TWI244869B - Data-on-demand digital broadcast system utilizing prefetch data transmission - Google Patents

Data-on-demand digital broadcast system utilizing prefetch data transmission Download PDF

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TWI244869B
TWI244869B TW91123507A TW91123507A TWI244869B TW I244869 B TWI244869 B TW I244869B TW 91123507 A TW91123507 A TW 91123507A TW 91123507 A TW91123507 A TW 91123507A TW I244869 B TWI244869 B TW I244869B
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dod
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TW91123507A
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Khoi Hoang
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Prediwave Corp
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Priority claimed from US10/054,008 external-priority patent/US20020175998A1/en
Priority claimed from US09/997,954 external-priority patent/US20020066100A1/en
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Abstract

A method and system for a decreased idle time scheduling matrix for a data file reduced into data blocks. A scheduling matrix is generated and idle time is filled with data blocks that appear later in the matrix, keeping with the original sequence of data blocks. This is then repeated, or equally a new decreased idle time scheduling matrix is created. A preloading stream can be configured from the decreased time scheduling matrix. Specially designed set-top boxes are then described which can received these data blocks. In a preferred embodiment the set-top box can store a continuously broadcast and optimized preloading stream for increased bandwidth savings.

Description

—4244869 五、發明說明(1) 【發明所屬之技術領域】 _隨夕選視:孔⑴心"“’,V0D)系統為隨選資料 (DOD)之一種。在v〇D系統中,視訊 現k貝 服器或伺服器網路提供給一個或多田二、基礎上由-伺 此技藝者所充分瞭解。 ^夕用戶。此系統為熟練 【先前技術】 在習用之VOD結構中,一個伺服器或 準層級之用戶及伺服器型態通信。例如 路二 康資料之請求給一健器(如一視訊以出:: 回應用戶之請求,.伺.服器送出該要求之檔給兩田準 個伺服器提供。該用戶可能儲存任侗j 為—個或/ w w仔任一個收到的資斜擴在# 揮叙性記憶體中供稍後使用。該用戶伺服器型能‘一雙 向j通信設施。目前雙向通信需建立新的設施因為現存 =電纜線僅能提供單向通信。雙向通信設施 叩貝的,且其服務效果未必能滿足大多數使用者。 菸提供者(例如電纜公司)欲提供v〇d服 imr广,標準用戶伺服器型態有許多限制,標 不斷收聽及滿足來自每一用戶之每—請求之機制。因^備 能收到服務之用戶數量係依此機制之容量而定。有一種 制係利用大量平行處理之電腦具有大而快之硬碟陣列作^ 部伺服器。然而,即使最快的現存局部伺服器僅能每—^— 4244869 V. Description of the invention (1) [Technical field to which the invention belongs] _ Sui Xi Xuan Shi: Kong Yixin " "', V0D) system is a kind of DOD. In the vOD system, The video server or server network is provided to one or Tada II. Based on the full understanding of the artisans. ^ Evening users. This system is skilled [prior art] In the conventional VOD structure, one Server or quasi-level user and server type communication. For example, Lu Erkang's data request to a health device (such as a video to output :: In response to the user's request, the server sends the requested file to Liangtian Provided by a quasi-server. The user may store the information received by any user or any of the users in a #storage memory for later use. The user-server model can be a bi-directional jCommunication facilities. Currently, new facilities need to be established for two-way communication because the existing cable cables can only provide one-way communication. Two-way communication facilities are not available, and their service effects may not satisfy most users. ) To provide v〇d service imr wide, There are many restrictions on the type of standard user server. The mechanism of continuously listening and satisfying every request from each user. Because the number of users who can receive services depends on the capacity of this mechanism. There is a system to use A large number of parallel processing computers have a large and fast hard disk array as the ^ server. However, even the fastest existing local server can only

第6頁 1244869^ 五、發明說明(2) ””ata stFeainS)給大約 1〇〇〇一2〇〇〇 用戶。因 a :服矛力更多用戶,局部伺服器之數目 服器需要更多高階飼服器以維持對局部飼服:: box ^ ί ί之另—限制為—個接收機頂盒(SeH〇p- Γ之i 影此服務視訊之前,②需下載所™服 ‘D而下: '即使具有甚快之下載速度,為放影所選之 0D而下載可觀之資料之延遲可能甚顯著而不可行。 己的ΪΪ用戶飼服器型態之另一限制為每一用戶需有並自 緩衝(c見二因Λ’需要之總頻寬與簽約之用戶數成正比。 pp ac e η己fe體配合局部飼服器曾經用於之 限制但因為緩衝記憶體亦為有限,故亦未能解決;見題之。 # # = ί為更能滿足更多用戶之隨選視訊服務,現存之服 ===加每一局部飼服器之用戶之比例至超過局部 給m。用:,一局部祠服器若能提供服務 約之用白 κ際上要未其服務.1 0,000用戶。若大多數簽 枯俯《 * = ^時訂購視訊服務,此技術也許可用。然而此 (例=1ΐ ί而架設,因為大多數用戶喜歡在同—時間接 荷。汽曰或週末)要看電視,因此使得局部伺服器過負 任何::媒口:f提供-種不必更換現有設施’而能經由 系統。某體k供1^選服務給大量之用戶而無顯著延遲之 第7頁 1244869 五、發明說明(3) 【發明内容】 本發明教示一種系統及方法以提供經由現存之DOD系 統提供減少存取時間之隨選資料(DOD)服務。本發明亦教 示一種系統及方法以提供經由減少頻寬之DOD服務。 在一例示性實施例中,於一伺服器端,傳送資料給一 用戶以提供隨選資料服務之方法,包含下列步驟:提供一 減少閒置時間、包含有一所選DOD服務之資料區塊(data block)之線性序列(liner sequence);自該減少閒置時間 乏資料區塊序列移除最常出現之資料區塊;放置該移除之 隶常出現之資料區塊於預取(pre fetch)實料串流(Stream) 中’使該預取資料串流包含對應於該選定之])〇D服務之預 取資料區塊;經由該傳送媒體傳送該預取資料串流;並經 由傳送媒體傳送其餘減少閒置時間之資料區塊序列,使接 ^設備可結合其餘減少閒置時間之資料區塊序列及該預取 貝料區塊以產生該選定之D〇D服務,因而減少傳送該D〇D服 務,需之頻寬。此D0D廣播伺服器方法進一步包含··自該 $ =閒置時間之資料區塊移除多數的其餘資料區塊;·放置 ‘ S ί f餘資料區塊於該預取資料串流使得該預取資料串 ‘ Ϊ二f最常出現之資料區塊及對應於所選D0D服務之其 料^堍區塊;並經由傳送媒體傳送其餘減少閒置時間之資 因而^序列及該預取資料區塊以產生該選定之D0D服務’ J滅少傳送該D0D服務所需之頻寬。 〜例示性實施例中,於用戶端,處理接收自一伺服Page 6 1244869 ^ V. Description of the invention (2) "" ata stFeainS) is given to about 1,000-2000 users. Because a: more users, the number of local servers requires more high-level feeders to maintain local feeding :: box ^ ί The other is limited to one receiver set-top box (SeH〇p- Γ 之 i Before the video of this service, ② download the server ™ down: 'Even with a fast download speed, the delay in downloading considerable data for playing the selected 0D may be significant and not feasible. Another limitation of the user's feeding device type is that each user needs to have and self-buffer (c see the total bandwidth required by the two factors Λ 'is proportional to the number of users signed up. Pp ac e η 自 fe body coordination The limitation of the local feeder was used but the buffer memory is also limited, so it cannot be resolved; see the title. # # = Ί For more on-demand video services to meet more users, the existing service == = Add the proportion of users of each local feeding device to more than the local supply m. Use: If a local shrine serving device can provide a service contract, it must be serviced. 10,000 users. If most of them sign When the video service is ordered when * = ^, this technology may be available. However, this (example = 1ΐ Er and set up Because most users like to pick up at the same time. (TV or weekend) to watch TV, so the local server is overloaded with any :: media port: f provides-a kind of system that can be passed without replacing existing facilities. k provides 1 ^ selected services to a large number of users without significant delay. Page 7 1244869 V. Description of the Invention (3) [Summary of the Invention] The present invention teaches a system and method for providing a reduction in access time through an existing DOD system. Data-on-Demand (DOD) service. The present invention also teaches a system and method to provide DOD services via reduced bandwidth. In an exemplary embodiment, on a server side, data is transmitted to a user to provide data-on-demand The service method includes the following steps: providing a linear sequence that reduces idle time and includes a data block of a selected DOD service; and removes the most frequently used data block sequence that reduces idle time Appearing data blocks; placing the removed frequently appearing data blocks in a pre-fetch material stream (make the pre-fetch data stream include In the selected]) 〇D pre-fetched data block of the service; transmitting the pre-fetched data stream via the transmission medium; and transmitting the rest of the data block sequence that reduces idle time via the transmission medium, so that the receiving device can combine the rest The data block sequence for reducing idle time and the pre-fetched shell material block to generate the selected DOD service, thus reducing the bandwidth required to transmit the DOD service. This DOD broadcast server method further includes ... Remove most of the remaining data blocks from the $ = idle time data block; · Place the 'S ί f remaining data blocks in the prefetched data stream so that the prefetched data string' appears most often The data block and its corresponding data block corresponding to the selected DOD service; and the rest of the data to reduce the idle time are transmitted via the transmission medium. Therefore, the sequence and the prefetched data block to generate the selected DOD service The bandwidth required to transmit the DOD service. ~ In the exemplary embodiment, at the client, the processing is received from a servo

1244869 五、發明說明(4) 器以提供隨選資料服務之方法,包含下列步驟:接收一對 應於選定之DOD服務包含預取數據區塊之預取數據串流; 儲存該預取數據區塊於一記憶位置;接收對應於選定之 DOD服務,包含原資料區塊之原始資料串流;並處理原始 資料區塊及預取資料區塊使一用戶能存取所選之D 〇 D服 務。此方法可進一步包含··接收用戶輸入之所^D〇I)服務 之指示;轉到對應於所選DOD服務之一個頻道以回應該用 =輸入,並自對應於所選D 〇 D服務之頻道接收原始.資料串 流。在一實施例,此法供處理接收自一伺服器之資料係為 用戶端之機頂盒所執行。 一隨選貧料系統包含一第一套頻道伺服器(Channel server )、, 一中.央控制伺服器以控制第一套頻道伺服器,一 f套增,斋耦合至第一套頻道伺服器,一結合器/放大 口至第一套增頻器,且一結合器/放大器適於經由一 送j體傳送育料。在一例示性實施例中,該隨選資料系 、)!進I f包含一頻道監視模組以監視此系統,一開關陣 ^ f 一套頻道伺服器及—第二套增頻器。.當系統發生 :守μ頻道監視模組係用以向中央控制伺服器報告。該 制伺服器,胃來自頻道監視模組之報告作出回應, 11二,陣列將第一套頻道伺服器之一故障頻道伺服器更 二套頻道伺服器内之-個頻道祠服器,並將第-頻=。為之故障增頻器置換為在第二套增頻器内之一增1244869 V. Description of the invention (4) The method for the device to provide on-demand data services includes the following steps: receiving a pre-fetched data stream corresponding to the selected DOD service including pre-fetched data blocks; storing the pre-fetched data blocks At a memory location; receiving the original data stream corresponding to the selected DOD service, including the original data block; and processing the original data block and the pre-fetched data block to enable a user to access the selected DOD service. This method may further include receiving an instruction from the user for all services provided by the user; go to a channel corresponding to the selected DOD service to respond to the use of the = input, and from the service corresponding to the selected service. The channel receives the raw data stream. In one embodiment, the method for processing data received from a server is performed by a set-top box on the client side. A lean-on-demand system includes a first set of channel servers, a central control server to control the first set of channel servers, an additional set of f, coupled to the first set of channel servers A combiner / amplifier port to the first set of boosters, and a combiner / amplifier is suitable for transmitting breeding material through a j-body. In an exemplary embodiment, the on-demand data is,)! The I f includes a channel monitoring module to monitor the system, a switching array ^ f a set of channel servers and a second set of frequency boosters. When the system occurs: The monitoring channel monitoring module is used to report to the central control server. The server responded to the report from the channel monitoring module of the system. 11 Second, the array changed one of the first channel server's faulty channel server into one of the two channel server's one channel server, and The first-frequency =. Replace the faulty frequency booster with one of the second frequency booster.

力一只施例教示一泛用STB,能接收並處理An example teaches a general STB, which can receive and process

1244869 五、發明說明(5) 一群數位伺服器如V0D及數位廣播。此 具有—高度彈性之架構能熟練地處理接=:料泛用 架構包含-資料匯流排,—第一通信 收之貝人枓。此 編信媒介,—記憶體典型地包含雙向 : 匯=久及暫存記憶體,一數位解碼器、,雙向輕 、άΓ非,以及一中央處理單元(CPU),雙向耦合至資料、陳+ &排。本發明此一實施例之cpu提供ST 、/、/ 例Γ;位解碼器及該解調器。細控1244869 V. Description of the invention (5) A group of digital servers such as V0D and digital broadcasting. This highly flexible architecture can handle the connection skillfully: the data is universally used The architecture includes-the data bus,-the first communication receiver. This letter medium—memory typically includes two-way: sink = long-term temporary memory, a digital decoder, two-way light, fast, and a central processing unit (CPU), which is bidirectionally coupled to data, and & Row. The cpu of this embodiment of the present invention provides ST, /, / examples Γ; a bit decoder and the demodulator. Fine control

程^能接收源於-減少'閒置時間排序矩.陣之^^制, 亦旎接收此類資料區塊之平行資料串流。 A 朴πίΐΓ實ί”,該複雜之STB架構允許一資.料區塊最 ΐ之STR之貝料串流(data Stream)被廣播並載入一閒 置ί ,々特定貧料區塊之預載入至一STB允許在有限時間 内即省頻見。該預載入(或預取)資料流,基於v〇d之要 求可被排程以預傳(predeHver)不同資料排序於 之 同時間。 卜 【實施方式】 圖1A顯不本發明之一實施例之例示性〇〇])系統丨〇 〇。在 此實施例中,DOD系統1 〇 〇提供被選之資料檔,例如視訊 檔’然而,DOD系統1 〇〇不限於提供被選之視訊檔,亦能提 供其他資料檔,例如隨選戲檔。D〇D系統包含一中央控制 伺服裔102。一中央儲存1〇3,一群頻道伺服器1〇“ —1〇4打 五、發明說明(6) ,一群增頻器106a-l〇6n,一結合器/放大器1〇8。中央控 制伺服器1 0 2控制頻道伺服器丨〇 4。中央儲存丨〇 3以數位格 式儲存資料檔。在一例示性實施例中,儲存在中央儲存 1 0 3之資料檔可為任何已授權之電腦例如連接到網路之中 央控制伺服器1 0 2經由標準網路介面(如乙太網路連結) 存取。每一頻道伺服器1 04被指定一頻道且耦合至」^曾頻 态1 0 6。頻道伺服器1 〇 4依據來自中央控制伺服器1 〇 2之指 示提供取自中央儲存103之資料檔。每一頻道伺服器1〇4之 輸出為四極調輻(QAM)調變之中頻(if)信號具有一適合之 頻率給湘應之增頻器106。該qAM調變之IF信號與所用σ之標 準有關,。在美國現行標準為數據在電纜系統上之介面規格 (DOCSIS)標準,該標準需求约43· 75MHz之IF頻率。該增頻 器106將接收自頻道伺服器;[04之IFMt號轉換為射頻^ ^ (RF信號)。該RF信號含有頻率及寬頻,係與所要之頻道及 所用標準有關。例如,在現行美國之有線電視頻道8 〇之穩 準’該RF信號具有約為5 59· 2 5MHz之頻率,且有一約⑽H2 之頻寬。增頻器1 0 6之輸出供給結合器/放大器丨〇8。該結 合器/放大器1 0 8調整並結合收到之RF信號,將信號輸^出 給傳送媒介11 〇。 b 在一例示性實施例,該中央控制伺服器丨0 2包含一緣 圖使用者介面(未顯示)使一服務提供者能以取放(drag曰 a n d (¾ r ο p )操作以排序數據分送。進一步,中央控制飼服 器1 0 2依據分送矩陣確認並控制頻道飼服器1 〇 4起動及停 止。在一例示性實施例,該中央控制伺服器丨〇 2自動選擇The program can receive the parallel-data stream that originates from-reduction of the idle time sorting matrix. It also does not receive such data blocks. A Pu π ΐ 实 ί ί ”, the complex STB architecture allows a data stream of the most STR of the data block to be broadcasted and loaded into an idle, preload of a specific lean block. Entering an STB allows frequency-saving to be seen within a limited time. The pre-loaded (or pre-fetched) data stream can be scheduled based on the requirements of vOd to predeHver different data at the same time. [Embodiment] FIG. 1A shows an exemplary system of one embodiment of the present invention. The system 丨 00. In this embodiment, the DOD system 100 provides a selected data file, such as a video file. The DOD system 100 is not limited to providing selected video files, but can also provide other data files, such as on-demand movie files. The DOD system includes a central control server 102. A central storage 103, a group of channel servos 10 "-104, five, invention description (6), a group of frequency boosters 106a-106n, a combiner / amplifier 108. The central control server 102 controls the channel server 丨 〇 4. Central storage 丨 〇 3 Save data files in digital format. In an exemplary embodiment, the data file stored in the central storage 103 may be any authorized computer such as a central control server connected to a network via a standard network interface such as an Ethernet link ) Access. Each channel server 104 is assigned a channel and is coupled to the ^ Zeng frequency 1 106. The channel server 104 provides a data file taken from the central storage 103 according to the instructions from the central control server 102. The output of each channel server 104 is a quadrupole modulated (QAM) modulated intermediate frequency (if) signal with a suitable frequency for the frequency response booster 106 of Xiangying. The qAM modulated IF signal is related to the σ standard used. The current standard in the United States is the DOCSIS standard for data on cable systems, which requires an IF frequency of about 43.75MHz. The frequency booster 106 will be received from the channel server; [04's IFMt number is converted to radio frequency ^ (RF signal). The RF signal contains frequency and broadband, which are related to the desired channel and the standard used. For example, in the current US cable television channel 80, the RF signal has a frequency of about 5 59 · 25 MHz and a bandwidth of about ⑽H2. The output of the frequency booster 106 is supplied to the combiner / amplifier 08. The combiner / amplifier 108 adjusts and combines the received RF signals and outputs the signals to the transmission medium 11. b In an exemplary embodiment, the central control server 丨 0 2 includes a edge map user interface (not shown) to enable a service provider to perform a pick and place (drag and and (¾ r ο p) operation to sort data Distribution. Further, the central control feeding device 102 confirms and controls the channel feeding device 104 to start and stop according to the distribution matrix. In an exemplary embodiment, the central control server 丨 02 is automatically selected

I244«69 五、發明說明(7) ' 一頻道並計算分送矩陣以在一選定之頻道傳送資料檔。該 中央控制伺服器1 〇 2提供離線增加、刪除及更新資料標資 訊(如持續期、種類、等級或/及簡述)。而且該中^控 制飼服器102以更新儲存其中之數據檔控制中央儲存1〇3' 在一例不性貫施例,一現存有線電視系統丨2〇會繼續饋送 信號給結合器/放大器108以提供非D0D服務給用戶。因 此,依本發明之實施例之D0D系統100並不中斷現有之有線 電視服務。I244 «69 V. Description of Invention (7) 'One channel and calculate distribution matrix to transmit data files on a selected channel. The central control server 102 provides offline addition, deletion, and update of information (such as duration, type, level, or / and brief description). In addition, the control feed 102 is controlled to update the data files stored therein, and the central storage 103 is controlled. In an inconsistent embodiment, an existing cable television system 20 will continue to feed signals to the coupler / amplifier 108 to Provide non-DOD services to users. Therefore, the DOD system 100 according to an embodiment of the present invention does not interrupt existing cable television services.

•圖1β顯不本發明另一實施例之例示性DOD系統1 00。除 了顯示於圖1Α之元件外,該D0D系統包含一開關矩陣112,、 一頻道監視模組11 4,一套備用頻道伺服器丨J h —丨丨6b,及 一套備用增頻器11 8a-118b。在一實施例中,該開關矩陣 112係實體上位於增頻器1〇6及結合器/放大器1〇8之間。FIG. 1β shows an exemplary DOD system 100 according to another embodiment of the present invention. In addition to the components shown in Figure 1A, the D0D system includes a switch matrix 112, a channel monitoring module 11 14, a backup channel server 丨 J h — 丨 丨 6b, and a backup frequency booster 11 8a -118b. In one embodiment, the switch matrix 112 is physically located between the frequency increaser 106 and the combiner / amplifier 108.

開關矩陣11 2被中央控制伺服器J 〇2所控制。該頻道監視模 組11 4包含一群有形的機頂盒,用以偵侧谮在之用戶,監 視DOD系統1〇〇之良窳。監視之結果為頻道監視器模組丨j * ,,給中央控制伺服器102。在頻道失效(如一頻道伺服 益損壞、一增頻道損壞、或一通信連接損壞)之情形,中 央控制伺服器1 〇 2經由開關矩陣i丨2斷開失效之元件並連上 一良好之備用元件11 6及/或j 8以恢復服務。 在一例示性實施例,為自D〇D系統丨〇 〇所廣播之資料檔 包含於電影專家群(MPEG)之檔案。每一評^:〇檔自動分成資 =區塊及次區塊,沿時間軸映射到資料檔之特定部分。此 貝料區塊及次區塊為中央控制伺服器】〇2所提供之三維分The switch matrix 112 is controlled by a central control server J 02. The channel monitoring module 114 includes a group of tangible set-top boxes for detecting users on the side and monitoring the goodness of the DOD system 100. The result of monitoring is the channel monitor module 丨 j *, which is provided to the central control server 102. In the case of a channel failure (such as a channel servo damage, an increased channel damage, or a communication connection damage), the central control server 102 disconnects the failed component via the switch matrix i 丨 2 and connects a good spare component 11 6 and / or j 8 to resume service. In an exemplary embodiment, the data file broadcast from the DOD system is included in the file of the film expert group (MPEG). Each comment ^: 0 files are automatically divided into data = blocks and sub-blocks, which are mapped to specific parts of the data file along the time axis. The shell material block and the sub-block are the three-dimensional analysis provided by the central control server.

第12頁Page 12

1244869 五、發明說明(8) 送矩陣所預定之時間内被送出。對於提供DOD服務之DOD系 統不再需要回饋頻道丨丨g '石W Hg川 道可用於其他用途,如收費或提供網際網路服務 圖2顯示依據本發明之一實施例之例示性頻道伺服 1 〇 4。該頻道伺服器1 〇 4包含一伺服控制器2 〇 2、CPU 2 0 4、 一QAM調變器20 6、一局部記憶20 8及一網路介面210。該伺 服控制器2 0 2指示C P U 2 0 4將資料檔分成區塊(更進一步分 成次區塊及資料包封(packets),選擇資料區塊依中央控 fJ器1 0 2提供之分送矩陣予以傳送,將選定之資料編碼, 壓縮編碼資料,然後將壓縮資料分送給QAM模組2〇6以控制 頻道飼服器104之整個運作。該QAM模組2〇6經由一匯流排 ^即PC I、CPU局部匯流排)或網際網路連結接收要傳送之 資料。在一例示性實施例中,該QAM調變器可包含一下游 C^^=ream) QAM調變器,一上游四極調幅/四極相移調 : Qpsk)具有正向錯誤改正解碼之猝發解調器,及/ ^ ^ 11 2 〇6 ^ ^ ^ ^ ^ ^ ^ ^ 1 04 2罔由路^介面21 〇將頻道伺服器1 04連接到其他頻道伺服哭 之排序及^^司人服器1〇2以執行來自中央控制飼服器^ 且自Ϊ央二η: ’將上況回報給中央控制伺服器102, 資料俨在次接收貝料檔。任何自中央儲存1 0 3取出之 可儲i於it私依據來自祠服控制器202之指令處理之前 例中,伺服裔1〇4之局部記憶208。在-例示性實施 ^員道飼服器m可依電境頻道之頻寬(如6 6 5或“1244869 V. Description of the invention (8) The matrix is sent out within the predetermined time. For a DOD system that provides DOD services, there is no longer a need to give back channels. G 'stone W Hg channel can be used for other purposes, such as charging or providing Internet services. Figure 2 shows an exemplary channel servo 1 according to an embodiment of the invention. 4. The channel server 104 includes a servo controller 20, a CPU 204, a QAM modulator 206, a local memory 208, and a network interface 210. The servo controller 2 0 instructs the CPU 2 0 4 to divide the data file into blocks (further divided into sub-blocks and data packets, and selects the data block according to the distribution matrix provided by the central control fJ device 102. It transmits, encodes the selected data, compresses the encoded data, and then distributes the compressed data to the QAM module 206 to control the overall operation of the channel feeder 104. The QAM module 206 passes a bus ^^ PC I, CPU local bus) or Internet link to receive the data to be transmitted. In an exemplary embodiment, the QAM modulator may include a downstream C ^^ ream) QAM modulator, an upstream quadrupole amplitude modulation / quadrupole phase shift modulation: Qpsk) a burst demodulator with forward error correction decoding , And / ^ ^ 11 2 〇 6 ^ ^ ^ ^ ^ ^ ^ ^ 1 04 2 罔 Yu Lu ^ Interface 21 〇 Connect the channel server 1 04 to other channel server crying sequence and ^^ Division Server 1〇 2 to execute from the central control feeder ^ and since the central control η: 'report the situation to the central control server 102, the data is received at the next time. Any of the data taken from the central storage 103 can be stored in the local memory 208 in accordance with the instructions from the temple controller 202 in the private example. In-exemplary implementation ^ the member feeding device m can be based on the bandwidth of the electrical channel (such as 6 65 or "

1244869 五、發明說明(9) 8MHz ) ’ QAM調變(如QAM64或“心“及—β⑽資料串流之壓 、^準^位元速率(即MPEG—〗或評以_2)之頻寬(如6.65 或8ΜΠζ )送出一或多個D〇])資料串流。 #以=2員示本發明之一實施例之泛用機頂盒(STB) 30 0。 =㈣〇包含__解調器3〇2、―cpu3〇4、一局部記憶 4、—級衝記憶,、一解碼器312具有視訊及音訊解碼 3及重豐模組314、一用戶介面318、-通信鏈 ^制'龍流排322如所示將這些元㈣合CP_2 貧料解碼,解壓縮已解碼之資料,重組已解 3U Λ 將已六解碼資料儲存於局•記憶308或緩衝記憶 中,/邮为送^ 資料給解碼器312。.在一例示性實施例 大号3m包含非揮發性記憶、(如硬碟),缓衝放 大為3 1 〇已含揮發性記憶。 列之在施例’QAM解調器302包含發射及接收模組及下 )力:多元件:加密/解密模組、正向錯誤改正解碼 電S cuJl谐控制、下游及上游處理器、CPU及記憶介面 號以再儲存資料。 月义之w號,取樣並解調該信 將至ί =性實施例巾,!存取已被認可,解碼器312 顯亍 】;:區塊解碼以將貧料區塊轉換為可供輸出銀幕 二Ϊ:像献解碼;312自簽約之用戶獲得指令,如放 Τ 、暫如、則進一段、重繞捲、向前等等。解碼器1244869 V. Description of the invention (9) 8MHz) 'QAM modulation (such as QAM64 or “heart” and —β⑽ data stream pressure, ^ quasi ^ bit rate (ie MPEG — or rated _2) bandwidth (Such as 6.65 or 8MΠζ) send one or more D0]) data stream. # 以 = 2members show a universal set-top box (STB) 300 according to an embodiment of the present invention. = ㈣〇 Contains __ demodulator 3〇2, ―cpu3〇4, a local memory 4, 4 grade memory, a decoder 312 with video and audio decoding 3, and Chongfeng module 314, a user interface 318 、 Communication chain system 'Long Liu Pai 322 decodes these elements into CP_2 as shown in the figure, decompresses the decoded data, reassembles the decoded 3U Λ stores the decoded data in the bureau • memory 308 or buffer memory Middle / post is to send ^ data to the decoder 312. In an exemplary embodiment, the large size 3m contains non-volatile memory (such as a hard disk), and the buffer magnification is 3 1 0, which already contains volatile memory. Listed in the example 'QAM demodulator 302 includes transmitting and receiving modules and below) Force: Multi-component: Encryption / Decryption Module, Forward Error Correction Decoding ScuJl Control, Downstream and Upstream Processors, CPU Remember the interface number to save the data. No. w Yueyi, sampling and demodulating this letter The access has been approved, and the decoder 312 is displayed.]: Block decoding to convert the poor blocks into output screen. Second: Xiangxian decoding; 312 obtains instructions from the user who signed the contract. , Then advance a section, rewind the roll, forward and so on. decoder

1244869 五、發明說明(10) 312提供解碼資料給一輸出元件3 24給用戶使用。輸出元件 3一2 4可為任何適合之元件,如電視、電腦,任何適合之顯 示監視器、VCR或其類。 ’、 、圖形重疊模組3 1 4增強顯示繪圖之品質,如提供扭轉“ 角或圖中圖之能力。在一例示性實施例,圖形重疊模組 3^4可用於在遊戲型恶時之繪圖加速,例如,當服務提供 者利用本發明之系統提供隨選遊戲時。 ’、 •一用戶介面318使使用者控制STB3〇〇,可為任何適合 凡件如遙控農置、键般、知接上 口之 之通作“ ^ 日慧卡等。通信鍵32 0提供另外 Z連、,ό。此可為耦合至另一電腦,或用以提供雔θ、f ^。資料匯流排32 2可用商品化之快 徒(、又向通 而要以執订育料之即時通信型態者較可取。 電視ΐ:例施例+,雖然資料播係廣播給所有有後 用隨選資料服務。在一例干性〜:用戶可以解碼並享 面318中之智慧卡獲得 、、甶用戶者介 當地商店或由服務提# 料槽。智慧卡可在 例示性實施例中,售機更新資料。在另一 用之資料檔。 糸、、充詖么、一用戶無限存取所有可 在較佳實施例中,隨谐咨 任何時間選擇一可用』a 7、互動功能允許—用戶於 選之資料標開始;:::::門=戶按下-選擇鍵至: 資源被分配(如頻寬、伺應時間。由於更多 服σσ之靶力)給DOD服務之用,1244869 V. Description of the invention (10) 312 provides decoded data to an output element 3 24 for use by the user. The output elements 3-4 can be any suitable elements, such as televisions, computers, any suitable display monitors, VCRs or the like. ',, Graphics overlay module 3 1 4 enhances the quality of display graphics, such as providing the ability to reverse the "angle" or the figure. In an exemplary embodiment, the graphics overlay module 3 ^ 4 can be used in the game-type evil Graphics acceleration, for example, when a service provider provides on-demand games using the system of the present invention. ', • A user interface 318 enables the user to control STB300, which can be any suitable device such as remote farming, keys, and The catch phrase is "^ Rihui Card, etc." The communication key 32 0 provides another Z connection, ό. This may be coupled to another computer or used to provide 雔 θ, f ^. The data bus 32 2 is available for commercialized fast-talkers (and those who want to use instant messaging in order to subscribe to breeding materials are preferred. TV ΐ: 例 例 +, although data broadcasting is broadcast to all after-use On-demand data service. In an example of dryness ~: The user can decode and enjoy the smart card in face 318 to obtain it, ask the user to refer to the local store, or provide the feed slot. The smart card can be sold in the exemplary embodiment. Update data in another machine. In another used data file. 糸 ,, charge ?, unlimited access for all users. In the preferred embodiment, any one can be selected at any time with the service. A7. Interactive function allowed—user Start at the selected data target; ::::: gate = user presses -select key to: resources are allocated (such as bandwidth, wait time. Due to more target force serving σσ) for DOD service,

第15頁 1244sm 五、發明說明(11) 反應時間就更短。在一例示性實施例中,反應時間可基於 評估資源之分配及所望服務品質而決定。當與將一第一資 料區塊置於一平行資料串流中之實施例結合時,反應時間 僅為接收並處理此第一資料區塊所需之時間。 在一實施例中,每一資料檔之資料區塊之數(NUM-OF -BLKS)可由下式計算: 估計BLK 大小= (DataFi le — Size*TS)/DataFile — Length (1) BLK —SIZE = (估計BLK 大小+CLUSTER —SIZE-lByte)/Page 15 1244sm 5. Description of the invention (11) The reaction time is even shorter. In an exemplary embodiment, the response time may be determined based on the evaluation of resource allocation and desired service quality. When combined with the embodiment where a first data block is placed in a parallel data stream, the response time is only the time required to receive and process this first data block. In one embodiment, the number of data blocks (NUM-OF-BLKS) of each data file can be calculated by the following formula: Estimated BLK size = (DataFi le — Size * TS) / DataFile — Length (1) BLK —SIZE = (Estimated BLK size + CLUSTER —SIZE-lByte) /

• (CLUSTER—SIZE) (2) BLK—SIZE一BYTES=BLK一SIZE*CLUSTER—SIZE (3) NUM—OF_BLKS = (DataFile —Si ze + BLK —SIZE—BYTES-1 Byte) /BLK—SIZE一BYTES (4) 於公式(1)及(4)中,估計之BLK大小為一估計之區塊 大小(以Bytes計);Dataf i le —Size為資料檔之大小(以 byte計);TS表示時間槽之長短(以秒計);Data —• (CLUSTER_SIZE) (2) BLK_SIZE_BYTES = BLK_SIZE * CLUSTER_SIZE (3) NUM_OF_BLKS = (DataFile_Si ze + BLK_SIZE_BYTES-1 Byte) / BLK_SIZE_BYTES (4) In formulas (1) and (4), the estimated BLK size is an estimated block size (measured in Bytes); Dataf le — Size is the size of the data file (measured in bytes); TS indicates time Length of slot (in seconds); Data —

Fi le —Length為資料檔之長短(以秒計);BLK-SIZE為每 一資料區塊所需群組(clusters)數;CLUSTER—SIZE為每一 頻道飼服器1 04 (如64Kby tes )局部記憶208之一群組之大 小’ BLK —SIZE —BYTES為以Byte計之區塊大小。在此實施例 中,區塊數(NUM —0T — BLKS)等於資料檔大小(以Byte計) 加上一資料區塊大小(以By t e計)減1 By t e並除以一資料 區塊大小(以By t e計)。公式(1 )到(4 )顯示一特殊實施Fi le — Length is the length of the data file (in seconds); BLK-SIZE is the number of clusters required for each data block; CLUSTER—SIZE is the feeder for each channel 1 04 (eg 64Kby tes) The size of one group of the local memory 208 'BLK —SIZE —BYTES is the block size in bytes. In this embodiment, the number of blocks (NUM — 0T — BLKS) is equal to the data file size (in Bytes) plus a data block size (in By te) minus 1 By te and divided by a data block size (In By te). Equations (1) to (4) show a special implementation

第16頁 1244869、 五、發明說明(12) 例。熟習此技蟄之人士應暸解其他方法於 檔之資料區塊數。例如區分一資料檔 异—資料 係估計之區塊大小及頻道伺服器之'、、貝'2 ^數主要 大小之函數。因此,本發明應不心群組 資料本?之一實施例產生-排序矩= 貝’、田私序。在一例示性實施例中,本笋明μ用# „ 割多工器⑴me dlvlslon multriple本;月”時間分 多工器(FDM)技術在伺服哭端分 ,M)及頻率分割 在-例示性實施^ /生—狀壓縮及排序。 -實施例中,每—資料陣給每-資料標。在 fa枓塊之數目而產生。典型地,一排序矩陣提供 料】二分送資料播之資料區塊之分送順序,致資Page 16 1244869, V. Explanation of the invention (12) Example. Those who are familiar with this technique should know the number of data blocks in other methods. For example, the difference between a data file—data is a function of the estimated block size and the main size of the channel server's, 、, and 22 ^ numbers. Therefore, the present invention should not care about group data books? In one embodiment, generation-ordering moment = P ', Tian private sequence. In an exemplary embodiment, the present application uses # „cut multiplexer ⑴me dlvlslon multriple; monthly” time division multiplexer (FDM) technology in the servo crying end, M) and frequency division in the -exemplary Implement ^ / raw-like compression and sorting. In the embodiment, each data array gives each data target. Generated in the number of blocks. Typically, a sorting matrix provides data] The distribution order of the data blocks of the two distribution data broadcast

^ a" A u ^壬一個想於任一時間存取資料檔之用戶所存 取0 U 一於步驟4〇2,幾個資料檔之資料區塊(χ)被接收,一第 又4數j被歸零(步驟4 〇 4 )。一參考陣列被清除(步驟 :、人’]與\比較(步驟4〇8)。若j小於X,第二變數i被歸 =^步驟412 )。其次i與X比較(步驟414 )。若i小於X, 貧料區塊儲存於排序矩陣内之行[(i + j)m〇dul〇(x)]被寫入 5亥參考陣列(步驟418 )。若參考陣列已有此資料區塊, 則T寫入此重覆資料。開始時,因排序矩陣尚無記載,此 步&可予省略。再來,參考陣列被檢查是否包含資料區塊 1 (步驟420 )。開始時,因所有在參考陣列中之記載已在^ a " A u ^ No. A user who wants to access the data file at any time is accessed by 0 U. In step 402, the data blocks (χ) of several data files are received, and the first and fourth numbers are received. j is zeroed (step 4 04). A reference array is cleared (step :, person '] and compared with (step 408). If j is less than X, the second variable i is returned = ^ step 412). Next, i is compared with X (step 414). If i is less than X, the row [(i + j) m0dul0 (x)] of the lean block stored in the sorting matrix is written into the reference array (step 418). If the reference array already has this data block, T writes this repeated data. At the beginning, this step & can be omitted because there is no record of the sorting matrix. Then, the reference array is checked whether it contains data block 1 (step 420). At the beginning, because all the entries in the reference

1244869 五、發明說明(13)1244869 V. Description of the invention (13)

步驟4 0 6被清除,在參考陣列中應無資料。若參考陣列未 含貢料區塊i ’貧料區塊i被加入排序矩陣之陣列位置 [(i + j ) modul o(x ),j ]及參考陣列中(步驟422 )。於資料 區塊i加進排序短陣及參考陣列後,i增加1,使卜i +1 (步 驟4 24 ),然後程序自步驟4 14重覆直至i>x。若參考陣列 包含資料區塊i,i增加1,使i = i +1 (步驟4 2 4 ) ·,然後程 序自步驟41 4重覆直至i = X ’當丨=X時,j增加1,使j = j + 1 (步驟416 )而程序自步驟4 0 6重覆直至j=x。當卜x時,整 個程序終止(步驟41 0 )。 在一例示性實施例,若一資料檔被分為6個資料區塊 (x = 6),該排序矩陣及參考陣列如下·· 排序矩陣(SM)Step 4 0 6 is cleared and there should be no data in the reference array. If the reference array contains no material block i ', the lean material block i is added to the array position [(i + j) modulo (x), j] of the ranking matrix and the reference array (step 422). After the data block i is added to the sorted short array and the reference array, i is increased by 1 to make i + 1 (step 4 24), and then the procedure is repeated from step 4 14 until i > x. If the reference array contains data block i, i is increased by 1 so that i = i +1 (step 4 2 4), and then the procedure is repeated from step 41 4 until i = X 'when 丨 = X, j increases by 1, Let j = j + 1 (step 416) and the procedure repeats from step 4 to 6 until j = x. When x is set, the entire program is terminated (step 4100). In an exemplary embodiment, if a data file is divided into 6 data blocks (x = 6), the sorting matrix and reference array are as follows: · Sorting matrix (SM)

TSO TS1 TS2 TS3 TS4 TS5 [0,0]blk0 [i,o]biki [2,0]blk2 [3,0]blk3 [4,0^^4 [5,0]blk5 [0,1] [l,l]blkO P,l] [3,1] Kl] [5,1] [0,2] [1,2] |;2,2]blk0 [3,2]blkl [4,2] [5,2] [0,3] [1,3] P,3] [3,3]blk0 [4,130] [5,3]blk2 [〇>4] [l,4]blk3 [3,4] [4,4]blkO [5,4]blkl [〇,5] [1>5] P,5] [3,5]blk4 [4,5] [5,5]blkOTSO TS1 TS2 TS3 TS4 TS5 [0,0] blk0 [i, o] biki [2,0] blk2 [3,0] blk3 [4,0 ^^ 4 [5,0] blk5 [0,1] [l , L] blkO P, l] [3,1] Kl] [5,1] [0,2] [1,2] |; 2,2] blk0 [3,2] blkl [4,2] [5 , 2] [0,3] [1,3] P, 3] [3,3] blk0 [4,130] [5,3] blk2 [〇 > 4] [l, 4] blk3 [3,4] [ 4,4] blkO [5,4] blkl [〇, 5] [1 > 5] P, 5] [3,5] blk4 [4,5] [5,5] blkO

參考陣列(RA)Reference Array (RA)

第· 18頁 TSO =>blkO TS1 =>blkO,blkl,blk3 TS2 =>blkO,blk2 TS3 =>blkO,blkl,blk3,blk4 TS4 =>blkO, blk4 TS5 =>blkO,blkl,blk2,blk5 在力一例不性 1244869 五、發明說明(14) space。 space 1 space2 space3 space.4 spaced TS0 bW bM blk3 blk4 blk5 TS1 blM bW blk2 blk3 blk4 TS2 blK bW blk3 blkl blk4 TS3 blK bUd tlkO blM blW blk2 TS4 blk4 bM) blk5 blt2 blkl TS5 bli5 bM bM blM blk3 blk4 附於此說明書之附錄A描述顯示於圖4之例示性程序, 一步一步之程序以產生上述排序矩陣及參考陣列。於此例 示性實施例’基於上述之排序矩陣,該資料檔之6個資料 區塊依下列次序送出: 、' ^ 則盾狂斤4用於呼管一』Page 18 TSO = > blkO TS1 = > blkO, blkl, blk3 TS2 = > blkO, blk2 TS3 = > blkO, blkl, blk3, blk4 TS4 = > blkO, blk4 TS5 = > blkO, blkl , blk2, blk5 An example of incapability 1244869 V. Description of the invention (14) space. space 1 space2 space3 space.4 spaced TS0 bW bM blk3 blk4 blk5 TS1 blM bW blk2 blk3 blk4 TS2 blK bW blk3 blkl blk4 TS3 blK bUd tlkO blM blW blk2 TS4 blk4 bM) blk5 blt2 b5 blk blk5 TS Appendix A of the description describes the exemplary procedure shown in FIG. 4, a step-by-step procedure to generate the above-mentioned sorting matrix and reference array. In this exemplary embodiment, based on the above-mentioned sorting matrix, the 6 data blocks of the data file are sent in the following order: ^ ^ Shield Kuangjin 4 is used for exhalation 1 "

排序矩陣在一預測之存取時間之前送出一資料2 ^ 一瓦 $之資料區塊數。例如,若一預先決定之前瞻二, 4間槽之長短(duration),對任何時間样^:間為一 時間槽,資料檔之第4資料區塊⑴k4)應曰、〜或等於I ^ 簽約之The sort matrix sends a number of data blocks 2 ^ one watt $ before a predicted access time. For example, if one determines in advance the duration of the two slots, for any time sample ^: a slot, the fourth data block of the data file (k4) should be, ~ or equal to I ^ Of

1244869 ----^ 五、發明說明(15) " 戶在或之别為一STB300所接收,但在TS4之前不會 被頌示。供產生一前瞻排序短陣之程序步驟係實質上與前 述圖4之_程序步驟相似,除了此實施例中之前瞻排序矩陣 ^ = 了前瞻時間排出一提早送出之次序而已。假設一資料 才:二為6個貢料區塊,一例示性基於—前瞻排序矩陣之分 送人序具有兩個時間槽之長短之前瞻時間,可表示如下: TSO TS1 TS2 TS3 TS4 TS51244869 ---- ^ V. Description of the invention (15) " The user was or was received by an STB300, but will not be honoured before TS4. The procedural steps for generating a look-ahead short matrix are substantially similar to the program steps of FIG. 4 described above, except that in this embodiment the look-ahead sort matrix ^ = the order of the look-ahead time is sent out early. Assume one data: two are six tribute blocks, an example is based on the look-ahead sorting matrix. The forward order has two time slots, and the look-ahead time can be expressed as follows: TSO TS1 TS2 TS3 TS4 TS5

->blkO-> blkO

=>blkO,blkl,blk3,blk4 ->blkO, blk2 =>blkO,blkl,blk3,blk4,blk5 =>blkO,blk5 = >blkO, blkl,blk2= > blkO, blkl, blk3, blk4-> blkO, blk2 = > blkO, blkl, blk3, blk4, blk5 = > blkO, blk5 = > blkO, blkl, blk2

—一彳固三維分送一套資料檔分送矩陣係基於此組資料檔 j母=資料檀之排序矩陣而產生。在此三維分送矩陣中, f第一、’、=產生包含此組資料檔之每一資料標之識別號碼 、、,/)。^,該三雄分送矩陣被計算以有效利用在每一頻道分 ί Ϊ多資料串流之可用頻寬。在一例示性實施例,一為此 八二所热知之迴旋(c〇nν 01 u t iοη )方法被採用以產生三維 ^ =矩陣來排序一組資料檔之有效率分送。例如,一迴旋 Γτς^可包含下列方針:(1)全數資料區塊在任一時間槽 墟^時間内送出者應保持最小之可能數目;且(2)若依 據方針(1)可用許多部分解決方案,最佳解決方案為其中—One solid 3D distribution of a set of data file distribution matrix is generated based on this set of data files j mother = data matrix ranking matrix. In this three-dimensional distribution matrix, f first, ', = generates the identification numbers of each data object containing this set of data files ,,, /). ^ The Sanxiong distribution matrix is calculated to effectively use the available bandwidth of multiple data streams distributed on each channel. In an exemplary embodiment, a method known as convolution (cónv 01 u t iοη), which is well known for this, is used to generate a three-dimensional ^ = matrix for efficient distribution of a set of data files. For example, a manoeuvre Γτς ^ may include the following policies: (1) All data blocks should be sent to the smallest possible number within any given time slot; and (2) Many partial solutions can be used in accordance with the policy (1) , The best solution is among them

第20頁 4244869 -*-----. 五、發明說明(16) 下列資料區塊相加後具最少資料區塊之和者··在任一參考 時間槽之時間内送出之資料區塊、在前一時間槽(以參考 時間槽為準)時間内送出之資料區塊及在次一時間槽(以 參考時間槽為準)時間内送出之資料區塊。例如,假如一 例示性系統送出二個短資料槽,Μ及N,其中每一資料棺被 分為6個資料區塊,則基於排序矩陣送出次序係如下述:Page 20 4244869-* -----. V. Description of the invention (16) The sum of the following data blocks with the least data block sum ... the data blocks sent within any reference time slot, Data blocks sent in the previous time slot (based on the reference time slot) and data blocks sent in the next time slot (based on the reference time slot). For example, if an exemplary system sends out two short data slots, M and N, each of which is divided into 6 data blocks, the order of submission based on the ranking matrix is as follows:

TSO 二〉blkO TS1 ->blkO,blkl,blk3,blk4 TS2 =>blk0,blk2 TS3 :>blkO, blkl,blk3, blk4, blk5 TS4 =>blkO,blk5 TS5 =>bikO, blkl,blk2 —一二Γ、、隹y刀送一套資料檔分送矩陣係基於此組資料檔 之母負料棺之排序矩陣而產生。在此三維分送矩陣中, 1第三維產_生包含此組資料檔之每一資料檔之識別號碼 #』該維分送矩陣被計算以有效利用在每一頻道分 ^ -A 4流之可用頻寬。在一例示性實施例,一為此 分送矩陣來排序'剛檑之有效率分:。用:如, t含下列方針:⑴全數資料區塊在任一時門一神 據方針⑴可料Λβ,束可能數目;且⑴若依 )了用4多部分解決方案’最佳解決方案為其中TSO II> blkO TS1-> blkO, blkl, blk3, blk4 TS2 = > blk0, blk2 TS3: > blkO, blkl, blk3, blk4, blk5 TS4 = > blkO, blk5 TS5 = > bikO, blkl, blk2 — A set of data file distribution matrices for one, two, Γ, and 隹 y knives are generated based on the sorting matrix of the female negative coffin of this set of data files. In this three-dimensional distribution matrix, the first three-dimensional distribution matrix contains the identification number of each data file of the set of data files. The "dimensional distribution matrix" is calculated to effectively use the ^ -A 4 stream of each channel. Available bandwidth. In an exemplary embodiment, a distribution matrix is used to sort the 'efficiency score: Use: For example, t contains the following policies: ⑴ All data blocks can be used at any one time. According to the policy, we can expect Λβ, the number of possible beams;

第21頁 1244869 五、發明說明(17) 下列資料區塊相加後具最少資料區塊之和者:在任一參考 柃間槽之時間内送出之資料區塊、在前一時間槽(以參考 日了間槽為準)時間内送出之資料區塊及在次一時間槽(以 蒼考時間槽為準)時間内送出之資料區塊。例如,假如一 7不性系統送出二個短資料檔,Μ及N,其中每一資料檔被 分為6個資料區塊,則基於排序矩陣送出次序係如下述:Page 21, 1244869 V. Description of the invention (17) The sum of the following data blocks with the least data block sum: the data block sent in any reference time slot, the previous time slot (reference Data blocks sent within the time slot) and data blocks sent within the next time slot (based on the Cangkao time slot). For example, if a 7-sex system sends two short data files, M and N, each of which is divided into 6 data blocks, the sending order based on the ranking matrix is as follows:

TSO 二〉b1kOTSO II> b1kO

. TSl =>blkO,blkl,blk3,blk4 TS2 =>blkO,blk2 TS3 =>blkO,blkl,blk3,blk4 TS4 =>blkO,blk4 TS5 =>blkO,blkl,blk2,blk5 應用如上述之例示性迴旋方法,分送矩陣之可能組合 如下述: 選擇1 ·以平移〇 T S分送視訊檔 總資料區塊. TSl = > blkO, blkl, blk3, blk4 TS2 = > blkO, blk2 TS3 = > blkO, blkl, blk3, blk4 TS4 = > blkO, blk4 TS5 = > blkO, blkl, blk2, blk5 Application example The above-mentioned exemplary convolution method, the possible combinations of distribution matrices are as follows: Option 1 · Distribute the total data block of the video file by shifting 0TS

—_____________ TS0=>M0,NO 2 TS1=>M0,Ml,M3,NO,Nl,N3 6 TS2=>M0,M2,NO,N2 4 TS3=>M0,Ml,M3,M4,NO,Nl,N3,N4 8 TS4=>M0,M4,NO,N4 a--_____________ TS0 = > M0, NO 2 TS1 = > M0, Ml, M3, NO, Nl, N3 6 TS2 = > M0, M2, NO, N2 4 TS3 = > M0, Ml, M3, M4, NO, Nl, N3, N4 8 TS4 = > M0, M4, NO, N4 a

第22頁 1244869 五、發明說明(18) TS5 = >M0,M1,M2,M5,N0,N1,N2,N5 8 總貢料區塊 選擇2 :以平移ITS分送視訊檔 TSO=>MO,NO,Nl5 N3 4 TS1二〉MO,Ml,M3,NO,N2 5 TS2二〉MO, M2, NO, Nl,N3, N4 6 TS3 = >M0, Ml,M3, M4, NO, N4 6Page 22, 1244869 V. Description of the invention (18) TS5 = > M0, M1, M2, M5, N0, N1, N2, N5 8 Total tributary block selection 2: Distribute video file by translation ITS TSO = > MO, NO, Nl5 N3 4 TS1II> MO, Ml, M3, NO, N2 5 TS2II> MO, M2, NO, Nl, N3, N4 6 TS3 = > M0, Ml, M3, M4, NO, N4 6

T.S4 = >M0, M4, NO, N1,N2, N5 6TS5 = > MO,Ml, M2,M5,NO 5 選擇3 :以平移2 TS分送視訊檔 總資料區塊 TS0二〉MO,NO,N2 3 TS1=>M0,Ml,M3, NO,Nl,N3,N4 7 TS2 = >M0, M2, NO, N4 4 TS3二〉MO, Ml, M3, M4, NO, Nl,N2, N5 8 TS4 = >M0, M4, NO 3T.S4 = > M0, M4, NO, N1, N2, N5 6TS5 = > MO, Ml, M2, M5, NO 5 Option 3: Distribute the video file total data block TS2 by Pan 2 TS , NO, N2 3 TS1 = > M0, Ml, M3, NO, Nl, N3, N4 7 TS2 = > M0, M2, NO, N4 4 TS3 2> MO, Ml, M3, M4, NO, Nl, N2, N5 8 TS4 = > M0, M4, NO 3

TS5=>M0,Ml,M2,M5,NO,Nl,N3 7 選擇4 :以平移3TS分送視訊檔 總資料區塊 TS0二〉MO,NO, Nl, N3,N4 5 TS1二〉MO,Ml,M3, NO, N4 5TS5 = > M0, Ml, M2, M5, NO, Nl, N3 7 Choice 4: Distribute the video file total data block TS0 2> MO, NO, Nl, N3, N4 5 Ml, M3, NO, N4 5

li ί\li ί \

第23頁 1244869 五、發明說明(19) TS2=>M0,M2, NO, N1,N2,N5 6 TS3 = >M0,Ml, M3, M4,NO, 5 TS4 = >M0? M4, NO, N1,N3 5 TS5二〉MO, Ml,M2, M5,NO,Nl, N2 6 選擇5 :以平移4TS分送視訊檔 總資料區塊 3TS1 二〉Page 24 1244869 V. Description of the invention (19) TS2 = > M0, M2, NO, N1, N2, N5 6 TS3 = > M0, Ml, M3, M4, NO, 5 TS4 = > M0? M4, NO, N1, N3 5 TS5 2> MO, Ml, M2, M5, NO, Nl, N2 6 Option 5: Distribute the video file total data block 3TS1 2 by panning 4TS

TS0 = >M0, NO, N4 M.0, Ml, M3, NO, Nl,N2, N5 TS2=>M0,M2,NO, TS3二〉MO, Ml,M3, M4,NO, Nl,N3 TS4=>M0,M4,NO, N2 TS5 = >M0,Ml, M2, M5,NO,Nl,N35 N4 選擇6 :以平移5TS分送視訊檔. 總資料區塊 TS0=>M0,NO, Nl, N2,N5 5 TS1二〉MO,Ml,M3, NO 4TS0 = > M0, NO, N4 M.0, Ml, M3, NO, Nl, N2, N5 TS2 = > M0, M2, NO, TS3 2> MO, Ml, M3, M4, NO, Nl, N3 TS4 = > M0, M4, NO, N2 TS5 = > M0, Ml, M2, M5, NO, Nl, N35 N4 Option 6: Distribute video files by panning 5TS. Total data block TS0 = > M0, NO, Nl, N2, N5 5 TS1> MO, Ml, M3, NO 4

TS2=>M0,M2,NO, Nl,N3 5 TS3=>M0,Ml,M3,M4,NO,N2 6 TS4=>M0,M4,NO,Nl,N3,N4 6 TS5=>M0,Ml,M2,M5,NO,N4 6 應用方針(1),選擇2、4及6具有最少資料區塊之最大TS2 = > M0, M2, NO, Nl, N3 5 TS3 = > M0, Ml, M3, M4, NO, N2 6 TS4 = > M0, M4, NO, Nl, N3, N4 6 TS5 = > M0, Ml, M2, M5, NO, N4 6 Application guideline (1), choose 2, 4 and 6 with the largest data block

第24頁 124486?Page 24 124486?

數(即6資料區塊)在任一時間槽内送出。應用方針(2), j例示^實施例之最佳分送矩陣為選擇4,因選擇4具有任 麥考枯間槽内之資料區塊加上相鄰時間槽之資料區塊 (如16 -資料具有區塊)具有最小之和。因此本實施例之最 佳化係貧料檔N之分送序列應平移3個時間槽。在例示性實 施例中三二,之分送矩陣產生給每一頻道伺服器1 0 4。 曰=每一貝料檔之資料區塊依據一分送矩陣而發送,大 ,之簽約用戶可在一隨機時間存取資料檔且資料檔之 貧料區塊將適時提供給每一簽約用戶。在上述提供之: 中,設一時間槽之長度等於5秒,該D〇D系統1〇〇為資1 及N發送資料區塊係依據最佳分送矩陣(即平移資料押田 分送序列3個時間槽)以下述方式發送: @之 時間00:00:00=>M0N0N1N3N4 時間00:00:05=>M0M1M3N0N4 時間00:00:10=>M0M2N0N1N2N5 時間00:00:15=>M0M1M3M4N0 時間00:00:20=>M0M4N0N1N3 時間00:00:25=>M0M1M2M5N0N2 時間00:00:30=>M0N0N1N3N4 時間00:00:35=>M0M1M3N0N4 時間00:00:40=>M0M2N0N1N2N5 時間00:00:45=>M0M1M3M4N0 時間00:00:50=>M0M4N0N1N3The number (that is, 6 data blocks) is sent in any time slot. Application guideline (2), j exemplified ^ The optimal distribution matrix of the embodiment is selection 4, because selection 4 has a data block in any slot of the micaucus plus a data block in an adjacent time slot (such as 16- The data has blocks) with the smallest sum. Therefore, the optimization sequence of lean material level N in this embodiment should be shifted by 3 time slots. In the exemplary embodiment, a distribution matrix of three or two is generated for each channel server 104. = The data block of each shell material file is sent according to a distribution matrix. Large users can access the data file at a random time and the lean material block of the data file will be provided to each contracted user in a timely manner. In the above provided: In the above, it is assumed that the length of a time slot is equal to 5 seconds, and the DOD system 100 is used to send data 1 and N. The data block is sent according to the best distribution matrix (that is, the panned data distribution sequence 3 Time slots) are sent as follows: @ 的 时间 00: 00: 00 = > M0N0N1N3N4 time 00: 00: 05 = > M0M1M3N0N4 time 00: 00: 10 = > M0M2N0N1N2N5 time 00: 00: 15 = > M0M1M3M4N0 time 00: 00: 20 = > M0M4N0N1N3 time 00: 00: 25 = > M0M1M2M5N0N2 time 00: 00: 30 = > M0N0N1N3N4 time 00: 00: 35 = > M0M1M3N0N4 time 00: 00: 40 = > M0M2N0N1N2N5 time 00: 00: 45 = > M0M1M3M4N0 time 00: 00: 50 = > M0M4N0N1N3

1244869 五、發明說明(21) 時間 0(Κ· 0 0 : 5 5 = >M0M1M2M5N0N2 設於時間00 ··〇〇··〇〇,用戶A選擇電,用户A2Stb 3 0 0接收’儲存,放影及拒絕資料區塊如下: 時間0 0 : 0 0 : 〇〇 = >接收M0 = >放影M0,儲存Μ〇· 時間0 0:0 0:0 5 =〉接收Μ1,Μ3二〉放影Ml儲左Mn… 〇〇:〇〇:1〇=>接收112=>放影心,儲存別,’4{1,们”,们.時間 時間0 0 : 0 0 : 1 5 = >接收M4 = >放影M3,儲在⑽/ · 時間 0 0 : 0 0:2 0=:> 無接收= > 放影M4,儲’1^2,^,^^· 時間〇〇:〇〇:25 = >接收廳= >放影心,儲在心’们,们,们^4· 〇, Ml,M2, M3, M4, M51244869 V. Description of the invention (21) Time 0 (K · 0 0: 5 5 = > M0M1M2M5N0N2 Set at time 00 ·· 〇〇 ·· 〇〇, user A selects electricity, user A2Stb 3 0 0 receives' store, put The shadow and rejection data blocks are as follows: Time 0 0: 0 0: 〇〇 = > Receive M0 = > Play M0, Store Μ0 · Time 0 0: 0 0: 0 5 => Receive Μ1, Μ3 2> Playback M1 and left Mn ... 〇〇: 〇〇: 1〇 = > Receive 112 = > Playback heart, store it, '4 {1, men', men. Time 0 0: 0 0: 1 5 = > Receiving M4 = > Playing M3, stored in ⑽ / · time 0 0: 0 0: 2 0 =: > No receiving = > Playing M4, storing '1 ^ 2, ^, ^^ · Time 〇〇: 〇〇: 25 = > Reception Hall = > Show your heart, store it in your heart 'men, men, men ^ 4 · 〇, Ml, M2, M3, M4, M5

設於時間0 0 ·· 〇 〇 ·· 1 〇,用戶B選擇電 30 0接收,儲存,放影及拒絕資料區塊^下’用戶B之STB 時間 00 :00: 1 〇 = > 接收M〇, M2 = > 放影M〇 時間 0 0 : 0 0 : 1 5 = > 接收M1,M3, M4 = > 故$ :存M〇, M2 MO, Ml, M2, M3, Μ4 心,餘存 儲存Μ〇, 儲存Μ〇, 储存Μ〇, 儲存M〇,Set at time 0 0 ·· 〇〇 ·· 1 〇, user B chooses to call 30 0 to receive, store, play and reject the data block ^ under 'STB time of user B 00: 00: 1 〇 = > receive M 〇, M2 = > Play M〇 time 0 0: 0 0: 1 5 = > receive M1, M3, M4 = > so $: save M〇, M2 MO, M1, M2, M3, M4 heart, I store M0, store M0, store M0, store M0,

Ml,Μ2,·Μ3, M4 Ml,M2, M3, Μ4, Μ5 Μΐ,M2, M3, Μ4, Μ5 Ml,M2, M3, Μ4, Μ5 時間0 0 : 0 0 : 2 〇 = >無接收= > 放影M2, 時間0 0 : 00 : 2 5 = >接收Μ5 = >放影M3, 時間0 0 : 0 0 ·· 3〇 = >無接收= > 放影μ4, 時間00:0 0:3 5 = >無接收= >放影撾5, 設於時間0 0 : 0 0 : 1 5, 用 戶C選擇電M1, M2, M3, M4 M1, M2, M3, M4, M5 M5, M2, M3, M4, M5 M1, M2, M3, M4, M5 time 0 0: 0 0: 2 〇 = > no reception = > Playback M2, time 0 0: 00: 2 5 = > Receive M5 = > Playback M3, time 0 0: 0 0 · 3〇 = > No reception = > Playback μ4, time 00 : 0 0: 3 5 = > no reception = > play Laos 5, set at time 0 0: 0 0: 1 5, user C selects electricity

用戶C之STBUser C's STB

影ΜShadow M

1244869 五、發明說明(22) 3 0 0接收,儲存 放 影及拒絕資料區塊如下 時間0 0 : 0 0·· 15 = >接收NO二〉放影N0,儲存Ν〇· 時間00 ·· 0 0 ·· 2 0二> 接收Ν1,Ν3二〉放影Ν1,儲存Ν 時間0 0 : 0 0 ·· 2 5 = >接收Ν2 = >放影Ν2,儲存㈣,^’^,1^3· 時間 00 ·· 0 0··30 = > 接收Ν4 = > 放影Ν3,儲存Ν〇,’ Ν ’ Ν2, Ν3· 時間 0 0 : 0 0 ·· 3 5 = > 無接收= > 放影Μ,儲存Ν〇, Ν ’ Ν2, Ν3, Ν4· 間 00 ·· 00 ·· 40 = > 接收 Ν5 = > 放影 Ν5,儲存Ν〇 ’ /2, Ν3, Ν4·時 , 1)Ν2^3,Ν4,Ν51244869 V. Description of the invention (22) 3 0 0 Receiving, storing, playing and rejecting the data blocks are as follows: 0 0: 0 0 ·· 15 = > Receiving NO 2> Playing N0, storing Ν〇 · Time 00 ·· 0 0 · 2 0 2 > Receive N1, N3 2> Play N1, store N time 0 0: 0 0 · 2 5 = > Receive N2 = > Play N2, store ㈣, ^ '^, 1 ^ 3 · Time 00 ·· 0 0 ·· 30 = > Receive Ν4 = > Play N3, store Ν〇, 'Ν' Ν2, Ν3 · Time 0 0: 0 0 · 3 5 = > None Receiving = > Play M, store Ν〇, Ν 'Ν2, Ν3, Ν4, 00 ... 00 = 40, > Receive Ν5 = > Play Ν5, store Ν〇' / 2, Ν3, Ν4 Hours, 1) Ν2 ^ 3, Ν4, Ν5

設於時間00:0 0:30,用戶D選擇電景4 300接收,儲存,放影及拒絕資料區塊如y下’:用戶D之STB 時間00:00:3 0 = &gt;接收腳,1^1,们,^ = &gt;放 N〇,NU3.K_:00:35=&gt; 無接收=&gt; N館 J NO, Nl,N3, N4 〜1,儲存 牯間00 · 00 ·· 4〇 = &gt;接收N2, N5 = &gt;放影付 Ν〇,Ν1,Ν2,Ν3,Ν4,Ν5 . 時間0 0 : 0 0 時間00 : 00 ^m〇〇:〇0:i, ^#Ν〇^Ν1^2^3,Ν4,Ν5 -&gt; ”、、接收=&gt; 放影 Ν4,儲存Ν0,ν1,Ν2,Ν3,ν4,ν5 如上述之例子'Ufl〉放影Ν5,儲存Ν〇,Ν1,Ν2,Ν3,Ν4,Ν5 擇並開p β、’且5之用戶可於一隨機時間獨立地選 = 服:提供者之資料槽。上述&quot;接收&quot;之名 流,但在任-已^ ::常接收—連續之由時間槽決定之串 已知之點’該接收之STB僅需某一 1244869 發明說明(23) 已、纟里接收之資粗 , 述”接收π所必[品塊亚儲存其他收到之資料區塊。此為上 正確。因此”接&quot;也許稱之為π非退回(n〇n re]ect)&quot;較 收&quot;較佳可稱為”入Α可謂π除了M4之外皆退回&quot;而π無接 冊钓全部退回”。 由上述例子Κ芒 別是,於至少某一士出可用頻寬未完全被充分利用。特 發生。此閒置時ή日::槽内有「閒置時間」,其中無傳送 選擇4為例,苴中潛在無效率的使用頻寬。以前述之 •。換言之,在:日”:望之時間槽内兩個資料區塊被傳送 個資料區塊之傳“二3有頻寬適於傳送6個資料區塊,4 重,但在資料:;;; =置:雖然在選擇4中不算嚴 使利用最佳化給合通::::料區塊大時即較為嚴重。即 空區塊空間之部分被^置,來組合育料’仍可能有可觀的 此空區塊空問★日隹· 寬,本發明之目的即Α :使用之頻寬’因此而浪費頻 之-實施例在排;能減少閒置時間’因此本發明 少閒置時間之=定後執行另一步驟,在此稱為減 一例示性減少閒置時間之排序矩陣之模型可夫 區塊排序輯來解釋哼上逑 二於傳送資料區塊之閒置時間為簡明起見以「V」表可 TSO-^blkO,&lt; &gt;,〈—〜〉〈—〜〉 TSl=&gt;blkO, blkl,blk3,〈--&gt;Set at time 0: 0 0:30, user D selects the electric scene 4 300 to receive, store, play, and reject the data block as y ': STB time of user D 00: 00: 3 0 = &gt; receiving foot, 1 ^ 1, ^ = &gt; Put No, NU3.K_: 00: 35 = &gt; No Reception = &gt; N Hall J NO, Nl, N3, N4 ~ 1, Stored in room 00 · 00 ·· 4〇 = &gt; Receiving N2, N5 = &gt; Playback payment 〇〇, Ν1, Ν2, Ν3, Ν4, Ν5. Time 0 0: 0 0 Time 00: 00 ^ m〇〇: 〇0: i, ^ # Ν〇 ^ Ν1 ^ 2 ^ 3, Ν4, Ν5-&gt; ", Receive = &gt; Play N4, store Ν0, ν1, Ν2, Ν3, ν4, ν5 As the above example, 'Ufl> Play Ν5, store Ν〇, Ν1, Ν2, Ν3, Ν4, Ν5 select and open p β, 'and 5 users can choose independently at a random time = server: the data slot of the provider. The above-mentioned "receiving" celebrities, but Incumbent-already ^ :: often received-continuous string of known points determined by the time slot 'The received STB requires only a certain 1244869 Invention Description (23) The received information of the received information has been described as "necessary to receive π [Pinya stores other received data blocks. This is correct. Therefore, "receiving" may be referred to as "π non-return (n〇n re) ect)", which is better than "receiving," which can be described as "returning π except M4". Return all. ”From the above example, K. Manbe is that the available bandwidth is not fully utilized in at least one taxi. This event occurs. This idle time :: There is" idle time "in the slot, no transmission option 4 As an example, the potentially inefficient use of bandwidth in 苴. Take the aforementioned •. In other words, two data blocks are transmitted in the time slot: "time", and the data block is transmitted "two of the three bandwidths are suitable for transmitting six data blocks, four times, but in the data: ;;; = Set: Although it is not strict to optimize the utilization in Choice 4: ::: It is more serious when the material block is large. That is, a part of the empty block space is placed to combine the breeding materials. 'There may still be a considerable amount of empty space in this empty block. ★ Sundial · Wide. The purpose of the present invention is Α: the bandwidth used.' -The embodiment is in the row; the idle time can be reduced. Therefore, the present invention reduces the idle time = after performing another step, it is referred to here as a model of an ordering matrix that reduces the idle time to reduce the idle time. Hum on the second, the idle time of the data block is transmitted. For the sake of brevity, the "V" table may be TSO- ^ blkO, &lt; &lt; <-&Gt;

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TS2 =〉blkO,blk2,&lt; — &gt;,&lt; — &gt; TS3 = &gt;blkO, blkl,blk3, blk4 TS“&gt;blkO, blk4, TS5二〉blkO, blkl,blk3, blk5 如圖5以表列形式所示,該排序矩陣明顯的在許多時間 槽内有以閒置時間表示之有未用之頻寬。本發明利用.時 槽與時間槽間之一定頻寬減少此閒置時間。達成經由利用 ^定頻寬減少閒置傳送時間之關鍵是要瞭解^資料區塊之 分送序列必需依附於’在資料區塊被傳送之所在之時間枰 並不切合,除非當時或之前必需被存取時資料區塊必二 Ϊ :據t ’ 一定頻寬之利用係以在每一時間槽之内依據由 來之分送序列傳送—定數量之資料區塊,並盘 排序矩陣指定之時間槽無關。 /、 門在二前面τί述之6個區塊排序矩陣中,有大量之閒置時 間在TS0、TS1、TS2 及TS4 中,盘 γ 與 ν 了 寬對應於傳送# P1擗4· &amp; α為了舉例假設所望之一定頻 塊被排序而傳送。此程序時間槽内4個資料區 之移至&quot;,:〗 次取其次之資料區塊,並將 塊被例子,在tsi,Mk〇中之第一區TS2 => blkO, blk2, &lt; — &gt;, &lt; — &gt; TS3 = &gt; blkO, blkl, blk3, blk4 TS "&gt; blkO, blk4, TS5 2> blkO, blkl, blk3, blk5 Figure 5 As shown in the form of a table, the sorting matrix obviously has unused bandwidth in idle time in many time slots. The present invention uses a certain bandwidth between the time slot and the time slot to reduce this idle time. Achieved The key to reducing idle transmission time by using ^ fixed bandwidth is to understand that the distribution sequence of ^ data blocks must be attached to 'when the data block is transmitted', which is not appropriate unless it must be accessed at the time or before The time data block must be two: According to t 'the use of a certain bandwidth is transmitted within each time slot according to the originating distribution sequence-a certain number of data blocks, and the time slot specified by the sorting matrix is irrelevant. / 、 The gate is in the 6 block sorting matrix described in the previous two, there is a large amount of idle time in TS0, TS1, TS2, and TS4. The widths of the disks γ and ν correspond to the transmission # P1 擗 4 · &amp; α For example, suppose that a certain frequency block is expected to be ordered and transmitted. The four data areas in the slot are moved to &quot;:〗 The next data block is taken, and the block is taken as an example. The first area in tsi, Mk〇

塊被移至TS0。在TSl,blko中夕4 一广A 後,因為TS0仍有一空白資料^ 一區塊亦被移前。然 移,TS0至肤拮、生 士、’區塊空間,由TS1之bl k3亦前 夕TS0至此填滿了所有空間,現在看起來有如:The block is moved to TS0. After TS1, blko 4 and 1 A, because TS0 still has a blank data ^ one block has also been moved forward. Of course, TS0 to the skin, faculty, and ‘block space’, bl k3 from TS1, and TS0 now filled all the space, it now looks like:

1244«份1244 «servings

TSO = &gt;blkO, blkO , blkl, blk3 現在TS1及大部分之TS2係空白,故自TS3之資料區塊 被移前。一旦此程序已完成,該矩陣看起來有如: TS0二〉blkO, blkO, blkl,blk3 TS1二〉blkO, blk2, blkO, blkl, TS2=&gt;blk3,blk4,blkO,blk4 TS3 = &gt;blkO, blkl,blk2,blk5 TS4 = &gt;TSO = &gt; blkO, blkO, blkl, blk3 Now TS1 and most of TS2 are blank, so the data block from TS3 is moved forward. Once this procedure has been completed, the matrix looks like: TS0 2> blkO, blkO, blkl, blk3 TS1 2> blkO, blk2, blkO, blkl, TS2 = &gt; blk3, blk4, blkO, blk4 TS3 = &gt; blkO, blkl, blk2, blk5 TS4 = &gt;

TS5二〉 , 此亦顯示於圖6之表格中。空白及不全之時間 本例中之TS4及TS5,很簡單的以重複原來之輕’如 ^3被右 滿,而仍充滿該時間槽。因此第一 6個時間槽形如· 凡 TSO = &gt;blkO, blkO, blkl, blk3 TSl=&gt;blkO, blk2,blkO,blkl, TS2-&gt;blk3, blk4,blkO,blk4 TS3 = &gt;blkO, blkl,blk2,blk5 TS4=&gt;blkO,blkO,blkl,blk3 TS5 = &gt;blkO, blk2,blkO,blkl 其次在此程序中之兩個時間槽以^及TS7,含有TS2 及TS5 2>, this is also shown in the table in Figure 6. Blank and incomplete time TS4 and TS5 in this example are very simple to fill the time slot by repeating the original light weight such as ^ 3. So the first 6 time slots are as follows: Where TSO = &gt; blkO, blkO, blkl, blk3 TSl = &gt; blkO, blk2, blkO, blkl, TS2- &gt; blk3, blk4, blkO, blk4 TS3 = &gt; blkO , blkl, blk2, blk5 TS4 = &gt; blkO, blkO, blkl, blk3 TS5 = &gt; blkO, blk2, blkO, blkl The two time slots in this program are ^ and TS7, which contain TS2 and

12448691244869

五、發明說明(26) TS3同樣的資料p抬 較短的排序矩陣。中▲此此耘序只際產生者,於-新的 示由充滿閒置圖7以表列顯 知,如果昭著肩的重稷矩陣。由上述之例子明顯可 分送相反的用广見在可以原排序矩陣即時 之中間槽並當收至,丨r 。一用戶甚至可進入系統 田 】起始區塊b 1 k 0時開始使用資料。 資料區塊變成id:大部分變成可計算之短篇,當 系統並開始接收資m?在該串流中一用戶可跳進此 ή單之步驟51〇,在4牛驟由41圖/可知’此額外之步驟為一較 在步驟41 〇之後執行。 一定值間㊁料ξι—10;前述敘述與所選頻寬被設定為 等於資料區塊之整::之整數個。然@,此-定頻寬不必 8所發展出來的欠^個、。而以簡單的條件取代,即附於圖 列產生之資料串=\\送序列。於圖8所發展出之分送序 硬體設備(如广^後棱供給一控制數位資料廣播之低階 m 1路卡或頻道伺服器)。 定給檔案料區塊’低階硬體設備不如在指 人不必擔心資所瞭解’在分送序列之抽取階段,吾 且低階硬體;送。而是,分送矩陣提供序列, 該指定大小之部分可充分利用。-旦 檔,直到頻寬再度可用:低階设備將暫停廣播此特定資料V. Description of the invention (26) The same data p for TS3 is a shorter sorting matrix. The middle and the bottom of this sequence are only the producers, and the Yu-Xin display is shown in the table of Figure 7 full of idleness. From the above example, it is obvious that the opposite can be sent to the middle slot of the original sort matrix in real time and when received, r. A user can even enter the system.] The starting block b 1 k 0 starts to use the data. The data block becomes id: most of it becomes a computable short story. When the system starts to receive data m? In the stream, a user can jump to step 51 of this price list, which is shown in Figure 4/4 in 4 steps This additional step is performed after step 41. A certain value between ξι-10; the foregoing description and the selected bandwidth are set to be equal to the whole number of data blocks :: an integer number. Of course, @ , 此-fixed bandwidth does not need to be developed due to ^^. Instead, it is replaced by a simple condition, that is, the data string attached to the chart = \\ send sequence. The distribution sequence hardware equipment developed in Figure 8 (such as the low-order m 1 card or channel server that provides a digital data broadcast control for wide-angle rear). Given the low-level hardware equipment of the archival material block, it is better to refer to the low-level hardware than the low-level hardware in the extraction phase of the distribution sequence. Instead, the distribution matrix provides the sequence, and the specified size portion can be fully utilized. -Once the bandwidth is available again: low-end devices will temporarily suspend broadcasting this particular data

第31頁 1244869 五、發明說明(27) ^----- ^ 於上述例子,在原排序矩陣中用到的最大頻寬係用於 該減少閒置時間矩陣。此係無論在那一點,—用戶開始接 收資料,最大等待間間並未改變。然而,對於熟習此技藝 者將可明白,頻寬可調整於時間成本或利潤間之一。重要 的疋’對於一隨選資料服務而言,所需最多的時間不會超 過執行該檔之時間。若此事發生,結果是排序矩陣之總時 間槽數大於原排序矩陣包含之時間槽數。若此為2小時之 電影,例如,也許需3小時來放影,將空檔留在電影的中 間。但有些應用’也許可利用甚至期望此功能,如資料串 流可以下載而不馬上使用。 單一資料串流具有一定、最佳化之頻寬可簡單地結合 在一起。其結果將是一群串.流頻寬,其總寬為該群單一串 流之和。 如所示,減少閒置時間為一單一資料串流計算排序矩 陣是很有用的。以此方式,當傳送一資料串流時一指定數 量之頻寬可被充分使用。然而如上述,本發明之一觀點為 產生一三維分送矩陣。用一三維分送矩陣,一減少閒置時 間之分送矩陣亦可在同樣的方式計算並建立。然而一旦以 減少閒置時間之分送矩陣用於單一串流,然後許多充分最 佳化之單一串流被結合在一起,在許多情形下此將傾向於 過度操作該三維矩陣。 罔此所具體展現的是以電腦為工具之方法以傳送一隨 選資料檔包含一動作,準備一分送矩陣規定一重複資料傳 送序列以利經由一種媒介廣播給在非特定方式之一群用Page 31 1244869 V. Description of the invention (27) ^ ----- ^ In the above example, the maximum bandwidth used in the original sorting matrix is used to reduce the idle time matrix. No matter at this point, the user starts to receive data, and the maximum waiting time has not changed. However, it will be apparent to those skilled in the art that bandwidth can be adjusted between time cost or profit. The important thing is that for an on-demand data service, the maximum time required will not exceed the time required to execute the file. If this happens, the result is that the total number of time slots in the sorting matrix is greater than the number of time slots contained in the original sorting matrix. If this is a 2 hour movie, for example, it may take 3 hours to show the movie, leaving the space in the middle of the movie. However, some applications ’may allow or even expect this feature, such as data streaming can be downloaded without immediate use. Single data streams with a certain, optimized bandwidth can simply be combined. The result will be a group of stream.stream bandwidths whose total width is the sum of the individual streams of the group. As shown, it is useful to reduce the idle time to calculate the sorting matrix for a single data stream. In this way, a specified amount of bandwidth can be fully utilized when transmitting a data stream. As mentioned above, however, it is an aspect of the present invention to generate a three-dimensional distribution matrix. With a three-dimensional distribution matrix, a distribution matrix that reduces idle time can also be calculated and established in the same way. However, once the distribution matrix is used to reduce the idle time for a single stream, and then many fully optimized single streams are combined, in many cases this will tend to over-manipulate the three-dimensional matrix. What is specifically shown here is the use of a computer as a tool to transmit an on-demand data file containing an action, preparing a distribution matrix to specify a repeating data transmission sequence for broadcasting via a medium to a group in an unspecified way.

1244869 五、發明說明(28) 戶。此準備分送矩陣之動作進_米 資料區塊,具有至少一第一區塊^包含將—資料檔縮減成 複之資料傳送序列。因此一使用^將貧料區塊排列為該重 送序列並在第一區塊收到之同時可接收此重複之資料傳 此資料檔。此重複資料傳送序列♦ I十斷之方式開始使用 最少之閒置時間來傳送重複資料=預設之頻寬,並且以 傳送需要一與用戶數無關之送序列。且隨選資料之 圖9以流程圖形式總結如何^ 序。首先於5 20,一原始排序矩陳^ =閒置時間之排序 肩始排序矩陣依據本發明之簡單之'&quot;、貝料播而產生。此 序矩陣。為清楚之目的而用「ϋ非減少閒置時間之排序 之重點不在矩陣之結構本身,而。二序;陣 陣中因“成為最重要。少間置時間之排序矩 5〇〇 Λ /此原始矩陣因而視為排序序列 時門Λ 可用以在原始矩陣中加入閒置 m旦頻寬已指定,資料區塊自矩陣被 直至所有閒置時間槽被移除。於55〇,智慧地,若切、 序列,閒置時間槽被消除。-旦此步驟完成二 :始排序矩陣之序列被重複。終止發生於有了新的重複序 ,此I視為一新的減少閒置時間排序矩陣56〇。 偉得注意的是,排序序列與減少閒置時間之排序序列 二t呈異是最主要的分辨。吾人可描繪為一線形順序而另 —為重複的矩陣,但結果是一使用者接收一穩定的串流而1244869 V. Description of invention (28) households. The action of preparing the distribution matrix is a data block, which has at least one first block ^ including a data transmission sequence that reduces the data file to a plurality. Therefore, one can use ^ to arrange the lean material block into the retransmission sequence and receive the duplicate data transfer to the data file while receiving the first block. This repeating data transmission sequence ♦ I start in ten-minute mode to transmit repeating data with the least idle time = default bandwidth, and the transmission requires a transmission sequence that is independent of the number of users. And Figure 9 on-demand information summarizes the sequence in the form of a flowchart. First at 5 20, an original sorting moment Chen ^ = idle time sorting. The shoulder-to-sequence sorting matrix is generated according to the simplicity of the present invention. This ordered matrix. For the purpose of clarity, "the priority of ordering instead of reducing idle time is not in the structure of the matrix itself, but in the second order; because of arrays," it becomes the most important. The sorting moment of the small idle time is 500 Λ / This original matrix is therefore regarded as a sort sequence. The gate Λ can be used to add idle m denier bandwidth to the original matrix. The data block has been specified from the matrix to all idle time slots. Removed. At 55 °, intelligently, if the cut, sequence, idle time slot is eliminated. -Once this step is completed 2: the sequence of the starting sort matrix is repeated. Termination occurs when there is a new repetitive sequence, this I is regarded as a new reduced idle time sorting matrix 56. It is important to note that the difference between the sorting sequence and the sorting sequence that reduces idle time is the main difference. We can be depicted as a linear sequence and the other-a repeated matrix, but the result is that a user receives a stable stream and

第33頁 K44869Page 33 K44869

五、發明說明(29) 非用戶特定之資料區塊 一種進一步減少D 〇 D資料區塊廣播排序矩陣所务 之方法係在一指定之頻道中不斷廣播所出現的資料%寬 自下列之例示性矩陣中可知,選定之資料區塊較在^、、$。 之其他資料區塊較常出現。此出現閒置時之原始分这Γ内 表示如下: TSO = &gt;blkO, TS1二〉blkO, blkl,blk3,&lt;--&gt;, T’S2 = &gt;blkO, blk2,〈--〉,〈--〉 TS3 = &gt;blkO, blkl, blk3, blk4 TS4 = &gt;blkO, blk4, blkO, blkl, blk2,blk5 由圖9之程序產生之具有減少閒置時間之分送矩卩車表 示如下 : TSO = &gt;blkO, blkO, blkl, blk3 TSl=&gt;blkO, blk2, blkO, blkl TS2 = &gt;blk3, blk4,blkO,blk4 TS3 = &gt;blkO, blkl,blk2,blk5 以線性重複資料串流代表之例示性減少閒置時間矩 係如下所示: 干V. Description of the invention (29) Non-user-specific data blocks A method to further reduce the data ordering matrix of broadcast data blocks is to continuously broadcast the data appearing in a specified channel. It can be seen from the matrix that the selected data blocks are more than ^, and $. Other data blocks appear more often. The original points in the idle when this occurs are represented within Γ as follows: TSO = &gt; blkO, TS1 2> blkO, blkl, blk3, &lt;-&gt;, T'S2 = &gt; blkO, blk2, <->, <-> TS3 = &gt; blkO, blkl, blk3, blk4 TS4 = &gt; blkO, blk4, blkO, blkl, blk2, blk5 The distribution moment car produced by the program of Fig. 9 with reduced idle time is expressed as follows: TSO = &gt; blkO, blkO, blkl, blk3 TSl = &gt; blkO, blk2, blkO, blkl TS2 = &gt; blk3, blk4, blkO, blk4 TS3 = &gt; blkO, blkl, blk2, blk5 Repeat the data stream linearly A representative example of the reduction in idle time is as follows:

第34頁 '12448的一一- 五、發明說明(30) streamO-blkO, blkO, blkl, blk3,blk0,blk2,blk0,blkl, blk3,blk4,blk0,blk4, blk0,blkl,blk2,blk5 自此例示性分送矩陣之線性表示法可見,b 1 kO比區塊 1 一 5更常傳送。在一所給之資料區塊内不同之分送矩陣會 導致不同之或多或少更常出現之資料區塊。在任何在序列 中更早出現之分送矩陣之資料區塊將更常出現,以區塊〇 最常出現。 因為b 1 k 0較其他資料區塊在資料串流中更常出現,在 一獨立之一較.密之傳送頻寬可以由資料串流中送出b丨k 〇及 其他常出現之區塊而獲得。此串流較原始串流可有大為減 少之頻寬,且指定給原始串流之頻寬亦可能以僅傳送較少 出現之資料區塊而減少。例如’若上述串流被拆為兩個分 開的串流,一預取串流僅帶著資料b 1 k 0而一原始串流帶著 餘下的貧料區塊’这兩個重複的串流會以下述形式出現·· primary Stream=&gt;blkl,blk3,blk2,blkl,blk3,blk4, blk4, blkl,blk2,blk5 因此在一獨立之資料串流傳送b 1 k 0,傳送原始資料串 流所窝之頻寬減少3 7 · 5 %。此減少量後因七1 k 0包含原始資 料串流中全部1 6個資料區塊中之6個,致有效移除b 1 k 〇而 減少6個在原始資料串流傳送的資料區塊。所需以傳送包Page 34-12 of 12448-Description of the invention (30) streamO-blkO, blkO, blkl, blk3, blk0, blk2, blk0, blkl, blk3, blk4, blk0, blk4, blk0, blkl, blk2, blk5 from The linear representation of this exemplary distribution matrix shows that b 1 kO is transmitted more often than blocks 1 to 5. Different distribution matrices within a given data block will result in different more or less frequently occurring data blocks. The data blocks of any distribution matrix that appear earlier in the sequence will appear more often, with block 0 appearing most often. Because b 1 k 0 appears more frequently in the data stream than other data blocks, a more independent transmission bandwidth can send b 丨 k 〇 and other frequently-occurring blocks from the data stream. obtain. This stream may have significantly reduced bandwidth compared to the original stream, and the bandwidth assigned to the original stream may also be reduced by transmitting only less frequently occurring data blocks. For example, "If the above stream is split into two separate streams, a pre-fetch stream only carries data b 1 k 0 and an original stream with the remaining lean blocks" these two repeated streams Will appear in the following form: primary stream = &gt; blkl, blk3, blk2, blkl, blk3, blk4, blk4, blkl, blk2, blk5 So send b 1 k 0 in a separate data stream, and send the original data stream Reduced nest bandwidth by 37.5%. After this reduction, 7 1 k 0 contains 6 of all 16 data blocks in the original data stream, which effectively removes b 1 k 0 and reduces 6 data blocks transmitted in the original data stream. Required to transmit packets

第35頁 1244869 五、發明說明(31) 含b 1 k 0之預取資料串流所增加之賴宫 由STB執行之缓衝量而定。此緩衝量以巳來是少的,係依 節之討論。 隻衝里 '下將以圖作更細 此傳送=法之明顯優點為減少存取—選定之隨選資料 服務所需之日守間。一旦一選定之])⑽服欠 _ &lt;υυΐ)服務之blkO被接收, 使用者即可開始使用所選之服務,然而若無預取串流,使 用者必須等到bl kO在原始資料串流中出現。因Mk〇在一獨 立串流上被傳送,一服務不必等候次—Mk〇自—包含許多 不同之資料區域而被連續傳送。當然,為了流暢地利用隨 選資料系統’ blkl在MkO完成之時必須被接收且立即可 用,此需要原始資料串流有大量足夠的頻寬以保證在b丨k 〇 已結束之剷接收b 1 k 1。圖1 0顯示依據本發明之一實施例之 程序60 0 ’供排序D0D資料區塊,以傳送一原始資料串流及 一預取資料串流。於步驟6 0 2,一減少閒置時間排序被以 線性排序資料區塊依要傳送之資料區塊之序列安排之。於 步驟6 0 4,最常出現之資料區塊自序列中移出,餘下較短 之序列資料區塊需要較窄之頻寬。如前討論者,b 1 k 0常常 最常出現。於步驟6 0 6,自減少閒置時間序列移除之資料 區塊被置於包含全部之b 1 k 0資料區塊之預取資料串流。然 後於步驟6 0 8,該截短之資料區塊序列被置於原始資料串 流中。然後於步驟6 1 0,該原始資料串流及該預取資料串 流經由傳送媒介11 0 (圖1 B )在不同之頻寬傳送兩個不同 之重複資料串流給接收之機頂盒。 圖11顯示一依一實施例之程序6 5 0,供排序D0D資料區Page 35 1244869 V. Description of the invention (31) The increase in the pre-fetched data stream containing b 1 k 0 depends on the amount of buffer performed by the STB. This amount of buffering is small, which is discussed in detail. The only advantage of the "Only Reaching" is to make the map more detailed. The obvious advantage of this transmission = method is to reduce access-the selected on-demand data service. Once a selected)) blkO service is received, the user can start using the selected service, but if there is no prefetch stream, the user must wait until bl kO is streaming in the original data Appears. Since Mk0 is transmitted on a separate stream, a service does not have to wait for time-Mk0-contains many different data areas to be transmitted continuously. Of course, in order to smoothly use the on-demand data system 'blkl must be received and available immediately when MkO is completed, this requires that the raw data stream has a sufficient amount of bandwidth to ensure that b 1 is received at the end of b 丨 k 〇 k 1. FIG. 10 shows a program 60 0 ′ for sorting a DOD data block to transmit an original data stream and a pre-fetched data stream according to an embodiment of the present invention. In step 602, a sort of reducing idle time is arranged in a linear order of data blocks according to the sequence of data blocks to be transmitted. At step 604, the most frequently occurring data blocks are removed from the sequence, and the remaining shorter sequence data blocks require a narrower bandwidth. As previously discussed, b 1 k 0 often occurs most often. At step 606, the data blocks removed from the reduced idle time series are placed in a pre-fetched data stream containing all b 1 k 0 data blocks. Then in step 608, the truncated data block sequence is placed in the original data stream. Then in step 6 10, the original data stream and the pre-fetched data stream send two different repeated data streams to the receiving set-top box at different bandwidths via the transmission medium 11 0 (Fig. 1B). FIG. 11 shows a procedure 6 50 according to an embodiment for sorting the D0D data area.

第36頁 1244869 五、發明說明(32) 塊以傳送一原始資料串、、ώ 一 DOD分送矩陣之許多D〇]) :取】料串流。*中來自- 傳送因而極小化傳送办°鬼,在—預取資料串流上被 減少間置時間之排序係:安;。料=-:驟652,- 資料區塊之線性序列# -排成貝枓區塊要傳送之順序之 fώ — 表不。於步驟654,最當出頰之眘枓 區塊,自序列中移出, 取㊉出現之貝科 需求較窄之頻寬。於步驟較知之原始資料區塊序列而 之資料區塊被置於—預取次自減〉、閒置時間序列移出 凡低直Α 預取賁料串流中。 於步驟6 58,做成一決仝·你… 始串流之頻甯Η 、疋·傳达較短之資料區塊之原 需::ΐί 縮減至一預定之門檀值。㈣驟可能 若原始資料串流所需:f :此技藝之人士所明白的。 驟660。於牛驄RRn :見在門板值以下,則程序繼續到步 由傳送媒、入H岡1’该原始資料串流及該預取資料串流經 介11G(圖1B)在不同之頻寬以兩個不同之重複資 枓串流傳送給接收之機頂盒。 貝 由孤右所而頻寬在門檻值以上,則程序回到步驟6 5 4,其 +驟之貪料區塊之最常出5見之資料區塊被移*。然後在 ’、移出之資料區塊被加入預取資料串流。程序繼 Ά 行重衩直至所需傳送頻寬於步驟6 5 8已判定低於門檻 值’此時程序停止於步驟66〇。在另一實施例中,其他^ 準Τ用以取代門檻值標準,如極小化原始及預取資料串流 一 ^所需之頻寬之總和。以此方式,所需傳送D〇D服務之 頻寬較黃奎(Khoi Hoang)於20 0 1年6月25日所申請之美國 申明案號9/982,017·名為11 Decreased Idle TimePage 36 1244869 V. Description of the invention (32) block to transmit a raw data string, a lot of DOD distribution matrix D]]: take] material stream. * The medium from-transmission is therefore minimized. The transmission is a ghost, and the ordering system is reduced in the -pre-fetch data stream: Ann ;. Material =-: Step 652,-Linear sequence of data blocks #-Arranged in the order in which the blocks are to be transmitted — Nothing. At step 654, the most cautious cautious block is removed from the sequence to obtain the narrow bandwidth required by the Beco that appears. The data block sequence of the original data block sequence that is better known in the step is placed in the prefetch time self-decreasing, and the idle time series is moved out of the Fan Zhi A prefetching material stream. At step 6 58, a match is made. You ... The frequency of starting the stream is better, and the original requirement of transmitting a shorter block of data is: ΐ ΐ Reduced to a predetermined threshold value. Steps may be required if the source data is streamed: f: as understood by those skilled in the art. Step 660. Yu Niu RRn: See below the door value, then the procedure continues to the transmission medium, the original data stream and the pre-fetched data stream through Hgang 1 'through the 11G (Figure 1B) in different bandwidths. Two different duplicate data streams are transmitted to the receiving set-top box. If the bandwidth is above the threshold, the program returns to step 6 5 4 and the most frequently seen data block of the + greedy block is moved *. Then, the removed data block is added to the prefetched data stream. The program repeats until the required transmission bandwidth is lower than the threshold value determined in step 6 5 8 ', and the program stops at step 66. In another embodiment, other thresholds are used instead of threshold standards, such as minimizing the sum of the bandwidth required for the original and pre-fetched data streams. In this way, the bandwidth required to transmit the DOD service is higher than the US Declaration No. 9/982, 017, named 11 Decreased Idle Time, which Khoi Hoang applied for on June 25, 2001.

