JPS6051178B2 - Record address recording and playback method - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention provides a system for recording signal data such as music, video, English conversation, etc. in multiple channels using methods such as time division multiplexing, frequency multiplexing, and mechanical channel division. By recording the address shown,
The present invention relates to a record address recording and reproducing method for the purpose of selecting target signal data more quickly and reliably.

The above-mentioned records include not only records made of magnetic tape or other tapes, but also disk-shaped records, such as the RCA method, which detects the unevenness of the signal groove as a change in capacity by bringing a stylus into contact with the disc. ), a method in which the stylus is brought into contact with the board and the unevenness of the signal groove is mechanically detected using a piezoelectric element (TED method), and a method in which the stylus is brought into contact with the board and the unevenness of the signal groove is detected mechanically using a piezoelectric element (TED method);
There is a method (Philips method) in which the information is reflected by a photodiode and optically detected by a photodiode, and the present invention particularly relates to an address recording and reproducing method for these records.

Conventionally, for example, when performing automatic tuning on a record with multiple songs recorded, if the record is a tape, there is no signal for about 4 seconds between songs, and the signal is switched off according to a pre-installed program. A method was used to sequentially detect signal areas.

In addition, when the record was a disc, a method was used in which a no-signal groove detector using reflected light was used to search for the gaps between songs from the outermost circumference to the innermost circumference of the disk and record them in memory. With such conventional methods, it is possible to locate the beginning of each song, but it is impossible to play multiple songs in sequence while skipping unnecessary songs or going backwards. However, unnecessary data may be skipped along the way,
This does not mean that there is no need to select multiple pieces of data sequentially while going backwards, for example: In the case of a video file in which multiple images are to be searched, an address is assigned to each image, and this is controlled by a microcomputer, etc., thereby allowing random access, including backward forwarding. The law has passed.

However, especially in the case of disc-shaped records, if you try to fast-forward or fast-reverse, you will be forced to skip over the signal groove and perform mistracking, so it is difficult to reproduce the recorded signal without making mistakes. Because it is not possible to
It was impossible to jump or go backwards unless you returned to the outer edge of the disc and then moved it again; for example, it was impossible to move from one song to another, or to move directly in the opposite direction.

The present invention was made to enable such automatic song selection and other signal data selection control, and it allows information to be recorded and a signal proportional to the remaining time of each song to be sent to the recording channel of the record. An address recording and reproducing method in which address information is multiplexed and sequentially added as address information,
Therefore, the remaining time of the song can always be known at the playback position (position of the hand).

The basic playback operation of a player for playing back a record recording the remaining time of a song is as follows.

Basically, the player always sets the playback state to normal/forward and detects the address information after moving the pickup in fast forward or fast backward, except when the performance ends.

After that, based on the address information detected at that location,
By entering the performance or following fast-forward or fast-reverse movement in the same way as described above, by confirming the address information in the playback state (normal forward), a mode that combines fast-forward or fast-reverse and normal forward without mistracking can be established. Basically, it is designed to allow automatic channel selection to be performed quickly. In other words, although the basic purpose of the present invention is to record a reverse address, the gist of the present invention is not simply to record a reverse address, but also to record a reverse address at a predetermined time, such as a master disc that is cut and played in advance. This means that something like a song whose time is known is subject to control.

By using this basic recording and playback method, the playback position of a song is detected each time, and it is possible not only to fast-forward or fast-reverse from the beginning or end of the song, but also to fast-forward or fast-reverse from the middle of the song. Automatic channel selection and other selection controls can be performed quickly and reliably. Assigning a reverse address can be easily performed without using any special technology by applying ordinary digital technology, for example, technology such as a digital watch to display the remaining time.

Various methods can be used for data processing, but by using a microcomputer as one of the best methods, operation is reliable, quick, compact, and easy to operate. You can also control jump display, track number display, remaining time display of the song, etc. An embodiment of the present invention will be described below with reference to the drawings.

