TWI223563B - Methods and systems for transmitting delayed access client generic data-on-demand services - Google Patents
Methods and systems for transmitting delayed access client generic data-on-demand services Download PDFInfo
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奶3563 108408_^ Ά_修正 五、發明說明(1) ~ - 1 ·發明之領域: 本毛月係有關於一般之數據隨(Data-on-demand, ◦D)系統,本發明特別是有關於數據傳輸排序。 2 ·發明之背景: 奴選視訊(Vi deo-on-demand, V0D)系統為隨選數據系 =一種型式,在v〇D系統中,視訊數據檔為由一種伺服 1 3種伺服網路在隨選基礎上所提供予一個或多個用戶^ 〜在種傳統V0D架構中,一伺服器或一伺服器網路以 〜種標準層級,用戶伺服器型態與用戶通信,舉例來說, tfi ^二發出、凊求給一伺服器,要求一數據檔(例如一視 t 檔),為回應該用戶之請求,該伺服器送出該請求 -”給該用戶’在該標準之用戶伺服器型態中,一用戶 有Γ Ϊ據ΐ之請求可由一個或多個伺服器來完成,該用戶 y忐有旎力去儲存任何一個已接收到的數據檔於非揮發 (non-v〇latile)記憶體中以供稍後使用,該標準之用戶 L服器型態為要一種雙向通信設施。目前,雙向通信需要 ,,新設施,因為現有之電纜僅能提供單向通信,雙向通 信設施例係混合光纖同轴電纜(hybrid f iber 〇ptics coaxial cables,HFC)或全光纖設施,取代現有之電纜設Milk 3563 108408_ ^ Ά_Revision V. Description of the Invention (1) ~-1 · Field of Invention: The present month relates to a general Data-on-demand (◦D) system, and the present invention is particularly related to Data transfer sequencing. 2. Background of the invention: The slave video (Vi deo-on-demand, V0D) system is an on-demand data system = a type. In the vOD system, the video data file is composed of one servo, one servo network, and three servo networks. Provided to one or more users on an on-demand basis. ^ In a traditional V0D architecture, a server or a server network communicates with the user at a standard level of user-server type, for example, tfi ^ Issue, request to a server, request a data file (such as a t-file), in response to the user's request, the server sends the request-"to the user" in the standard user server type In the state, a user's request can be completed by one or more servers. The user has no power to store any received data file in non-volatile memory. For the later use, the standard user server type is a two-way communication facility. Currently, two-way communication requires new facilities because the existing cable can only provide one-way communication. 1. hybrid fiber coaxial cable er 〇ptics coaxial cables (HFC) or all-fiber facilities, replacing existing cable installations
第6頁 〜 發明說明(2) ~^~ i μ 义昂貴然而所促成之服敌+ 負擔。 战之服務未必能由大部分使用者何 另 〜電纜 ^ ’該 提供〜 每〜項 的能力 列之大 11 °然 訊數據 > ’因 力口,支曾 器之控 務提供 戶時有 服務提 來自每 數量與 且快速 當作局 一次僅 1 0 0 0至 器之數 以維持 者(例係 許多限 供者必須 一用戶之 此一機制 的磁碟陣 部伺服 能送出視 2 0 0 0個用 量必須增 局部伺服 外,該標準用戶伺服器型離卷 二J,圖提供V0D服務給::i 不準用戶伺服器型態之_ 機制以在該網路中不斷 1f該 ^ * 1卜®^揽聽及滿足 二求,因此,可收取到服 其中-機制利用具有大;二 =並聯 uasSiVely_parallel)電腦 法,即使既有最快的局部伺服器每 =(video-on-demand streain)給約 此’為了服務更多用戶,局部伺服 加局部伺服器需要更多上層伺服器 制。 " 的頻ΐ準Ξ C伺服器3態之另:限制為每-用戶需要自己 抑 匕,此總需求頻寬係正比於用戶的數量,於届 ;:Γ ί中之緩衝記憶體(cache memory)已經用來提昇頻 仍是ΐί ί用緩衝記憶體並未解決該問題因為緩衝記憶體 目前’為了使視訊隨選服務更能讓用戶負擔得起,現 存的服務提供者正在局部伺服器的能力上,增加每一局部Page 6 ~ Description of the invention (2) ~ ^ ~ i μ is expensive, but it contributes to adversary + burden. The service of war may not be provided by most users. ~ Cable ^ 'It should be provided ~ The capacity of each item is 11% larger than that of the data.>' Because of the power of the mouth, the control of the support device provides services from time to time. The number from each number and quickly regarded as a round only 1 0 0 0 to the device to maintain (for example, many suppliers must be a user of this mechanism of the disk array servo can send video 2 0 0 0 The amount must be increased in addition to the local servo. The standard user server type is off the second volume. The map provides V0D services to: i The user server type is not allowed to _ mechanism to continuously 1f the ^ * 1bu in the network. ® ^ Listen and satisfy the two requirements, therefore, can be charged to the service-the mechanism has a large use; two = parallel uasSiVely_parallel) computer method, even if the fastest local server has About this, in order to serve more users, local servos plus local servers need more upper-level server mechanisms. " The frequency of the C server is different from the three states of the C server: the limitation is that each user needs to suppress the dagger. This total required bandwidth is proportional to the number of users. Yu:; Γ Buffer memory memory) has been used to improve the frequency. Buffer memory is not used to solve the problem because buffer memory is currently 'in order to make video on-demand services more affordable for users, existing service providers are using local server capabilities. Add each part
第7頁 1223563 Λ 修正 曰 案號 9110841 五、發明說明(3) 伺服器的用戶比例,一般夾% ^ ^ H况’有能力提供服務給1 0 0 0個 用戶的夕局1】服益’事實上可用來服務1 0 0 0 0個用戶, 5 ί:不在同—個時間中點選視訊的話’此技 術也許可盯’:而’此技術卻是 的用戶有可能在同一個時間中相面*1、 口乃穴夕數 、两士、 ^ E. y 心要觀看視δίΐ (例如傍晚與 週末),而&成該局部伺服器在如此 因此,本發明希望提供一有能力於n出負載 供應隨選服務給大量用戶的夺轉、 可传傳輸媒介上 施,進-步來說,本發明希望提供—且有血存的設 關的傳輸頻寬之用戶通用廣播系统,;^:者數量無 其有能力來供應要求縮小傳輸頻寬的用戶系統, 選服務。 I用-人佳資料隨 3·發明之目的及概述 本發明提供一種可於一狹窄頻寬下同時 個虽作縮小頻寬用戶通用資料區塊排 =二個或夕 戶之DOD廣播系統,而沒有雙向通信之貝,枓^給大量用 步提供—視訊接收器,其可在資料檔被&用 發明進一 靡 間中,透過一用戶通用格式開始進 的短暫時 一步提供—更大噸t对率的方车卡丁貝科褚廣播,更進 遲完玲拉 寬羊的方法來下载資料檔案,刹田μPage 7 1223563 Λ Amendment Case No. 9110841 V. Description of the Invention (3) Proportion of users of the server, generally% ^ ^ H condition 'the ability to provide services to 1 0 0 users 1] Benefits' In fact, it can be used to serve 1 0 0 0 0 users, 5 ί: If you do n’t click on the video at the same time, 'this technology is also allowed to stare': while 'this technology is the same user may be at the same time Face * 1, the number of mouths and holes, two persons, ^ E. y want to watch as δίΐ (such as evening and weekends), and & into the local server in this way, therefore, the present invention hopes to provide an ability to The load-on-demand service is provided to a large number of users, and the transferable transmission medium is applied. Further, the present invention hopes to provide a universal broadcast system for users with a well-established transmission bandwidth; ^: The number of users has no ability to provide user systems that require a reduction in transmission bandwidth and choose services. I-Personal Information with 3. Objects and Overview of the Invention The present invention provides a DOD broadcasting system that can simultaneously reduce the bandwidth of a user ’s general data block arrangement = two or even under a narrow bandwidth, and There is no two-way communication, and it provides a large number of uses-a video receiver, which can be provided in a short time when the data file is & On the side of the square car Kart Beco Chu broadcast, even more Chi Wanling widen the way to download the data file, brake field μ
遲客戶接收時間來允許一智慧機一 ▲ ’、利用I 1223563 案號 91108408 五、發明說明(4) 簡略地說 稽案的數據隨 當作一縮小的Allow customers to receive a smart phone one time later ▲ ’, use I 1223563 case number 91108408 V. Description of the invention (4) Briefly, the data of the audit case is treated as a reduced
來廣播 廣播伺 廣播伺 不斷地 輸媒體 被選擇 用來接 的時間 者,在 料檔案 大量資 服器連 服器連 將大量 傳給大 之資料 收此授 之後用 此段被 之前儲 ,本發明之 選選廣播系 客戶通用資 料檔案的數 結的傳輸媒 結的大量接 的資料檔案 量的接收器 檔案一致的 權資訊;並 一觀點 統來實 料區塊 據隨選 · 、妥 體,遷 收器; 以縮小 ;在此 授權資 來顯示一部份的被 指定的時間之中能 存一部份的資料檔 可由一用來傳輸大量資料 現,其中每一資料檔案被 序列來傳輸,包含:一用 (D〇D)廣播伺服器;一與 過此傳輸媒體與數據隨選 在數據隨選廣播伺服器中 的客戶通用格式透過此傳 接收器中會用來要求與一 訊;此接收器會更進一步 收器更進一步在一段指定 選擇之資料檔案給使用 讓接收器在開始顯示此資 案。 序列之 資料區 資料區 土鬼序列 > ’於 %。更 間反應 佳資料 〜時間 發明之另 數據隨 塊之次 塊是完 構成之 此一接 進一步 出至少 傳輸序 間隔; 侧貫施例為 ⑺木得輸次佳的 選(D0D)廣播方法,包含下列動作 佳資料傳輸序列,在序列中沒有相 全相同的;以非用戶指定方式 案根據次佳傳輸序列傳輸給 戶將於指定得時段中開始接 二^定的時段有-持續時間,而 組資料區塊序列的資訊,此 列的動作包含:準備一資料柃 根據此時間間隔將此資料檔‘;析 準備一 鄰的兩個 由資料區 大量用 收資料檔 此持續時 一準備次 明確指定 成大量資Those who come to broadcast, broadcast, and broadcast continuously broadcast the time that the media is selected to pick up. After receiving a large amount of data from the server, the server and the server will send a large amount of data to the server. The selected broadcast is the client ’s general data file, the number of transmission media, the number of data files, and the number of data files in the receiver. The receiver ’s files are consistent with the right information. Receiver; to shrink; in this authorized resource to show a part of the specified time can store a part of the data file can be used to transfer a large amount of data, where each data file is transmitted serially, including : One-use (DOD) broadcast server; one with the transmission media and data on-demand in the data-on-demand broadcast server. The general format of the client will be used to request and send a message through this receiver; this receiver The receiver will take the receiver one step further and specify a selected data file for use by the receiver to display the item at the beginning. Sequence Data Area Data Area Dungeon Sequence > ’in%. Better response data ~ Time-invented another data with the block. The block is completed. This one at a time further transmits at least the transmission sequence interval. The lateral example is the best-selected (D0D) broadcast method of Tochigi, including The following action data transmission sequences are not exactly the same in the sequence; transmission in a non-user-specified manner according to the next-best transmission sequence to the user will begin in the specified time period and then have a duration of- The information of the data block sequence. The actions in this row include: prepare a data, and then prepare this data file according to this time interval; analyze and prepare two neighboring data files that are used by the data area in large quantities. Large amount of money
1223563 —一__案號 91108408 _^ 月 五、發明說明(5) 料區塊使得每一資料區塊在該時間間隔中成為可顯示的; ,定一所需要的時區(time slots)數來傳送數據檔案,在 每一個時間狹縫中擁有等同於時間間隔之一持續時間;將 至少一個此大量數據區塊中之一的數據區塊配置給每一個 時間狹縫。 一數據隨選系 伺服器用來控制此 -組頻道伺服器相 器相連結, 據。在一實 組用來監視 器,以及一 統錯誤發生 服器會對頻 一組頻道伺 道伺服器中 之一增頻器 而一結 施例中 此系統 第二組 時用來 道監視 服器中 之一頻 切換成 統包含第一組頻道伺服器,一中央控制 苐一組頻道伺服器’第一組增頻器與第 連,,一結合器/放大器與第一組增頻 合器/放大器可透過傳輸媒體來傳送數 ’此數據隨選系統更包含一頻道監視模 :一個開關矩陣,一第二組頻道伺服 增頻器。此頻道監視模組是安裝來在系 回報給中央控制器之用。此中央控制伺 模、、且回報採取反應’指示開關矩陣將第 =缺陷之一頻道伺服器切換成第二組頻 2伺服器,並將第一組增頻器中有缺陷 一組增頻器中之一增頻器。 一用來接收以士 法,包含下列動作^ ^數據區塊序列傳輸之數據檔案的 案;在指定的時段收一使用者輸入指示選擇的數據 大量數據區塊之〜秘存至少一個此次佳數據區塊序列= :記憶體位置;在指定的 田案的第一部份給使用者;至少1223563 —One __Case No. 91108408 _ ^ Fifth, the description of the invention (5) The data block makes each data block displayable in the time interval;, determine the number of time slots required The data file is transmitted, and each time slot has a duration equal to one of the time intervals; at least one data block of one of the plurality of data blocks is allocated to each time slot. A data-on-demand server is used to control this-a set of channel server phase links. In a real group, it is used for monitoring, and a unified error occurrence server will serve one of the frequency increasers in a group of servers, and in the embodiment, the second group of the system is used to monitor the server. The first frequency switching system includes a first group of channel servers, a central control and a group of channel servers. The first group of frequency increasers is connected to the first, a combiner / amplifier and the first group of frequency increasers / amplifiers. The data can be transmitted through the transmission medium. This data on-demand system also includes a channel monitoring module: a switch matrix, and a second group of channel servo boosters. This channel monitoring module is installed for reporting to the central controller. This central control servo mode, and returns to take a response 'instructs the switch matrix to switch the channel server of the first defect to the second group of frequency 2 servers, and the defective group of frequency increasers in the first group of frequency increasers. One of the upconverters. One is used to receive the data file of Ethiopia, which contains the following actions: ^ ^ Data block sequence data file transmission; Receive a user input instruction to select a large number of data blocks within a specified period of time ~ keep at least one Data block sequence =: memory location; to the user in the first part of the designated field plan; at least
段過去後至少顯At least
第10頁 1223563 案號 91108408 Λ:_η 曰 修正 五、發明說明(6) 收一個此次佳數據區塊序列之大量數據區塊中之一額外數 據區塊;並將數據檔案的第二部份以結合至少一個此被儲 存數據區塊之一與此額外的數據區塊的方式顯示給使用 者。 此的視訊接收器並不打算被限制成屬於電視機的裝 置,而也許包含任何能夠用來以與本發明一致的方法接收 廣播傳輸的裝置包括更進步的電視系統與電腦。 ί.發明詳細說明 本發明提供一種能在狹窄頻寬中同時將一個或多個數 據檔案以次佳數據區塊序列的形式傳輸給大量用戶之數據 隨選(DOD)廣播系統,而不需要雙向通訊,本發明更進一 步提供一種能在用戶選取一數據檔案的短暫時間中開始撥 放此數據檔案之機頂盒(STB),更進一步地,提供一更大 頻寬效率的方法來下載數據檔案,利用延遲客戶接收時間 來允許一智慧機頂盒在開始撥放數據檔案之前承載部分數 據檔案。Page 10 1223563 Case No. 91108408 Λ: _η Revision V. Description of the invention (6) Receive one additional data block from a large number of data blocks of the best data block sequence; and the second part of the data file Displayed to the user by combining at least one of the stored data blocks with this additional data block. This video receiver is not intended to be limited to a device belonging to a television, but may include any device capable of receiving broadcast transmissions in a manner consistent with the present invention, including more advanced television systems and computers. DETAILED DESCRIPTION OF THE INVENTION The present invention provides a data-on-demand (DOD) broadcasting system capable of simultaneously transmitting one or more data files in the form of a sub-best sequence of data blocks to a large number of users in a narrow bandwidth, without the need for two-way Communication, the present invention further provides a set-top box (STB) that can start to release a data file within a short period of time when the user selects the data file. Furthermore, it provides a more bandwidth-efficient method to download the data file. Delay the client's reception time to allow a smart set-top box to carry part of the data file before it starts playing the data file.
