JPH01155555A - Disk recorder - Google Patents

Abstract

PURPOSE:To gain the transfer speed by providing a memory of >=3 clusters, rearranging then data addresses of individual channels for accessing a disk to be in ascending order and accessing them in turn in near order by the previously accessed address. CONSTITUTION:The memory 4 is provided with >=3 clusters, and in case of reproducing, a request counter 5 is increased by each channel whenever the memory 4 has one vacant cluster. On the other hand, a routine for accessing a disk 2a1 is always monitoring the request counter 5, and when one remaining cluster is achieved, the addresses are rearranged to be in ascending order, so as to access in turn the disk 2a1 in near order by an address for the previously accessed track. When the access order of the disk 2a1 is controlled by such a method as stated above, the max. seek is only one time even in the worst condition, while the remainder is under an average seek time. By this method, the transfer speed can be gained.

Description

[Detailed description of the invention] [Industrial application field] The present invention relates to a disc recorder.

[Prior art]

In a digital disk recorder that uses a hard disk device as a storage medium and accesses it randomly, when trying to record/play back audio on multiple channels, the number of channels is limited by the transfer speed of the disk.

When playing back or recording, if the speed at which the digitized audio is read or written from the disk is slower than the speed at which it is sampled, it becomes impossible to continue the recording/playback.

For example, when digitizing with a sampling frequency of 50 KHz and a quantization bit rate of 16 bits, the amount of data is 100 Kbytes/sec, and if there are 4 channels, 400 bytes can be read onto the disk at a speed of 7 seconds. must be heard. FIG. 3 shows a case in which a conventional two-channel device is expanded to four channels. Each channel has two memories and writes and reads are executed sequentially. Let's say the rotation speed of day, south and ri is 3482 rpms.
l track capacity is 16384 bytes, maximum seek time is 50m5. -The amount of data read/written at a time is 3276
Assuming 8 bytes (2 tracks), the transfer rate obtained by randomly accessing this disk is obtained by the following equation. The transfer speed S is 5=32768/(3XT+0.05)...where T is the time required for one rotation, which is 0.017 seconds in this example.

The reason why Tx3 is used is that it takes two revolutions to read data, and it is necessary to wait at most one revolution because it is not known where the head is on the track when reading data.

If S (transfer rate) is calculated in this way, it will be approximately 322 bytes and 7 seconds, which means that it is impossible to record/play back audio of 4 channels without interruption. (In some cases, it may not be possible to make it in time.) [Problems to be solved by the invention] - If a block of data that is read and written at one time is called a cluster, normally two clusters worth of memory is provided for each channel. While one side is playing, data is read from the disk on the other side, and when one side is finished playing, the memory is replaced and the same operation is repeated. With this, the transfer speed cannot be increased as mentioned above.

[Failure to solve the problem]

Therefore, a memory with one or more 3 clusters (any number is fine as long as there are 3 or more) is provided, and then the addresses of the data of each channel accessed to the disk are sorted in ascending order, and the data of each channel is sequentially accessed in order of the address closest to the address accessed at the destination. The data is stored in memory and read from the disk in an access order that minimizes the seek time of the pick-up and print heads.

[Effect]

Assuming that the memory capacity is three clusters, in the case of playback, each channel increments the request counter each time one cluster becomes available in the memory, as shown in FIG.

The value of the instant request counter is the number of free areas in memory. The routine that accesses the disk always monitors the request counters, and when one of them reaches 2 (one cluster remaining), it checks the other request counters to sort the addresses in ascending order. Sort the addresses on the disk that are requested (with free space) in ascending order. If the address of the track accessed first is closest to the largest address or the smallest address, the disk is accessed in ascending order, otherwise the disk is accessed in ascending order. The request counter of the channel that accessed the disk should be decremented.

If you control the disk access order in this way,
It is no longer necessary to always consider the maximum seek time for each access, and even in the worst case, the maximum seek is only done once and the remaining seek time is less than the average seek time, so the transfer speed can be increased.

S=C×32768/((3×T+0.05)+'(c
-1)X(3XT+0.023) )...
・・・・・・・・・・・・(1) Here, C is the number of channels,
0.05 is the maximum seek time, 0.023 is the average seek time, and T is the time required for one rotation.

If four channels are processed using this method, the transfer rate will be about 4o2.8 bytes, 7 seconds, and data for four channels can be covered.

〔Example〕

The present invention will be explained with reference to the drawings. FIG. 1 is a block diagram of a disc recorder according to the present invention. This is an A/D for a computer control system 1 that uses a disk 2a as an external storage device 2.

This includes a D/A converter 3.