FIG. 1 shows, as an example, a state in which signal data converted into binary codes is recorded in a recording channel of a record such as a tape or a disk by a time-determined multiplexing method.

In this figure, the signal data to be reproduced includes, for example, first channel signal data (1ch), second channel (2ch), third channel (3ch), fourth channel (4ch), and error detection code (CRC). ) are provided respectively, and a frame (n
)(n+1)... are inserted respectively. 1st channel (1ch) and 2nd channel (2c)
h) to compose a stereo song, and the third channel (3ch
) and the fourth channel (4ch) constitute other stereo songs, a plurality of songs 1 and 2 as shown in FIG. 3 are recorded.

In this case, the area where the song is actually recorded is area 10 within the song, and the blank area between the songs (usually 4
(about seconds) is called an inter-track area 11, and the first blank part of the recording channel is called a head area 12. For example, every other odd-numbered frame (n+1) (n+3)
... is used for address information to achieve the present invention.

The odd-numbered frames (n+1), (n+3), etc. are assumed to consist of 13 bits, and FIG. 2 shows a state in which these are extracted and rearranged in parallel. In this figure 2, even numbered frames (n) (n+2)
(n+4)(n+6) synchronization signals are also shown. Odd numbered frames (n+1) (n+3) (n15) (n+
7), one address information for 1Ch (52Ch) is recorded with 26 bits including 2 frames (n+1) (n+3), and 3Ch and 4Ch with 26 bits including 2 frames (n+5) (n+7). One address information is recorded for each frame (n+1) (n+3)... More specifically, the first two bits of each frame (n+1) (n+3)... are for the channel number, for example 1Ch 0F, 2Ch
is "10゛, 3Ch is ゜゜11゛, 4Ch is "゜00゛, and the first half frame (n+1) (n+5)...
The third 1 bit indicates whether it is area 10 during a song or area 11 between songs. For example, during a song, it is “゜1゛,” between songs is “0゛,” and the remaining 10 bits are is the song number out of the total number of songs (frame (n
+1), the fifth song, and frame (n+5), the first song.
0th song) is hail. Also, the second half frame (n+3
)(n+7)... The remaining 11 bits are used to write an address proportional to the remaining time of the signal data. That is,
If the remaining time of the song from the actual playback position to the start position of the next song is set as an address that is subtracted by 1 every second, then in frame (n+3) in Figure 2, the next signal data from the playback position is 183 hoes to the beginning position, frame (n+
7), it is 15 seconds. These address information are
Recording is performed sequentially over the entire recording channel through the mid-song area 10 and the inter-song area 11 in the figure. Note that the head area 12 includes the odd frames (n+1) (n+3),
Address information such as the total number of songs, song number and performance time for each song may be further recorded in synchronization with...

An example of a circuit for reproducing records recorded in the above manner will be explained with reference to FIG.

20 is an audio disc player, 21 is a decoder, 22 is an output amplifier, and 23 is a speaker.

In addition, 24 is an operation keyboard for program performance;
25 is a display section, and 26 is a microcomputer (hereinafter referred to as microcomputer) as a control circuit. The microcomputer 26 has an 8-bit CPU1R. Ar! 4.RO
It is composed of a control section 27 consisting of M, and input/output sections 28, 29, and 30 for connection with other devices.

In the operation keyboard 24, RO and R9J are keys used to specify the song number and number of times, 1PLAY is a play start command key, and 1SUC is a key that causes the song to be played sequentially from that song to the last song on the record. The Rx key is a key used to specify the number of repeat performances, and the 1+ key is a key used to connect songs.

Also, Rpause/JumpJ'NextJ are used only during performance, and among these. RPause is a key that pauses when pressed once and continues playing when pressed again. RJUnlPj is a key that jumps to the next song on the program and then plays as programmed. 1N
extJ is a key that jumps to the next song on the record and then plays normally. ノ1C1earJ is a key that returns to the initial settings (when the power is turned on) when pressed at any time, 1Tim.
If you press this anytime, the time will be displayed on the display section 25.

In the display section 25, 31 is a display element such as a liquid crystal that displays the song number, number of times, and time, RPROGJlMU.
Is TEJrPLAYJ currently operating the program?
An indicator light that indicates whether muting is in progress or playing.
RAMJrPMJ is an indicator light that indicates whether it is morning or afternoon when displaying the time, and 1 song number/number of times is an indicator light that indicates the number and number of songs among the display elements 31. .

Next, the input/output section 28 of the microcomputer 26 is connected to the operation keyboard 24 and the display section 25, the input/output section 29 is connected to the address output terminal of the decoder 21, and the input/output section 30 is connected to the audio 1 fast forward of disc player 20. J. rFast rewind J. It is connected to the r playback and 1 end terminals, the RCh selection and RMUTEJ terminals of the decoder 21, and the display section 25.

Next, an outline of the reproduction of a record recorded according to the method of the present invention will be explained with reference to the flowchart of FIG.

First, when a music selection program is written using the operation keyboard 24, its contents are stored in the RAM of the control section 27 of the microcomputer 26 via the input/output section 28.

Next, when automatic music selection is started, the data sequentially written in the program is read out from the memory (RAM) of the control unit 27 using the target address. Then, it is determined whether the target address is in memory. If it does not exist, it is determined that the program has ended, and the program exits. If the target address exists, the address is read from the record by playback. Here, it is determined whether the target address matches the address read from the record. If they do not match, the time T to reach the target address is calculated, and based on this calculation, fast-forwarding or fast-reversing is performed for a predetermined time T. This operation is repeated several times until the time difference with the target address falls within the allowable value, and gradually converges to the target value. Then it enters the normal playback state. Note that if only the first address information extracted from the record is adopted as is, there is a risk of malfunction due to errors in writing or reading.

Therefore, if the same address information is obtained several times in succession, malfunctions can be prevented by determining that it is correct information and using it. Next, as a standard operation example, )R3JXr2J+R5J+RlJrsUcoRPLA
Assuming that the Y keys are pressed in sequence, this operation will be explained in detail based on the flowcharts in FIG. 6 and subsequent figures.

Note that during programming, the microcomputer 26 selects 1 of r1, r
1-2J. 1 of r2, 2 of 12...1 of RN. j.
rN's 2yo... ・R5O's 1J. s150 to each memory of 2J and press the RPLAYj key, this content is saved to 2J of LNr5l of R5l. ．． r52 L. ．． r52's 2J・・RN(7)1J. ．． 2J of rN・1 of RlOO. j. 2 of r100. If you want to run the above program again during a performance, the program will move to J. Until you press the RPLAY key,
This allows you to continue playing the previous performance.

(1) First, in the write flowchart of FIG. 6, when starting (power is turned on), ROJ is placed in memory Z, r1 is placed in memory (N), and 10 is written in memory (Y).
The operation of inserting the function is performed.

Here, (Z) is a memory that temporarily stores the keyed-in numbers, and (N) is a memory that stores the program number.
Y) is a memory for storing whether the number of repetitions of 1SUCJ is 1 or not.

(2) At this point, R3J is keyed in first, so RPL
AY? No. ．． lSUC? Is it NOll×? yo is R
No, 1+? No. ．． rO, l, 2...9? The answer is YES.

And is the content of 1 memory (Z) O? Since the answer is YES, enter the keyed R3J into the memory (Z). (3) Next, 1×yo is keyed in, so RPL
AY? No. ．． lSUC? Is it NOll×? Yo is Y
ES, is the content of 1 memory (Y) O? Yes, yes
Since the memory (1 of N) is N=1 at this point, that is, the content of the memory (Z) is R3J in the memory (1 of 1).

After this, RO. Insert J, memory (N's 2
), that is, set (1 of 2).