第11頁 I223563Page 11 I223563
曰 修正 ~ _ ^_ 案號 91108408_^ 五、發明說明(7) =二當這些方法提供用戶通用數據隨選(D0D)服務時’更 步減少了所舄要的傳輸頻寬來交換接收時間的延遲 第1A圖說明一與本發明之一實施例一致的典型隨選視 ^統1 00,在此實施例中,此視訊隨選系統i 〇〇並不只被 於提供視訊檔案隨選,而是也能夠來提供其他的數據 :案。舉例來說,隨選遊戲檔案。此視訊隨選系統100包 括一中央控制伺服器102, 一中央儲存器1〇3,大量 服器104a-10 4n,大量增頻器106a—106n,以及一結合器/ 放大器108,此中央控制伺服器102控制頻道伺服器 此中央儲存器103以數位格式儲存數據檔案,在一°典型的 實施例中,數據檔案儲存在中央儲存器1〇3中,可^經由 一標準的網路介面(例如,乙太網路連結)讓任何經授權的 電腦所取得,使得中央控制伺服器1 〇 2連結到網路上。每 一頻道伺服器1 04都被指定到一個頻道,並且連結到一個 增頻器106,此通道伺服器104提供由中央儲存器1〇3根據 中央控制伺服器1 0 2之指令所取回之數據檔案,每一通道 伺服器之輸出均是一種直角相位差放大調變(quadrature amplitude modulation,QAM)之已調變的中間頻率 (intermediate frequency, I F )訊號,擁有適當的頻率來 對應增頻器1 0 6,此QAM已調變I F訊號取決於已被採用之標 準,目前美國最新的採用標準為電缦上數據系統界面規τ (data-over-cable-systems-interface-specification DOCS IS)標準,此標準需要一大約43. 5 7MHz的IF頻率,此Amendment ~ _ ^ _ Case No. 91108408_ ^ V. Description of the invention (7) = Second, when these methods provide users with universal data on demand (DOD) services, 'the required transmission bandwidth is further reduced to exchange reception time. Delaying FIG. 1A illustrates a typical on-demand video system 100 consistent with one embodiment of the present invention. In this embodiment, the video-on-demand system i 00 is not limited to providing video files on-demand, but Can also provide other data: case. For example, on-demand game files. The video-on-demand system 100 includes a central control server 102, a central storage 103, a large number of servers 104a-10 4n, a large number of boosters 106a-106n, and a combiner / amplifier 108. The central control servo The central storage 103 stores the data files in a digital format. In a typical embodiment, the data files are stored in the central storage 103, which can be accessed through a standard network interface (such as , Ethernet connection) to be obtained by any authorized computer, so that the central control server 102 is connected to the network. Each channel server 104 is assigned to a channel and is connected to an up-converter 106. This channel server 104 provides it to be retrieved by the central storage unit 103 according to the instructions of the central control server 102. Data files. The output of each channel server is a modulated intermediate frequency (IF) signal of quadrature amplitude modulation (QAM). It has an appropriate frequency to correspond to the booster. 1 0. This QAM has modulated the IF signal depending on the adopted standard. The latest adopted standard in the United States is the data-over-cable-systems-interface-specification DOCS IS standard. This standard requires an IF frequency of about 43. 5 7MHz.
第12頁 1223563 厂--0^_JU〇84〇8___ 年月曰 修正 五、發明說明(8) 增頻裔1 0 6將由頻道伺服器丨〇 4接收的丨F訊號轉換成無線電· 頻率訊號(RF signal ),此包含頻率與頻寬的RF訊號,取 決於一需要的頻道與所採用的標準。舉例說明,目前美國 i 有線電視頻道80的最新標準下,此RF訊號的頻率大約55 9. j5MHz,而頻寬大約6MHz,該增頻器106的輸出則送到結合 器/放大器1 08,該結合器/放大器丨〇8放大、整理、結合該 接收到的RF訊號在輸出至傳輸媒體1丨〇。 在一典型實施例中,該中央控制伺服器1 〇 2包括一圖 形使用者介面(未顯示出)讓一服務提供者利用取放(drag一暑 and- drοp)彳呆作來進行數據傳送之排序,更進一步,此中 央控制伺服器1 0 2授權並控制該頻道伺服器丨0 4根據傳送矩 陣來啟動或停止’在一典型實施中,該中央控制伺服器 102自動選擇一頻道並在所選擇的頻道中計算傳送矩陣來 傳輸數據檔案,此中央控制伺服器1 〇 2提供離線的數據檔 案資訊的增加、刪除以及更新(例如:持續時間、類別、 額訂功率、以及/或簡短的敘述),更進一步,此中央控制 伺服器1 0 2利用更新數據檔案以及儲存數據庫在其中來控 制中央控制器1 0 3。 • 在一典型的實施例中’一存在的有線電視系統1 2 0也 許持續供應訊號到該結合器/放大器1 0 2來提供非隨選視訊 服務給客戶,因此與本發明一致之隨選視訊系統1 0 0並不 - 中斷該有線電視服務。 1Page 12 1223563 Factory--0 ^ _JU〇84〇8 ___ Years and months of revision V. Description of the invention (8) The frequency-increasing source 1 0 6 converts the F signal received by the channel server 丨 〇4 into a radio frequency signal ( RF signal), which includes an RF signal of frequency and bandwidth, depending on a required channel and the standard used. For example, under the latest standard of the US i cable channel 80, the frequency of this RF signal is about 55 9. j5MHz, and the bandwidth is about 6MHz. The output of the booster 106 is sent to the combiner / amplifier 108. The combiner / amplifier 8 amplifies, organizes, and combines the received RF signal and outputs it to the transmission medium 1. In a typical embodiment, the central control server 102 includes a graphical user interface (not shown) for a service provider to perform data transmission using drag and drop operations. Sequencing, further, the central control server 1 0 2 authorizes and controls the channel server 丨 0 4 to start or stop according to the transmission matrix. In a typical implementation, the central control server 102 automatically selects a channel and The transmission matrix is calculated in the selected channel to transmit the data file. This central control server 10 provides offline data file information addition, deletion and update (for example: duration, category, subscription power, and / or short description) ) Furthermore, the central control server 10 uses the updated data file and stores the database to control the central controller 103. • In a typical embodiment, 'an existing cable television system 120 may continue to supply signals to the combiner / amplifier 102 to provide non-on-demand video services to customers, so on-demand video in accordance with the present invention System 1 0 0 does not-interrupt the cable TV service. 1
第13頁 1223563 -年月日 修正 五、發明說明(9) 第1 B圖說月—與本發明之另一實施例一致的典 視訊系統1 0 0,除了 A,ΛΕ;Ι 士 α %⑽认一主 '、 、 于Τ在弟1 a圖中所說明的7〇素之外,該隨 選視訊系統1 0 0還句人 叫M &访! ! 〇 ^ _ 疋巴含一開關矩陣1 1 2,一頻道監視模組 114’ 一組備用頻道伺服器116a-116b,一組備用增頻器 11 8a— 11 8b ’在—實施例中,該開關矩陣11 2被置放在增頻 器1 0 6以及結合器/放大器1 0 8之間,該開關矩陣1丨2由中央 控制伺服器所控制,該頻道監視模組11 4則由大量已裝配 的模擬潛在用戶的視訊接收器(s e t — t 〇 p _ b 〇 χ e s )所組成, 用來監視隨選視訊系統1 ο 0的正常狀態,監視結果利用頻 道監視模組11 4傳送到中央控制伺服器i 〇 2,如果發生錯誤 (例如:一頻道伺服器失效、一增頻器失效、或者一通訊 連結失效)’該中央控制伺服器1 〇 2則透過開關矩陣1丨2解 除故障的元件,並接合正常的備份元件n 6以及/或n 8來 衝重新恢復服務。 在一典型的實施例中,由數據隨選系統(DOD)1〇〇廣播 的數據檀案動慇影像壓縮標準(motion pictures expert group,MPEG),每一動態影像壓縮標準檔案都被動態分割 成數據區塊以及子區塊’並隨著一時間轴繪製到一特定的 數據檔案’這些數據區塊以及子區塊依照由中央控制伺服 器1 0 2所提供的三維傳送矩陣在一預定的時間内被送出, 一回授頻道並不需要由數據隨選系統1 〇 〇來提供數據隨選 服務,然而,如果一回授頻道可利用,則此回授頻道就無Page 13 1223563-Year Month Day Amendment V. Description of the Invention (9) Figure 1 B illustrates the month—a typical videoconferencing system 100 consistent with another embodiment of the present invention, except for A, ΛΕ; Ιαα% recognition One master ',, and Yu are described in the picture of brother 1a, and the video-on-demand system 100 also calls M & visit! !! 〇 ^ _ 疋 Bar contains a switch matrix 1 12, a channel monitoring module 114 'a set of backup channel servers 116a-116b, a set of backup boosters 11 8a-11 8b' in-embodiment, the switch Matrix 11 2 is placed between up-converter 106 and combiner / amplifier 108, the switch matrix 1 2 is controlled by a central control server, and the channel monitoring module 11 4 is assembled by a large number of Video receiver (set — t 〇 p _ b 〇 χ es) to simulate potential users, used to monitor the normal state of the on-demand video system 1 ο 0, the monitoring results are transmitted to the central control using the channel monitoring module 114 Server i 〇2, if an error occurs (for example: a channel server failure, a booster failure, or a communication link failure) 'the central control server 1 〇 2 through the switch matrix 1 丨 2 to remove the faulty component , And join the normal backup element n 6 and / or n 8 to restore the service. In a typical embodiment, data is broadcasted by the Data On Demand System (DOD) 100. Motion picture expert group (MPEG), each motion picture compression standard file is dynamically divided into The data blocks and sub-blocks are drawn to a specific data file along with a time axis. These data blocks and sub-blocks follow a three-dimensional transfer matrix provided by the central control server 102 at a predetermined time. Is sent out, a feedback channel does not need the data on-demand service provided by the data on-demand system 1000, however, if a feedback channel is available, this feedback channel will be
第14頁 1223563 ______PM 911〇84〇g_±__Ά_Θ______—修正_ 五、發明說明(10) 法用來進行其他用途,如廣告或提供網際網路服務。 第2圖說明一與本發明之一實施例一致的典型頻道伺 服器1 0 4,該頻道伺服器1 0 4包含一伺服控制器2 0 2,一中 央處理機(CPU) 2 04,一直角相位差放大調變器(QAM) 20 6,Page 14 1223563 ______PM 911〇84〇g_ ± __Ά_Θ ______— Amendment_ V. Description of Invention (10) The law is used for other purposes, such as advertising or providing Internet services. FIG. 2 illustrates a typical channel server 104 consistent with an embodiment of the present invention. The channel server 104 includes a servo controller 202, a central processing unit (CPU) 204, and a right angle. Phase Difference Amplifier Modulator (QAM) 20 6,