The figure shows the data bus during playback, but during recording, the seek process is the same except that the direction is reversed.

The D/A converter reads data from the memories (A, B, O) of each channel (four channels in the figure) of the memory 4 in synchronization with the sampling frequency. Memory 4 consists of three blocks for each channel, and if we call them A, B, and C, the D/A converter will
While reading data from the memory block of
CPL] 1a prepares to read subsequent data from the disk into the next memory block through the disk interface 6.

It increases this transfer rate and enables random access of 4 channels. First, the three blocks of memory (A, B, C) of memory 4 and a counter 5 that shows how many blocks of memory are free. establish. When this counter 5 is reset, it is set to 3, that is, the number of all memories (A, B, 0). When starting playback, each time the booter is read from disk 2al, it is incremented, and the D/A converter records one block (
Each cluster is read out, and is incremented every time the memory 4 is switched by the switch Snb.

In the conventional method, when one side of the memory becomes free, the next data is simply fetched from the disk. Instead of doing this for each channel independently, the amount of data for each channel is measured by a counter 5. At the same time, the addresses to be read next are arranged in ascending order while monitoring, and when any counter 5 becomes 2 or more, that is, when there is only one memory 4 left in any channel, the addresses are arranged in ascending order. Then, an operation (reading or reading) from the disk 2al of the disk device 2a is caused.

For example, if the address of the disc to be played on each channel is as follows, CHI: l-2-3
-4-5-6-7-8-9-10CH2: 1000-
1001-1002-1003-0H3: 11-12
-13-14-15-16-17-CH4: 1006
-1007-1008-1009-Sound like this is IC
If playback is performed using two memories for each channel in the order of U to 40H, the accessed disk addresses will be as follows.

1.2,1000,1001,11.12,1006°1007.3,1002,13,1008,4,100
3゜14.1009... In this case, the transfer rate cannot be increased because it is necessary to alternately seek very far places each time reading and folding.

If you sort these in ascending order before reading and writing and access them all at once, you won't have to alternately seek far away locations, and you will still remember the address that was accessed first.
The total seek time can be minimized by accessing the addresses in ascending order starting from the closest address.

In the above example, it is accessed as follows.

1.2,3,1000,1001,1002,11,1
2゜13.1006,1007,1008゜Up to this point, if we sequentially read from channel 1 to empty memory 4 and line up the addresses to be read next, we get 4゜1003.14.10
09, rearrange this in ascending order, 4, 14, 1003.
1009 and compare it with the previously accessed address 1008, access from the nearest 1009 in ascending order, and 1
009゜Data on disk 2al from the one closest to 4
゜15. Accesses addresses 1004 and 1010 in order, and stores data in IB, 3B, 2B, and 4B of memory 4 in order.

As mentioned above, by sorting the access order in ascending order of addresses and accessing in the order closest to the address accessed at the destination, in the worst case, the maximum seek is only one time, and the rest is much shorter than that and the average seek time is less than the average seek time. As a result, the transfer rate can be increased.

(According to the υ formula, if C=4.T3=0.023, then S=
It takes about 402.8 bytes in 7 seconds, which is 1.25 times the conventional time, and it becomes possible to support 4-channel random access.

The reason why each channel has three memories is because it is necessary for multiple channels to issue read requests in order to rearrange addresses.

If there are two memories, the memory of the first channel that made a request will be empty before multiple channels issue a read request, so as long as there are three or more memories, you can have as many memories as you like, but you can use less. Then there's nothing better than that.

〔Effect of the invention〕

According to the present invention, each channel has three memories, so each channel can always have an empty memory, so data can be accessed in the order closest to the destination address in ascending order of addresses, and seek time can be shortened.

[Brief explanation of the drawing]

FIG. 1 is a block diagram of the present invention, FIG. 2 is an embodiment of a data reading routine, and FIG. 3 is a block diagram of a conventional example.

Claims (1)

[Claims] In a disk recorder that records and plays back four or more channels, the disk device that plays the disk and the disk that sequentially accesses the data on the disk have at least three channels on each channel.
A memory means for sequentially storing data in the memory of the cluster, a means for rearranging the addresses of the data to be stored in the memory of each channel in the memory means in ascending order, and a means for rearranging the addresses of the data to be stored in the memory of each channel in ascending order, and data on the disk is stored in the order of addresses from the address closest to the address accessed first. The apparatus is equipped with an access means for accessing, and a DA converter for sequentially converting the data stored in the memory means for each channel into an audio signal for each channel, thereby shortening seek time and continuously reproducing audio signals for a plurality of channels. A disc recorder characterized by: