JPS63213089A - Pattern signal processing circuit - Google Patents

Abstract

PURPOSE:To accelerate processing by executing the processing of the center of gravity operation with the aid of a hardware. CONSTITUTION:The data aij on a picture element (i and j) is temporarily held by a flip flop 10. This data is transmitted to three arithmetic circuits which respectively operate X (i and j), Y (i and j) and Z (i and j). The multiplication of column numbers (j) and (aij) is executed by the X arithmetic circuit and the result of that is held by a first register 1. The partial sum X (i and j-1) up to the picture element (i and j-1) is held by a fourth register 4. The multiplication of row numbers (i) and (aij) is executed by the Y arithmetic circuit and the result of that is held by a second register 2. The partial sum Y (i and j-1) is held by a fifth register 5. The data is integrated as it is by the Z arithmetic circuit. The partial sum Z (i and j-1) up to the just before picture element (i and j-1) is held by a sixth register 6. When the scanning of one picture is finished, the total sum X, Y and Z are respectively inputted to a seventh to a ninth registers 7-9. The centers of gravity of an (x) direction and a (y) direction are obtained by these X, Y and Z.

Description

DETAILED DESCRIPTION OF THE INVENTION (7) Technical Field The present invention relates to a system that uses hardware to perform high-speed calculation of the center of gravity position, which is frequently used in the field of two-dimensional pattern recognition.

In the field of image processing, the technical field of analyzing contact patterns produced by planar contact sensors, etc., it is often necessary to calculate the position of the center of gravity of a pattern. The pattern here is
It is a collection of binarized data of pixels arranged in a matrix.

FIG. 1 shows an example of one image pattern. The entire screen is divided into NxN pixels vertically and horizontally.

A pixel is the smallest unit that makes up an image. As with a regular matrix, number the rows vertically and number the columns horizontally. Let the row number be 11 and the column number be j. The position of a pixel on the screen can be expressed by (i, j).

It is assumed that i and j take values of 0, 1.2, . . . , (N-1). The pixel in the i-th row and j-th column is expressed by (iej).

Let i=0 be the O-th row, and j=o be the θ-th column.

The center of a pixel is called a grid point. The position of a pixel can be expressed by (iej), but it can also be expressed by the position of a grid point.

Let all grid points of the uppermost leftmost pixel (i=o, j=o) on the screen be the origin. With this as the origin, take the X coordinate to the right and the X coordinate to the bottom. The grid spacing is set to 1 for simplicity.

Then, the X coordinate of the grid point of pixel (i, j) is j.

The X coordinate is i.

The grid point is the center of the pixel. A pixel has one piece of data. In the case of a black and white binary image, this data is either O or 1.

One piece of data corresponds to that pixel. Therefore, data for pixel (ilj) will be written as aij. This is either O or 1. A pixel is the smallest unit of an image that has an area, but a grid point is a point that does not have a total area. In terms of location, there is a one-to-one correspondence. Therefore,
In the operation of determining the center of gravity, pixels and grid points may be used without distinction.

Generally, the X coordinate of the grid point of pixel (i ej) is x(i, j
), and the X coordinate is written as y(i, j).

As already mentioned, if the lattice spacing is 1, then x(i.

j)=j, y(i, j)=i.

The figure in Fig. 1 can be any pattern, for example,
The sound of the object to be grasped by the robot hand is captured by a camera,
Alternatively, it is captured by a planar contact sensor. Binarization processing is performed, and the data of each pixel is either 1 or 0.
It is given in the binary code.

In this figure, it is assumed that the data of the pixels in the white background part is 1, and the data of all the pixels in the black background part is O. Assume that the white part represents the shape of the object. In order for the robot hand to grasp this object, it is necessary to obtain position information of the object.

In order to understand the shape of an object and find its position, it is also necessary to find the position of the outline formed by the black and white boundary line. There are several ways to do this.

Rather, what is important in determining the position of an object is the position of its center of gravity. If the shape of the object is already known or the shape of the object is extremely simple, the most important information is the position of the center of gravity of the pattern representing the object.