一局部記憶體2 0 8,以及一網路介面2 1 0。此伺服控制器 2 0 2利用命令中央處理為20 4分割數據槽案成區塊來控制頻 道伺服器1 0 4的全部動作(更進一部分分割成資區塊以及數 據封包),根據由中央控制伺服器1 〇 2所提供的傳送矩陣來 選擇數據區塊,將所選擇的數據編碼,將以編碼的數據壓 縮,然後傳送壓縮數據到QAM調變器20 6,該QAM調變器206 接收數據再透過匯排流(例如,PC I、CPU局部匯排流)或乙 太網路連結來傳輸,在一典型的實施例中,該QAM調變器 2 0 6也許包括一下游QAM調變器,一帶有往前錯誤修正解碼 器的上游 QAM/QPSK(quadrature phase shift keying)爆 破式解調變器,以及/或一上游調節器,該qAM調變器2〇 6 的輸出則是一能直接送到一頻1 〇 6的I F訊號。該網路介面A local memory 2 0 8 and a network interface 2 1 0. This servo controller 2 uses the command central processing to divide the data slots into 20 4 blocks to control the entire operation of the channel server 104 (a further part is divided into data blocks and data packets). According to the central control The transmission matrix provided by the server 102 selects data blocks, encodes the selected data, compresses the encoded data, and transmits the compressed data to the QAM modulator 20 6 which receives the data And then transmitted through a bus (for example, PC I, CPU local bus) or an Ethernet connection. In a typical embodiment, the QAM modulator 206 may include a downstream QAM modulator , An upstream QAM / QPSK (quadrature phase shift keying) blast demodulator with a forward error correction decoder, and / or an upstream regulator, the output of the qAM modulator 206 can be directly IF signal sent to a frequency of 1.06. The web interface
1 0 2連接此頻道伺服器1 〇 4到其他頻道伺服器1 〇 4以及中央 控制饲服β 1 0 2來執行排序並控制從中央控制伺服器丨〇 2來 的才曰不’回報狀態給中央控制伺服器丨〇 2,並從中央儲存 2 l〇f接收數據資料’任何從中央儲存器1〇3接收的數據資 二Ϊ此,在數據資料根據伺服控制2 0 2的指令被進行處理 誉月:ϋ I在頻道飼服11 10 4的局部記憶體20 8中。在一典型 % ? 此頻道飼服器1 04可能根據有限頻道的頻寬送1 0 2 Connect this channel server 1 〇 04 to other channel servers 1 〇 4 and the central control feeding service β 1 0 2 to perform sorting and control the status from the central control server 〇 02. The central control server 丨 〇2, and receive data from the central storage 2 l0f 'any data received from the central storage 103 data, so the data is processed according to the instructions of the servo control 202 Yuyue: ϋI is in the local memory 20 8 of the channel feeding 11 10 4. In a typical%? This channel feeder 1 04 may be sent according to the bandwidth of a limited channel
•第15頁 1223563 案號 91108408 年 月 曰 修正 五、發明說明(11)出一個或多個數據隨選資料串(例如:6、6 · 5、或8奸 QAM調變(例如QAM64或QAM2 5 6,以及壓縮標準/隨選數I) ’ 流之位元率,即是MPEG-1或MPEG-2)。 康串 第3圖說明一與本發明之一實施例一致的典型機了貝4 300,此 STB30 0 包括一 QAM 解調變器 302,一 CPU304,二 條件存取模組3 0 6 (例如一智慧卡片系統),一局部記恢^ 3 0 8’ 一緩衝記憶體30 9,一 ST胳制器310,一解碼器〜— 3 1 2,以及一圖形重疊模組3丨4,此ST她制器3 1 0控制所有 3丁83 0 0的工作,利用控制〇?113 0 2以及卩入河解調變器3 0 2來選 擇數據以回應用戶的要求,解碼所選擇之數據,解壓所已 解碼之數據,重新組合解壓縮數據,儲存解壓縮數據到局 部記憶體3 0 8或暫存記憶體3 0 9,並傳送儲存數據至解壓縮 器3 1 2,在一典型的實施例中,此ST鹏制器3丨〇根據從傳 輸媒體11 0接收到的數據封包的數據封包端控制所有 STB3 0 0的工作。在一典型的實施例中,此局部記憶體3〇8 包含非揮發性(non —volati le)記憶體(如一硬體驅動),以 及包含非揮發性記憶體之暫存記憶體3 0 9。 在此一實施例中,QAM解碼器30 2包含傳送器以及接收 器模組,以及一個或多個如下之元件:加密/解密模組、 前錯誤修正解碼器/編碼器、調節器控制、下游與上游處 理器、cpu與記憶體介面電路,此qAM解調變器3〇2接收已 調變I F訊號,取樣並解調變此訊號成原來的數據。• Page 15 1223563 Case No. 91108408 Rev. V. Invention Description (11) One or more data on-demand data strings (for example: 6, 6, 5, or 8 QAM modulation (for example, QAM64 or QAM2 5 6, and the compression standard / on-demand number I) 'The bit rate of the stream, which is MPEG-1 or MPEG-2). Fig. 3 illustrates a typical machine in accordance with an embodiment of the present invention. The STB 300 includes a QAM demodulator 302, a CPU 304, and two conditional access modules 3 0 6 (for example, a Smart card system), a local memory ^ 3 0 8 ', a buffer memory 30 9, a ST controller 310, a decoder ~-3 1 2 and a graphics overlay module 3 丨 4, this ST her The controller 3 1 0 controls all the operations of 3 3 3 0 0, and uses the control 0 1 3 2 0 2 and the Lu River demodulator 3 2 2 to select data in response to the user's request, decode the selected data, and decompress the data. The decoded data is recombined with the decompressed data, the decompressed data is stored in the local memory 308 or the temporary storage memory 309, and the stored data is transmitted to the decompressor 3 1 2 in a typical embodiment. The ST Peng controller 3 controls the operation of all STBs 300 according to the data packet end of the data packet received from the transmission medium 110. In a typical embodiment, the local memory 308 includes non-volatile memory (such as a hard drive), and temporary memory 309 including non-volatile memory. In this embodiment, the QAM decoder 302 includes a transmitter and receiver module, and one or more of the following components: an encryption / decryption module, a previous error correction decoder / encoder, a regulator control, and a downstream With the upstream processor, CPU and memory interface circuit, this qAM demodulator 302 receives the modulated IF signal, samples and demodulates this signal into the original data.
第16頁 1223563 -1^ 91108408 年月曰 修正 五、發明說明(12)Page 16 1223563 -1 ^ 91108408 Date of Amendment V. Description of Invention (12)
在授權以及/或已收取一適當費用後,存取受到同意 時’此有條件之存取模組3 〇 6允許一解碼程序,存取條件 則由服務提供者所規定。After authorization and / or an appropriate fee has been charged, and when access is granted, the conditional access module 306 allows a decoding process, and the access conditions are specified by the service provider.
授權的方法包括直接插入訂閱等級以及預告等級到傳 輸隨選數據货標中,機頂盒(STB ) 3 0 0以此方法讀取訂閱等 級並比較儲存在STB3〇〇中之訂閱等級,如果儲存的訂閱等 級與隨著數據隨選服務傳來的訂閱等級相同,則客戶即被 授權接收此服務,至於預告等級,一 STB3〇〇讀取一隨著數 據隨選服務傳來的預告等級,並顯示與預告等級相符之訊 息’此授權方法由黃奎的專利應用所提出,專利標題為 「數據隨選客戶存取控制」(Controlling Data-〇n-Demand Client Access)^ 2 0 0 1 年 7月 9日提出, 所申請之美國申請案號09/902,503,其已納入參考文獻 中 〇 在一實施例中,當存取一旦被同意,該解碼器3 1 2解 碼至少一數據區塊轉換成可在輸出螢幕上顯示的影像,此 解碼Is 3 1 2支援由盯閱客戶端來的指令,如撥放、停止、 暫停、步進、倒轉,前轉等。 該圖形重疊模組3 1 4利用,舉例說明,提供阿法混合 (alp ah blending)或圖中圖(pictur-in - picture)來增加The authorization method includes directly inserting the subscription level and advance notice level into the on-demand data tag. The set-top box (STB) 3 0 0 reads the subscription level and compares the subscription level stored in STB 300. If the stored subscription The level is the same as the subscription level that came with the data on-demand service, then the customer is authorized to receive this service. As for the notice level, one STB300 reads the notice level that comes with the data on-demand service and displays the same as A message that matches the notice level 'This authorization method was proposed by Huang Kui's patent application, and the title of the patent is "Controlling Data-On-Demand Client Access" ^ July 2011 It was filed today that the applied US application number 09 / 902,503 has been incorporated in the reference. In one embodiment, once access is granted, the decoder 3 1 2 decodes at least one data block into Output the image displayed on the screen. This decoding Is 3 1 2 supports the instructions from the viewing client, such as play, stop, pause, step, reverse, forward, etc. This graphics overlay module 3 1 4 uses, for example, to provide alp ah blending or pictur-in-picture to increase
第17頁 !223563 —-m^rnm_日 修正 五^明說明'——- =不的圖形品質’在一實施例中,此圖形重疊模組3"能 =用來在遊戲執行模式時進行圖形加速,例如當服務提供 者利用本發明提供隨選遊戲服務時。 二在一實施例中,即使數據資料被廣播到所有有線電視 二f者,唯一只有隨選數據訂閱者擁有相容的STB3 0 0而能 =碼並享受數據隨選服務’在實施例中,可經由:; =制模中之智慧卡系統來獲得數據標案隨選条之牛 =思柘,一智慧卡也許能在一般商店取得,也可能 棱供者提供並安裝,在另一實施例中,統一 訂閱者對可取得的數據槽案進行無限制的存:糸、、先“、 在一實施例中,隨選數據互動的優 時間選擇可取得的數據稽案,從客戶按;在任何 一直到所選取的數據構案開始撥放的時,5 ,的時刻 稱之為反應時間,當越多的資源(例如.B 、B、間總數 *上:皮配置來提供隨選數據服務,則反應時間合::器: 一實施例中,一反應時間可由資源配置邀曰 ,在 的估算來決定。 、所而的服務品質 在一實施例中,一已選擇的反應時 時間,此時區(TS)的持續時間是讓客戶j :時區的持續 區塊的時間間隔。在一實施例中,—數 =速度放數據 案,被分成大量的數據區塊,使得每—愈^;如視訊檔 裝據區塊可在此時Page 17! 223563 —-m ^ rnm_Day Correction 5 ^ Describes '——- = No graphics quality' In one embodiment, this graphics overlay module 3 " can = be used in game execution mode Graphics acceleration, such as when service providers use the invention to provide on-demand game services. In an embodiment, even if the data is broadcast to all cable TV users, only the on-demand data subscribers have compatible STB3 0 0 and can == code and enjoy data on-demand services. 'In the embodiment, Can be obtained through :; = Smart card system in the mold to obtain the data of the optional item of the bull = thinking, a smart card may be available in general stores, or may be provided and installed by the supplier, in another embodiment In the case of unified subscribers, unrestricted storage of available data slots is available: 糸 ,, first ", in one embodiment, the preferred time for on-demand data interaction is to select available data audits, and press from customers; in Any time until the selected data structure starts to be released, the time of 5 is called the response time. When more resources (such as .B, B, the total number of rooms * upper: skin configuration to provide on-demand data services Then, the response time is :: In one embodiment, a response time can be determined by the resource allocation request, and the estimated service quality. Therefore, in an embodiment, a selected response time, this Time zone (TS) duration The time is the time interval between the continuous blocks of the client j: time zone. In one embodiment, the number of data files is divided into a large number of data blocks, so that each of them becomes more and more; Available at this time