(4) Next, R2 is keyed in, so as above,
RO, l,...9? YES becomes YES, and 1 memory (Z
) is the content 0? YES is YES, so memory (Z)
to R2. j enters.

(5) Next, 1+ is keyed in, so 1+? YES, and is the content of 1 memory (Y) O? Yo is Y
It becomes ES.

Since memory (2 of 1) is set in (3) above, is memory 1 (1 of N) set? The result is NO, and the content R2J of the memory (Z) is stored in the memory (N of 2), that is, (1 of 2).

After this, put RO in the memory (Z), add r1 to N in the memory (N), and now the contents of the memory (N), that is, the program number, are R2. J, and sets L in memory RN, that is, 1 in memory R2.

(6) Next, R5J is keyed in, so RO, l,
...9? Is the content of 7 memory (Z) 0? YES is YES, so R5 is stored in the memory (Z).

(7) Next, 1+ is keyed in, so 1+? The answer is YES.

Is the content of 1 memory (Y) 0? YES becomes 1
Is memory (N of 1) set? Since the answer is YES, the content R5 of the memory (Z) is stored in the memory (N's 1), that is, (2's 1), and the memory (N's 2), that is, (2's 1) is stored.
RlJ is entered in 2-2).

After this, put 10 in the memory (Z), set the contents of the memory (Z) to R3J, and save the memory (1 of N), that is, (1 of 3)
Set.

(8) Next, RL is keyed in, so RlJ is similarly stored in the memory (Z). (9) Next, Rsuco is keyed in, so 1SUC? YES and 1
Is the content of memory (Z) 0? Since 2 is NO, the content 1 of memory (Z) is stored in memory (1 of N), that is, (1 of 3).
1-1 in memory (2 of N), that is, (2 of 3)
J enters.

Furthermore, put 11J into the memory (Z) and r into the memory (Y).
Insert 1.

(10) Next, RPLAYJ is keyed in, so R
PLAY? YES becomes YES.

Then, is it programmed to select one station? Yes, yes.
．． Is the content of r memory (Y) O? Because yo is NO,
The product of the content Rlj of memory (Z) and the content 1-1 of memory (N's 2), that is, memory (3's 2), is 1 x (-
1)=-1 to new memory (2 of N), that is (2 of 3)
Put it in.

Hereinafter, the process will move on to the readout flowchart shown in FIG. In the flowchart of FIG. 7, when the RPLAY key is pressed, the memory address changes to r1, 1, r1, 2Jr2, 1Jr2, 2J-JN.
Move to the address added by 50 addresses from 1arN, 2yo....

Then, put R5L in memory RNJ, and put R in memory RKJ.
Add O.J. Note that RKyo is a memory that indicates how many songs were continuously played during RsucJ. Next, decoder 21
The command RMUTEJ is issued to J (2 of N) to read memory (2 of N) (in this example, 2J of memory 1N = 2).
Half 0? Since the answer is YES, is the beginning of the song in memory 1 (1 of N) stored in memory? judge the situation.

Since the answer is NO at first, I issue a command to the player 20 to send it normally in 1 play, and it asks if it can read the 1 address. If the paper is readable (YES), it is readable.
), is the first song number correct? judge the situation. Here, when the needle first falls on the outermost edge of the disc, the song number is number 1 and the song number written is number 3, so the answer is NO. As a result, until the beginning of the next song (1, 2 Ch or 3, 4
Distance calculation is performed for whichever channel is closest. That is, the distance to the beginning of the next song (corresponding to the remaining time of the song) is calculated based on the address information added in the direction opposite to the playback direction. Next, 1 feed distance t≧0? For example, for sending from No. 1 to No. 3, etc., t>0, so YES. On the other hand, for sending from No. 5 to No. 1, etc., since t<O, the answer is NO. Then, the player 20 is instructed to fast-forward t or fast-reverse t. As you repeat the above operations, does it gradually converge and the first song number is correct? YES becomes YES.
Next, one song? If NO in , fast rewind in the same way.
is commanded to the player 20, and is repeated until the result becomes YES, and the result gradually converges. The song number and song start location determined in this way are stored in memory. Next, r1, 2Ch? If YES, r1, 2ch (7) MUTEOFFJ
is commanded to the decoder 21 and displayed on the display unit 25.