However, the center of gravity here refers to the center of gravity of the pattern in the binarized image. It is not the center of gravity of a real object.

Data aij′f: 1 is associated with a pixel where an object exists, and Ot is associated with a pixel where an object is not present. The X coordinate of the center of gravity of this object pattern is "gsy coordinate is y"
If g, then i=o j=.

The center of gravity is calculated by

By calculating the center of gravity, the center position of the obtained pattern figure can be found.

B) Prior art The calculation of the center of gravity of a binarized image Bakun using equations (1) and (2) has conventionally been generally performed by a computer. Computers can be easily used in combination with other processes, so the range of uses can be expanded. It is widely used because of its excellent versatility.

When processing six images using a computer, all matrix data is stored on an image memory called a frame memory, and then the data is read out pixel by pixel and all centroid calculations are performed.

(b) Problems with the Prior Art Since the data for determining the entire image is stored in the frame memory, various calculations and processes other than the calculation of the center of gravity can be performed.

However, since image processing by a computer involves memory, it has the fatal drawback of requiring a long processing time. If there is one image, it is scanned using some method, A/D converted, binarized, and written to frame memory. Writing is completed when scanning is completed. After writing is complete, read the data and calculate the center of gravity.

In many cases, the screen is updated in a short period of time.

It is necessary to calculate the center of gravity faster than the screen is updated. This is difficult. The calculation of the center of gravity by a computer must be performed in a short period of time between the end of scanning one screen and the start of scanning the next screen. Calculating the center of gravity takes time because the calculation is performed while reading data from memory.

If this happens, the screen update cycle becomes too long.

In recent years, there has been a severe demand for higher speeds in the automation field. For this reason, increasing the speed of pattern processing is also an important issue.

2) Purpose It is an object of the present invention to calculate the entire center of gravity of an image pattern by hardware, without relying on a computer.

At the same time as the data on the screen is scanned, the center of gravity calculation progresses.
A second object of the present invention is to enable high-speed processing in which the center of gravity calculation is also completed when scanning is completed.

Not relying on a computer means that there is no double operation of writing image data into a frame memory and reading it out for calculation.

(a) Configuration It is an object of the present invention to obtain equations (1) and (2) regarding the center of gravity, which have been calculated by a computer, by using hardware instead of using a computer.

The values that must be found in (1) and (2) are. In both cases, double integration ΣΣ must be performed.

Since N is a finite value, no matter how the order of integration is changed, the result will not change.

Various methods can be considered for scanning the screen. The simplest one is rask-order scanning. This starts from the upper left pixel (i=0.j=0),
Scan one line from left to right (j = 0 to N-1), then
Go down one line (i=1) and scan from left to right, changing lines.

Such a scan is performed by adding jvo to N-1 and adding this to i from 0 to N-1 (3) to (5).
) strictly corresponds to the double multiplication of

Therefore, when scanning a TV screen etc. in rask order, (
The operations 3) to (5) can be performed in parallel with scanning.

However, the possibilities of the invention are not dependent on the scanning method.

Other scanning methods may be used. In this case, the order of double integration can be changed to match the scanning method.

Now, consider the partial sum up to the i-th row and the j-th column.

This is calculated as X(i, j), Y(i
, j) Z(i, j).

As already mentioned, X(i,j)=j1y(i,j)−
1, so (8) and (4) can be written as follows.

On the other hand, the partial sum is defined as. At the end of the scan, the partial sums are equal to the total sum x1y,z.

X (N-1, N-1) = X (11
)Y(N-1,N-1) =Y(12) Z(N-1,N-1) =4 (13)
Comparing the values of row i, column j and the previous row i, column (j-1), this is jaij, 1aij, aij
is simply added to the previous value.

X(i, j) = X(i, j-1) + jaij
(14) Y(i, j)=Y(i, j-1)+1
aij(15)Z(1-j) = Z(1-j-1)
+ aij (16) By counting the clock pulses while scanning, j and i are immediately known. The reason for scanning is to obtain the data aij Yt. Once the data aij is known, the value obtained by multiplying it by 11 jTi'' is immediately obtained.

This is not a multiplication, but an addition. Since aij is O or 1, jaij is either 0 or j. You can tell whether it is 0 or j without multiplying by j x aij. If multiplication is involved, this value cannot be obtained in the scanning time of one pixel. However, since j is known, there is no need to calculate it.

The same applies to 1aij. This is i or 97i.

Regarding aij, it is simple: it is either O or 1.

In the present invention, each time a pixel in the i-th row and j-th column is scanned and data aij is obtained, jaij, 1aij, aij
is added to the previous partial sum, and the partial sum up to (i, j) is
(i, j), Y (i, j), and Z (i, j) are obtained, and upon completion of scanning, the total sums X, Y, and Zi are obtained. Then, the center of gravity is found by xg=l(17) yg=7(1f3). x, y parallel to the scan
, z are found. After completion of scanning (17), (18
) is all you need to do. This can be done in a short amount of time.

Therefore, the coordinates of the center of gravity of the pattern can be determined at the same speed as the screen is scanned.

This will be explained using drawings.

FIG. 3 shows the matrix structure of the screen.

Column numbers are numbered from left to right and row numbers are numbered from top to bottom. The θth row is aoo, aol, a02, ... a0
7. This is a matrix with 8 rows and 8 columns.

In other words, this is an example where N=8. This is an example where N is extremely small. In practice, such a small N is sometimes used, but in many cases N takes a large value such as 256.512.1024.

In FIGS. 3 and 4, N=8 is used to simplify the explanation.

FIG. 4 shows the flow of data by rask type scanning. aoo s..., leads to ao7, then al next line below
Move to o. The scanning of this row ends at a17, and the next row a20
Move to.

In this way, a series of data is obtained.

From these data series, along with scanning, partial sums X(i, j) of j aij, partial sums Y(i, j) of 1at3, a
Find the partial sum Z(i, j) of ij.

FIG. 2 shows a pattern signal processing circuit for performing such calculations. The whole consists of flip-flops, registers, counters, multipliers, dividers, etc.

However, although the multipliers and dividers will be explained later, they are not ordinary analog or digital multipliers and dividers.

Because high-speed calculations are required, special circuits are used for multiplication and division.

In FIG. 2, the flip-flop 1o has a pixel (iej
) is temporarily held. aij is 1
or O, so a flip-flop is sufficient. The time for holding data is the scanning time for one pixel. In other words, it is for one clock. For example, this is 100nsec
It is a short time of ~150 nsec.

This data is sent to three arithmetic circuits.

X(i,j)'e operation"r 7) Tame(7) X operation circuit, Y operation circuit for calculating Y(i,j),
This is a Z calculation circuit for calculating Z(i,j)'e.

The X operation circuit has a first count that generates all column numbers j. This is counted up every clock generated for each pixel, and when one row of data is input, t j = o...
・, N-1), cleared. In other words, for each row of pixels, the column number jt of the pixel currently being scanned is generated.

A multiplier 13 multiplies j and aij. This must be done in less than one clock, so
It's not a normal multiplier. This will be explained later. The result of the multiplication is jaij. This is temporarily held in the first register.

The fourth register 4 stores the partial sum X(
i, j-1) are retained. The output of this register and the output of the first register are added by an adder 15. This result is circulated and placed in register 4. In other words, by register 1.4 and adder 15, X(i, j-1) + jaij = X(i, j)
The entire calculation (19) is performed. This is (14
) is the same as the expression.

The fourth register always stores partial sums up to X(i,j). The same thing applies when the line number changes. Partial sums are always calculated and stored in the same order as the scan order.

When the line number changes, X(i, N-1) + (N-1) at N-1= X(
i+1,0).