1223563 --- 案號91108408_年月日 絛正 五、發明說明(14) 區的持續時間中讓數據檔案得以播放。 在一實施例中,每一數據檔案之數據區塊的數量 (NUM-OF-BLKS)可由以下的方式來計算:1223563 --- Case number 91108408_ 年月 日 绦 正 5. The description of the invention (14) allows the data file to be played during the duration of the zone. In one embodiment, the number of data blocks (NUM-OF-BLKS) of each data file can be calculated in the following ways:
Estimate-BLK-Size= (DataFile-Size*TS)/DataFile-Length (1) BLK-SIZE=(Estimate BLK-Size +CLUSTER-SIZE-1 Byte) /CLUSTER-SIZE (2)Estimate-BLK-Size = (DataFile-Size * TS) / DataFile-Length (1) BLK-SIZE = (Estimate BLK-Size + CLUSTER-SIZE-1 Byte) / CLUSTER-SIZE (2)
BLK-SIZE-BYTES:BLK-SIZE*CLUSTER-SIZE (3) NUM-OF-BLKS=(DataFile-Size+BLK-SIZE-BYTES-lByte) /BLK —SIZE一BYTES ( αλ 在方程式(1)到(4)中,其£31^11^七6一31^-8126是一區塊預 測大小(以位元組為單位);其D a t aF i 1 e — S i z e是數據標案 大小(以位元組為單位);TS為時區的持續時間(以秒為單 位);DataF i le — Length是數據檔案持績時間(以秒為單 位);BLK — SIZE是每一數據區塊所需的叢集數量, CLUSTER一S I ZE是在局部記憶體2 〇 8中給每一頻道伺服器所 用的叢極大小(如64K位元組),bLK_SIZE_BYTES以位元組 顯示的區塊大小’在此實施例中,區塊的數量 〇龍一〇『一61^3)等於數據檔案大小(以位元組)加上_以位 元組顯示之數據區塊大小減一個位元組在除以以位元組顯 示之數據區塊大小,方程式(丨)到(4 )說明了 一特有的實施BLK-SIZE-BYTES: BLK-SIZE * CLUSTER-SIZE (3) NUM-OF-BLKS = (DataFile-Size + BLK-SIZE-BYTES-lByte) / BLK —SIZE_BYTES (αλ in equation (1) to ( 4), where £ 31 ^ 11 ^ seven 6-31 ^ -8126 is the predicted size of a block (in bytes); its D at aF i 1 e — S ize is the size of the data item (in bits) Tuples); TS is the duration of the time zone (in seconds); DataF i — Length is the data file performance time (in seconds); BLK — SIZE is the cluster required for each data block The number, CLUSTER-SI ZE is the size of the cluster (such as 64K bytes) for each channel server in local memory 2.08, and bLK_SIZE_BYTES is the block size in bytes. In this embodiment, , The number of blocks 〇long one 『『 61 ^ 3) is equal to the size of the data file (in bytes) plus _ the size of the data block displayed in bytes minus one byte in the divided by the bytes The size of the displayed data block, equations (丨) to (4) illustrate a unique implementation
第19頁 1223563 ___案號年月日 修正__ 五、發明說明(15) 例’一熟悉此技藝之人會承認有其他方法來計算一數據檔 案之數據區塊數量,如將數據檔案分成一定量的數據區塊 唯一區塊預測大小及頻道伺服器1 〇 4上局部記憶體2 0 8之叢 集大小的主要功能,因此本發明並不貝特定限制於以上所 提出之實施例。 第4圖說明一根據本發明之一實施例來產生用來送出 數據檔案的序列矩陣之典型程序,在實施例中,本發明使 用時間分割多工器(time division multiplexing,TDM)以 及頻率分割多工器(frequency division mutiplexing,FDM)技術在伺服器端進行壓縮與序列數據傳 送,在一實施例中,每一數據檔案被分成一定數量的數據 區塊’並根據數據區塊的數量產生序列矩陣,典型來說, 依序列矩陣提供一送出的順序從伺服器傳送數據檔案的區 塊給客戶,而這些數據區塊可由任何想要在任意的時間中 存取這些數據檔案的客戶依序存取。 “Page 19, 1223563 __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ The main functions of a certain amount of data block unique block prediction size and the cluster size of the local memory 208 on the channel server 104 are not limited to the above-mentioned embodiments. FIG. 4 illustrates a typical procedure for generating a sequence matrix for sending data files according to an embodiment of the present invention. In the embodiment, the present invention uses time division multiplexing (TDM) and frequency division Frequency division mutiplexing (FDM) technology performs compression and sequence data transmission on the server side. In one embodiment, each data file is divided into a certain number of data blocks and a sequence matrix is generated according to the number of data blocks. , Typically, the sequence matrix provides a sent sequence of blocks of data files from the server to the client, and these data blocks can be sequentially accessed by any client who wants to access these data files at any time . "
在步驟4 0 2 ’數據槽案的數據區塊數量(χ)被接收,一第一 個變數,j,被設定成零(步驟404),一參考陣列被清除 (步驟406 ),此參考陣列為了内部的管理目的持續追蹤數 據區塊,接著,j與X比較(步驟408),如果j小於χ,一第 二變數i被設定=零==412),接下來,丨與义比較 414),如果i小於X’則被儲存在排序矩陣中行 (X)]的數據區塊責備寫到參考陣列中(步驟41 考陣列已有此數據區塊’則不重複寫A,最開始時,:為The number of data blocks (χ) in the data slot case is received in step 4 2 ', a first variable, j, is set to zero (step 404), and a reference array is cleared (step 406). This reference array Data blocks are continuously tracked for internal management purposes. Next, j is compared with X (step 408). If j is less than χ, a second variable i is set = zero = = 412). Next, 丨 is compared with meaning 414) If i is less than X ', the data block of row (X)] stored in the sorting matrix is blamed and written to the reference array (step 41, the test array already has this data block', then A is not repeatedly written. At the beginning, for
1223563 _案號91108408_年月曰 修正_ 五、發明說明(16) 排序矩陣尚未有入口,此一步驟可以被省略,接下來,參 考陣列被檢查是否含有數據區塊i (步驟4 2 0 ),起初因為所 有參考陣列中的入口都在步驟4 0 6中被清除,所以參考陣 列中沒有任何資料,如果參考陣列沒有包含數據區塊i, 數據區塊i則被加入排序陣列中[(i + j )對(X ),j ]的位置以 及參考陣列中(步驟4 2 2),在數據區塊i加入排序陣列以 及參考陣列中之後,i被加1,如此i = i + 1 (步驟4 2 4 ),接著 程序重複步驟4 1 4直到i = X,如果參考陣列包含數據區塊 i,i則被加1,如此i = i + 1 (步驟4 2 4 ),接著程序重複步驟 4 1 4直到i = X,當i = X,則j被加1,如此j = j + 1 (步驟4 1 6 ), 接著程序重複步驟4 0 6直到j = X,整個程序在j = X時結束(步 驟 4 1 0 )。 在一實施例中,如果一數據檔案被分成六個數據區塊 (x = 6 ),則此排序矩陣以及參考陣列即如下所示: 排序矩陣(SM) TSO TS1 TS2 TS3 TS4 [0,0]blk0 [l,0]blkl [2,0]blk2 [3,0]blk3 [4,0]blk4 TS5 [5, 0]blk5 [0,1] [1,1 ]blk0 [2, 1] [3,1] [4,1]1223563 _Case No. 91108408_ Year Month Amendment_ Five. Description of the invention (16) There is no entry for the sorting matrix. This step can be omitted. Next, the reference array is checked to see if it contains data block i (step 4 2 0) At first, because all the entries in the reference array are cleared in step 406, there is no data in the reference array. If the reference array does not contain data block i, data block i is added to the sorting array [(i + j) the position of (X), j] and the reference array (step 4 2 2), after the data block i is added to the sorting array and the reference array, i is incremented by 1, so i = i + 1 (step 4 2 4), then the program repeats step 4 1 4 until i = X. If the reference array contains data block i, i is incremented by 1, so i = i + 1 (step 4 2 4), and then the program repeats step 4 1 4 until i = X, when i = X, j is incremented by 1, so j = j + 1 (step 4 1 6), then the program repeats step 4 0 6 until j = X, the entire program is at j = X End (step 4 1 0). In an embodiment, if a data file is divided into six data blocks (x = 6), the sort matrix and reference array are as follows: Sort matrix (SM) TSO TS1 TS2 TS3 TS4 [0,0] blk0 [l, 0] blkl [2,0] blk2 [3,0] blk3 [4,0] blk4 TS5 [5, 0] blk5 [0,1] [1,1] blk0 [2, 1] [3 , 1] [4,1]
1223563 案號 91108408 年月曰 修正 五、發明說明(17) [5, 1] [0,2] [5, 2] 1,2] [2, 2]blk0 [3, 2]blkl [4, 2] [2, 3] [〇, 3] [1,3] [5, 3]blk2 [3, 3]blk0 [4, 3] [0,4] [1,4]blk3 [2, 4] [5, 4]blkl [3, 4]1223563 Case No. 91108408 Rev. V. Description of the Invention (17) [5, 1] [0,2] [5, 2] 1, 2] [2, 2] blk0 [3, 2] blkl [4, 2 ] [2, 3] [〇, 3] [1,3] [5, 3] blk2 [3, 3] blk0 [4, 3] [0,4] [1,4] blk3 [2, 4] [ 5, 4] blkl [3, 4]
[4, 4]b1kO[4, 4] b1kO
[2, 5] [0,5] [1,5] [5, 5]blk0 [3, 5]blk4 [4, 5] 參考陣列(RA)[2, 5] [0,5] [1,5] [5, 5] blk0 [3, 5] blk4 [4, 5] Reference Array (RA)
spaceO space 1 space2 space 3 space4 space5 TSO blkO blkO blkO blkO blkO blkO TS1 blkl blkl blkl blkl blkl blkl TS2 blk2 blk2 blk2 blk2 blk2 blk2 TS3 blk3 blk3 blk3 blk3 blk3 blk3 TS4 blk4 blk4 blk4 blk4 blk4 blk4 TS5 blk5 blk5 blk5 blk5 blk5 blk5 第22頁 1223563 _室號91108408_年月 _日 修正__ 五、發明說明(18) 在此典型的實施例中,基於以上之排序陣列,此數據檔案 的六個數據區塊被以下列的排序送出: TS0 = > b1k0 TS1 => blkO, blkl, blk3 TS2 => blkO, blk2 TS3 => blkO, blkl, blk3, blk4 TS4 => blkO, blk4 TS5 => b 1 k0,b1k1,b 1 k2,b1k5 泰 在另一典型的實施例中,一預視程序能用來在預測存 取時間之前計算出一個預視排序矩陣以送出一指定的數據 槽案之數據區塊,舉例來說,一預先指定的預視時間是一 時區’對任何時區大於或等於第4時區來說,數據檔案的 數據區塊(b lk) 4應該在時區(TS) 3時或之前被訂閱客戶處 的>STB3 0 0所接收,但直到TS4時Mk4都不應該被撥放,此 ,來產生一預視排序矩陣的程序步驟將一提早傳輸序 二,序、’此程序步驟相當類似於在以上所敘述圖&之spaceO space 1 space2 space 3 space4 space5 TSO blkO blkO blkO blkO blkO blkO TS1 blkl blkl blkl blkl blkl blkl TS2 blk2 blk2 blk2 blk2 blk2 blk2 TS3 blk3 TS4 blk 4 blk4 blk3 blk3 blk3 blk3 TS4 blk4 blk3 Page 22 1223563 _ Room No. 91108408_ Year Month _ Day Amendment __ V. Description of the Invention (18) In this typical embodiment, based on the above-mentioned sorted array, the six data blocks of this data file are sorted by the following Sort out: TS0 = > b1k0 TS1 = > blkO, blkl, blk3 TS2 = > blkO, blk2 TS3 = > blkO, blkl, blk3, blk4 TS4 = > blkO, blk4 TS5 = > b 1 k0, b1k1, b 1 k2, b1k5 In another typical embodiment, a preview program can be used to calculate a preview matrix before the access time is predicted to send a data block of a specified data slot. For example, a pre-specified preview time is a time zone. For any time zone greater than or equal to the 4th time zone, the data block (b lk) 4 of the data file should be subscribed at or before the time zone (TS) 3 customer Received at the user's station> STB3 0 0, but Mk4 should not be released until TS4. Therefore, the program steps to generate a preview matrix will be transmitted early. The sequence and sequence are similar. Figures & described above
二預視排序矩陣的程序步驟,在此實施例中根據_預^ 】=、,假言史一數據構,被分成六個*據區塊,一典型= ==排序陣列來傳送並在兩個時區之間擁 傳延序列能被表示如下: 卞間iThe procedural steps of the two preview matrix are, in this embodiment, according to _ 预 ^] =, the hypothesis history is a data structure, which is divided into six * data blocks, a typical === sorting array to transfer and transfer in two The time delay between the time zones can be expressed as follows: 卞 间 i
I223563 _ 案號91108408 年月日 修正 五、發明說明(19) TS0 二〉blkO TS1 => blkO, blkl, blk3, blk4 TS2 => blkO, blk2 TS3 => b 1 k0, b1k1,b 1 k3,b 1 k4,b1k5 TS4 => blkO, blk5 TS5 => blkO, blkl, blk2 一用來傳 排序矩陣所產 用,在此三維 檔案所用之三 每一頻道中有 -典型 被用來 行排序 所有數 小可能 所同意 而數據 -典型 案皆被 序如下 實施例 產生一 ,舉例 量的數 數量, ’則優 區塊則 系統送 分成六 送一組數據檔案的三維傳送矩陣 _ _______ _ 生來讓該數據檔案組中之每一個數據檔案所 傳送矩陣中產生一讓數據檔案組之每一數據 維識別證,此三維傳輸矩陣是被計算出來在 效利用可用之頻寬來傳輸多個數據串流,在 中,一迴旋方法,其在此技術中相當有名, 二維傳輸矩陣來對數據檔案組的有效傳輸進 说明,一迴旋方法也許包含下列策略:(1 ) 據區塊在任何時區得持續期間都應保持在最 以及(2 )如果有多個部分解決方案被策略(】 先的方案即是擁有最小數據區塊的那一個, 在之前的時區期間被送出,舉例來說,假設 出兩個簡短的數據檔案Μ以及N,每一數舉 個數據區%,根據排序矩陣所形成的傳送:I223563 _ Case No. 91108408 Amendment Month and Day V. Invention Description (19) TS0 II> blkO TS1 = > blkO, blkl, blk3, blk4 TS2 = > blkO, blk2 TS3 = > b 1 k0, b1k1, b 1 k3, b 1 k4, b1k5 TS4 = > blkO, blk5 TS5 = > blkO, blkl, blk2 are used to transmit the ordering matrix produced. In this three-dimensional file, three channels are used-typically used for Row ordering all numbers may be agreed and the data-typical cases are sequenced to generate one in the following example, for example, the number of numbers, 'The optimal block is sent by the system into a three-dimensional transfer matrix of six data files _ _______ _ It is generated that each data file in the data file group transfer matrix generates a data file ID for each data file group. This three-dimensional transmission matrix is calculated to make use of the available bandwidth to transmit multiple data files. Data streaming. Among them, a convolution method, which is quite famous in this technology. A two-dimensional transmission matrix is used to explain the effective transmission of data file groups. A convolution method may include the following strategies: (1) According to the block in any Time zone support Both periods should be kept at the top and (2) if there are multiple partial solutions that are strategized (), the first solution is the one with the smallest data block and is sent during the previous time zone. For example, suppose Two short data files M and N, each with a data area%, according to the transfer formed by the sort matrix:
TS0 => b 1 k0TS0 = > b 1 k0
第24頁 1223563 案號 91108408 年月曰 修正 五、發明說明(20)Page 24 1223563 Case No. 91108408 Revised V. Description of Invention (20)
TS1 => blkO, blkl, blk3, blk4 TS2 => blkO, blk2 TS3 => blkO, b 1 k 1, blk3, blk4, TS4 二> blkO, blk4 TS5 => blkO, b 1 k 1, blk2,blk5 應用 上述 所提 出的 典型 迴旋 方法 ,傳 輸矩 陣的 丨可 能 結 合 如 下: 選擇 1 :在轉移 二0時 區(TS)時 送出 視訊 檔案 總 數 據 區 塊 TSO MO, NO 2 TS1 => MO, Ml, M3, NO, Nl, N3 6 TS2 => MO, M2, NO, N2 4 TS3 => MO, Ml, M3, M4, NO, Nl, N3, N4 8 TS4 MO, M5, NO, N4 4 TS5 => MO, Ml, M2, M5, NO, Nl, N2, N5 8 選擇 2 :在轉移I TS時送 出視 訊槽 案 總 數 據 區 塊 TSO => MO, NO ,Nl, N3 4 TS1 => MO, Ml, M3, NO, N2 5 TS2 => MO, M2, NO, Nl, N3, N4 6 TS3 => MO, Ml, M3, M4, NO, N4 6 .第25頁 1223563 _案號91108408_年月日 修正 五、發明說明(21) TS4 => M0, M4, NO, Nl, N2, N5 6 TS5 二> M0, Ml, M2, M5, NO 5 選擇[ 丨:在轉移2TS時送 出視 訊槽 案 總數據區塊 TS0 => M0, NO, N2 3 TS1 二> M0, Ml, M3, NO, Nl, N3, N4 7 TS2 => M0, M2, NO, N4 4 TS3 => M0, Ml, M3, M4, NO, Nl, N2, N5 8 TS4 => M0, M4, NO 3 TS5 二> M0, Ml, M2, M5, NO, Nl, N3 7 選擇4 ··在轉移3TS時送出視訊檔案 總教據區塊 TSO MO, NO, Nl, N3, N4 5 TS1 => MO, Ml, M3, NO, N4 5 TS2 => MO, M2, NO, Nl, N2, N5 6 TS3 => MO, Ml, M3, M4, NO 5 TS4 => MO, M4, NO, Nl, N3 5 TS5 => MO, Ml, M2, M5, NO, Nl, N2 6 選擇 5 :在轉移4TS時送 出視 訊槽 案 總數據區塊 TSO => MO, NO, N4 3TS1 = > blkO, blkl, blk3, blk4 TS2 = > blkO, blk2 TS3 = > blkO, b 1 k 1, blk3, blk4, TS4 two > blkO, blk4 TS5 = > blkO, b 1 k 1 , Blk2, blk5 Apply the above-mentioned typical roundabout method, the transfer matrix 丨 may be combined as follows: Option 1: Send the video file total data block TSO MO, NO 2 TS1 = > MO when the time zone 2 (TS) is transferred , Ml, M3, NO, Nl, N3 6 TS2 = > MO, M2, NO, N2 4 TS3 = > MO, Ml, M3, M4, NO, Nl, N3, N4 8 TS4 MO, M5, NO, N4 4 TS5 = > MO, Ml, M2, M5, NO, Nl, N2, N5 8 Option 2: Send the total data block of the video slot when transferring I TS TSO = > MO, NO, Nl, N3 4 TS1 = > MO, Ml, M3, NO, N2 5 TS2 = > MO, M2, NO, Nl, N3, N4 6 TS3 = > MO, Ml, M3, M4, NO, N4 6 .page 25 1223563 _ Case No. 91108408_ Revised Year, Month, and Day V. Description of the invention (21) TS4 = > M0, M4, NO, Nl, N2, N5 6 TS5 II > M0, Ml, M2, M5, NO 5 Select [丨: Send video slot total data block TS0 when transferring 2TS => M0, NO, N2 3 TS1 2 > M0, Ml, M3, NO, Nl, N3, N4 7 TS2 = > M0, M2 , NO, N4 4 TS3 = > M0, Ml, M3, M4, NO, Nl, N2, N5 8 TS4 = > M0, M4, NO 3 TS5 two > M0, Ml, M2, M5, NO, Nl , N3 7 select 4 ·· Send video file general data block TSO MO, NO, Nl, N3, N4 5 TS1 = > MO, Ml, M3, NO, N4 5 TS2 = > MO, when transferring 3TS M2, NO, Nl, N2, N5 6 TS3 = > MO, Ml, M3, M4, NO 5 TS4 = > MO, M4, NO, Nl, N3 5 TS5 = > MO, Ml, M2, M5, NO, Nl, N2 6 Option 5: Send the total data block of the video slot when transferring 4TS TSO => MO, NO, N4 3
第26頁 1223563 _案號91108408_年月日_ 五、發明說明(22) TS1 M0, Ml, M3, NO, Nl, N2, N5 7 TS2 二> M0, M2, NO 3 TS3 => M0, Ml, M3, M4, NO, Nl, N3 7 TS4 => M0, M4, NO, N2 4 TS5 M0, Ml, M2, M5, NO, Nl, N3,N4 8 選擇 6 :在轉移5TS時送 出視 訊槽 案 總數據區塊 TS0 -> M0, NO, Nl, N2, N5 5 TS1 M0, 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 應用策略(1 ),選擇2,4,與6擁有在任何時區間傳送 的最小的數據區塊之極大值(例如,6個數據區塊),應用 策略(2 ), 在典型的實施例中,最佳的傳輸矩陣為選擇 4,因為選擇4擁有任何參考時區之數據區塊與鄰近之時區 的數據區塊之最小總和(例如,1 6個數據區塊),因此就本 實施例之最佳化來說,數據檔案N隻傳送序列鷹平移3個時 區,在一實施例中,三維之傳送矩陣由每一個頻道伺服器 1 0 4所產生。Page 26 1223563 _Case No. 91108408_ Year Month Date_ V. Description of the invention (22) TS1 M0, Ml, M3, NO, Nl, N2, N5 7 TS2 II > M0, M2, NO 3 TS3 = > M0 , Ml, M3, M4, NO, Nl, N3 7 TS4 = > M0, M4, NO, N2 4 TS5 M0, Ml, M2, M5, NO, Nl, N3, N4 8 Option 6: Send when transferring 5TS Video slot total data block TS0-> M0, NO, Nl, N2, N5 5 TS1 M0, 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 Apply strategy (1), select 2 , 4, and 6 have the maximum value of the smallest data block transmitted in any time interval (for example, 6 data blocks), and apply the strategy (2). In a typical embodiment, the optimal transmission matrix is selected 4. Because 4 has the smallest sum of data blocks in any reference time zone and data blocks in adjacent time zones (for example, 16 data blocks), the data file N is optimized for the purpose of the optimization of this embodiment. Only teleports the Eagle to pan 3 time zones, Embodiment, a three-dimensional matrix of transmission channels generated by each embodiment of a server 104.