第37頁 1244869 五、發明說明(33)Page 37 1244869 V. Description of the invention (33)

Constant Bandwidth Data-〇n~Demand Broadcast Delivery Matrices11 (中華民國9〇年9月21日申請案案號 9 0 1 2 3 2 8 3號)之方法更進一步減少。 自原始資料串流移除常出現之資料區塊具有大為減少 包含於原始資料串流中之資料區塊總數之效果,因而減少 傳达原始育料串流所需之頻寬。下表描述自原始資料串流 移除選定之資料區塊所節省之樣本頻寬: 資料區塊一 C預戴入〕 資料氐塊中預估之出現頻寬 C於6個區塊序列中〕^ 預佔節省之頻寬 BlkO., 丄0〇心 Blkl^ 〜5 09^ ^SBW- Blk2- -3396*, Blk3^ ~25心 25BW^ Blk4v- R 1 h-c, ^ 〜16知 表一:於6區塊序列中為特殊之預載入資料區塊節省頻寬 為了立刻存取以上述方法傳送之D〇D服務,接收之 Q T R v / 笔 須預载入並儲存一部或全部包含於預取資料串流中 之二料。因為該STB必須先載入並儲存任一D0D服務要顯示 之blkO ’在節省傳送頻寬與存取延遲時間之間有所取 捨以預載入該預取資料串流並儲存該預取資料區塊,用 可用極少延遲存取所選D0D服務。一接收之STB需 要有足夠之記憶儲存包含於預取資料串流之資料。 各圖1 2顯示依據本發明之一實施例之預載入程序之機頂 A 00 °於步驟7 02,機頂盒被置於一間置或被動模式。典The method of Constant Bandwidth Data-On ~ Demand Broadcast Delivery Matrices11 (Application No. 9 0 1 2 3 2 8 3 of the Republic of China on September 21, 1990) has been further reduced. Removing frequently occurring data blocks from the original data stream has the effect of greatly reducing the total number of data blocks contained in the original data stream, thus reducing the bandwidth required to communicate the original breeding stream. The following table describes the sample bandwidth saved by removing selected data blocks from the original data stream: Data block 1 C pre-worn in] Data block estimated occurrence frequency C in 6 block sequences] ^ Preserved bandwidth BlkO., 丄 0〇 heart Blkl ^ ~ 5 09 ^ ^ SBW- Blk2- -3396 *, Blk3 ^ ~ 25 heart 25BW ^ Blk4v- R 1 hc, ^ ~ 16 Know Table 1: 6 Block sequence saves bandwidth for special pre-loaded data blocks. In order to immediately access the DOD service transmitted by the above method, the received QTR v / pen must be pre-loaded and stored one or all of which are included in the pre-load. Take the second one from the data stream. Because the STB must first load and store the blkO to be displayed by any D0D service. There is a trade-off between saving transmission bandwidth and access delay time to preload the prefetched data stream and store the prefetched data area Block, with selected delays to access the selected DOD service. A received STB needs to have enough memory to store the data contained in the prefetched data stream. Each of Figures 12 shows a set-top A 00 of a pre-loaded program according to an embodiment of the present invention. At step 702, the set-top box is placed in an interposed or passive mode. Canon

Ϊ244869 五、發明說明(34) 型地此模式應係所有用戶之機頂盒之預設(default)模 式。於步驟7 04,機頂盒接收在指定頻寬上之預取資料串 流。在一例示性實施例,預取資料串流自電子節目指引頻 道(programming guide channel)上被接收。於步驟 7〇6,、 機頂盒判定收到之預取資料區塊是否較早已儲存之資料區 塊為新。一在預取資料區塊内之識別碼指示機頂盒之資 區塊有多新。若收到之資料區塊較新,機頂盒即儲最德 的預取資料區塊於一内部記憶3〇8中(圖3),此為步驟 708,典型地係一硬碟驅動之磁性儲存設備。典型地 之^料區塊為較新之預取資料區塊所蓋(〇verwitte 然舊版因各種理由也許仍予保留。 ^ # ί ί Γ# V ^^11 ^ ^ ^ ^ ^ ^ ^ — H成貝孤接收柃自動切換到主動模式。為卜卜一 貫施例,此機頂盒留在主動模式 間 3一 到被動模式。 了1之後它會回 於步驟712,該機頂盒接收到一Ϊ244869 V. Description of invention (34) This mode should be the default mode for all users' set-top boxes. At step 704, the set-top box receives a pre-fetched data stream at a specified bandwidth. In an exemplary embodiment, the pre-fetched data stream is received from an electronic programming guide channel. In step 706, the set-top box determines whether the pre-fetched data block received is earlier than the previously stored data block is new. An identification code in the prefetched data block indicates how new the set-top box's asset block is. If the received data block is newer, the set-top box stores the most pre-fetched data block in an internal memory 308 (Figure 3). This is step 708, which is typically a hard disk drive magnetic storage device. . Typically, the ^ data block is covered by the newer prefetched data block (〇verwitte, although the old version may still be retained for various reasons. ^ # Ί ί Γ # V ^^ 11 ^ ^ ^ ^ ^ ^ — H Chengbei ’s receiver automatically switches to the active mode. As a consistent example, this set-top box stays in the active mode between 3 and passive mode. After 1 it returns to step 712, the set-top box receives a

服務。此時由一使用者鈐Α ρ 7去放〜所述之D0D 自-電子二相應於所選_服務之碼或 自電子即目扣引服務内之節目單 驟m機頂盒顯示自儲存於步驟7 於步 D0D服務之第-資料區塊(b_。於牛驟塊所選 至適當之頻道並接收、儲存 _=6該機頂盒調譜 流於記憶3。8(圖3)。在一例示性趣:服矛力之原始資料串 714平行發生到使用者停止 Λ t 1,4 乂驟與步驟 颂不所4D0D服務為止。於步驟 124486, 五、發明說明(35) 718 ’若依所選D〇D服務之所有預取資料區塊,已經播放, 程序繼續到步驟72〇。於步驟72〇,自先前於步驟716接收 之原始串流資料區塊之D0D服務之其餘部份被播放。此步 驟許當依據所選DOD服務之原始資料區塊之串流自一伺 服器下載時所儲存之預取資料區塊2D0D服務之第一資料 區塊(blkO)或前幾個D〇D服務之資料區塊被播放。此系統 允T無接缝之D0D服務之觀看而具極小之存取時間延遲且 ^ Y頻寬之需求。為極佳化功能起見,足夠的預取資料區 ^必需被儲存及播放以允許由原始資料串流之資料區塊及 供使用。此時可能需要增加預取串流之頻寬或原 始貝科串流之頻寬。 經尚=置入預取資料串流且最終載入閒置之機頂盒之資料 t 天、一週或一月的運作後會改變以反映使用者的直 一 ί ί生一最大的頻寬節省。例如,在秋天的週一晚上了 殊,運動事件會在預取資料串流中較為流行,反之在 最务的Ϊ新的家庭特徵的電影發行會在預取資料串流中在 取先的序列中被送出。 在 遲,盒使用者可於任何時間開始所望的節目而無延 stb中此預序列#此盒在閒置模式時已經儲存在 反映使用者對節目作預先改t (例 ; 序列之更ί )或基約之D0D使用者要求預置 雖然本發明已以較佳實施例揭露如上,然其並非用以service. At this time, it is put by a user 钤 Α ρ 7 ~ The D0D self-electronic two corresponding to the selected service code or the electronic program is deducted from the program list. The set-top box displays the self-storage in step 7. In step D0D service, the -data block (b_. Select the appropriate channel in Niu Su block and receive and store _ = 6. The set-top box tunes the stream to memory 3.8 (Figure 3). An example of sexual interest: The raw data string 714 of the spear force occurs in parallel until the user stops the Λ t 1,4 steps and the step 4D0D service. At step 124486, V. Description of the invention (35) 718 'If the selected DOD All pre-fetched data blocks of the service have been played, and the process continues to step 72. At step 72, the rest of the DOD service from the original streaming data block previously received at step 716 is played. This step allows Prefetched data block stored in the source data block of the selected DOD service when downloaded from a server. The first data block (blkO) of the 2D0D service or the data area of the first several DOD services The block is played. This system allows the viewing of seamless D0D services with minimal access The requirement of time delay and ^ Y bandwidth. In order to optimize the function, sufficient prefetched data area ^ must be stored and played to allow data blocks and for streaming from the original data. At this time may need to increase The bandwidth of the pre-fetch stream or the original Beco stream. Survival = The data of the pre-fetched data stream that is eventually loaded into the idle set-top box will be changed to reflect the operation after a day, week or month. The user ’s savings are the biggest bandwidth savings. For example, in the fall on Monday night, sports events will be more popular in the pre-fetch data stream, but the most important thing is the new family characteristics. The movie release will be sent in the pre-fetched sequence in the pre-fetched data stream. At the end, the box user can start the desired program at any time without delay. Stb This pre-sequence # This box is already stored in idle mode In response to the user's prior modification of the program (for example, the sequence is more) or the user's request for presetting, although the present invention has been disclosed above in a preferred embodiment, it is not intended to

IHIH

第40頁 1244869Page 1244 44869

限制本發明,任 和範圍内,當可 範圍當視後附之 何熟習此技藝者 作些許之更動與 申請專利範圍所 在不脫離本發明 調節,因此本發明 界定者為準。 之精神 之保護Limiting the present invention, within the scope and scope, when the scope is attached, depending on what is attached to the person skilled in the art to make a few changes and the scope of patent application does not depart from the adjustment of the present invention, so the definition of the present invention shall prevail. Protection of spirit

附錄A 減少 閒置時間及定量頻 寬之隨選資料廣播分送矩陣 下述係一步一步敘述顯示於圖4之例示性程序 排序矩陣給具有6個資料區塊之資料檔: 開始 設資料區塊 (步驟4 0 2 )接收一群資料區塊給一資料槽(χ ) 數為 6 ( X = 6 )。 (步驟404)置j = 〇。 (步驟4 06)清除一參考陣列(ra)。 (步驟4 0 8 ) j與X比較。 (步驟 412)j 少於 χ(〇&lt;6)使 i = 〇。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(〇&lt;6),於⑽讀取位於行[0]之矩陣並寫 入RA ;開始時,該SM為空的故無資料寫入RA。(步驟42〇) 是否含有資料區塊i *blk〇? (步驟422)RA不含有任何資料因其係空的。在SM及RA中寫Appendix A The On-Demand Data Broadcast Distribution Matrix to Reduce Idle Time and Quantitative Bandwidth The following is a step-by-step description of the exemplary program ranking matrix shown in Figure 4 for a data file with 6 data blocks: Start setting data blocks ( Step 4 0 2) Receive a group of data blocks to a data slot (χ) with a number of 6 (X = 6). (Step 404) Set j = 0. (Step 4 06) Clear a reference array (ra). (Step 4 0 8) j is compared with X. (Step 412) j is less than χ (0 &lt; 6) such that i = 〇. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (0 &lt; 6), and the matrix at row [0] is read and written into RA; at the beginning, the SM is empty and no data is written into RA. (Step 42) Does it contain data block i * blk〇? (Step 422) RA does not contain any data because it is empty. Write in SM and RA

1 1 I 1! I Ϊ 1 II I I 麗 1 I 1 II 第41頁 1244869 五、發明說明(37) 入blkO於位置[〇, 〇 ]。 (步驟424)加1於i(i = l)以導入位置[1,〇]之值,回到步驟 414 〇 (步驟414)i與X比較。 (步驟418)i少於χ(1&lt;6),於SM讀取位於行[1]之矩陣並寫 入RA ;開始時,該SM為空的故無資料寫入RA。 (步驟420 )RA是否含有資料區塊i或blkl? (步驟422)RA不含有blkl。在SM及RA中寫入blkl於位置 [1,0] ° Γ步驟4 24)加1於i(i = 2)以導入位置[2, 〇]之值,回到步驟 414 〇 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(2&lt;6),於SM讀取位於行[2]之矩陣並寫 入RA ;開始時,該SM為空的故無資料寫入RA。 (步驟4 20 ) RA是否含有資料區塊i或bl k2? (步驟422)RA不含有blk2。在SM及RA中寫入blk2於位置 [2,0] ° (步驟424)加1於i( i =3)以導入位置[3, 〇]之值,回到步驟 414 〇 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(3&lt;6),於SM讀取位於行[3]之矩陣並寫 入RA ;開始時,該SM為空的故無資料寫入RA。(步驟42 0 ) RA是否含有資料區塊i或blk3? (步驟422)RA不含有blk3。在SM及RA中寫入blk3於位置1 1 I 1! I Ϊ 1 II I I Li 1 I 1 II Page 41 1244869 V. Description of the invention (37) Enter blkO at position [〇, 〇]. (Step 424) Add 1 to i (i = 1) to introduce the value of position [1, 0], and return to step 414 (step 414) i is compared with X. (Step 418) i is less than χ (1 &lt; 6), the SM reads the matrix located in row [1] and writes it to RA; at the beginning, the SM is empty and no data is written to RA. (Step 420) Does RA contain data block i or blkl? (Step 422) RA does not contain blkl. Write blkl at position [1,0] ° in SM and RA. Step 4 24) Add 1 to i (i = 2) to import the value of position [2, 〇] and return to step 414 〇 (Step 4 1 4) Compare i with X. (Step 418) i is less than x (2 &lt; 6), and the SM reads the matrix located in row [2] and writes it to RA; at the beginning, the SM is empty and no data is written to RA. (Step 4 20) Does the RA contain data block i or bl k2? (Step 422) The RA does not contain blk2. Write blk2 at position [2,0] ° in SM and RA (step 424) plus 1 to i (i = 3) to import the value of position [3, 〇], and return to step 414 〇 (step 4 1 4 ) i is compared to X. (Step 418) i is less than x (3 &lt; 6), and the SM reads the matrix located in row [3] and writes it to RA; at the beginning, the SM is empty and no data is written to RA. (Step 420) Does the RA contain data block i or blk3? (Step 422) The RA does not contain blk3. Write blk3 in position in SM and RA

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第42頁 1244869 五、發明說明(38) [3, 0]。 (步驟4 2 4 )加1於i ( i = 4 )以導入位置[4,0 ]之值,回到步驟 414 ° (步驟4 1 4) i與X比較。 (步驟418)i少於χ(4&lt;6),於SM讀取位於行[4]之矩陣並寫 入RA ;開始時,該SM為空的故無資料寫入RA。 (步驟420 )RA是否含有資料區塊i或blk4? (步驟422)RA不含有blk4。在SM及RA中寫入blk4於位置 [4,0] ° (步驟4 2 4 )加1於i ( i = 5 )以導入位置[5,〇 ]之值,回到步驟 414 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(5&lt;6),於SM讀取位於行[4]之矩陣並寫 入RA ;開始時,該SM為空的故無資料寫入RA。 · (步驟420)RA是否含有資料區塊i或 (步驟422)RA不含有blk5。在SM及RA中寫入blk5於位置 [5, 0] 〇 (步驟424)加1於i(i = 6)。回到步驟414。 (步驟4 1 4 ) i與X比較。 (步驟416)i等於x(6 = 6)。j增加1 ( j = l)。回到步驟4〇6。 .........................................(^ 406 )清除一參考陣列(RA)。 (步驟4 0 8 ) j與X比較。 (步驟412) j 少於χ(1&lt;6)使i = 0。Page 42 1244869 V. Description of the invention (38) [3, 0]. (Step 4 2 4) Add 1 to i (i = 4) to introduce the value of position [4,0] and return to step 414 ° (Step 4 1 4) i and X are compared. (Step 418) i is less than χ (4 &lt; 6), the SM reads the matrix located in row [4] and writes it to RA; at the beginning, the SM is empty, so no data is written to RA. (Step 420) Does RA contain data block i or blk4? (Step 422) RA does not contain blk4. Write blk4 at position [4,0] ° in SM and RA (step 4 2 4) plus 1 to i (i = 5) to import the value of position [5, 〇], and return to step 414 (step 4 1 4) Compare i with X. (Step 418) i is less than x (5 &lt; 6), and the SM reads the matrix in row [4] and writes it to RA; at the beginning, the SM is empty and no data is written to RA. (Step 420) Whether RA contains data block i or (Step 422) RA does not contain blk5. Write blk5 to position [5, 0] in SM and RA (step 424) and add 1 to i (i = 6). Return to step 414. (Step 4 1 4) i is compared with X. (Step 416) i is equal to x (6 = 6). j is increased by 1 (j = l). Return to step 4〇6. ............... (^ 406) Clear a reference array ( RA). (Step 4 0 8) j is compared with X. (Step 412) j is less than χ (1 &lt; 6) so that i = 0.

第43頁 n44mPage 43 n44m

(步驟4 1 4 ) i與x比較。 (步驟418)i少於χ(〇&lt;6),於SM讀取位於行Π]之矩陣並寫 入RA。位置[1,〇]含有blkl ;因此,blkl被寫入RA。所有 其他位置是空的。 (步驟420)RA是否含有資料區塊][或blkO? (步驟422)RA不含有blkO。在SM及RA中寫入blkO於位置 [1,1 ]。RA 現在有blkl 及blkO。 (步驟424)加1於i (卜1 )以導入位置[2, 1 ]之值,回到步驟 414 〇 (步驟414)i與X比較。 (步驟418)i少於x(l&lt;6),於SM讀取位於行[2]之矩陣並寫 入RA。位置[2, 0 ]含有bl k2 ;所有其他位置是空的。ra現 在有blkl、blkO 及blk2。 (步驟4 20 )RA是否含有資料區塊i或blkl? (步驟424)RA含有blkl。因此,無資料寫入位置[2,1]。加 1於i (i = 2 )以導入位置[3,1 ]之值,回到步驟41 4。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(2&lt;6),於SM讀取位於行[3]之矩陣並寫 入RA。位置[3, 0]含有blk3 ;所有其他位置是空的。RA現 在有blkl、blkO、blk2 及blk3。 (步驟4 20 )RA是否含有資料區塊i或blk2?(步驟424)RA是含 有blk2。因此,無資料窝入位置[3,1]。加1於i(i = 3)以導 入位置[4,1 ]之值,回到步驟4 1 4。 (步驟4 1 4 ) i與X比較。(Step 4 1 4) i is compared with x. (Step 418) i is less than χ (0 &lt; 6), and the matrix at row Π] is read at SM and written into RA. Position [1, 0] contains blkl; therefore, blkl is written to RA. All other locations are empty. (Step 420) Does RA contain a data block? [Or blkO? (Step 422) RA does not contain blkO. Write blkO at position [1,1] in SM and RA. RA now has blkl and blkO. (Step 424) Add 1 to i (Bu1) to introduce the value of position [2, 1], and return to Step 414. (Step 414) i is compared with X. (Step 418) i is less than x (l &lt; 6), reads the matrix in row [2] at SM and writes it to RA. Position [2, 0] contains bl k2; all other positions are empty. ra now has blkl, blkO, and blk2. (Step 4 20) Does RA contain data block i or blkl? (Step 424) RA contains blkl. Therefore, no data is written to location [2, 1]. Add 1 to i (i = 2) to import the value of position [3, 1], and return to step 41 4. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (2 &lt; 6), read the matrix in row [3] at SM and write it to RA. Position [3, 0] contains blk3; all other positions are empty. RA now has blkl, blkO, blk2, and blk3. (Step 4 20) Does RA contain data block i or blk2? (Step 424) RA contains blk2. Therefore, there is no information on the nesting position [3, 1]. Add 1 to i (i = 3) to import the value of position [4, 1], and return to step 4 1 4. (Step 4 1 4) i is compared with X.

1244869 —— 五、發明說明(40) (步驟418)i少於x(3&lt;6),於SM讀取位於行[4]之矩陣並寫 入RA。位置[4, 0 ]含有bl k4 ;所有其他位置是空的。RA現 在有 blkl、blk0、blk2、blk3 及 blk4。 (步驟4 20 )RA是否含有資料區塊i或blk3? (步驟424)RA是含有blk3。因此,無資料寫入位置[4,1]。 加1於i(i=4)以導入位置[5,1]之值,回到步驟414。 (步驟4 14) i與X比較。 (步驟418)i少於x(4&lt;6),於SM讀取位於行[5]之矩陣並寫 入RA。位置[5, 0 ]含有blk5 ;所有其他位置是空的。RA現 各有blkl、blk0、blk2、blk3、blk4 及blk5。 (步驟4 20 )RA是否含有資料區塊i或blk4? (步驟424)RA是含有blk4。因此,無資料寫入位置[5, 1 ]。 加1於i (i = 5)以導入位置[〇, 1 ]之值,回到步驟414。 (步驟4 1 4) i與X比較。 (步驟418)i少於χ(5&lt;6),於SM讀取位於行[〇]之矩陣並寫 入RA。位置[0, 0 ]含有blkO ;所有其他位置是空的。ra現 在有b 1 k 0 ;因此,b 1 k 0被搬棄。 (步驟4 20 )RA是否含有資料區塊j或b 1 k5? (步驟424)RA是含有blk5。因此,無資料寫入位置[〇,ι]。 加1於i (i = 6 )。回到步驟41 4。 (步驟414)i與X比較。(步驟416) i等於χ(6 = 6) 〇 j增加1 (卜 2)。回到步驟40 6。1244869 —— V. Description of the invention (40) (step 418) i is less than x (3 &lt; 6), read the matrix in row [4] in SM and write it into RA. Position [4, 0] contains bl k4; all other positions are empty. The RA now has blkl, blk0, blk2, blk3, and blk4. (Step 4 20) Does RA contain data block i or blk3? (Step 424) RA contains blk3. Therefore, no data is written to location [4, 1]. Add 1 to i (i = 4) to import the value of position [5, 1], and return to step 414. (Step 4 14) i is compared with X. (Step 418) i is less than x (4 &lt; 6), read the matrix in row [5] at SM and write it to RA. Position [5, 0] contains blk5; all other positions are empty. The RA now has blkl, blk0, blk2, blk3, blk4, and blk5. (Step 4 20) Does RA contain data block i or blk4? (Step 424) RA contains blk4. Therefore, no data is written to location [5, 1]. Add 1 to i (i = 5) to import the value of position [0, 1], and return to step 414. (Step 4 1 4) i is compared with X. (Step 418) i is less than χ (5 &lt; 6), and the matrix at row [0] is read at SM and written into RA. Position [0, 0] contains blkO; all other positions are empty. ra now has b 1 k 0; therefore, b 1 k 0 is removed. (Step 4 20) Does RA contain data block j or b 1 k5? (Step 424) RA contains blk5. Therefore, no data is written to the location [0, ι]. Add 1 to i (i = 6). Go back to step 41 4. (Step 414) i is compared to X. (Step 416) i is equal to χ (6 = 6) and j is increased by 1 (B 2). Go back to step 40 6.

第45頁 1244869 五、發明說明(41) (步驟4 0 6 )清除一參考陣列(R A )。 (步驟4 0 8 ) j與X比較。 (步驟 412)j 少於 x(2&lt;6)使 i = 0。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(〇&lt;6),於SM讀取位於行[2]之矩陣並寫 入RA。位置[2, 0 ]含有bl k2 ;所有其他位置是空的。RA現 在有blk2。 (步驟420 )RA是否含有資料區塊i或bikO? (步驟422)RA不含有blkO。在SM及RA中寫入blkO於位置 Γ2,2] ° RA 現在有blk2 及blkO。 (步驟4 2 4 )加1於i ( i = 1 )以導入位置[3,2 ]之值,回到步驟 414。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(l&lt;6),於SM讀取位於行[3]之矩陣並寫 入RA。位置[3, 0 ]含有bl k3 ;所有其他位置是空的。ra現 在有blk2、blkO 及blk3。 (步驟4 20 )RA是否含有資料區塊i或blkl? (步驟422)RA不含有blkl。在SM及RA中寫入blkl於位置 [3,2] °RA 現在有 blk2、blkO、blk3 及 blkl。 (步驟424)加1於i(i = 2)以導入位置[4,2]之值,回到步驟 414。(步驟414) i與X比較。 (步驟418)i少於x(2&lt;6),於SM讀取位於行[4]之矩陣並寫 入RA。位置[4, 0 ]含有blk4 ;所有其他位置是空的。ra現 在有blk2、blkO、blk3、blkl 及blk4。Page 45 1244869 V. Description of the invention (41) (step 4 0 6) Clear a reference array (RA). (Step 4 0 8) j is compared with X. (Step 412) j is less than x (2 &lt; 6) such that i = 0. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (0 &lt; 6), read the matrix in row [2] at SM and write it to RA. Position [2, 0] contains bl k2; all other positions are empty. RA now has blk2. (Step 420) Does RA contain data block i or bikO? (Step 422) RA does not contain blkO. Write blkO in SM and RA at position Γ2,2] ° RA now has blk2 and blkO. (Step 4 2 4) Add 1 to i (i = 1) to import the value of position [3, 2], and return to step 414. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (l &lt; 6), and the matrix at row [3] is read at SM and written into RA. Position [3, 0] contains bl k3; all other positions are empty. ra now has blk2, blkO, and blk3. (Step 4 20) Does RA contain data block i or blkl? (Step 422) RA does not contain blkl. Write blkl at position [3,2] in SM and RA ° RA now has blk2, blkO, blk3, and blkl. (Step 424) Add 1 to i (i = 2) to introduce the value of position [4, 2], and return to step 414. (Step 414) i is compared to X. (Step 418) i is less than x (2 &lt; 6), read the matrix in row [4] at SM and write it to RA. Position [4, 0] contains blk4; all other positions are empty. ra now has blk2, blkO, blk3, blkl, and blk4.

第46頁 1244869 一 丨_ 丨丨· ' ' ·_·Ν· _ __丨__ 五、發明說明(42) (步驟4 20 )RA是否含有資料區塊i或blk2? (步驟4 24)RA是含有blk2。因此,無資料寫入位置[4,2]。 加1於i ( i = 3 )以導入位置[5,2 ]之值,回到步驟4 1 4。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(3&lt;6),於SM讀取位於行[5]之矩陣並寫 入RA。位置[5, 0]含有blk5 ;所有其他位置是空的。RA現 在有 blk2、blk0、blk3、blkl、blk4 及 blk5 ° (步驟4 20 )RA是否含有資料區塊i或blk3? (步驟424)RA是含有blk3。因此,無資料寫入位置[5,2]。 加1於i (i =4)以導入位置[0, 2]之值,回到步驟414。 (步驟414) i與X比較。 (步驟418)i少於x(4&lt;6),於SM讀取位於行[0]之矩陣並寫 入RA。位置[0,0]含有blkO ;所有其他位置是空的。因 此,b 1 k 0被拋棄。 (步驟4 20)RA是否含有資料區塊i或blk4? (步驟4 24)RA是含有blk4。因此,無資料寫入位置[0,2]。 加1於i (i = 5)以導入位置[1,2 ]之值,回到步驟4 1 4。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於χ(5&lt;6),於SM讀取位於行[1]之矩陣並寫 入RA。RA已經含有blkl及blkO ;因此,blkl及blkO被拋 棄。所有其他位置是空的。 (步驟4 20 )RA是否含有資料區塊i或blk5? (步驟4 24)RA是含有blk5。因此,無資料寫入位置[1,2.]。 加1於i (i = 6 )回到步驟41 4。Page 46 1244869 A 丨 _ 丨 丨 · '' · _ · Ν · _ __ 丨 __ 5. Explanation of the invention (42) (Step 4 20) Does the RA contain data block i or blk2? (Step 4 24) RA contains blk2. Therefore, no data is written to location [4, 2]. Add 1 to i (i = 3) to import the value of position [5, 2] and return to step 4 1 4. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (3 &lt; 6), read the matrix in row [5] at SM and write it to RA. Position [5, 0] contains blk5; all other positions are empty. The RA now has blk2, blk0, blk3, blkl, blk4, and blk5 ° (step 4 20) Does the RA contain data block i or blk3? (Step 424) The RA contains blk3. Therefore, no data is written to location [5, 2]. Add 1 to i (i = 4) to import the value of position [0, 2], and return to step 414. (Step 414) i is compared with X. (Step 418) i is less than x (4 &lt; 6), read the matrix at row [0] at SM and write it to RA. Position [0,0] contains blkO; all other positions are empty. Therefore, b 1 k 0 is discarded. (Step 4 20) Does RA contain data block i or blk4? (Step 4 24) RA contains blk4. Therefore, no data is written to location [0,2]. Add 1 to i (i = 5) to import the value of position [1, 2] and return to step 4 1 4. (Step 4 1 4) i is compared with X. (Step 418) i is less than χ (5 &lt; 6), and the matrix at row [1] is read at SM and written into RA. RA already contains blkl and blkO; therefore, blkl and blkO are discarded. All other positions are empty. (Step 4 20) Does RA contain data block i or blk5? (Step 4 24) RA contains blk5. Therefore, no data is written to location [1, 2.]. Add 1 to i (i = 6) and return to step 41 4.