And REND? Is it NO and RENT? No
While this is the case, the player 20 is instructed to play normally after one play, and continues playing.

REND? If YES becomes YES, 1 memory (2 of N) (
In this example, 2)〉0? If the answer is YES, subtract r'1yo from the memory (N's 2) and select RMUTE near the starting point.
Return to J, command MUTE, then 1 memory (2 of N
)Half 0? The result of J ((RN's 2J-1) half 0 in this embodiment) is YES, and the third song is repeated again.

If you play the third song twice, you will get 1 memory (N of 2) and half 0.
? The result is NO, and the memory (51 no 2) is (51 no 1)
Then, (52 no 1) becomes (51 no 2), etc. Move on to the fifth song and play it once in the same way.

Next, in the same way, move on to the first song, play sequentially until the last song, and when the last song is finished playing, 1 memory (
N's 1) Half 0? The answer is YES, and the player 20 is instructed to perform 1 return to REND. It becomes U.

Next, some other representative examples of engravings are as follows.

Of course, both examples are based on the above-described flowchart. However, it is a well-known fact that even if the purpose and operation are the same, the flowcharts will generally be different depending on the programmer handling the flowchart, and the above flowchart is only one example.

(Ii) R6JrXJr3jr+yo.

4yo1+Jr2jrXJr2IPLAY! Assume that each key is pressed in sequence.

In this case, play the 6th song 3 times, then play the 4th song 1 time.
They will perform twice, and then the second song will be played twice. (
111) When only the 1PLAY key is pressed, the songs are played sequentially from the first song (normal play).

(IV) When 13JrSUCJr+/2yo RPLAYJ is pressed, the third song is played to the end, and then the second song is played once.

(V) When pressing 13/SUC/×R2JrPLAYJ, the third to last song is played twice.

The following is an example of operating the keyboard during a performance.

(Vi)1C1ear/3/×/2Jr+JrPLAY
When you press , press RclearJ to interrupt the performance,
Next, RPLAYJ began playing the third song twice and the fifth song once.

(Vii) R3yo.

×yoR2Jr+Jr5JrPLAY. When you press J, the previous performance continues until you press 1PLAY, and the RPL
Pressing AYj starts playing the new rewritten program. (Vji) When only the RNeXtJ key is pressed, the playback jumps to the next song on the record, and after that, the playback becomes normal.

(IX) When only the RJUmPJ key is pressed, the program jumps to the next song on the program and plays, and thereafter the program continues as per the program.

(x) If you press only the 1PLAYJ key, normal performance will start from the first song.

(Xi) When r+R4/PLAY is pressed, the fourth song is added after the preset program.

(Xii) 1 x yo R5/+yo 12. Press J'PLAY to play the current song 5 times in a row, then play the second song
Perform once, and then perform as per the program.

(Xiii) 1SUC/×. When you press Ul2yo1PLAYJ, the song being played will be played normally twice.

In the above embodiment, the case where the signal data as the information to be reproduced is music, but other videos,
It can also be applied to English conversation, various industrial data, etc.

For example, in English conversation, recitation, etc., chapters, sections, etc. correspond to the songs in the above-mentioned music. Therefore, by recording address information proportional to the remaining time for each chapter and section, the present invention can be used. The method can be applied as is. Furthermore, in the embodiment described above, address information is time-divisionally inserted at a rate of one word every two frames, but the invention is not limited to this.
If only some of the channels are being used on a disc or TR, the remaining channels can be used, and on a VTR, the audio track can also be used.

