JPS63200367A - Digital signal recording and modulating system - Google Patents

Abstract

PURPOSE:To increase window width, by converting a data of eight bits of one unit out of two units of data to a data of nine bits, and converting a remaining one unit to a data of 11 bits, when a data is converted to a data of 20 bits setting two units of digital data in which eight bits are set as one unit as a pair. CONSTITUTION:The data of one byte unit inputted to an input terminal 1 is modulation-processed at every two byte units, and one byte of front eight bits is latched at a latch circuit 2, and one byte of rear eight bits at a latch circuit 3, respectively. Next, outputs from those circuits are inputted to an 8 9 conversion circuit 4 constituted of a ROM, and they are converted to a data of nine bits, and it is supplied to a parallel-serial conversion circuit 5, and the output of the circuit is added on a laser device 8, and the data of first half one byte of the data of two byte units is recorded on a disk 9. Also, the data of latter half one byte is allocated 8 11 conversion circuits 11-13 constituted of the ROMs, and is converted to a parallel-serial signal at a parallel-serial conversion circuit 14, then, it is recorded on the disk 9 similarly.

Description

DETAILED DESCRIPTION OF THE INVENTION [Field of Industrial Application] The present invention relates to a recording modulation method that is commonly used and suitable for recording digital data on optical recording media such as optical disks and magneto-optical disks.

[Summary of the invention]

This invention is a modulation method that converts 8-bit data, that is, 1-byte data, into 20-bit data in 2-byte units during recording, and the number of "1" bits among the 20 bits is 9+II, 10, or When converting in units of 2 bytes, each byte is first converted into 9 bits, and the remaining 1 byte is converted into 11 bits. According to the present invention, when reproducing data from an optical recording medium, the margin in the time axis direction becomes large, and data can be reproduced with high reliability.

[Conventional technology]

Optical disks and magneto-optical disks that can be written to and read from, that is, recorded and played back, have been proposed, but when recording on these disks, data is processed using so-called channel coding, which converts 8-bit data into data with a larger number of bits. will be held.

In this case, when the bits of the signal recorded on the disk are called channel bits, m (m < n) of n channel bits (a modulation method that divides the data of rlJ into the bits of tlil) is used. is used as one of this channel coding.

During demodulation, the reproduced high-frequency output from the disk as shown in Figure 7 is sampled as shown by the arrow in the figure using a channel clock synchronized with the channel focus formed in synchronization with the reproduced data. Sampling is performed n times, and the number of "1" bits from among the n sampling values, that is, m sampling points are selected in order from the highest level to the bits of the selected m sampling points. is determined as “1” and fJ[1
1. This is called a differential detection method.

As mentioned above, in this method, when channel coding, the number of channel bits converted per byte is n.
Since the number of "1"s among them is determined to be m, the determined m positions are detected by a differential method.

The following methods are conventionally known as modulation methods for this differential detection.

One of them is called 4-11 modulation, which is a modulation in which 1 byte of data is muffed into 11 channel bits, and the number of rlJ bits is always 4 among the 11 channel bits. It is a method.

An example of this modulation method is shown in FIG. 8A. In this case, the recorded data will be such that a pulse appears at the "1" bit, and the recorded bits on the disk will be as shown in the figure.

In addition to this method, there is also a method called 4-15 modulation. This maps 1 byte of data to 15 channel bits, and one of the 15 channel bits is "1".
However, if there is a ``0'' bit between ``1'' and ``1'' bits, it will be ``0''.
This is a modulation method that ensures that there are always two or more bits.

An example of a modulation pattern of this 4-15 modulation is shown in FIG. 8B. At this time as well, the recording pulses to the disk will be as shown in Figure B, and the recording bits will also be as shown in Figure B.

[Problem that the invention seeks to solve]

During demodulation using the Rudy and chikujo method during the differential arrangement described above, the channel clock is locked to the reproduced signal so as to sample exactly the peak point of the envelope of the reproduced high-frequency output, as shown by the arrow in Figure 7. If so, demodulation can be performed correctly.

However, the channel clock is not necessarily locked in that way, and the sampling point in Figure 7 is not at the peak of the high-frequency output envelope, but at the shoulder, as shown by the dotted arrow. Even in this case, if the interval between channel bits is wide, it will not have much of an effect, and the longer the shortest distance between bits, the better this clock will be. This makes it resistant to shifts.

That is, as a modulation method for the differential detection method, it is better that the width of one channel bit, which is called the window width, and the shortest inter-bit distance are large.

By the way, if the recording wavelength of an optical disc is set to 7.6 μm per byte, then 4
In the case of -11 modulation, the window width is 0.69 .mu.m as shown in FIG. 8A, whereas in the case of 4-15 modulation, the window width is narrower to 0.51 .mu.m. Therefore, when looking only at the window width, 4-11 modulation is preferable. However, in the case of 4-11 modulation, "1" and "
The shortest distance between pits is 1.38 μm when the number of O atoms present between “1” is 1. On the other hand, as shown in FIG. 8B, in the case of 4-15 modulation, the shortest pit-to-pit distance is 1.52 μm because it is determined that there are two rOJs between “1” and “1”. becomes. Therefore, when looking at the shortest distance between bits, 4-15 modulation is superior.