第27頁 1223563 _案號9Π08408_ 年月曰___ 五、發明說明(23) 當每一個數據檔案的數據區塊被根據傳送矩陣送出 時,大量的訂閱的客戶可以在隨機的時間中存取這些數據 檔案,而此數據檔案之數據區塊能夠在同時被每一個訂閱 客戶所取得,如上所舉的範例中所述,假設時區的持續時 間為5秒,隨選數據(DOD )系統1 0 0則根據最佳傳送矩陣在 來送出數據檔案Μ與N之數據區塊,如以下所述的方法: 時間 0 0 : 0 0 : 0 0 => MO NO N1 Ν3 Ν4 時間 0 0 : 0 0 : 05 二〉MO Ml M3 NO N4 時間 00 : 00 : 10 二〉MO M2 NO N1 N2 N5Page 27 1223563 _ Case No. 9Π08408_ Year Month ___ V. Description of the Invention (23) When the data block of each data file is sent out according to the transfer matrix, a large number of subscribed customers can access these at random times Data file, and the data block of this data file can be obtained by each subscriber at the same time, as described in the example above, assuming the duration of the time zone is 5 seconds, the data on demand (DOD) system 1 0 0 Then, the data blocks of the data files M and N are sent according to the optimal transmission matrix, as described below: Time 0 0: 0 0: 0 0 = > MO NO N1 Ν3 Ν4 Time 0 0: 0 0: 05 2> MO Ml M3 NO N4 time 00: 00: 10 2> MO M2 NO N1 N2 N5
時間 0 0 : 0 0 : 15 => MO Ml M3 M4 NO 時間 0 0 : 0 0 : 2 0 => M0 M4 NO N1 N3 時間 00:00:25 => M0 Ml M2 M5 NO N2 時間 00 : 0 0 : 3 0 => M0 NO N1 N3 N4 時間 0 0 : 0 0 : 3 5 => M0 Ml M3 NO N4Time 0 0: 0 0: 15 = > MO Ml M3 M4 NO time 0 0: 0 0: 2 0 = > M0 M4 NO N1 N3 time 00:00:25 = > M0 Ml M2 M5 NO N2 time 00 : 0 0: 3 0 = > M0 NO N1 N3 N4 time 0 0: 0 0: 3 5 = > M0 Ml M3 NO N4
時間 00:00:40 => M0 M2 NO N1 N2 N5 時間 0 0 : 0 0 : 45 => M0 Ml M3 M4 NO 時間 0 0 : 0 0 : 5 0 => M0 M4 NO N1 N3 時間 00:00:55 => M0 Ml M2 M5 NO N2 如果時間0 0 : 0 0 ·. 0 0有一個客戶A選擇影片M,則此客戶A端 的例(STB) 30 0接收、儲存、撥放、退回這些數據區塊之動 作如下:Time 00:00:40 = > M0 M2 NO N1 N2 N5 Time 0 0: 0 0: 45 = > M0 Ml M3 M4 NO Time 0 0: 0 0: 5 0 = > M0 M4 NO N1 N3 Time 00 : 00: 55 = > M0 Ml M2 M5 NO N2 If time 0 0: 0 0 ·. 0 0 there is a client A who selects movie M, then the client A's example (STB) 30 0 receives, stores, plays, The action of returning these data blocks is as follows:
第28頁 1223563 _案號911Q8408_年月曰 修正_ 五、發明說明(24) 時間0 0 : 0 0 : 0 0 = >接收Μ 0 = >撥放Μ 0,儲存Μ 0 . 時間 0 0 : 0 0 : 0 5 =>接收 Ml,Μ3 = >撥放 Ml,儲存 MO, Ml, M3. 時間 00 : 00 : 10 =>接收 M2 = >撥放 M2,儲存 MO,Ml, M2, M3· 時間 00 : 00 : 1 5 =>接收 M4 = >撥放 M3,儲存 M0, Ml,M2, M3, M4. 時間 0 0 : 0 0 : 2 0 = >無接收= >撥放 M4,儲存 M0,M1,M2,M3,M4. 時間 00 : 00 : 25 =>接收 M5二〉撥放 M5,儲存 M0, Ml, M2, M3, M4, M5. 如果時間0 0 : 0 0 : 1 0有一個客戶B選擇影片Μ,則此客戶B端 的視訊接收器(STB) 3 0 0接收、儲存、撥放、退回這些數據 區塊之動作如下: 時間 0 0 : 0 0 : 1 0 = >接收 M0, M2 = >撥放 M0,儲存 M0,M2. 時間 0 0 : 0 0 ·· 15 = >接收 Ml,M3, M4 = >撥放 M1,儲存 M0, Ml, M2, Μ3,Μ4· 時間 0 0 : 0 0 : 20 = >無接收撥放 M2,儲存 Μ0, Ml, M2, M3, Μ4· 時間 00 : 00 : 25 = >接收 Μ5 = >撥放 M3,儲存 Μ0, Ml, M2, M3, Μ4, Μ5· 時間 00 : 0 0 : 3 0 = >無接收=>撥放 Μ4,儲存 Μ0, Ml,M2, M3, Μ4, M5. 時間 0 0 : 0 0 : 3 5 = >無接收= >撥放 M5,儲存 M0,Ml,M2,M3,M4, Μ5· 如果時間0 0 : 0 0 : 1 5有一個客戶C選擇影片Ν,則此客戶C端Page 28 1223563 _Case No. 911Q8408_ Year Month Revision_ V. Description of the Invention (24) Time 0 0: 0 0: 0 0 = > Receive Μ 0 = > Play Μ 0, store Μ 0. Time 0 0: 0 0: 0 5 = > Receive M1, M3 = > Play M1, store MO, Ml, M3. Time 00: 00: 10 = > Receive M2 = > Play M2, store MO, Ml , M2, M3 · Time 00: 00: 1 5 = > Receive M4 = > Play M3, store M0, Ml, M2, M3, M4. Time 0 0: 0 0: 2 0 = > No receive = > Play M4, store M0, M1, M2, M3, M4. Time 00: 00: 25 = > Receive M5 2> Play M5, store M0, Ml, M2, M3, M4, M5. If time 0 0: 0 0: 1 0 A client B selects the video M, then the video receiver (STB) 3 0 0 of this client B receives, stores, plays, and returns these data blocks as follows: Time 0 0: 0 0: 1 0 = > Receive M0, M2 = > Play M0, store M0, M2. Time 0 0: 0 0 ·· 15 = > Receive M1, M3, M4 = > Play M1, store M0 , Ml, M2, M3, M4 · Time 0 0: 0 0: 20 = > No receive play M2, store M0, Ml, M2, M3, M4 · Time 00: 00: 25 = > Receiving M5 = > Playing M3, storing M0, M1, M2, M3, M4, M5 · Time 00: 0 0: 3 0 = > No receiving = > Playing M4, storing M0, M1, M2 , M3, M4, M5. Time 0 0: 0 0: 3 5 = > No reception = > Play M5, store M0, M1, M2, M3, M4, M5. If time 0 0: 0 0: 1 5There is a client C who selects the movie N, then this client C
第29頁 1223563 _案號91108408_年月日__ 五、發明說明(25) 的視訊接收器(STB) 3 0 0接收、儲存、撥放、退回這些數據 區塊之動作如下: 時間0 0 : 0 0 : 1 5 = >接收N 0 = >撥放N 0,儲存N 0 時間 00 : 00 : 20 =>接收 Nl,N3 = >撥放 N1,儲存 NO, Nl,N3 時間 00 : 00: 2 5 =>接收 N2 = >撥放 N2,儲存 NO, Nl,N2, N3, N4 時間 0 0 : 0 0 : 3 0 =>接收 N4 = >撥放 N3,儲存 NO, Nl,N2, N3, N4 時間 0 0 : 0 0 : 35 =>無接收撥放 N4,儲存 NO, Nl,N2, N3, N4, N5 時間 00 : 00 : 40 =>接收 N5 = >撥放 N5,儲存 NO, Nl,N2, N3, N4, N5 如果時間0 0 : 0 0 : 3 0有一個客戶D選擇影片N,則此客戶D端 的例(STB) 3 0 0接收、儲存、撥放、退回這些數據區塊之動 作如下: 時間 0 0 : 0 0 : 30 =>接收 NO, Nl,N3, N4 = >撥放 N0,儲存 NO, N1, Ν3,Ν4· 時間0 0 : 0 0 : 35 =>無接收=>撥放Ν1,儲存NO, Nl,Ν3, Ν4. 時間 00 : 00 : 40 =>接收 N2, N5 = >撥放 N2,儲存 NO, Nl,N2, N3, Ν4,Ν5· 時間 0 0:0 0:45 =>無接收=>撥放 Ν3,儲存 NO, Ν1,Ν2, Ν3, Ν4, N5. 時間 00 : 00 : 50 =>無接收=>撥放 N4,儲存 NO, Nl,N2, N3, N4,Page 29 1223563 _Case No. 91108408_Year Month and Day__ V. The video receiver (STB) of the invention description (25) 3 0 0 The actions of receiving, storing, playing, and returning these data blocks are as follows: time 0 0 : 0 0: 1 5 = > Receive N 0 = > Play N 0, store N 0 time 00: 00: 20 = > Receive Nl, N3 = > Play N1, store NO, Nl, N3 time 00: 00: 2 5 = > Receive N2 = > Play N2, store NO, Nl, N2, N3, N4 time 0 0: 0 0: 3 0 = > Receive N4 = > play N3, store NO, Nl, N2, N3, N4 time 0 0: 0 0: 35 = > No receive play N4, store NO, Nl, N2, N3, N4, N5 time 00: 00: 40 = > receive N5 = > Play N5, store NO, Nl, N2, N3, N4, N5 If time 0 0: 0 0: 3 0 there is a client D to select movie N, then the client D side example (STB) 3 0 0 receives, The operations of storing, playing, and returning these data blocks are as follows: Time 0 0: 0 0: 30 = > Receive NO, Nl, N3, N4 = > Play N0, store NO, N1, Ν3, Ν4 · time 0 0: 0 0: 35 = > No reception = > Play N1, store NO, Nl, Ν3, Ν4. Time 00: 00: 40 = > Receiving N2, N5 = > Playing N2, storing NO, Nl, N2, N3, Ν4, Ν5 · Time 0 0: 0 0:45 = > No receiving = > Playing N3, storing NO, Ν1, Ν2 , Ν3, Ν4, N5. Time 00: 00: 50 = > No reception = > Play N4, store NO, Nl, N2, N3, N4,
第30頁 1223563 _SR_lll〇8408 五、發明說明(26) N5. 曰 修正 時間0 0 : N5. 如 獨 上面的例子所示,杯彳 立地對服務提供者所搵戶的組合皆可在任意的時間中 放 不 ,如果用來傳輪上述實旖仞 > ,仃選擇並開始播 _ 足時,仍有可能用據檔m 時間延遲下來傳送相杳大e A:人專輸程序在極小的 法提供客戶通用隨選數棱 上之的時間最佳 客戶能更進一步縮小所需 』序&出了提供 戶通用P #嫱良¥ T而頭寬以父換存取時間之延遲之客 尸逋用數據廣播服務的方法。 、各 弟5圖說明一與本發明夕 杳A Ysl , , 列矩陣的i型程序在Λ ;/ V 處理序 工w八w序在實施例中,本發明使用時間分割多 工=time division multipiexing,TDM)以及頻率分割多 (frequency division mutiplexing,FDM)技術在伺^ =螭進行壓縮與序列數據傳送,在一典型實施例中,一次 一排序陣列為了每一數據檔案而產生,在一實施例中,^ 數f稽案被分成一定數量的數據區塊,並根據數據區塊 的數里產生序列矩陣’典型來說,依序列矩陣提供一送出 、順序從伺服器傳送數據檔案的區塊給客戶,而這此 T由任何想要在任意的時間中存取這些數據檔案的交 戶依序存取。 各Page 30 1223563 _SR_lll〇8408 V. Explanation of the invention (26) N5. Revision time 0 0: N5. As shown in the example above, the combination of the cup and the service provider can be combined at any time. No, if it ’s used to pass the above-mentioned practice. When you select and start broadcasting, it is still possible to use the file m time delay to transmit the relative e. A: The human input process is in a very small way. Provide customers with universal on-demand time. The best customers can further reduce the need. "Order & out of the general customer P #P 良 ¥ T and the head width is changed from the parent to the access time delay. Method of using data broadcasting service. Each figure 5 illustrates an example of the Y-type program of the column matrix in accordance with the present invention. The i-type program of the column matrix is in Λ; / V processing sequence w eight w sequence In the embodiment, the present invention uses time division multiplexing = time division multipiexing (TDM) and frequency division mutiplexing (FDM) technologies perform compression and sequence data transmission. In a typical embodiment, an array is generated one at a time for each data file. In one embodiment, In the case, the number f is divided into a certain number of data blocks, and a sequence matrix is generated based on the number of data blocks. Typically, a sequence matrix is provided to send out and sequentially send data file blocks from the server to Customers, and these are accessed sequentially by anyone who wants to access these data files at any time. each
第31頁 1223563 ϋΙ〇Β408Page 31 1223563 ϋΙ〇Β408
五、發明說明(27) 第-:ίΠ ί的數據區塊數量(X)被接收,- 除(步驟506),此ΪΪϊίί f'5〇4^,—參考陣列被清 數據區塊,接著/ i '"、内αΡ的官理目的持續追蹤 第二變數二為j;s=,5:8),如…X,-驟5 1 4 ),如果i小Μ 7 ^ 接下來,1與X比較(步 ^果小於X,則被儲存在次佳排序矩 :Omodulo(x)]的數據區塊責備 5⑻,如果參考陣列已有此數據= (步驟 開始時’因為次佳排序矩陣尚未2 ::重J寫入’: 省略,接下來,夂去曈別4认士百 此一步驟可以被 520),起初因為;有參考;二;數據區塊i(步驟 清除,所以參考陣列中沒有任何資料口如m 5 06二被 2數據區塊i,則檢查次佳排序一>行p歹,二 2i=i_521),如果次佳排二;; 矩塊)^數據區塊1即被加入次㈣ 數據區塊i則被加入次佳排序^ (X),J]的位置以及參考陣列中(步 J) ^ · 加入次佳排序矩陣以及參考陣列中之、在數據區塊1 1==ι + 1 (步驟5 24),接著程序重複步驟1?加―1,如此 考陣列包含數據區塊i或次佳排序矩 1 · 11,如果f 524/^lt 1223563 案號 91108408 年月曰 修正 五、發明說明(28) X時 516),接著程序重複步驟5 0 6直到j =x,整個程序在 結束(步驟5 1 0 )。 在一實施例中,如果一數據檔案被分成六個數據區塊 (x = 6 ),則此次佳排序矩陣以及參考陣列即如下所示: 次佳預定矩陣(SSM) TSO TS1 TS2 TS3 TS4 [0,0]blk0 [l,0]blkl [2,0]blk2 [3,0]blk3 [4,0]blk4 TS5 [5,0]blk5 [〇, 1] [5, 1] [〇, 2] [5, 2] 1,1] [2, 1] [3, 1] [4, 1] [1,2] [2, 2]blk0 [3, 2]blkl [4, 2] [0,3] [1,3] [5, 3]blk2 [2, 3] [3, 3] [4, 3] [0,4] [1,4]blk3 [2, 4 [3,4] [4, 4]blk0V. Description of the invention (27) Number-: ίΠ ί The number of data blocks (X) is received,-divided (step 506), this ΪΪϊί f'5〇4 ^,-reference array cleared data blocks, then / i '", the official purpose of the internal αP continues to track the second variable j = s =, 5: 8), such as ... X,-step 5 1 4), if i small M 7 ^ Next, 1 Compare with X (If the step is less than X, it is stored in the second-best sorting moment: Omodulo (x)]. If the reference array already has this data = (at the beginning of the step 'because the second-best sorting matrix is not yet available) 2 :: Heavy J write ': Omit, next, go to Farewell 4 steps. This step can be 520), initially because; there are references; two; data block i (steps are cleared, so refer to the array If there is no data port such as m 5 06, 2 is the data block i, then check the next best order 1> row p 歹, 2 2 = i_521), if the second best row is 2 ;; moment block) ^ data block 1 is The data block i is added to the second-best sort ^ (X), J] and the reference array (step J) ^ · The second-best sort matrix and the reference array are added to the data block 1 1 == ι + 1 ( (Step 5 24), then the program repeats step 1? Plus -1, so that the array contains the data block i or the next-best sorting moment 1.11, if f 524 / ^ lt 1223563 (28) X 516), then the program repeats step 5 6 until j = x, and the entire program is ended (step 5 1 0). In one embodiment, if a data file is divided into six data blocks (x = 6), the best-order matrix and reference array are as follows: Sub-best predetermined matrix (SSM) TSO TS1 TS2 TS3 TS4 [ 0,0] blk0 [l, 0] blkl [2,0] blk2 [3,0] blk3 [4,0] blk4 TS5 [5,0] blk5 [〇, 1] [5, 1] [〇, 2 ] [5, 2] 1,1] [2, 1] [3, 1] [4, 1] [1,2] [2, 2] blk0 [3, 2] blkl [4, 2] [0, 3] [1,3] [5, 3] blk2 [2, 3] [3, 3] [4, 3] [0,4] [1,4] blk3 [2, 4 [3,4] [4 , 4] blk0