In addition, address information is made into one data in 2 frames (26 bits), but it is not limited to this, and can be 1 frame (13 bits) or 3 frames (39 bits).
etc., you can allocate any bits you need. In the above embodiments, the case of a recording method using time division multiplexing or mechanical channel division was explained, but the present invention is not limited to this, and the present invention can be applied to a recording method using other multiplexing methods such as frequency division multiplexing. can be used.

Furthermore, in the case of a disc-shaped record, the present invention is not limited to the case where the pickup during performance moves from the outer circumference to the inner circumference, but conversely, the pickup may be played while moving from the inner circumference to the outer circumference. For example, the present invention can be used even in such a case, and the present invention can also be used even when the speed during performance changes continuously or in stages. Since the present invention employs the recording and playback method described above, the current location can be determined at any playback position on the record, so fast forwarding and rewinding can be done quickly and reliably, and it is easy to jump to the next song in the middle of a performance. It has extremely excellent effects such as being able to display the number of the song currently being played, being able to display the remaining time of the song being played, and being able to display the required time of the automatic performance program when it is inserted into the keyboard. .

With disc-shaped records, it is necessary to search by randomly skipping over addresses when fast-forwarding or fast-reversing, and this is particularly effective in such cases.

[Brief explanation of the drawing]

Fig. 1 is an explanatory diagram of the recording channels of a record when recording is performed using the time division multiplexing method, and Fig. 2 is an illustration of the present invention in which only the synchronization signal and address information are extracted from the recording channels of Fig. 1 and rearranged in parallel. An explanatory diagram of the recording and playback method of
FIG. 3 is an explanatory diagram of a recording channel of a record, FIG. 4 is a block diagram showing an embodiment of a reproducing apparatus according to the present invention, FIG. 5 is a flowchart for explaining the general operation, and FIG. A flowchart for explaining the write operation, FIG. 7 is a flowchart for explaining the read operation, and FIG. 8 is a flowchart for explaining the read operation.
Figures A and b are explanatory diagrams of memory addresses. (n) (n+1) (n+4) ~... synchronization signal, (n+
1) (n+3) (n+5) ~ Address information, (1
ch) (2ch) ~...data signal, 1,2~...
Song number, 10... area during the song, 11... area between songs,
12...Head area, 20...Audio disc player, 21...Decoder, 22...Output amplifier, 23..."Speaker, 24...Operation key, 25...
・Display section, 26... Microcomputer, 27... Control section (CP
U,. RAM. ．． ROM, etc.), 28, 29, 30...
Input/output section, 31...display element.

Claims (1)

[Claims] 1. In a record in which a plurality of signal data to be reproduced are recorded in a plurality of recording channels, a channel number for each signal data and a Address information indicating a numerical value approximately proportional to the remaining time of the signal data is sequentially recorded in predetermined units based on the starting position of the signal data, and an address for determining the presence or absence of signal data is recorded, and this information is recorded. This player includes a player for playing records recorded sequentially, and this player sets the pickup to fast-forward or fast-backward movement and then normal forward playback mode except when the performance ends, and detects correct address information. , a record address recording and playback method that enables automatic tuning of the record. 2. Record the number of all signal data to be reproduced, the number of each signal data to be reproduced, and each address of the reproduction time in the first recording channel part of the plurality of signal data to be reproduced,
2. The record address recording and reproducing method according to claim 1, wherein an address of a channel number is recorded for each signal data of a recording channel. 3 In addition to a player that detects signal data, plays records, fast-forwards, fast-reverses, etc., there is also a decoder that decodes signals detected by this player, a playback device that plays back signals from this decoder, and operations for program control. The microcomputer is equipped with a keyboard and a microcomputer that controls these devices, and the microcomputer records and calculates address information from the decoder and operation signals from the operation keyboard, and gives the player play, fast forward, fast reverse, and start commands. 3. A record address recording and reproducing method according to claim 1 or 2, wherein commands such as , return, pause, etc. are given to the decoder, and channel selection, mutating, etc. are given to the decoder.