As mentioned above, each conventional modulation method has its advantages and disadvantages, and there has been no optimal modulation method for the differential detection method.

The present invention aims to provide a modulation method in which the shortest bit distance is equal to that of 4-15 modulation, and the window width is wider than that of 4-11 modulation.

[Means for solving problems]

In this invention, when two units of digital data each consisting of 8 bits are converted into 20-bit data, there are 9, 10, or 11 "1" bits among the 20 bits. In this recording modulation method, one unit of 8-bit data out of two units of data is converted to 9-bit data using a first conversion table, and the remaining one unit of 8-bit data out of the two units is converted into 9-bit data. The bit data is 11-bit data, and when added to the number of "1" bits in the output of the first conversion table, the result is 9, 10, or 11 "1" bits.
” 11-bit data including the number of bits in the second
This is a recording modulation method for digital data that is obtained using a conversion table.

[Effect]

Since each 2 byte is converted into 20 channel bits, what is the window width? , 6X 2 + 20 = 0.76μ
ta, which is wider than in the case of the 4-11 variable (0.69μ bond). Also, in this case, the shortest bit distance is twice the window width, so 0.76X 2 = 1.5
2μ information, which is almost the same as in the case of 4-15 modulation.

〔Example〕

An embodiment of the present invention will be described with reference to the encoder and decoder shown in FIG.

In FIG. 1, (1) is an input terminal into which 1-byte data is input. Data in units of 1 byte through this input terminal (1) is modulated in units of 2 bytes, and the data in units of 2 bytes is latched by latch circuits (2) and (3). In other words, input data is divided into 2-byte units, the latch circuit (2) receives the first 1 byte (8 bits) of the 2-byte data, and the latch circuit (3) receives the following 1 byte (8 bits). bits) are each latched. The output of the latch circuit (2) is supplied to an 8-9 conversion circuit (4). This 8-9 conversion circuit (4) is an RO having a conversion table as shown in Figure 3.
M, and has 256 modulation patterns consisting of 3, 4, and 5 cases in which the number of focuses of ``1'' among 9 bits is 3, 4, and 5. This 9-bit data is supplied to the parallel-serial conversion circuit (5) and converted into serial data, which is then sent to the laser device (
8) and the laser onto the disk (9).
The first 1 byte of data in byte units is recorded.

Next, the latter 1 byte of data is supplied from the latch circuit (3) to the selector (10). Selector (10) has 3
8-11 conversion circuits (11), (12) and (
13) are connected, and as described later, these three
The output of the latch circuit (3) is supplied to one of the three conversion circuits (11), (12) and (13).

Then, the 8-11 conversion circuit (11)
It consists of a ROM that stores 256 modulation patterns in which the number of "1" bits is 4, and the 8-11 conversion circuit (12) similarly has 5 "1" bits out of 11 bits. It consists of a ROM that stores 256 modulation patterns, and a conversion circuit (13) that stores 256 modulation patterns.
It consists of a ROM that stores 256 modulation patterns in which the number of 1 bits is 6.

In the 8-9 conversion circuit (4), there are three cases where the number of "1" bits included in the 9 bits is 3, 4, and 5 as shown in Figure 3 above. Yes, the number of "1" bits out of 20 bits as a whole of 2 bytes is 9
Therefore, the 9-bit output converted by the 8-9 conversion circuit (4) contains "1".
” conversion circuit (11), (1
This allows the user to select either of 2) and (13).

Therefore, the 2-bit signal from the 8-9 conversion circuit (4) controls the selector (10) depending on whether the number of "1"s is 3, 4, or 5. If the number of rlJOs is 3, the conversion circuit (13) is selected, if the number of rlJOs is 4, the conversion circuit (12) is selected, and if the number of rlJOs is 5, the conversion circuit (11) is selected. The selector (10) is controlled. It is also possible to select the ROM to be read by controlling the chip select terminals of the conversion circuits (11, 12) and (13) without providing the selector (10).

The last 1 byte of the 2-byte unit converted into 11-bit modulation pattern data is parallel-
The serial converter circuit (14) converts the signal into a serial signal, which is then passed through the parallel-serial converter circuit (5), the laser drive circuit (6), and the switch circuit (7).
is supplied to the laser device (8) through the disk (9).
) will be recorded following the above.

In this way, there are 20 data in 2-byte units on the disk.
It is recorded as bit data, and moreover, it is recorded as data including nine "1" bits among the 20 bits. So to speak, 9-20 modulation is performed.

8-11 conversion circuits (11), (12) and (13)
) examples of conversion tables are shown in FIGS. 4, 5, and 6, respectively.