mm 第33頁 1223563 _案號91108408_年月曰 修正_ 五、發明說明(29) [5, 4]blkl [0,5] [1,5] [2,5] [3, 5]blk4 [4, 5] [5, 5] 參考陣列(RA) spaceO space 1 space2 space3 space4 space5 TS0 blkO blkl blk2 blk3 blk4 blk5 TS1 blkl blkO blk2 blk3 b 1 k4 blk5 TS2 blk2 blkO blk3 blkl blk4 blk5 TS3 blk3 blkl blkO blk4 blk5 blk2 TS4 blk4 blkO blk5 blk2 blkl blk3 TS5 blk5 blk2 blkl blkO blk3 b 1 k4 在此典型的實·施例中,基於以上之次佳排序陣列,此數據 檔案的六個數據區塊被以下列的排序送出: TSO => blkO TS1 => b lk 0 , blk3 TS2 二> blkO, blk2 TS3 二> blkl, blk3, blk4 TS4 blkO, blk4mm Page 33 1223563 _Case No. 91108408_ Year and Month Revision_ V. Description of the Invention (29) [5, 4] blkl [0,5] [1,5] [2,5] [3, 5] blk4 [ 4, 5] [5, 5] Reference array (RA) spaceO space 1 space2 space3 space4 space5 TS0 blkO blkl blk2 blk3 blk4 blk5 TS1 blkl blkO blk2 blk3 b 1 k4 blk5 TS2 blk2 blkO blk3 blkl blk4 blk5 TS3 blk3 blkl blk blk2 TS4 blk4 blkO blk5 blk2 blkl blk3 TS5 blk5 blk2 blkl blkO blk3 b 1 k4 In this typical embodiment, based on the next-best sorting array, the six data blocks of this data file are sent out in the following order : TSO = > blkO TS1 = > b lk 0, blk3 TS2 second > blkO, blk2 TS3 second > blkl, blk3, blk4 TS4 blkO, blk4
第34頁 1223563 曰 修正 案號 91108408 五、發明說明(30) TS5 => blkl, blk2, blk5 當與第4圖之典型”最佳"排列順序比較 次佳”排列順序較其少傳送了 3個數據區塊,這個社典型之” 在利用此典型之次佳排序傳輸時少i 8 · 7 5 %之頻 導敎 佳排序需要一個接收之例(STB)來延遲一時區以而此次 選取之數據檔案給一使用者。 〜$〜所 上列實施例之次佳排序僅是無數可能以本技蓺 巧以加以具體化實現之排序組合之一,要在用來^ ,技 檔案之頻寬縮減得越多,就必須在STB#取時間上數據 更多的延遲以達成,在接收的STB能顯示一所選取之 樓案之刚,範例應包含2個或更多時區延遲之傳輸。 第6圖說明一與本發明之一實施例一致之利用次佳序 列矩陣傳輸來顯示資料檔案的典型視訊接收器(STB)程 序,此程序6 0 0在步驟6 0 2開始,此處的STB30 0 (第3圖)從 數據隨選(DOD)廣播系統1〇〇 (第1A圖)接收一電子節目指引 (electronic guide program,EPG),此 EPG列出可由 DOD系 統1 00取得之所有槽案,在步驟603中,一使用者在EPG上 透過按下一訂購按鈕來選取一數據檔案,此STB等待一定 之時區來允許足夠所選取檔案之數據區塊來儲存,使得此 S T B可以撥放數據檔案而不需要何中斷,根據一實施例, 此CPU3 0 4 (第3圖)能夠指出需要多少延遲能保證所選取的Page 23, 1223563, Amendment No. 91108408 V. Description of the invention (30) TS5 = > blkl, blk2, blk5 When compared with the typical "best " better order of arrangement" in Figure 4, the arrangement order is transmitted less than it. 3 data blocks, this company is typical "when using this typical second-best sorting transmission, i 8 · 75% less frequency-guided best sorting requires a reception case (STB) to delay a time zone and this time The selected data file is given to a user. ~ $ ~ The second-best sorting of the above embodiments is just one of the countless sorting combinations that may be implemented by this technique, which can be implemented concretely. The more the bandwidth is reduced, the more time it takes to get the data in STB # to achieve it. The received STB can show the rigidity of a selected building case. The example should include two or more time zone delayed transmissions. Figure 6 illustrates a typical video receiver (STB) procedure for displaying data files using a sub-optimal sequence matrix transmission consistent with one embodiment of the present invention. This procedure starts at step 602, where STB30 0 (Figure 3) Wide selection from data (DOD) The system 100 (Figure 1A) receives an electronic guide program (EPG). This EPG lists all the slots that can be obtained by the DOD system 100. In step 603, a user clicks on the EPG by pressing Next order button to select a data file. The STB waits for a certain time zone to allow enough data blocks for the selected file to be stored, so that the STB can play the data file without any interruption. According to an embodiment, the CPU3 0 4 (Figure 3) can indicate how much delay is needed to guarantee the selected
第35頁 1223563Page 1223563
修正 五、發明說明(31) 檔案得以順利無中斷地, 地隨著從_系統100 J據::不2演算法或簡單 遲來達成,在這樣據=頭中韓入所需要的延 ”运矩陣以使延遲所需 j更次佳排 相關聯。 興可取侍之頻寬得以 一旦STB以延遲了 & g ^ ,動開始播放選案時間一來:= 束,根據兩實施例之_, - 一寻間延遲結 選取之數據檔案。 STBk不一使用者開始播放一所 -般操作: -服務提供者能安排在廣播前傳送 :視訊槽案)到頻道词服器Π4,該中央二案(例 异並达出三維傳送矩陣(識別、時區、以工5益1 02計 順序)到頻道伺服器丨〇4,H 數據區塊傳送 考該三維傳送矩陣以適者順j,,頻道伺服器104參 -數據檔案被分成數據區塊使得 ς =據區塊,每Amendment V. Description of the invention (31) The file can be smoothly and uninterruptedly, with the slave system 100 J data: it is not reached by 2 algorithms or simply late, so the data = the delay required by China and South Korea In order to make the delay j more optimal, the preferred bandwidth can be used. Once the STB is delayed by & g ^, the time to start playing the election will come: = bundle, according to the two embodiments, _,- The data file selected by a search delay node. STBk different users start to play a general operation:-The service provider can arrange to transmit before broadcast: video slot case) to the channel word server Π4, the central two case ( Exemplify the difference and arrive at the three-dimensional transmission matrix (identification, time zone, and order based on labor and benefits) to the channel server. 〇4, H data block transmission test This three-dimensional transmission matrix is to be suitable for the j, channel server 104. The parameter-data file is divided into data blocks such that ς = according to the block, each
據檔案之數據區塊的數量與所選擇:區:檔案’-數 據檔案之數據串流位元率相la,舉例持二時間以及數 率之MPEG數據串流中,各一個數摅 在一固定位元 u午〔百4位元組/秒〕According to the number of data blocks of the file and the selected: Area: File '-The data stream bit rate of the data file is la. For example, in the MPEG data stream that holds two times and the data rate, one number is fixed at a time. Bit u afternoon [One hundred and four bytes / second]
第36頁 1223563 案號 91108408 五、發明說明(32) X時間(秒)/ 8 ( 1 )。Page 36 1223563 Case No. 91108408 V. Description of the invention (32) X time (seconds) / 8 (1).
修正 在一典型貫鉍例中,一數據區塊一 多倍的頻道伺服器1 04之局部纪卜興 ’、"整成高出| (1)所計算出的長度為72萬位元纪,上面所舉之方程式 叢集大小為6萬4千位元組,則數’果局部記憶體2 0 8$ 千位元組,在此實施例中,數摅:塊的長度應為7 6萬8 個子區塊,每一子區塊都擁有與^應更進一步被分成多In a typical example of bismuth, a local block of Channel Server 1 04 that is more than one times as much as a data block is "rounded up" (1) The calculated length is 720,000 bits The size of the cluster of equations mentioned above is 64,000 bytes, so the number of local memory is 208 $ kilobytes. In this embodiment, the number of blocks: the length of the block should be 76,000 8 sub-blocks, each sub-block has and should be further divided into multiple
施例中,數據區塊擁有1 2個6葸4斗I、相同的大小’在此負 雨4卞位元組之子區塊。 一子區塊能再被分成數據封白 ^ 包頭以及封包資料,此封包資料声,一數據封包含一套 cpu傳送數據去的自然層之最大4長 ':、每一頻道伺服器纪 ΓΠ ,MTU)相關,在此所提出的實施例中’封白 一、^ 、匕^料之總大小應小於MTU,然而,對最大效率來 說’封包資料長度則應盡可能地長。In the embodiment, the data block has 12 6-by-4 buckets of the same size, and the sub-blocks of the 4-byte byte are rained here. A sub-block can be further divided into a data packet ^ packet header and packet data. This packet data sounds, a data packet contains a set of maximum lengths of 4 natural layers that the cpu sends data to: ', each channel server period ΓΠ, MTU). In the embodiment proposed here, the total size of the 'Bai Baiyi, ^, and Diao materials should be smaller than the MTU. However, for maximum efficiency,' the packet data length should be as long as possible.
在一典型實施例中,在封包頭中之資料包含允許訂閱 客戶之S Τ Β 3 0 0解碼所有接收數據之訊息’並指出數據封包 是否屬於所選擇的數據檔案(如協定簽署、版本、識別 證、或封包型式資訊),此封包首標也許亦包含其他訊 息,就如所需的”次佳”延遲時間,區塊/子區塊/封包數 量、封包長度、環狀剩餘檢查(Cycl redundancyIn a typical embodiment, the data in the packet header includes the message S TB 3 0 0 which allows the subscribing customer to decode all received data messages and indicate whether the data packet belongs to the selected data file (such as agreement signature, version, identification Certificate, or packet type information), this packet header may also contain other information, such as the required "second best" delay time, block / sub-block / packet number, packet length, ring remaining check (Cycl redundancy
第37頁 1223563 _案號91108408_逢二月 日 修正 ^ 五、發明說明(33) check, CRC)以及子區塊中之補償,與/或編碼訊息。 一旦被頻道伺服器1 04所接收,數據封包即被送到QAM 調變器2 0 6,其處另一個封包首標被加入數據封包以產生 一 Q AM調變IF輸出訊號,而給QAM調變器2 0 6的最大輸出位 元率則與可取得之頻寬相關,舉例來說,給一 QAM調變器 2 0 6六百萬位元組之頻寬,其最大位元率為5 · 0 5 (位元/符 號)X 6 (百萬位7〇組)=3 0 · 3百萬位疋/秒。Page 37 1223563 _Case No. 91108408_Every February Day Amendment ^ V. Description of the Invention (33) check, CRC) and compensation in sub-blocks, and / or encoded messages. Once received by the channel server 104, the data packet is sent to the QAM modulator 2 06, where another packet header is added to the data packet to generate a Q AM modulated IF output signal, and the QAM is modulated. The maximum output bit rate of the transformer 2 0 6 is related to the available bandwidth. For example, for a QAM modulator 2 6 6 million bytes, the maximum bit rate is 5 0 5 (bits / symbol) X 6 (7 million sets of bits) = 3 0 3 million bits per second.