^244869 、發明說明(43) 步驟4 1 4 ) i與X比較。 回到步驟4 06 步騍4 1 6 ) i等於X ( 6二6 )。j增加1 (卜3 ) (步驟4 0 6 )清除一參考陣列(RA)。 (步驟4 08) j與X比較。 (步驟4l2)j 少於x(3&lt;6)使i = 0。 (步驟414)i與X比較。 (步驟418)i少於x(0&lt;6),於SM讀取位於行[3]之矩陣並寫^ 244869, Description of the Invention (43) Step 4 1 4) i and X are compared. Return to step 4 06. Step 4 1 6) i is equal to X (6 2 6). Increase j by 1 (Bu 3) (step 4 0 6) to clear a reference array (RA). (Step 4 08) j is compared with X. (Step 4l2) j is less than x (3 &lt; 6) so that i = 0. (Step 414) i is compared to X. (Step 418) i is less than x (0 &lt; 6), read the matrix located in row [3] at SM and write

人RA。位置[3,0]含有blk3且位置[3,2]含有blkl;blk3及 blkl被寫入RA。所有其他位置是空啲。 (步驟420)RA是否含有資料區塊i或blkO? (步驟422)RA不含有blkO。在SM及RA中寫入blkO於位置 [3,3] 〇RA 現在有 blk3、blkl 及 blkO。 (步驟4 24)加1於i(i = l)以導入位置[4, 3]之值,回到步驟 414 〇 (步驟4 1 4 ) i與X比較。People RA. Location [3,0] contains blk3 and location [3,2] contains blkl; blk3 and blkl are written to the RA. All other locations are empty. (Step 420) Does RA contain data block i or blkO? (Step 422) RA does not contain blkO. Write blkO in position [3,3] in SM and RA. RA now has blk3, blkl, and blkO. (Step 4 24) Add 1 to i (i = 1) to introduce the value of position [4, 3], and return to step 414 (step 4 1 4) i and X are compared.

(步驟418)i少於x(l&lt;6),於SM讀取位於行[4]之矩陣並寫 入RA。位置[4, 0 ]含有b 1 k4 ;所有其他位置是空的。ra現 在有blk3、blkl、blk0及blk4。(步驟42〇)ra是否含有資 料區塊i或blkl? 、 (步驟424) RA是含有blkl。因此,無資料寫入位置[4, 3]。 加1於i(i = 2)以導入位置[5, 3]之值,回到步驟414。 (步驟414)i與X比較。(Step 418) i is less than x (l &lt; 6), reads the matrix in row [4] at SM and writes it to RA. Position [4, 0] contains b 1 k4; all other positions are empty. ra now has blk3, blkl, blk0, and blk4. (Step 42) Does ra contain the data block i or blkl? (Step 424) RA contains blkl. Therefore, no data is written to the location [4, 3]. Add 1 to i (i = 2) to import the value of position [5, 3], and return to step 414. (Step 414) i is compared to X.

1244869 五、發明說明(44) (步驟418) i少於x( 2 &lt;6 ),於SM讀取位於行[5]之矩陣並寫 入RA。位置[5, 0 ]含有bl k5 ;所有其他位置是空的。RA現 在有 blk3、blkl、blkO、blk4 及 blk5。 (步驟4 20 )RA是否含有資料區塊i或blk2? (步驟4 22)RA不含有blk2。在SM及RA中寫入blk2於位置 [5,3] °RA 現在有 blk3、blkl、blkO、blk4、blk5 及 blk2。 (步驟4 2 4 )加1於i ( i = 3 )以導入位置[Ο,3 ]之值,回到步驟 414 〇 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(3&lt;6),於SM讀取位於行[0]之矩陣並寫 入RA。位置[0, 〇]含有blkO ;所有其他位置是空的。RA已 經含有blkO ;因此,拋棄blkO。 (步驟4 20 )RA是否含有資料區塊i、blk3? (步驟424)RA是含有blk3。因此,無資料寫入位置[〇, 3 ]。 加1於i ( i =4)以導入位置[1,3 ]之值,回到步驟4 1 4。 (步驟41 4) i與X比較。 (步驟4 18) i少於χ(4&lt;6),於SM讀取位於行[1 ]之矩陣並寫 入RA。位置[1,〇 ]含有blkl且位置[1,1]含有blkO ;所有其 他位置是空的。RA已經含有blkl及blkO ;不重複寫入。 (步驟420 )RA是否含有資料區塊i或blk4?(步驟424)RA是含 有MM。因此,無資料寫入位置[1,3]。加1於i ( i = 5)以導 入位置[2,3 ]之值,回到步驟41 4。 (步驟414)i與X比較。1244869 V. Description of the invention (44) (step 418) i is less than x (2 &lt; 6), read the matrix located in row [5] in SM and write it into RA. Position [5, 0] contains bl k5; all other positions are empty. RA now has blk3, blkl, blkO, blk4, and blk5. (Step 4 20) Does RA contain data block i or blk2? (Step 4 22) RA does not contain blk2. Write blk2 at position [5,3] in SM and RA. RA now has blk3, blkl, blkO, blk4, blk5, and blk2. (Step 4 2 4) Add 1 to i (i = 3) to introduce the value of position [0, 3], and return to step 414 (step 4 1 4) i and X are compared. (Step 418) i is less than x (3 &lt; 6), and the matrix at row [0] is read at SM and written into RA. Position [0, 〇] contains blkO; all other positions are empty. RA already contains blkO; therefore, blkO is discarded. (Step 4 20) Does RA contain data blocks i, blk3? (Step 424) RA contains blk3. Therefore, no data is written to location [0, 3]. Add 1 to i (i = 4) to import the value of position [1, 3] and return to step 4 1 4. (Step 41 4) i is compared with X. (Step 4 18) i is less than χ (4 &lt; 6), read the matrix in row [1] at SM and write it to RA. Position [1, 0] contains blkl and position [1, 1] contains blkO; all other positions are empty. RA already contains blkl and blkO; no rewrites. (Step 420) Does RA contain data block i or blk4? (Step 424) RA contains MM. Therefore, no data is written to location [1, 3]. Add 1 to i (i = 5) to import the value of position [2, 3], and return to step 41 4. (Step 414) i is compared to X.

1244869 五、發明說明(45) (步驟418)i少於χ(5&lt;6),於SM讀取位於行[2]之矩陣並寫 入RA。位置[2, 0]含有blk2且位置[2, 2]含有blkO ;所有其 他位置是空的。RA已經含有blk2及blkO,不重複寫入。 (步驟4 20 )RA是否含有資料區塊i或blk5? (步驟424)RA是含有blk5。因此,無資料寫入位置[2,3]。 加1於i(i = 6),回到步驟414。 (步驟414)i與X比較。 (步驟416)i等於χ(6 = 6)增加l(j = 4)。回到步驟4〇6。 (步驟4 06)清除一參考陣列(RA)。 (步驟4 08 ) j與X比較。 (步驟412) j 少於x(4&lt;6)使 i = 0。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(0&lt;6),於SM讀取位於行[4]之矩陣並寫 入RA。位置[4, 0 ]含有blk4 ; blk4被寫入RA。所有其他位 置是空的。 (步驟420 )RA是否含有資料區塊i或blkO? (步驟422)RA不含有blkO。在SM及RA中寫入blkO於位置 [4, 4]。RA 現在有blk4 及blkO。 (步驟424)加1於i(i = l)以導入位置[5, 4]之值,回到步驟 414 ° (步驟414) i與X比較。 (步驟418)i少於x(l&lt;6),於SM讀取位於行[5]之矩陣並寫1244869 V. Description of the invention (45) (step 418) i is less than χ (5 &lt; 6), reads the matrix located in row [2] in SM and writes it to RA. Position [2, 0] contains blk2 and position [2, 2] contains blkO; all other positions are empty. RA already contains blk2 and blkO and is not rewritten. (Step 4 20) Does RA contain data block i or blk5? (Step 424) RA contains blk5. Therefore, no data is written to location [2,3]. Add 1 to i (i = 6) and return to step 414. (Step 414) i is compared to X. (Step 416) i is equal to χ (6 = 6) increased by l (j = 4). Return to step 4〇6. (Step 4 06) Clear a reference array (RA). (Step 4 08) j is compared with X. (Step 412) j is less than x (4 &lt; 6) so that i = 0. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (0 &lt; 6), read the matrix in row [4] at SM and write it to RA. Position [4, 0] contains blk4; blk4 is written to RA. All other locations are empty. (Step 420) Does RA contain data block i or blkO? (Step 422) RA does not contain blkO. Write blkO in position [4, 4] in SM and RA. RA now has blk4 and blkO. (Step 424) Add 1 to i (i = l) to introduce the value of position [5, 4], and return to step 414 ° (step 414) i is compared with X. (Step 418) i is less than x (l &lt; 6), read the matrix at row [5] and write at SM

第50頁 1244869 五、發明說明(46) 入RA。位置[5,0]含有blk5且位置[5,3]含有blk2 ;所有其 他位置是空的。RA現在有blk4、blkO、blk5及blk2。 (步驟420)RA是否含有資料區塊i或blkl? (步驟422)RA不含有blkl。在SM及RA中寫入blkl於位置 [5,4] °RA 現在有 blk4、blkO、blk5、blk2 及 blkl。 (步驟4 2 4)加1於i (i = 2 )以導入位置[〇,4 ]之值,回到步驟 414 〇 (步驟414)i與X比較。 (步驟4 1 8 ) i少於X( 2&lt;6 ),於SM讀取位於行[〇 ]之矩陣並寫 入RA。位置[0, 0 ]含有bl kO ;所有其他位置是空的。ra已 經含有blkO,因此,不重複寫入。 (步驟420 )RA是否含有資料區塊i或blk2? (步驟424)RA不含有blk2。加1於i(i=3)以導入位置[1,4] 之值,回到步驟41 4 〇 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(3&lt;6),於SM讀取位於行[1 ]之矩陣並寫 入R A。位置[1,〇 ]含有b 1 k 1 ;所有其他位置是空的。R A現 在已經含有blkl及blkO ;不重複寫入。 (步驟420 )RA是否含有資料區塊i或blk3? (步驟422)RA不含有blk3。在SM及RA中寫入blk3於位置 [!,4]。1?八現在有 blk4、blk0、blk5、blk2、blkl 及 blk3。(步驟424)加1於i (i = 4)以導入位置[2, 4]之值,回 到步驟41 4 〇 (步驟414)i與X比較。Page 50 1244869 V. Description of Invention (46) Enter RA. Position [5,0] contains blk5 and position [5,3] contains blk2; all other positions are empty. The RA now has blk4, blkO, blk5, and blk2. (Step 420) Does RA contain data block i or blkl? (Step 422) RA does not contain blkl. Write blkl at position [5,4] in SM and RA. RA now has blk4, blkO, blk5, blk2, and blkl. (Step 4 2 4) Add 1 to i (i = 2) to introduce the value of position [0, 4], and return to step 414 (step 414) i and X are compared. (Step 4 1 8) i is less than X (2 &lt; 6), read the matrix in row [0] at SM and write it to RA. Position [0, 0] contains bl kO; all other positions are empty. ra already contains blkO, so no duplicates are written. (Step 420) Does RA contain data block i or blk2? (Step 424) RA does not contain blk2. Add 1 to i (i = 3) to introduce the value of position [1, 4] and return to step 41 4 0 (step 4 1 4) i is compared with X. (Step 418) i is less than x (3 &lt; 6), read the matrix in row [1] at SM and write R A. Position [1, 0] contains b 1 k 1; all other positions are empty. R A now contains blkl and blkO; no duplicate writes. (Step 420) Does RA contain data block i or blk3? (Step 422) RA does not contain blk3. Write blk3 at position [!, 4] in SM and RA. 1 to 8 now have blk4, blk0, blk5, blk2, blkl, and blk3. (Step 424) Add 1 to i (i = 4) to import the value of position [2, 4], and return to step 41 4 0 (step 414) i is compared with X.

第51頁 1244869 五、發明說明(47) (步驟418)i少於x(4&lt;6),於SM讀取位於行[2]之矩陣並寫 入RA。位置[2,0]含有blk2且位置[2,2]含有blkO ;所有其 他位置是空的。RA已經含有blk2及blkO,不重複寫入。 (步驟420)RA是否含有資料區塊i或]3lk4? (步驟424)RA是含有blk4。因此,無資料寫入位置[2,4]。 加1於i (i = 5 ),以導入位置[3,4 ]之值。回到步驟41 4。 (步驟4 1 4) i與X比較。 (步驟4 1 8) i少於x( 5&lt;6),於3^1讀取位於行[3]之矩陣並寫 入RA。位置[3, 0]含有blk3、位置[3, 2]含有blkl且位置 [3,3 ]含有b 1 k 0 ;所有其他位置是空的。r a已經含有 blk3、blkl及blkO;不重複寫入。 (步驟4 20 )RA是否含有資料區塊i或blk5? (步驟424)RA是含有bl k5。因此,無資料寫入位置[3, 4 ]。 加1於i ( i = 6 ),回到步驟41 4。 (步驟4 1 4 ) i與X比較。 (步驟416)i等於χ(6 = 6)。j增加1 ( j = 5)。回到步驟4 0 6。 (步驟4 06)清除一參考陣列(RA)。 (步驟4 08) j與X比較。 (步驟412)j少於x(5&lt;6)使i = 0。(步驟414Μ與X比較。 (步驟418)i少於χ(〇&lt;6),於SM讀取位於行[5]之矩陣並寫 入RA。位置[5,0]含有blk5位置[5,3]含有blk2,且位置、 [5,4]含有blkl ;blk5及blk2及blkl被寫入RA。所有其他Page 51 1244869 V. Description of the invention (47) (step 418) i is less than x (4 &lt; 6), read the matrix in row [2] in SM and write it into RA. Position [2,0] contains blk2 and position [2,2] contains blkO; all other positions are empty. RA already contains blk2 and blkO and is not rewritten. (Step 420) Does RA contain data block i or 3lk4? (Step 424) RA contains blk4. Therefore, no data is written to location [2,4]. Add 1 to i (i = 5) to import the value of position [3, 4]. Go back to step 41 4. (Step 4 1 4) i is compared with X. (Step 4 1 8) If i is less than x (5 &lt; 6), read the matrix in row [3] at 3 ^ 1 and write it to RA. Position [3, 0] contains blk3, position [3, 2] contains blkl and position [3, 3] contains b 1 k 0; all other positions are empty. r a already contains blk3, blkl, and blkO; do not write repeatedly. (Step 4 20) Does RA contain data block i or blk5? (Step 424) RA contains bl k5. Therefore, no data is written to location [3, 4]. Add 1 to i (i = 6) and return to step 41 4. (Step 4 1 4) i is compared with X. (Step 416) i is equal to χ (6 = 6). j is increased by 1 (j = 5). Go back to step 4 0 6. (Step 4 06) Clear a reference array (RA). (Step 4 08) j is compared with X. (Step 412) j is less than x (5 &lt; 6) so that i = 0. (Step 414M is compared with X. (Step 418) i is less than χ (0 &lt; 6), and the matrix at row [5] is read at SM and written to RA. Position [5,0] contains blk5 position [5, 3] contains blk2 and position, [5,4] contains blkl; blk5 and blk2 and blkl are written to RA. All others

一《44869 五、發明說明(48) 位置是空的。 (步驟4 20 )RA是否含有資料區塊i或blkO? (步驟422)RA不含有blkO。在SM及RA中寫入blkO於位置 [5,5] °RA 現在有 blk5、blk2、blkl 及 blkO。 (步驟4 2 4 )加1於i ( i = 1 )以導入位置[0,5 ]之值,回到步驟 414 〇 (步驟414) i與X比較。I. 44869 V. Description of Invention (48) The position is empty. (Step 4 20) Does RA contain data block i or blkO? (Step 422) RA does not contain blkO. Write blkO at position [5,5] in SM and RA. RA now has blk5, blk2, blkl, and blkO. (Step 4 2 4) Add 1 to i (i = 1) to introduce the value of position [0,5], and return to step 414. (Step 414) i is compared with X.

(步驟418)i少於x(l&lt;6),於SM讀取位於行[〇]之矩陣並寫 入RA。位置[0, 0 ]含有blkO且所有其他位置是空的。RA現 在含有blk5、blk2、blkl 及blkO。 (步驟4 20 )RA是否含有資料區塊i或blkl? (步驟424)RA是含有blkl。加1於i(i=2)以導入位置[1,5] 之值,回到步驟41 4。 (步驟414) i與X比較。(Step 418) i is less than x (l &lt; 6), and the matrix at row [0] is read at SM and written into RA. Position [0, 0] contains blkO and all other positions are empty. RA now contains blk5, blk2, blkl, and blkO. (Step 4 20) Does RA contain data block i or blkl? (Step 424) RA contains blkl. Add 1 to i (i = 2) to import the value of position [1, 5] and return to step 41 4. (Step 414) i is compared with X.

(步驟418)i少於x(2&lt;6)。於SM讀取位於行[1]之矩陣並寫 入RA。位置[1,0]含有blkl、位置[1,1]含有blkO且位置 [1,4]含有blk3 ;所有其他位置是空的。ra已經含有blkO 及blkl ;因此,不重複寫入。將blk3寫入RA。RA現在含有 blk5、blk2、blkl、blk0 及 blk3。(步驟 420 )RA 是否含有 資料區塊i或blk2? (步驟424)RA是含有blk2。加1於i(i=3)以導入位置[2,5] 之值/回到步驟41 4。 (步驟4 1 4 ) i與X比較。 (步驟418)i少於x(3&lt;6)。於SM讀取位於行[2]之矩陣並寫(Step 418) i is less than x (2 &lt; 6). The matrix at row [1] is read at SM and written to RA. Position [1,0] contains blkl, position [1,1] contains blkO and position [1,4] contains blk3; all other positions are empty. ra already contains blkO and blkl; therefore, no duplicate writes. Write blk3 to RA. The RA now contains blk5, blk2, blkl, blk0, and blk3. (Step 420) Does RA contain data block i or blk2? (Step 424) RA contains blk2. Add 1 to i (i = 3) to import the value of position [2,5] / return to step 41 4. (Step 4 1 4) i is compared with X. (Step 418) i is less than x (3 &lt; 6). Read the matrix in row [2] and write at SM

第53頁 1244869 五、發明說明(49) 入RA。位置[2,0]含有blk2、位置[2,2]含有blkO ;所有其 他位置是空的。RA已經含有blk2及blkO ;不重複寫入。 (步驟420)RA是否含有資料區塊i或blk3? (步驟4 2 4)RA是含有blk3。加1於i(i=4)以導入位置[3, 5] 之值,回到步驟41 4。 (步驟4 1 4) i與X比較。 (步驟418)i少於χ(4&lt;6),於SM讀取位於行[3]之矩陣並寫 入RA。位置[3, 0 ]含有blk3、位置[3, 2]含有blkl、位置 [3, 3]含有blkO。所有其他位置是空的。RA已經含有 b’lk3、blkl及blkO ;不重複寫入。 (步驟4 20 )RA是否含有資料區塊i或blk4? (步驟4 22 )RA不含有blk4。在SM及RA中寫入blk4於位置 [3,5] 'RA 現在有 blk5、blk2、blkl、blkO、blk3 及 blk4 〇 (步驟424)加1於i(i = 5)以導入位置[4,5]之值,回到步驟 414 〇 (步驟414)i與X比較。 (步驟418)i少於x(5&lt;6),於3^1讀取位於行[4]之矩陣並寫 入RA。位置[4, 0 ]含有blk4、位置[4, 4]含有blkO。所有其 他位置是空的。RA已經含有blk4及blkO ;不重複寫入。 (步驟420)RA是否含有資料區塊i或blk5? (步驟422&gt;RA是含有blk5。因此,無資料寫入位置[3, 4 ]。 (步驟424)加1於i(i = 6)。回到步驟414。 (步驟4 1 4) i與X比較。___________________Page 53 1244869 V. Description of Invention (49) Enter RA. Position [2,0] contains blk2, position [2,2] contains blkO; all other positions are empty. RA already contains blk2 and blkO; no repeated writes. (Step 420) Does RA contain data block i or blk3? (Step 4 2 4) RA contains blk3. Add 1 to i (i = 4) to import the value of position [3, 5] and return to step 41 4. (Step 4 1 4) i is compared with X. (Step 418) i is less than χ (4 &lt; 6), and the matrix at row [3] is read at SM and written into RA. Position [3, 0] contains blk3, position [3, 2] contains blkl, and position [3, 3] contains blkO. All other positions are empty. RA already contains b'lk3, blkl, and blkO; no rewrites. (Step 4 20) Does RA contain data block i or blk4? (Step 4 22) RA does not contain blk4. Write blk4 at position [3,5] in SM and RA. 'RA now has blk5, blk2, blkl, blkO, blk3, and blk4 0 (step 424) plus 1 (i = 5) to import position [4, 5], return to step 414 (step 414) i and X. (Step 418) i is less than x (5 &lt; 6), and the matrix at row [4] is read at 3 ^ 1 and written into RA. Position [4, 0] contains blk4 and position [4, 4] contains blkO. All other positions are empty. RA already contains blk4 and blkO; no rewrites. (Step 420) Does RA contain data block i or blk5? (Step 422> RA contains blk5. Therefore, no data is written to location [3, 4]. (Step 424) Add 1 to i (i = 6). Return to step 414. (Step 4 1 4) i is compared with X. _________________

第54頁 1244869 五、發明說明(50) (步驟4 1 6 ) i等於X ( 6 = 6 )。j增加1 (j二5 )。回到步驟4 0 6。 (步驟4 0 6 )清除一參考陣列(RA)。 (步驟4 0 8 ) j•與X比較。 (步驟410) j 等於x(6 = 6); 完畢。Page 54 1244869 V. Description of the invention (50) (step 4 1 6) i is equal to X (6 = 6). j is increased by 1 (j = 5). Go back to step 4 0 6. (Step 4 0 6) Clear a reference array (RA). (Step 4 0 8) j • Compare with X. (Step 410) j is equal to x (6 = 6); Done.

第55頁 1244869Page 1212 44869

器 圖ΙΑ 圖IB 圖2 顯示本發明之一 顯不本發明另一 顯不依據本發明 具施例之例示 践施例之例示 之一實施例之 性DOD系統 性D〇D系統 例示性頻道伺服 陣 圖3 圖4 圖5 顯示本發明之 顯示本發明之 顯示本發明之 貫施例之泛用機頂盒(gTB) 貫施例產生一排序矩陣之程序 具知例之6個資料區塊排序矩 圖6 顯示圖5中排序矩 置時間槽被充滿 陣之資料區塊被上移直到全 圖7 排序矩陣 顯示本赉明之一實施例之新的減少閒置頻寬之 圖8 描述額外之減少閒置時間之實施例 圖9 以流程圖开》式描述減少間置時間之排序之實施 例如何實現圖1 〇 係一流程圖顯示依據本發明之一實^: 之程序供排序DOD資料區塊,以傳送一原始資料串流及一1 預取 資料串流 圖11 係一流程圖顯示依據本發明之另一實施例之程 序供排序D0D資料區塊,以傳送一原始資料串流及一預王 取資料串流 圖1 2 係一流程圖顯示依據本發明之一實施例之機頂 盒預戴入程序 、 符號說明:Device Figure IA Figure IB Figure 2 shows one display of the present invention, another display of the present invention, according to an example of the embodiment of the present invention, an example of an embodiment, an exemplary DOD system, a DoD system, an exemplary channel servo Matrix map 3 Figure 4 Figure 5 shows the present invention showing a universal set-top box (gTB) of the present invention showing a consistent embodiment of the present invention. The sequence of generating a sort matrix is a known example of six data block sorting moments. 6 The data block in the sorting moment slot that is filled in the array shown in FIG. 5 is moved up to the whole figure 7. The sorting matrix shows a new embodiment of the present invention to reduce the idle bandwidth. FIG. 8 describes the additional idle time reduction. Embodiment Figure 9 describes the implementation of the sorting method for reducing the interposition time in the form of a flowchart. Figure 10 is a flowchart showing a procedure according to one embodiment of the present invention for sorting DOD data blocks to transmit a block of data. Raw data stream and a pre-fetched data stream Figure 11 is a flowchart showing a procedure according to another embodiment of the present invention for sorting DOD data blocks to transmit a raw data stream and a pre-fetched data stream Flow diagram 1 2 It is a flowchart showing a pre-installation procedure of a set-top box according to an embodiment of the present invention. Symbol description:

1244869 圖式簡單說明 80 有線電視頻道 100 D 0 D系統 102 中央控制伺服器 103 中央儲存 104a-1 04η 頻道伺服器 106a -106η增頻器 108 結合器/放大器 110 傳送媒介 112 開關矩陣 118a -118b 備用增頻器 120 有線電視糸統 102 中央控制伺服器 202 伺服控制器 2 04 CPU 206 QAM調變器 208 局部記憶 210 網路介面 302 QAM解調器 304 CPU 308 局部記憶 310 緩衝記憶 312 解碼器 314 圖形重疊模組 318 用戶介面· 32 0 通信鏈 322 快速資料匯流排 324 輸出元件 402, 40 4____......... 422, 424 步驟 5 1 0, 52 0, 530, 54 0, 550, 5 60, 4 1 0 步驟1244869 Brief description of the diagram 80 Cable TV channel 100 D 0 D system 102 Central control server 103 Central storage 104a-1 04η Channel server 106a-106η Frequency booster 108 Combiner / amplifier 110 Transmission medium 112 Switch matrix 118a-118b Spare Inverter 120 Cable TV system 102 Central control server 202 Servo controller 2 04 CPU 206 QAM modulator 208 Local memory 210 Network interface 302 QAM demodulator 304 CPU 308 Local memory 310 Buffer memory 312 Decoder 314 Graphics Overlay module 318 user interface · 32 0 communication chain 322 fast data bus 324 output element 402, 40 4 ____......... 422, 424 step 5 1 0, 52 0, 530, 54 0, 550, 5 60, 4 1 0 steps

第57頁Page 57

Claims (1)