Next, decoding during reproduction of data recorded in this manner will be described.

That is, data extracted by the laser device (8) is supplied to the waveform equalization circuit (21) via the switch (7). In this waveform equalization circuit (21), reproduction equalization is performed by a so-called analog filter, and its output is supplied to an A/D conversion and differential detection circuit (22) as well as a clock reproduction circuit (23).

The clock regeneration circuit (23) has a PLL and generates a channel clock synchronized with the regeneration signal. This channel clock is supplied to the A/D conversion and differential detection circuit (22). As mentioned above, this differential detection circuit (22) selects 9 reproduced data for every 20 channel clocks in descending order of level, sets the positions of the 9 as rlJ, and outputs 20 bits. The first 9 bits of the 20 bits are a 9-8 conversion circuit (24)
The latter 11 bits are respectively supplied to the 11-8 conversion circuit (25). Then, in the 9-8 conversion path (24), the conversion table shown in FIG.
Lunch will be served. On the other hand, in the 11-8 conversion circuit (25), the 11-bit data is returned to 8-bit data corresponding to FIGS. ) is latched. Then, each latch output is supplied to a combining circuit (28), and a 2-byte signal is correctly demodulated by this combining circuit (28) and obtained from an output terminal (29).

In this way, so-called 9-20 modulation is performed in units of 2 bytes, making it possible to realize a modulation method with a wide window width and shortest inter-bit distance.

In other words, Fig. 2 shows an example of the modulation pattern of this 9-20 modulation method. Since the conversion is performed in units of 2 bytes, each byte is converted into 7.6 μm, which is within 15.2 μm, that is, the time length. Two bytes of data are recorded as 20 bits of data in 2.7 μsec.

If we consider 1 byte of data, this is an 8-10 conversion (
This corresponds to conversion of the number of bits) and is converted to 10-bit data, so the channel bit width, that is, the window width, is 0.00000.
It is better than 69 μm and can realize a window width of 0.76 μm.

Also, the shortest distance between pits is "0" between rlJ and "1".
However, when considered in units of 1 byte, it is equivalent to 8-10 conversion, so the shortest focus distance is also 0.
．． The result is 76X 2-1.52 μ tide, which is approximately the same value as 4-11 modulation. Therefore, compared to the two modulation methods described as conventional examples, a modulation method with a wider window width and a wider minimum distance between pits can be realized.

By the way, just as a 2-byte unit, 16 bits - 2
Even if you prepare a 0-bit conversion table, the 9
-20 modulation can be performed. However, in that case, 2 bytes of data, or 16 bits of data,
Since it is necessary to convert data into 0-bit data at once, it is necessary to prepare a map for encoding and decoding all 65,536 patterns, and this map becomes a very large-scale circuit. However, as shown in Figure 1, if byte-by-byte mapping is performed, such a huge number of maps is not necessary.
In the case of FIG. 1, 256×4 mapping patterns are sufficient.

Note that the above example is based on the case where 2-byte data is expressed by 9 "1"s out of 20 channel bits, but the number of bits in rlJ is 10t[I or 11(
This is also possible with IN. However, 2 bytes of data is 2
It cannot be expressed by 81 "1's" in the 0 channel pit, nor can it be expressed by 12 or more "1's" in the 20 channel bit.

Although the above description has been made with reference to the case where the recording medium is a disk, the present invention can be applied to cases in which digital signals are recorded on various recording media and reproduced from the recording medium.

〔Effect of the invention〕

According to this invention, it is possible to realize a modulation method with a wide minimum pit distance and a wide window width, resulting in highly reliable and stable data reading. J[t)lil can be performed.

Furthermore, even when trying to increase the recording density of the current device, according to the present invention, since both the shortest pit distance and the window width are wide, there is ample margin, and it is possible to improve the recording density. It is.

[Brief explanation of the drawing]

FIG. 1 is a diagram for explaining one embodiment of the present invention, and FIG.
The figure is a diagram for explaining the main part, Figures 3 to 6 are diagrams showing an example of a conversion table of the main part, Figure 7 is a diagram for explaining the differential detection system, and Figure 8 is a diagram for explaining the differential detection system. FIG. 2 is a diagram for explaining an example of a conventional modulation method. (4) is an 8-9 conversion circuit which is an example of the first conversion table, and (11), (12) and (13) are 8-11 conversion circuits which are examples of the second conversion table.

Claims (1)

[Claims] When converting two units of digital data each consisting of 8 bits into 20-bit data, 9, 10, or 11 of the 20 bits are "1". A recording modulation method that converts one unit of 8-bit data of the two units of data to 9-bit data using a first conversion table, and converts the remaining one of the two units of data to 9-bit data. The 8-bit data of the unit is 11-bit data, and when added to the number of "1" bits in the output of the first conversion table, the result is 9, 10, or 11. A digital signal recording modulation method that obtains 11-bit data including the number of 1 bits using a second conversion table.