此Q AM調變I F訊號被送到增頻器1 〇 6被轉換成適用於特 定頻道的RF訊號(例如,給CATV頻道80,559.25 0百萬赫 灶’以及6百萬位元組之頻寬),舉例來說,如果一有線網 路具有高頻寬(或位元率),每一頻道能用來提供一個以上 之數據串流,而每一數據串流皆佔有一虛擬頻道,又舉例 來說,三MPEG1之數據串流能利用QAM調變放進(f i t ) 6百萬 位元組之頻寬中,此增頻器1 0 6之輪出連結到送出已結合 之訊號到傳輸媒體11 0之結合器/放大器1 〇 8。This Q AM modulated IF signal is sent to an upconverter 106 which is converted into an RF signal suitable for a specific channel (for example, to CATV channel 80,559.25 0 million Hz 'and a frequency of 6 million bytes). Wide). For example, if a cable network has high bandwidth (or bit rate), each channel can be used to provide more than one data stream, and each data stream occupies a virtual channel. It is said that the three MPEG1 data streams can be put into the 6 megabyte bandwidth using QAM modulation. The round-trip of this frequency booster 106 is connected to send the combined signal to the transmission medium 11 0's combiner / amplifier 1.08.
在一典型實施例中,用來傳輪π Νπ數據串流的全部系 統頻寬(BW)為BW = N X bw,其中bw為每一數據串流所需要 的頻寬,舉例來說,三MPEG-1之數據串流可在同一時間由 具有系統頻寬每秒30· 3百萬位元組之DOCS IS有線頻道來傳 輸,因為每一 MPEG- 1之數據串流佔有每秒9百萬位元組之 系統頻寬。In a typical embodiment, the total system bandwidth (BW) used to transmit round π π data streams is BW = NX bw, where bw is the required bandwidth of each data stream. For example, three MPEGs A data stream of -1 can be transmitted at the same time by a DOCS IS cable channel with a system bandwidth of 30 · 3 million bytes per second, because each MPEG-1 data stream occupies 9 million bits per second The system bandwidth of the tuple.
第38頁 1223563 _______案说 91108408_ 月__g_—修正 _ 五、發明說明(34) 典型來說,頻寬不在乎數量地被實際存取DOD服務之 訂閱客戶所用盡,因此即使沒有訂閱客戶正在使用DOD月艮 務’頻寬仍被用盡以確保系統的隨選能力。 此STB30 0—旦開啟,便持續不斷地接收與更新儲存在 一 S T B 3 0 0之局部記憶體3 0 8的節目指引,在一典型實施例 中’此STB30 0顯示包含電視螢幕上最新節目指引之數據檔 案資訊’數據檔案資訊,如視訊檔案資訊,也許包含影片 識別證、影片標題、描述(以多國語言)、類型(如動作片 或兒里片)、等級(如r、p G1 3 )、有線公司政策(如價格、 可預覽長度)、訂購期間、影片海報,以及影片預告。在 一典型實施例中,數據檔案資訊透過一預約頻道被傳出, 如此頻道被預約來讓韌體更新廣告的以及/或緊急資訊, 在另一實施例中,資訊由與其他數據串流分享之頻道所送 出。 一訂閱客戶能觀看顯示在電視螢幕上依照種類排列之 可提供的數據檔案清單,當客戶選擇該數據檔案中之一個 數據檔案時,此STB3〇〇控制其硬體轉入相對應之頻道及/ 或虛擬子頻道以開始接收該數據檔案之數據封包,此 STB30 0指定每一個封包首標,解碼數據封包中之數據,並 指出接收的數據封包是否應該被保留,如果STB3〇〇指出接 收的數據封包不應該被保留,則該數據封包即被丟棄,否Page 23, 1223563 _______ case 91108408_ month __g_—correction_ V. Description of the invention (34) Typically, the bandwidth is not used up by the subscription customers who actually access the DOD service, so even if no subscription customers are DOD bandwidth is still being used up to ensure on-demand capabilities of the system. Once the STB300 0 is turned on, it will continuously receive and update the program guide stored in the local memory 308 of a STB 300. In a typical embodiment, the STB30 0 display contains the latest program guide on the TV screen. Data file information 'Data file information, such as video file information, may include video identification cards, video titles, descriptions (in multiple languages), types (such as action movies or children's films), and grades (such as r, p G1 3 ), Cable company policies (such as price, previewable length), subscription period, movie posters, and movie trailers. In a typical embodiment, the data file information is transmitted through a reserved channel, so that the channel is reserved for the firmware to update the advertising and / or emergency information. In another embodiment, the information is shared with other data streams From your channel. A subscription customer can view the list of available data files arranged on the TV screen according to the type. When the customer selects one of the data files, this STB300 controls its hardware to transfer to the corresponding channel and / Or virtual subchannel to start receiving data packets of the data file. This STB300 specifies the header of each packet, decodes the data in the data packet, and indicates whether the received data packet should be retained. If STB300 indicates the received data, The packet should not be retained, then the data packet is discarded, no
第39頁 1223563 _______案號91108408_年月曰 修$ 五、發明說明(35) 則該數據封包即被放到局部記憶體3 0 8以待不久後取回, 或是暫存到記憶體暫存器3 0 9直到被送至解碼器3 1 2為止。 在一典型實施例中,此STB3 0 0使用一種,,滑動視窗 (slide window)"的預測技術來觀察在記憶體暫存器3〇9中 之預期數據區塊’以避免過頻繁之讀取/寫入局部記憶體 3 0 8來提昇性能效率。如果預測視窗產生一預測狀態時, 數據區塊被直接從記憶體暫存器3 〇 9傳出至解碼器3丨2,如 果預測沒有產生,則數據區塊在從記憶體暫存器3 〇 9傳送 至解碼器31 2前,會先由局部記憶體3〇8讀進記憶體暫存器 3 0 9中。 α 在一典型實施例中,該STB3 0 0透過紅外線(infrared 人控早7^按鍵、一紅外線(IR)鍵盤、或前面板按鍵包 ::停藤慢動作撥放、倒轉、畫面伸縮、單步 鈕來回應訂購客戶之指令,在一實施例中,如果一= :d 1:中沒有輸入任何動作(例如捲動節目單、Page 39 1223563 _______ Case No. 91108408_ Year and Month Revision $ 5. Description of the Invention (35) The data packet is then placed in local memory 3 0 8 for retrieval soon, or temporarily stored in memory The register 3 0 9 is sent to the decoder 3 1 2. In a typical embodiment, the STB 300 uses a prediction technique of "slide window" to observe the expected data block in the memory register 309 to avoid reading too frequently. Fetch / write local memory 308 to improve performance efficiency. If the prediction window generates a prediction state, the data block is directly transmitted from the memory register 3 009 to the decoder 3 丨 2; if the prediction is not generated, the data block is from the memory register 3 〇 9 is transmitted to the decoder 31 2 before being read from the local memory 308 into the memory register 309. α In a typical embodiment, the STB3 0 0 through infrared (infrared human control early 7 ^ key, an infrared (IR) keyboard, or front panel key package :: stop rattan slow motion playback, reverse, screen zoom, single Step button to respond to the ordering customer's instructions. In one embodiment, if one =: d 1: no action is entered (such as scrolling the program list,
此排定的,告會在訂閱客戶提供一動作“下= 控早疋上之按鍵)時會自動停止撥放, 該STB30 0會自動在一視 另實靶例中, #者可以#金# # 進仃播放時插入廣告,該服務恭For this schedule, the subscriber will automatically stop playing when the subscriber provides an action "Down = Control the key on the morning button." The STB300 will automatically see the target in another view, # 者 可以 # 金 # # Insert ads during playback, the service respectfully
頻ίΚ:;斷:格方針來規定在視訊進行播放時應有IFrequency ίΚ :; break: grid policy to specify that there should be I
第40頁 1223563Page 1223563
年 月 曰 fJt_911p8408 五、發明說明(36) 如果在數據封包首標中發現有緊急訊息位元時,此 STB3 0 0將暫停任何數據接收動作,並將其硬體轉至該預定 用來接收該數據檔案資訊的頻道,以取得以及解碼任何緊 急訊息並顯示到輸出螢幕上。在一典型實施例中,當此 =首〇閒2主將會被轉至預定用來接收該數據檔案資訊 訊息^ B夺準備#沒有延遲地接收及顯示任何的緊急 的範例說明某種本發明的典型實施例 的實施例、調整以及變形都是在本技蓺中 /且,、他 本發明不因此而被此上述討論特;斤而易見的, 確切地由下面之直二U 實施例所限制,而更 田下面之專利申請範圍來定義:Year and month fJt_911p8408 V. Description of the invention (36) If an emergency message bit is found in the data packet header, this STB3 0 0 will suspend any data receiving action and transfer its hardware to the intended use to receive the Data file information channel to get and decode any emergency messages and display them on the output screen. In a typical embodiment, when this = the first time, the host will be transferred to receive the data file information message ^ B 夺 备 # Receiving and displaying any urgent examples without delay to illustrate some of the invention Exemplary embodiments, adjustments, and variations are all in the present technology, and the present invention is not specifically discussed by the above; therefore, it is easy to see, exactly by the following straight two U embodiments Restrictions are defined by the scope of the following patent applications:
第41頁 1223563 _案號91108408_年月日__ 圖式簡單說明 第1 A圖說明一與本發明之一實施例一致的典型隨選視 訊系統; 第1 B圖說明一與本發明之另一實施例一致的典型隨選 視訊系統; 第2圖 說明一與本發明之一實施例一致的典型通道伺 服器; 第3圖 說明一與本發明之一實施例一致的典型視訊接 收器; 第4圖 說明一與本發明之一實施例一致之處理序列矩 陣的典型程序; 第5圖 說明一與本發明之一實施例一致之次佳處理序 列矩陣的典型程序;以及 第6圖 說明一與本發明之一實施例一致之利用次佳序 列矩陣傳輸來顯示資料檔案的典型視訊接收器(STB)程 序。 符號說明:Page 41 1223563 _Case No. 91108408_Year Month Day__ Brief Description of Drawings Figure 1 A illustrates a typical video-on-demand system consistent with one embodiment of the present invention; Figure 1 B illustrates one and another of the present invention. A typical video-on-demand system consistent with an embodiment; FIG. 2 illustrates a typical channel server consistent with an embodiment of the present invention; FIG. 3 illustrates a typical video receiver consistent with an embodiment of the present invention; FIG. 4 illustrates a typical procedure for processing a sequence matrix consistent with one embodiment of the present invention; FIG. 5 illustrates a typical procedure for processing a second-best sequence matrix consistent with an embodiment of the present invention; and FIG. One embodiment of the present invention is a typical video receiver (STB) program that uses a sub-optimal sequence matrix transmission to display data files. Symbol Description:
1223563 案號 91108408 年 月 曰 修正 圖式簡單說明 100 隨選視訊系統 102 中央控制伺服i 103 中央儲存器 104 頻道伺服器 106 增頻器 108 結合器/放大器 110 傳輸媒體 112 開關矩陣 118a, 118b 116a ,116b 備用頻道 120 有線電視系統 202 伺服控制器 204 中央處理機 206 調變器(QAM) 208 局部記憶體 210 網路介面 300 機頂盒 302 QAM解調變器 304 CPU 306 存取模組 308 局部記憶體 309 暫存記憶體 310 STB控制器 312 解碼器 314 圖形重疊模組 400 程序 402- 424 步驟 500 程序 5 0 2- 524 步驟 600 程序 6 0 2- 6 0 8 步驟 第43頁1223563 Case No. 91108408 Revised diagrams Brief description 100 On-demand video system 102 Central control servo i 103 Central storage 104 Channel server 106 Inverter 108 Combiner / amplifier 110 Transmission medium 112 Switch matrix 118a, 118b 116a, 116b Spare channel 120 Cable TV system 202 Servo controller 204 Central processor 206 Modulator (QAM) 208 Local memory 210 Network interface 300 Set-top box 302 QAM demodulator 304 CPU 306 Access module 308 Local memory 309 Temporary memory 310 STB controller 312 decoder 314 graphics overlay module 400 program 402-424 step 500 program 5 0 2- 524 step 600 program 6 0 2- 6 0 8 step page 43
Claims (1)
Applications Claiming Priority (5)
Application Number | Priority Date | Filing Date | Title |
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US09/892,017 US20020026501A1 (en) | 2000-05-31 | 2001-06-25 | Decreased idle time and constant bandwidth data-on-demand broadcast delivery matrices |
US09/902,503 US20020049980A1 (en) | 2000-05-31 | 2001-07-09 | Controlling data-on-demand client access |
US09/933,696 US20020059620A1 (en) | 2000-05-31 | 2001-08-20 | Selective inactivation and copy-protection |
US10/003,642 US20030084461A1 (en) | 2001-10-25 | 2001-10-25 | Method and apparatus for transmitting non-VOD services |
US09/997,968 US20020138845A1 (en) | 2000-05-31 | 2001-11-28 | Methods and systems for transmitting delayed access client generic data-on demand services |
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US9148628B2 (en) | 2007-08-16 | 2015-09-29 | Yahoo! Inc. | Intelligent media buffering based on input focus proximity |
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US9148628B2 (en) | 2007-08-16 | 2015-09-29 | Yahoo! Inc. | Intelligent media buffering based on input focus proximity |
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