1244869 六、申請專利範圍 1 · 一種以電腦為工具之泛用廣播方法,包含為一群 用戶準備一分送矩陣(deliery matrix)之動作,以定義適 於廣播之資料傳送序列(data transmission seqUence)之 非用戶特定之方式之隨選資料(on demand data),其中傳 送包含具有第一次集(subset)之資料傳送序列(sequence) 之第一串流(stream)及具有第二次集之資料傳送序 二串流之隨一 2·如申喷專利耗圍第1項之以電腦為工且 方法’其中該第一次集及第二次集不包括任何:有乏= 區塊。 3.如申請專利範圍第2項之以電腦為工具之 方法’其中該第-次集包含一最佳化資料區塊集,心續 第二次集為該資料傳送序列之其餘部#圭; 料區塊集與資料區塊分配之頻率有關。 '-竑佳化貝 4·如申請專利範圍第1項之以雷腦為工目 方法,其中該第一串流更包括一預;;泛用廣播 方法5:其如中 以預載入H碎Λ訊f予一、泛用機頂盒 子儲存器中。6如申过^ 串洲·於該泛用機頂盒之電 泛用廣播:法:其中該;專用=圍入第二頁之以電腦為工具之 機頂益S該泛用機頂盒為閒置 第58頁 1244869 〜- 六、申請專利範圍 時可載入該第一串流。 7. 如申請專利範圍第1項之以電腦為工具之泛用廣播 方法,其中該第一次集包括一需要以起始一節目之最小量 之資料區塊,因此該最小量係與可用之頻寬有關。 8. 如申請專利範圍第1項之以電腦為工具之泛用廣播 方法,其中在該第一次集及第二次集之間保留給傳送序列 之比係依據可用之傳送頻寬之量作最佳化規劃。 9. 如申請專利範圍第1項之以電腦為工具之泛用廣播 方法,其中用以傳送該隨選資料檔需要一與一群用戶之數 量無關之傳送頻寬之量。 10. 如申請專利範圍第9項之以電腦為工具之泛用廣 播方法,其中產生一分送矩陣之動作包含以下動作:準備 一第一排序矩陣以適於第一資料檔之傳送,該第一資料檔 以第一群資料區塊代表,該第一排序矩陣提供一第一序列 以在一時間槽内依序傳送該第一群資料區塊,其方式是任 一用戶接收依據該第一排序矩陣傳來之該第一資料檔可開 始存取在一時間槽内之該第一資料檔。11.如申請專利範 圍第10項之以電腦為工具之泛用廣播方法,其中該第一排 序矩陣為一定量頻寬之排序矩陣。 ___圓 第59頁 1244869 六、申請專利範圍 12. 如申請專利範圍第1 0項之以電腦為工具之泛用廣 播方法,其中該第一排序矩陣為一可變頻寬之排序矩陣。 13. 如申請專利範圍第1 1項之以電腦為工具之泛用廣 播方法,其中來自該第一群資料區塊一定量資料係排序以 在預留之頻寬傳送。 14. 如申請專利範圍第1 3項之以電腦為工具之泛用廣 播方法,其中控制預留頻寬之傳送係由一低階硬體設備執 存者。 15. 如申請專利範圍第1項之以電腦為工具之泛用廣 播方法,其中該第一串流為預取串流,該預取串流由一泛 用機頂盒以一套預取接收腳本存取。 16. 如申請專利範圍第1 5項之以電腦為工具之泛用廣 播方法,其中套預取接收腳本包括一使用者預載入一命令 於該機頂盒之動作。 17. 如申請專利範圍第1項之以電腦為工具之泛用廣 播方法,其中該套預取接收腳本包括一來自使用者之預訂 節目0 18. 一種以電腦為工具之方法,供產生一定量頻寬,1244869 VI. Scope of patent application1. A general-purpose broadcast method using a computer as a tool, including the action of preparing a distribution matrix for a group of users to define the data transmission sequence (data transmission seqUence) suitable for broadcasting. On-demand data in a non-user-specific manner, in which a first stream including a data transmission sequence having a first subset and a second data transmission are transmitted The second stream of the second sequence is as follows: 2. The application of the patent as described in the first patent application method is computer-based and the method 'where the first and second episodes do not include any: there is a shortage = block. 3. The method of using a computer as a tool according to item 2 of the patent application, wherein the first-time set includes an optimized data block set, and the second set is the remaining part of the data transmission sequence. The data block set is related to the frequency of data block allocation. '-竑 佳 化 贝 4. If the brainstorm method is used as the first item in the scope of the patent application, the first stream further includes a preview; the universal broadcast method 5: it is preloaded as it is Broken Λ message f to a, universal set-top box storage. 6 If applied ^ Chuanzhou · Electric Universal Broadcasting on the Universal Set-Top Box: Method: Of which; Dedicated = Set-top benefits using a computer as a tool enclosed on the second page S The universal set-top box is idle page 58 1244869 ~-6. The first stream can be loaded when applying for a patent. 7. For the general broadcasting method using a computer as a tool in the scope of patent application item 1, wherein the first set includes a minimum amount of data blocks required to start a program, the minimum amount is related to the available amount. Bandwidth related. 8. For the general-purpose broadcast method using a computer as a tool in the scope of patent application item 1, wherein the ratio reserved for the transmission sequence between the first episode and the second episode is based on the amount of available transmission bandwidth. Optimization planning. 9. For example, the general-purpose broadcasting method using a computer as a tool in the scope of patent application, wherein the transmission of the on-demand data file requires an amount of transmission bandwidth that is independent of the number of users. 10. For a computer-based universal broadcast method using item 9 of the scope of the patent application, the action of generating a distribution matrix includes the following actions: preparing a first sorting matrix to be suitable for the transmission of the first data file, the first A data file is represented by a first group of data blocks, and the first sorting matrix provides a first sequence to sequentially transmit the first group of data blocks within a time slot, in a manner that any user receives The first data file transmitted from the sorting matrix can start to access the first data file within a time slot. 11. The general-purpose broadcasting method using a computer as a tool according to item 10 of the patent application range, wherein the first sorting matrix is a sorting matrix with a certain amount of bandwidth. ___Circle Page 59 1244869 6. Scope of Patent Application 12. For the general broadcasting method using a computer as a tool in item 10 of the scope of patent application, the first sorting matrix is a sorting matrix with variable frequency. 13. For example, a computer-based universal broadcast method using item 11 in the scope of patent application, in which a certain amount of data from the first group of data blocks is sorted for transmission at a reserved bandwidth. 14. For the general broadcast method using a computer as a tool in the scope of patent application No. 13 in which the transmission of the reserved bandwidth is controlled by a low-level hardware device holder. 15. For example, the universal broadcast method using a computer as a tool in the scope of patent application, wherein the first stream is a prefetch stream, and the prefetch stream is stored by a universal set-top box with a set of prefetch receiving scripts take. 16. For the general broadcasting method using a computer as a tool in the scope of patent application No. 15, the pre-fetching receiving script includes a user pre-loading a command to the set-top box. 17. The general-purpose broadcasting method using a computer as a tool in the scope of patent application item 1, wherein the set of prefetching receiving scripts includes a reserved program from a user 0 18. A method using a computer as a tool for generating a certain amount bandwidth, 第60頁 1244869 申讀專利範圍 八〜間置時間 刀送隨選資料 產生一排〜1 ΐ 良 第一群 ΐ歹,J以在-時 1 〜用戶接 ϋ開始存取在 決 於依據 決 料區塊 機頂盒19 含: a. b· 流, c. 流内; d· 而找回 流放影 e. 第一部 之排序矩 ’該方法 序矩陣使 資料區塊 間槽内依 收依據該 一時間槽 望之定量 排序矩陣 確分配給 之串流; 頂盒不傳 一種泛用機頂 陣’使在一非用戶特定之格式適於 至少包含 適於傳送 表示,該 序傳送該 以下動作 第一資料 第一排序 第一群資 權,該 矩陣提 料區塊 第一排序矩陣傳來之該第 第一資料 供一第一 ,其方式 資料檔 定一所 該第一 定一正 為分開 ,而機 轉換到 當在間 於電腦 且 於自該 該資料 一第一 自動地 分與該 内之該第 傳送頻寬 之次序依 依據該定 且載入至 送到一顯 盒存取隨 一資料檔 定量頻寬則被用 資料區 區分成 之串流之一至 ,其中該 序串流該 量頻寬而 少該分開 示設備。 選視訊之方法,至少包 塊; 串流之資 一閒置模式; 置模式時之某一時間存取一預載入傳送串 可讀媒介上儲存資料區塊於該預載入傳送串 ,置模式轉換到一主動模式且為一用戶指示 ,,之一部分,並自該儲存之預載入傳送串 卽目之第一部份; ;二,正確之資料區塊,致在該節目之該 第一郎目之其餘部份之間無中斷; 1244869 六、申請專利範圍 郎目而不必進 其中該找回步驟允許該 —步存取廣播資料。 2 0·如申請專利範圍第1 9項之方法,具有能在該閒置 模式中執行週期性地更新該預取串流之額外功能。 21如申請專利範圍第2 〇項之方法,其中該週期性地 更新步驟係依據一使用者之指令。 22·如申請專利範圍第20項之方法,其中該週期性地 籲 更新步驟係依據一隨選視訊之喜好度。· 23·如申請專利範圍第20項之方法,其中該週期性地 更新步驟係依據一使用者之歷史記錄。 24·如申請專利範圍第19項之方法,其中該轉換到一 閒置模式之步驟係在一組閒置模式腳本之下被執行。 25·如申請專利範圍第24項之方法,其中該閒置模式 腳本包括使用者不活躍。 供預載入一有廣播資 26· 一種以電腦為工具之方法 料之機頂盒(STB),包含以下動作 轉換該S T B為一閒置模式;Page 60 1244869 Applicable patent scope Eight ~ Intermediate time Knife feed on-demand data produces a row ~ 1 良 Good first group J, J starts at-1 用户 User access starts and depends on the decision The block set-top box 19 contains: a. B · stream, c. In stream; d · and find reflow and projection e. The ordering moment of the first part 'the method order matrix makes the data in the slot between data blocks according to the time The desired sorting matrix is assigned to the stream; the set-top box does not pass a general-purpose set-top array, which makes a non-user-specific format suitable for at least containing suitable representations. The sequence sends the following actions: The first order is the first group of assets. The first data sent from the first order matrix of the matrix feed block is provided for a first. The order of switching to the current computer and automatically dividing the first transmission bandwidth from the data according to the order and loading it to a display box for access is quantified with a data file. Bandwidth is divided by data area One of the streams to, where the sequence stream is the amount of bandwidth and less the separate display device. Video selection method, at least block; Streaming data idle mode; At a certain time when setting the mode, access a preloaded transmission string stored data block on the readable medium in the preloaded transmission string, and set the mode Switch to an active mode and it is a user instruction, a part, and the first part of the pre-loaded transmission string from the stored; 2. the correct data block, to the first in the program There is no interruption between the other parts of Langmu; 1244869 VI. The scope of patent application for Langmu without having to enter the recovery step allows the one-step access to broadcast data. 2 0. The method of item 19 in the scope of patent application has the additional function of periodically updating the prefetch stream in the idle mode. 21 The method of claim 20 in the scope of patent application, wherein the periodic updating step is based on a user's instruction. 22. The method of claim 20 in the scope of patent application, wherein the periodic call for update step is based on a preference of an on-demand video. · 23 · The method of claim 20, wherein the step of periodically updating is based on a user's historical record. 24. The method of claim 19, wherein the step of converting to an idle mode is performed under a set of idle mode scripts. 25. The method of claim 24, wherein the idle mode script includes user inactivity. Pre-loaded with a broadcast resource 26. A computer-based method of set-top box (STB), including the following actions, to convert the STB to an idle mode; 第62頁 I244S69Page 62 I244S69 六、申請專利範圍 自動調諧該STB至該第—頻道, 擇資料,該第一頻道包含—預取資料〜串了^預先安排該STB選 當該使用者轉換该S T B到一活動模式 群資料區塊,該處理至少包含以下步驟&quot;处理該收到之一 將該收到之資料區塊解碼; ’ 解壓縮該收到之資料區塊; 依需要重新組合該收到之資料區塊;且 儲存該收到之資料區塊至一存在該STB 憶;且 u又局部吕己 由該泛用STB該使用者提供該第—資料稽給_㈣設 備 其中2:輪:=為㈣;:第26項之,為工具之方法, 之方法 甘士 ϋ认如申睛專利範圍第2 6項之以電腦為工具 &quot;ν μ雨出設備為一顯示監視器。 2 9 . . 其中該輪出°:=利範圍第26項之以電腦為工具之方法, 專利範圍第視訊卡帶錄影機(VCR)。30.如申請 為-電腦系統:M電腦為工具之方法’其中該輪出設備 31· 如申睛專利範圍第1 8項之以電腦為工具之方 法6. The scope of the patent application automatically tunes the STB to the-channel, selects data, and the first channel contains-prefetched data ~ strung ^ pre-arranged the STB selection when the user converts the STB to an activity mode group data area Block, the process includes at least the following steps: &quot; processing one of the received to decode the received data block; 'uncompress the received data block; reassemble the received data block as needed; and Store the received data block to a STB memory; and u partly provide the first data record to the _㈣ device by the user of the universal STB. 2: round: == ㈣; Among the 26 items, it is a tool method. The method Gan Shizheng recognizes as the patent application No. 26 of the patent scope is to use a computer as a tool &quot; ν μ rain out device as a display monitor. 2 9.. Among them, the round out °: = the method of using the computer as a tool in the 26th scope of the patent, the patented video cassette recorder (VCR) in the scope of the patent. 30. If applied for-computer system: M computer is a method of tool ’, where the wheel-out device 31. The method of using a computer as a tool in item 18 of the patent scope —f244869—F244869 六、申請專利範圍 其中該正確分配係依據可用之頻寬 32·一一種資料分送方法,至少包人: 一資料擋被分為一資料區塊群· 3 · 該資料區塊群被下列步驟安丄 、Γ-、 文徘成為一藉次庠: a)區分該資料檔為資料區塊群; 種斤 b )將第一變數歸零; c )清除一參考陣列; 第一變數與資料區塊群之總數比較; ,e右該第一變數少於資料區塊群之總數,將一第二變 數歸零, ^將該第二變數與該資料區塊群之總數比較; g)若該第二變數少於資料區塊群之總數,將一個或更 多儲存在一排序矩陣之一行之資料區塊寫入該參考陣列, 該行以[(HDmodU)]決定,其中i為該第二變數,]·為該 第一變數’ X為該資料區塊之數; h )若該參考陣列已經至少有該儲存之資料區塊之一 時,不寫入另一版本; i )檢查該參考陣列是否包含一相應於該第二變數之區 塊; j )若该參考陣列不包含相應於該第二變數之該資料區 塊,該資料區塊被加入該參考陣列及該排序矩陣於矩陣 於[(i + j)mod(x),j]之位置,且第二變數增加1 ; k)若該參考陣列包含相應於該第二變數之該資料區Sixth, the scope of patent application where the correct allocation is based on the available bandwidth 32. A method of data distribution, at least including: a data file is divided into a data block group · 3 · The data block group is the following Steps An 丄, Γ-, and Wen become a borrowing time: a) Differentiate the data file as a data block group; b) Reset the first variable to zero; c) Clear a reference array; First variable and data Compare the total number of block groups; e, the first variable is less than the total number of data block groups, zero a second variable, and ^ compare the second variable with the total number of data block groups; g) if The second variable is less than the total number of data block groups. One or more data blocks stored in one row of a sort matrix are written into the reference array. The row is determined by [(HDmodU)], where i is the first Two variables,] is the first variable; X is the number of data blocks; h) if the reference array already has at least one of the stored data blocks, do not write another version; i) check the reference Whether the array contains a block corresponding to the second variable; j) if the parameter The test array does not include the data block corresponding to the second variable. The data block is added to the reference array and the ranking matrix at the position of the matrix at [(i + j) mod (x), j]. Two variables increase by 1; k) if the reference array contains the data area corresponding to the second variable 12 桃 69 六、申請專利範圍 塊,則該第二變數增加1 ; 1)重複步驟g)到k)直到該第二變數等於該資料區塊總 數; m )該第一變數增加1 ; η)重複步驟c)到m)直到該第一變數等於該資料區塊總 數;且 〇 )重新安排該排序矩陣為一群串流; 其中該次序經由一具有指定給該資料檔之頻寬之媒介 傳送一重複訊號,且其中該頻寬被該重複訊號完全利用。 3 3.如申請專利範圍第3 2項之資料分送方法,其中該 重新安排該排序矩陣為一群串流之步驟包括決定指令給該 資料檔之頻寬之大小,並決定一第一分送串流之最佳化大 小,使第一分送串流包含該資料檔之起始序列。 34. 如申請專利範圍第33項之資料分送方法,其中決 定指定給資料檔之頻寬之大小使該頻寬為最小。 35. 如申請專利範圍第33項之資料分送方法,其中決 定指定給資料檔之頻寬之大小使該頻寬為最大。 36. 如申請專利範圍第32項之資料分送方法,其中該 資料分送串流群得自一包含一不同之資料權之資料分送矩 陣。12 桃 69 6. For the patent application range block, the second variable is increased by 1; 1) Repeat steps g) to k) until the second variable is equal to the total number of data blocks; m) the first variable is increased by 1; η) Repeat steps c) to m) until the first variable is equal to the total number of data blocks; and 0) rearrange the sorting matrix into a group of streams; wherein the order is transmitted via a medium with a bandwidth assigned to the data file- Repeated signals, and wherein the bandwidth is fully utilized by the repeated signals. 3 3. The data distribution method according to item 32 of the scope of patent application, wherein the step of rearranging the sorting matrix into a group of streams includes determining the size of the bandwidth given to the data file and determining a first distribution The stream is optimized in size so that the first distribution stream contains the starting sequence of the data file. 34. If the data distribution method of item 33 of the patent application scope is adopted, the size of the bandwidth assigned to the data file is determined to minimize the bandwidth. 35. For the data distribution method of item 33 in the scope of patent application, which determines the size of the bandwidth assigned to the data file to maximize the bandwidth. 36. The data distribution method according to item 32 of the patent application scope, wherein the data distribution stream group is obtained from a data distribution matrix including a different data right. 第65頁 1244869 六、申請專利範圍 37. 一種以電腦為工具之泛用資料廣播方法,至少包 含下列步驟: 於一泛用資料廣播系統,執行下列動作: 準備一分送矩陣,定義一適於廣播之資料傳送序列給 一群用戶,非用戶特定之方式之隨選資料,其中傳送該隨 選資料包含一第一串流含有一第一次集之該資料傳送序列 及一第二串流含有一第二次集之該資料傳送序列;其中該 第一串流係預載入傳送資料; ’ 提供一第一頻道伺服器適於經由第一頻道傳送該預載 入傳送資料; 在資料廣播之前,準備一第二頻道伺服器以傳送一電 子節目指引,該電子節目指引(EPG)包含在一有至少該第 二串流之第三頻道上之訊息;且自該第一頻道及該第二頻 道傳送資料;且 在一泛用STB,執行下列動作: .當該泛用STB係在被動模式時,於該第一頻道之該第 一串流接收一數位資料; 當泛用STB係在被動模式時,儲存該第一串流内之該 數位資料於一電腦可讀取媒介中; 接收該EPG之數位資料; 提供該EPG資料給泛用STB之一使用者; 自泛用STB之一位使用者接收資料處理指示;且 自泛用S T B之一位使用者使用該指示。Page 65 1244869 VI. Application scope 37. A general-purpose data broadcasting method using a computer as a tool, including at least the following steps: In a general-purpose data broadcasting system, perform the following actions: Prepare a distribution matrix and define a suitable Broadcast data transmission sequence to a group of users, non-user-specific on-demand data, where the on-demand data includes a first stream containing a first stream of the data transmission sequence and a second stream containing a The data transmission sequence of the second episode; wherein the first stream is preloaded transmission data; 'providing a first channel server adapted to transmit the preloaded transmission data via the first channel; before the data broadcast, Preparing a second channel server to transmit an electronic program guide, the electronic program guide (EPG) includes a message on a third channel having at least the second stream; and from the first channel and the second channel Transmitting data; and performing a general STB, performing the following actions: when the general STB is in a passive mode, receiving a digital data stream on the first stream of the first channel When the universal STB is in the passive mode, the digital data in the first stream is stored in a computer-readable medium; the digital data of the EPG is received; the EPG data is provided to one of the universal STBs for use One of the users of the self-used STB receives the data processing instruction; and one of the users of the self-used STB uses the instruction. 第66頁 4244869Page 66 4244869 六、申請專利範圍 申凊專利範圍 ,其中產生〜 第37項之以電腦為工具之泛用資 分送矩陣之動作至少包含下列動 3 8· 如 料廣播方法 作: 準備一第一排序矩陣 資料檀以一第一群資料 第一序列以在一時間槽 用戶接收依 其方式是任 資料檔可開 組數係依據 其中該 同資料區塊 始存取在一時 在該預載入串 第一資料檔不 適用於第一資料檔之傳送,該第 區塊表示,該第一排序矩陣提供 内依序傳送該第一群資料區塊, 據5亥弟'排序矩陣傳來之該第^ _ 間槽内之該第一資料檔,其中該 流之區塊數; 包括任何包括在談載入串流之共 39.如申請專利範圍第38項之以電腦為工且之泛 料廣播方法’纟中進_步包含在資料廣播之前 ^ 備一第三頻道伺服器供隨選資料資訊之】= ,頻道祠服器包括-尋回(retrieving)該分送 經第三頻道伺服器及㈣包括☆由排序矩陣為分送= 之§亥第一串流於该第三頻道伺服器之記憶内之 該第三頻道傳送資料。 &quot;動作’且自 4.0.如申請專利範圍第39項&lt; 以電腦為工 料廣播方法,其中該第一排序矩陣A +曰4 ^ 用 陣。 併斤矩陣為一定量頻寬排序矩6. Scope of patent application: The scope of patent application, which generates ~ Item 37, the general use of computer as a tool to send a general-purpose material distribution matrix, including at least the following actions 3 8 · As expected broadcast method: Prepare a first sort matrix data The first sequence of the first group of data is received by the user in a time slot. In this way, any data file can be opened according to the number of data blocks in the pre-loaded string. The file is not suitable for the transmission of the first data file. The first block indicates that the first sorting matrix provides the first group of data blocks to be transmitted in sequence. According to the first ^ _ The first data file in the slot, in which the number of blocks of the stream; including any of the 39 included in the talk stream, such as the computer-based and general-purpose broadcast method such as the 38th patent application scope Progressive _ step included before the data broadcast ^ Prepare a third channel server for on-demand data information] =, the channel server includes-retrieving the distribution via the third channel server and ㈣ includes ☆ Distribution by sorting matrix = § The first channel transmits data on the third channel in the memory of the third channel server. &quot; Action &quot; and since 4.0. For example, the 39th scope of the patent application &lt; computer-based broadcasting method, wherein the first ranking matrix A + 4 ^ is used. Sorting matrix for a certain amount of bandwidth 1244讎1244 雠 41. 如申請專利範圍第40項之以電腦為工 料廣播方法,其中一定數量之該第一資料區具之泛用資 在每一時間槽傳送。 、’叩鬼群被排序以 42. 如申請專利範圍第40項之以電腦為工且 料廣播方法,其中該第一排序矩陣為一可 =泛用貧 陣。 1見排序矩 43. 一種隨選資料(D〇D)廣播伺服器方法,接_ 服務以適於以一種非用戶特定之型態,經由— 播給一群用戶,至少包含下列動作: 、迗.、;|廣 提供一第一資料串流包括一群第一.資料區塊及— 外之育料區塊,該第一資料區塊及該額外資料區 、勺 含一選定之DOD服務; ^ ^ 自該第一資料串流移出該第一資料區塊群; 將至少一個該第一資料區塊置於一第二資料串流;且 經由該傳送媒介傳送該第二資料串流;且 經由該傳送媒體,傳送該第一資料串流,使一使用者 接收該第一及第二資料串流可結合該第一及第二資料串流 之至少一部分以便存取該選定之D〇D服務。 44·如申請專利範圍第43項之隨選資料(DOD)廣播伺 服器方法,其中該第一及第二資料串流被安排為一機頂盒41. For example, if a computer is used as a broadcasting method for item 40 in the scope of the patent application, a certain amount of the universal data of the first data area is transmitted in each time slot. The “叩 ghost group” is ranked as 42. For example, the computer-based broadcast method is used in item 40 of the scope of the patent application, wherein the first ranking matrix is a universal matrix. 1 See sorting moment 43. An on-demand data (DOD) broadcast server method that accesses services suitable for broadcasting to a group of users via a non-user-specific type, including at least the following actions:, 迗. 、; | Guangzhou provides a first data stream including a group of first. Data blocks and-outside breeding blocks, the first data block and the additional data area, including a selected DOD service; ^ ^ Removing the first data block group from the first data stream; placing at least one of the first data block in a second data stream; and transmitting the second data stream via the transmission medium; and via the The transmission medium transmits the first data stream so that a user receiving the first and second data streams can combine at least a part of the first and second data streams in order to access the selected DOD service. 44. The on-demand data (DOD) broadcast server method according to item 43 of the patent application, wherein the first and second data streams are arranged as a set-top box 第68頁 1244869 'Page 68 1244869 '' 所接收。Received. 4 5·如申請專利範圍第4 3項之π左 服器方法,其中該額=料資 及一群第三資料區塊。 括一群弟一貧料區塊 46·如申請專利範圍第45項之隨 服器方法,其中進一步包含下列動;^貝料(_廣播飼 .2!二資料區塊移出該群第二資料區塊,· =育料區塊之至少之一置於該第二資料串流 一、第中一;用者接收該第-及第二資料串流可將該第 之剛服:貧料區塊之至少一部分結合以存取該選定 7.如申請專利範圍第45項之隨^ ^ 服器方法,苴中在敌兮、g 遠貝科()廣播伺 服務 之至 々八中存取該遠疋之D〇D服務包括觀看該DOD脂, —部分之影像表演。 •一種卩过遥^料(000)廣播接待(recepfi〇n)方法, t、、、二由〜DOD接收器存取自一非用戶特定之D〇D廣播伺服器 系統傳送之DOD服務,至少包含下列動作: ° 、 經由一傳送媒介接收含有一個選定之DOD服務之一部 分之第一資料串流;且 緩由一傳送媒介接收含有該選定之D〇D服務之一部分4 5. If the method of π server is used in item 43 of the scope of patent application, where the amount = material and a group of third data blocks. Including a group of young people, a poor material block 46. If the patent application scope of the 45th method of the server, which further includes the following actions; ^ shell material (_broadcast feed. 2! Second data block moved out of the second data area Block, · = At least one of the breeding blocks is placed on the second data stream one and the first one; the user receives the first and second data streams and can use the first data stream: the lean material block At least a part of the combination is used to access the selection. 7. If the server method of the 45th scope of the patent application is applied, the remote server accesses the remote server, the remote remote server, and the remote remote server. The DOD service includes viewing the DOD, a part of the video performance. • A remote receiving (000) broadcast reception (recepfi) method, t ,, and two are accessed by the ~ DOD receiver from The DOD service transmitted by a non-user-specific DOD broadcast server system includes at least the following actions: °, receiving a first data stream containing a part of a selected DOD service via a transmission medium; and slowly transmitting it through a transmission medium Receive part of the service that includes the selected DOD 第69頁 1244869Page 69 1244869 六、申請專利範圍 之第二資料串流;且 結合該第一及第二資料串流以存取該選定之D〇D服 務。 49·如申請專利範圍第48項之隨選資料(D〇D)廣播接 待(reception)方法,進一步至少包含: 八 預裁入=第二資料串流之至少一部分於一記憶區; 將遠預載入資料串流之至少一部分與該第一資料串流之至 少一部分結合以放影該D〇D服務之至少一部分。 50·如申請專利範圍第48項之隨選資料(D〇D)廣播接 待(reception)方法,進一步至少包含接收一電子 ^0EPG),該EPG允許該D0D接收器選择一被存取之d〇d服曰 51.—種隨選資料(D0D)廣播飼服器方法,以提 合於以㈣戶,,之方式經傳送媒介廣播剛服務給 一群用戶,至少包含下列動作: 提供-減少閒置時間之資料區塊線性序列含有包括一 選定D0D服務之資料; 自該減少閒置時間之資料區塊序列移出一最常出現之 資料區塊; 將該移出之最常出現之資料區塊置於一預取資料串流 使該預取資料串流包括對應於該選定之D〇D服務之預取資6. The second data stream within the scope of the patent application; and combining the first and second data streams to access the selected DOD service. 49. If the on-demand data (DOD) broadcast reception method of item 48 of the scope of patent application, further includes at least: eight pre-cuts = at least a part of the second data stream in a memory area; At least a portion of the loaded data stream is combined with at least a portion of the first data stream to project at least a portion of the OD service. 50. If the on-demand data (DOD) broadcast reception method of item 48 of the patent application scope further includes receiving at least an electronic ^ 0EPG), the EPG allows the DOD receiver to select a accessed d 〇d service 51.—A kind of on-demand data (DOD) broadcast feeder method, in order to provide users with a way to broadcast a service to a group of users via a transmission medium, including at least the following actions: provide-reduce idle A linear sequence of data blocks of time contains data that includes a selected DOD service; one of the most frequently occurring data blocks is removed from the sequence of data blocks that should reduce idle time; the most frequently occurring data block is moved to one Prefetched data stream so that the prefetched data stream includes prefetched funds corresponding to the selected DOD service 12448691244869 申請專利範圍 料區塊, 經由該傳送媒介傳送該預取資料串流;以及 經由該傳送媒介傳送該其餘減少間置吟間之貝料區塊 序列使一接收設備可將該其餘減少閒置時間之資料區塊序 列及該預取資料區塊以產生該遠定之d〇d服務’因而減少 傳送該DOD服務之頻寬需求。 52 ·如申請專利範圍第5 1項之隨選資料(D0D)廣播伺 服器方法,其中該預取資料串流包括一電子節目指引 (EPG),該EPG包括指出何種服務可自該DOD廣播伺服器提 供之訊息。 53·如申請專利範圍第52項之隨選資料(DOD)廣播飼 服器方法,其中該EPG允許一使用者選擇該選定之D〇D服務 供觀看。 54·如申請專利範圍第51項之隨選資料(DOD)廣播伺 服器方法,進一步至少包含:Applying for a patent range of material blocks, transmitting the pre-fetched data stream via the transmission medium; and transmitting the remaining sequence of reduced material blocks between the intermediary groin via the transmission medium, so that a receiving device can reduce the remaining idle time by The sequence of data blocks and the pre-fetched data blocks to generate the distant dod service, thus reducing the bandwidth requirements for transmitting the DOD service. 52 · If the on-demand data (DOD) broadcast server method of item 51 of the patent application scope, wherein the prefetched data stream includes an electronic program guide (EPG), the EPG includes indicating which services can be broadcast from the DOD Information provided by the server. 53. The on-demand data (DOD) broadcast feeder method of claim 52, wherein the EPG allows a user to select the selected DOD service for viewing. 54. If the on-demand data (DOD) broadcast server method of item 51 of the patent application scope further includes at least: 自該減少閒置時間之資料區塊序列移出一群額外資另 區塊, 將該群額外資料區塊之至少一部分置於該預取資料; 流使該預取資料串〉άk #具+山?目4 t 、 认#—nnn J 包括該取吊出現之資料區塊及該對》 於忒&amp;疋DOD服務之額外資料區塊;以及 經由該傳送媒介傳送該其餘減少閒置時間之資料區夫Remove a group of additional data blocks from the data block sequence that reduces idle time, and place at least a part of the group of additional data blocks in the prefetched data; stream the prefetched data string> άk # 具 + 山? Item 4—nnn J includes the data blocks that appear in the lifting and the additional data blocks of the pair in the 忒 &amp; 疋 DOD service; and the remaining data blocks that reduce the idle time through the transmission medium. 1244869 六、申請專利範圍 序列使一接收設備可將該其餘減少間置時間之資料區塊序 列及該預取資料區塊結合以產生該選定之J}〇J)服務,因而 進一步減少傳送D 〇 D服務之頻寬需求。 印55·如申請專利範圍第54項之隨選資料(D〇D)廣播伺 ^ = f法,其中該群額外資料區塊包括一群第二最常出現 貝料區塊,因而進一步減少傳送該D〇D服務之頻寬需 5 6·如申請專利範圍第5 5 服=方法,其中該群額外資料 之資料區塊,因而進一步減少 求。 項之隨選資料(DOD)廣播肖 區塊包括一群第三最常出現 傳送該DOD服務之頻寬需 57· 服器方法 之資料區 求。58· 服器方法 之資料區 求。 如申晴專利範圍第56項之隨選資料(D〇D)廣播伺 ,其中該群额外資料區塊包括一群第四最常出 塊,因而進一步減少傳送該DOD服務之頻寬需 如申請專利範圍第57項之隨選資料(D〇D)廣播 ,其中該群額外資料區塊包括一群第五最常出 塊因而進一步減少傳送該DOD服務之頻寬需1244869 VI. The patent application sequence allows a receiving device to combine the remaining sequence of data blocks with reduced intermission time and the pre-fetched data blocks to generate the selected J} 〇J) service, thereby further reducing transmission D 〇 Bandwidth requirements for D services.印 55 · If the on-demand data (DOD) broadcast method of the 54th scope of the patent application is applied, the group of additional data blocks includes a group of the second most frequently used shell material blocks, thus further reducing the transmission of The bandwidth of the DOD service needs 56. · If the scope of patent application is the 5th service = method, the data block of this group of additional data, so the demand is further reduced. The on-demand data (DOD) broadcast block of the item includes a group of the third most frequently requested data area for transmitting the bandwidth of the DOD service. 58 · Information area of server method. For example, the on-demand data (DOD) broadcast service of the 56th scope of Shen Qing's patent, where the group of additional data blocks includes a group of the fourth most frequently produced block, so further reducing the bandwidth for transmitting the DOD service requires patent application. On-demand data (DOD) broadcast of scope item 57 where the group of additional data blocks includes a group of the fifth most frequently produced block thus further reducing the bandwidth requirement for transmitting the DOD service 124486ft124486ft 廣播伺服器傳送並 該EPG包括指示])0I)服務為該D〇D 用者接收之資訊。 可由使 6〇· —種隨選資料(D0D)廣播接待方 DOD接收裔自一非用戶特定之廣播伺 服務,至少包含下列動作·· 法,供存取經由一 服器系統傳送DOD 接收一預取資料串流包括對應於一 取資料區塊; 選定DOD服務之預 • 儲該預取資料區塊於一記憶區中; 接收一包括對應於選定之D〇D服務之 原始資料串流;以及 原始資料區塊之 處理該原始資料區塊及該預取資料 者存取該選定之DOD服務。 區塊以允許一使用 ,6 1 ·如申請專利範圍第6 0項之隨選資料(D0D)廣播接 待方法,進一步至少包含接收一電子節目指引(EPG),包 括指示可提供DOD服務之訊息。62·如申請專利範圍第6〇 項之隨選資料(DOD)廣播接待方法,進一步至少包含: 接收使用者之輸入,指示該選定之DOD服務; 轉換對應於該選定之DOD服務之頻道以響應該使用者 之輸入;以及 自該對應於選定之D 0 D服務之頻道接收該原始資料串 流0The broadcast server transmits and the EPG includes instructions]) 0I) the service receives the information for the DOD user. It is possible to make a 60-degree on-demand data (DOD) broadcast receiving party DOD receiver from a non-user-specific broadcast server, including at least the following actions ... methods for accessing the DOD transmitted through a server system receive a prefetch The data stream includes a pre-fetched data block corresponding to the selected DOD service. The pre-fetched data block is stored in a memory area. A raw data stream including the corresponding DOD service is received. Data block processing The original data block and the pre-fetched data access to the selected DOD service. The block is allowed to be used, 61. If the on-demand data (DOD) broadcast receiving method of item 60 of the patent application scope, further includes at least receiving an electronic program guide (EPG), including a message indicating that a DOD service is available. 62. If the on-demand data (DOD) broadcast reception method of item 60 of the patent application scope further includes at least: receiving input from a user, instructing the selected DOD service; switching a channel corresponding to the selected DOD service to respond Should be input by the user; and receive the raw data stream 0 from the channel corresponding to the selected D 0 D service 12448691244869 y 63·如申請專利範圍第62項之隨選資料(DOD)廣播接 待方法,包括接收一電子節目指引(EPG)。 y 64·如申請專利範圍第63項之隨選資料(D0D)廣播接 待方法’其中該電子節目指引提供訊息給使用者,指出自 該D0D廣播伺服器可提供之D〇D服務。 ^ 65·如申請專利範圍第64項之隨選資料(DOD)廣播接 待方法’其中該使用者可由選擇一接收自該對應於該選定 之D0D服務之EPG之訊息觀看一選定之D〇D服務。 ^ 66·如申請專利範圍第63項之隨選資料(D〇D)廣播接 待方法’進一步至少包含經由一視訊放影裝置放影該D0D 服務。67· —種泛用機頂盒(STB),能接收及處理一群數 位服務’如D0D及數位廣播,其中該數位服務之至少一個 係傳送一群獨立之資料串流,該泛用STB至少包含: 一資料匯流排; 一第一通信設備適於耦合至一數位廣播通信媒介;該 第一通信設備可操作以接收包括一原始資料串流及一預取 資料串流之數位資料; 雙向輕合至該資料匯流排之記憶; 一數位資料解碼器雙向编合至該資料匯流排;以及 一中央處理單元(CPU)雙向耦合至該資料匯流排,該cpu作 一STB控制程序之工具以控制該記憶,該數位解碼器及該y 63. The on-demand data (DOD) broadcast receiving method according to item 62 of the patent application scope includes receiving an electronic program guide (EPG). y 64. If the on-demand data (DOD) broadcast receiving method of item 63 of the scope of patent application 'is used, the electronic program guide provides information to the user, indicating the DOD service that can be provided from the DOD broadcast server. ^ 65. If the on-demand data (DOD) broadcast reception method of item 64 of the patent application scope ', wherein the user can view a selected DOD service by selecting a message received from the EPG corresponding to the selected DOD service . ^ 66. The on-demand data (DOD) broadcast receiving method according to item 63 of the patent application scope further includes at least playing the DOD service through a video playback device. 67 · —A general-purpose set-top box (STB) that can receive and process a group of digital services, such as DOD and digital broadcasting. At least one of the digital services transmits a group of independent data streams. The general-purpose STB includes at least: a data A bus; a first communication device adapted to be coupled to a digital broadcast communication medium; the first communication device operable to receive digital data including a raw data stream and a pre-fetched data stream; bi-directionally tapping to the data Memory of the bus; a digital data decoder bidirectionally combined to the data bus; and a central processing unit (CPU) bidirectionally coupled to the data bus, the cpu acts as a tool of the STB control program to control the memory, the Digital decoder and the 第74頁 1244869 &gt;、申請專利範圍 解調器,該STB控制程序可操作以處理接收自該第一通信 設備之數位資料,該CPU進一步可操作以處理該原始資料 串流及預取資料串流。 68. ^申請專利範圍第67項之泛用機頂盒(STB),其 中該預取^料係儲存在該記憶中。 6 9、ί Γ專利範圍第6 7項之泛用機頂盒(STB ),其 中該資料匯k排為一高速資料g _適於執行在—即時方 式ίΐΪϋί二以供完成即時處理接收自該第-通信設備 之該數位貧枓之需。 ^ ·孤二、Ϊ二專利^圍第Μ項之泛用機頂盒(STB),其 中該-貝料匯k排為一USB資料匯流排。 由妨L!·斜U二專利f圍第67項之泛用機頂盒(以8),其 中以貝7 ’々丨L為一高速(f i rewire)資料匯流排。 &gt; 請專利範圍第67項之泛用機頂盒(STB),其 中该§己憶υ括一大量儲存裝置。 73·如=請專利範圍第72項之泛用機頂盒 ,苴 中該大量儲存袭置係一硬碟。 八Page 74 1244869 &gt; Patent-pending demodulator, the STB control program is operable to process digital data received from the first communication device, and the CPU is further operable to process the raw data stream and pre-fetched data stream flow. 68. The general-purpose set-top box (STB) of item 67 of the scope of patent application, wherein the prefetched material is stored in the memory. 6 9. Γ The general set-top box (STB) of item 6 of the patent scope, wherein the data bank k is a high-speed data g_ suitable for performing in the-real-time mode til 2 for completion of real-time processing received from the- This digital need for communication equipment. ^ The "Second and Twenty-Second Patent" refers to the universal set-top box (STB) around item M, in which the -K material bank is a USB data bus. The universal set-top box (item 8) of item 67 in the patent F! Of the oblique U2 can be used. Among them, the high-speed (f i rewire) data bus is taken as the frame 7′々 丨 L. &gt; Please refer to the general-purpose set-top box (STB) of item 67 of the patent, in which the §memory includes a large number of storage devices. 73. If = please use the universal set-top box in item 72 of the patent, where the mass storage device is a hard disk. Eight 第75頁· 1244869 六、申請專利範圍 其 t ι,ΐ11 ; J ; J J f6; ^ ^ ^ ^ ^^STB) l#v^ 中該ΐ.速第74項之泛用機頂盒⑽)’其 c u係隨機存取記憶。 巾嗲數彳ΐ ^ :明專利範圍第67項之泛用機頂盒(STB),其 中該數位貝料解瑪器具有視頻及聲頻解碼能力。 中今7Ht f專利範圍第67項之泛用機頂盒(stb),其 中該數位貪料解碼器係一 MPEG解碼器。 占7如申请專利範圍第67項之泛用機頂盒(STB),其 ^ ^輕^育料解碼器支援各種數位資料命令包括放影、停 止、暫停、前進—段及向前。 Τ + 如申請專利範圍第67項之泛用機頂盒(STB),复 中該弟—通信設備包括一解調器。 八 中該Κ請專利範圍第67項之泛用機頂盒(STB),其 中解凋盗係一QAM解調器。 ’如申請專利範圍第67項之泛用機頂盒(STB),其 通“ 5又傷為一有線通信介面(Μ 〇 d e m) 〇Page 75 · 1244869 Sixth, the scope of the patent application is t ΐ, 11; J; JJ f6; ^ ^ ^ ^ ^^ STB) l # v ^ The general-purpose set-top box for speed item 74)) 'its cu It is a random access memory. Number of frames ^: The universal set-top box (STB) of item 67 of the Ming patent, in which the digital shell decoder has video and audio decoding capabilities. The universal set-top box (stb) of item 67 of the 7Htf patent scope in China, wherein the digital decoder is an MPEG decoder. The universal set-top box (STB), which accounted for item 67 in the scope of the patent application, has a ^^^^ feedstock decoder that supports a variety of digital data commands including play, stop, pause, forward-segment and forward. T + If the universal set-top box (STB) of the 67th area of the patent application is applied, the communication device includes a demodulator. The eighth of this patent claims the universal set-top box (STB) of item 67 of the patent scope, in which the anti-theft system is a QAM demodulator. ’For example, the general-purpose set-top box (STB) in the 67th area of the patent application, which is commonly used as a wired communication interface (Μ 〇 d e m). -4-24486^ 六、申請專利範圍 ^~ - 82·如中請專利範圍第67項之泛用機頂盒(stb),其 中該CPU用作使用者介面驅動器工具,適於解釋接收自一 耦合至該資料匯流排之使用者介面之命令。 δ3·如申請專利範圍第82項之泛用機頂盒(STB),其 中該使用者介面包括一遙控設備。 84·如申請專利範圍第82項之泛用機頂盒(STB),其 中該使用者介面包括一鍵盤。 85·如中請專利範圍第82項之泛用機頂盒(STB),其 中該使用者介面包括一電腦系統。8 6·如申請專利範圍/第 67項之泛用機頂盒(STB),其中該STB控制程序係能決定經 由一電子節目指引提供之資訊判定在一群頻道内接收之^ 料之性質。 、 87·如申請專利範圍第86項之泛用機頂盒(STB),其中 該STB控制程序為可操作以同時調諧頻道群之兩個或多 個,並同時處理自二個或多個該頻道群之資料。 88·如申請專利範圍第87項之泛用機頂盒(STB),其 中該STB控制程序為可操作以提供EPG資料給該泛用STB之 使用者。-4-24486 ^ VI. Scope of patent application ^ ~-82 · If you request the general set-top box (stb) of item 67 of the patent scope, the CPU is used as a user interface driver tool, which is suitable for explaining the reception from a coupling to User interface commands for the data bus. δ3. A general-purpose set-top box (STB) such as item 82 of the patent application, wherein the user interface includes a remote control device. 84. A general-purpose set-top box (STB) as claimed in item 82 of the patent application, wherein the user interface includes a keyboard. 85. The general-purpose set-top box (STB) in item 82 of the patent, wherein the user interface includes a computer system. 8 6. If the scope of patent application / item 67 is a general-purpose set-top box (STB), the STB control program can determine the nature of the materials received in a group of channels based on the information provided by an electronic program guide. 87. The general-purpose set-top box (STB) of item 86 in the scope of patent application, wherein the STB control program is operable to tune two or more of the channel group at the same time, and simultaneously process from two or more of the channel group. Information. 88. A general-purpose set-top box (STB), such as the 87th in the scope of patent application, wherein the STB control program is operable to provide EPG data to users of the general-purpose STB. 第77頁 1244869 六、申請專利範圍 8 9.如申請專利範圍第88項之泛用機頂盒(STB),其 中該STB控制程序為可操作以接收並自該泛用STB之使用者 提供指示。 90.如申請專利範園第87項之泛用機頂盒(STB),其 中該STB控制程序為可操作以調諧該STB到一第一頻道以選 擇由使用者要求之資料,判定該選定之資料之性質,解碼 該選定之貧料’解壓縮該選定之貧料’重組該解碼貧料5 儲存該選定之資料於該記憶,並在一適當處理過之後提供 該選定之資料至一輸出設備。Page 77 1244869 VI. Scope of patent application 8 9. If the general-purpose set-top box (STB) of the scope of patent application No. 88, the STB control program is operable to receive instructions from the user of the general-purpose STB. 90. The general-purpose set-top box (STB) of item 87 of the patent application park, wherein the STB control program is operable to tune the STB to a first channel to select data requested by the user, and determine the selected data. By nature, decoding the selected lean material 'decompresses the selected lean material' and reassembles the decoded lean material 5 stores the selected data in the memory and provides the selected data to an output device after proper processing. 第78頁Page 78
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