Now, in the Y calculation circuit, there is a second counter 12. This is a counter that generates line numbers. This remains unchanged while one line of data is input, and is counted up by one when one line of data is completed. In other words, the value increases by one every N clocks. Then, when one screen is completely scanned, it is cleared. In other words, it is O at the start of scanning, and is counted up by one for each row, so the row number i is output.

Multiplier 14 multiplies data aij by row number i=2.

The second register 2 temporarily holds this value 1aijk.

The fifth register 5 is a register that stores the entire partial sum Y(i, j-1) up to the immediately preceding pixel when scanning a pixel (itj). The output of this register and the second register 2
An adder 16 adds the output of the . Then, this result is input to the fifth register 5. This operation is Y(i, j-1) +1aij = Y(i, j) (
21). This is the same as equation (15).

In this way, the fifth register 5 calculates all the partial sums Y(i, j) up to the pixel being scanned (i I j) and always holds the partial sum Y(i, j).

Since the Z calculation circuit integrates the data as is, the column number 11 and row number i are unnecessary. Therefore, there is no counter. Data aij of pixel (ilj) is held in register 3.

The sixth register 6 is a register that stores the partial sum Z (i, j-1) of Z up to the immediately preceding pixel (i, j-1). An adder 17 adds the output of the sixth register 6 and the data of the third register.

The added result is input to the sixth register 6. The calculation here is Z(i, j-1) + atj= Z(i, j)
(22) can be written. This is the same as equation (16).

In this way, the sixth register always holds the partial sum zci, . . . i)v of the data up to the pixel being scanned.

In this way, while one screen is being scanned, the fourth register 4, fifth register 5, and sixth register 6 contain partial sums x(i, j), Y(i, j), Z(i, j) is maintained. While scanning continues, no latch signal is applied to the seventh register 7, the eighth register 8, and the ninth register 9. In other words, do not enter data in the middle.

When all matrix elements have been scanned, that is, when one screen has been scanned, a latch signal is applied to the seventh register 7, the eighth register 8, and the ninth register 9.

At this time, the previous stage register 4.5.6 contains the total sum x1y,
z is required.

The seventh register inputs the total sum X.

The eighth register 8 inputs the total sum Y″ft.

The ninth register 9 is operated by the total sum z2.

The value X1Z of the seventh register 7 and the ninth register 9 is divided by the divider 18 to obtain the center of gravity Xg in the X direction. In other words, perform the division xg=7(28). This is the same as equation (17).

From the value Y12 of the eighth register 8 and the ninth register 9, the center of gravity yg in the X direction is obtained by performing all divisions by the divider 19. In other words, 7g = -z (Sae) (2o
). This is the same as equation (18).

What should be noted is that the total sum x1y,z is determined when the scanning of the pixel matrix of one screen is completed. Although the time and height required for the calculation of the divider 18 and 19 will vary, the center coordinates (xg
e yg ) can be found.

There is almost no delay between the completion of scanning and the calculation of the center of gravity.

This is an excellent effect of the present invention.

The problems in the circuit configuration shown in FIG. 2 are the multipliers and the dividers. Extremely fast multiplication and division are required.

When processing all video signals, the clock is 100nsec
~150 nsec. For example, 30 per second
Assuming that one screen contains 512 x 512 pixels, simply distribute and one clock is 127 nsec.
becomes.

Multiplication and division must be executed between 100nsec and 150nsec. However, regarding division,
Since scanning is performed only once for one screen, there is some leeway.

With normal methods, it is impossible to perform all multiplications in this time interval.

However, in the present invention, although it is called multiplication, since the data aij is a binary code of 0 or 1, all of this can be realized with a simple configuration. FIG. 5 shows an example of the configuration of a multiplier.

Flip-flop 10 holds data aijk. The counter corresponds to counter 11 or 12 in FIG. In the case of the counter 11, the clocks are directly counted and the column number j is output. In the case of counter 12, 1/
The frequency is divided by N and the row number i is output.

Buffer IC 20 is connected to counter 11.12. Buffer IC21 is connected to the value 0.

The buffer ICs 20 and 21 have a select terminal, and when the voltage is 1", the input data is output as is. If the select terminal is NO", no data is output (the output becomes high impedance).

The data aij held by the flip-flop 10 is 0
If so, it is inverted by the inverter 22 and the buffer xc21
will output the value Ot.

If the data aij is 1, the buffer IC 20 is selected. Value j or il held by counter, buffer I
C20 outputs.

In this way, if the data is 0, 0 is output, and if the data is 1, j is output, and this is held in register 1 or 2.

The buffer ICs 20 and 21 can be configured with one selector IC. This is not a multiplication because it simply selects one of two inputs and outputs one. Computation time is extremely short. It takes less than 100nsec.

In this way, the multiplications jaij, 1aij can be performed by selectors.

Although the divider is not required to process as quickly as this, it is still desirable to be able to process in a short time.

It is not possible to use a digital divider.

Therefore, we will use the table reference method using ROM''fc.

The case where xg=X/Z'f is determined will be explained.

Divide address input into two groups. It is designated by one address input group 1kx. The other address input group is designated by Z. Here, X is ΣΣjaij, Z
is ΣΣaij.

The entire address is specified by XZ. The value of X/Z will be stored in the memory address specified by this address.

In this way, the address xZ is designated and the center of gravity Xg is obtained as the output of the memory.

This is provided as a fixed memory in ROM, and all of this is read out, and division is not performed each time. The division is performed in advance and all X/Z values are stored.

The sum of the Z and XQ bit numbers becomes the ROM address bit number. For example, if N=8 as shown in FIGS. 3 and 4, Z and X are 6-bit numbers.

The address bits of the ROM are 12 bits.

If N = 512, ZlX will be an 18-bit number, so the address bits of the ROM will be 32 bits.

(To) Effect The centroid calculation process that is frequently performed in the field of pattern processing is not performed by a computer, but by hardware. According to this hardware configuration, partial sums are obtained at the same speed as the scanning of the sensor, and the center of gravity is obtained upon completion of scanning one screen. It is a high-speed calculation.

Real-time processing, which was impossible when using a computer, becomes possible.

It can be suitably used in the field of pattern processing, particularly in the field of image processing and data processing of planar contact sensors that can obtain sensing data in a matrix form.

[Brief explanation of the drawing]

FIG. 1 is a binary image showing an example of a pattern consisting of a set of pixels divided into a matrix. FIG. 2 is a configuration diagram of a pattern signal processing circuit according to the present invention. FIG. 3 is a plan view of an 8×8 matrix image. FIG. 4 is a data string diagram when an 8×8 image is scanned in rask order. FIG. 5 is a circuit configuration diagram of a multiplier used in the circuit of FIG. 2. FIG. 6 is a circuit configuration diagram of a divider used in the circuit of FIG. 2. 1 to 9...Register 10...Flip-flop 11.12...
・Counter 13.14... Multiplier 18.19... Divider

Claims (1)

[Claims] The screen is divided into a large number of rectangular pixels arranged vertically and horizontally, and pixels (i, j) are specified by row number i and column number j, and each pixel (i, j) binary data a_i_j
The position of the center of gravity of the pattern created by the pixel with a_i_j=1 is xg=Σ_iΣ_jja_i_j/Σ_iΣ_ja_i
＿jyg=Σ_iΣ_jia_i_j/Σ_iΣ_ja
This is a circuit for calculating by the formula _i_j, which scans the screen, inputs data a_i_j as time series data together with a clock, and inputs the column number j and row number i, which are coordinate data in the x direction and y direction, with the clock. The data a_i_j is generated using a counter that operates by counting
When is 0, select 0, and when data a_i_j is 1, select column number j or row number i to find ja_i_j or ia_i_j, ja_i_j, ia_i_j,
a_i_j are added by a cyclic adder in synchronization with the clock, and as soon as one screen is scanned, the total sum X=
ΣΣja_i_j, Y=ΣΣia_i_j, Z=ΣΣa
_i_j is obtained, the combination of
, Y, and Z values.