JP2744229B2 - Image processing device - Google Patents

Description

Description: TECHNICAL FIELD The present invention relates to an image processing apparatus for processing image data.
You. <Conventional technology> Conventionally, high-density color image data is converted or edited.
The processing equipment is Cytec's Response 30
0 Series, Crossfield STUDIO-800 Series
And other printing systems are known. However, these devices are large and require
There was a disadvantage that the processing time was long. <Purpose> The present invention has been made in view of the above points, and has been made in consideration of the following.
Transfer processing with a simple configuration, without loss of image information, and
The purpose is to enable high-speed execution.
Compressed data that stores compressed data obtained by compressing m by m pixels
A data memory, and compressed data from the compressed data memory.
Reading means for reading the pressure, and the pressure read by the reading means.
Decodes compressed data and outputs image data every m lines
Decoding means, and m lines decoded by the decoding means
Image data without scaling
Processing means for performing a conversion process for tilting
Mxm image data of m lines converted from
Compression means for compressing in elementary units and compression by the compression means
Storing compressed data in the compressed data memory.
Storage means, and the compressed data
Reading compressed data in the first direction from the memory;
Decoding by the decoding means, and
Forward the image without scaling the image by the processing means
Performing a first conversion process inclining in the first direction;
The converted image data is compressed by the compression means.
And storing the compressed data in the compressed data memory by the storage means.
A first image tilting process, and
The compressed data in a second direction orthogonal to the first direction;
The compressed data from the data memory, and
Decrypting the decoded image data to the processing means.
The image in the second direction without scaling the image
A second conversion process of tilting is performed, and the second conversion process is performed.
Compressing the image data by the compression unit;
The second image tilt stored in the compressed data memory
Processing, and the compressed data memory by the reading means
Read compressed data in the first direction from
Decoding by the decoding means, and
Forward the image without scaling the image by the processing means
Performing a first conversion process inclining in the first direction;
The converted image data is compressed by the compression means.
And storing the compressed data in the compressed data memory by the storage means.
The third compressed image memory
To the image represented by the compressed data stored in
To provide an image processing apparatus for performing rotation processing by
You. <Example> (Overview) Generally, the functions of an image processing apparatus include: The above two editing functions are required. The former is generally
Called a pipeline processor with hardware
This device has a certain high speed with image editing function
Execute for items that require. Depends on the latter CPU
Processing is performed interactively with humans
Execute (may take some time). In other words, the former pipeline processor
Affine transformation (enlarge / reduce / transfer)
Motion / rotation) and spatial filtering (image enhancement / smoothing)
Etc.) and images such as color conversion processing by look-up table (LUT)
The sequential processing of images is mainly performed. The latter CPU processing is generally complicated processing and hardware.
Performs processing that cannot be easily converted to air. Here, the image can be any shape
Or copy the cropped image to another location.
Processing, such as correcting part of an image. these
Is generally a creative process created by the operator.
It takes some time and it is acceptable. However
This function needs to be sophisticated. Performing the above two editing functions with maximum performance
In order to apply it, the system architecture of the editing device
It is necessary to think from. In other words, both processes are fully functional
To enable high-speed execution, configure the system
System and how image data is handled (formmatsu
G), signal flow, function analysis, etc. need to be considered.
You. As a result of various studies, the system as a color image processing device
The following conclusions can be drawn as
Was. (1) To perform image editing, the image data must be compressed data
Have as. (2) As a compression method, m × m blocks are represented by one code.
Block coding or vector quantization. (3) In order to perform high-definition image editing,
Image data needs to be rewritten. In (1), high-resolution and high-gradation image editing processing is performed.
Therefore, the image data capacity becomes extremely enormous.
For example, color reading of A4 1page with 16pel / mm each 8bit / PIXEL
Data capacity of about 48 Mbytes for R, G, and B colors.
You. The above-mentioned image editing is interactive and
Compression of such color image data for high performance
The key technology is to make it easy to edit. others
The block coding or vector quantization method of (2)
Was determined to be optimal. This means that m × m pixels of the image are one fixed-length code
High compression ratio and image data
Data can be easily edited because the position information of the
It is. If you have image data using such a compression method,
Is required to perform high-resolution image editing as shown in (3).
To the pixel data within m × m of the compression unit
I noticed that I needed to rewrite it. That is, the vector quantity
A set of code data held in m × m units by the childization method
As a result of editing, depending on the editing item,
We need to change it. Hereinafter, the present invention will be described in detail based on embodiments. (Detailed Description of the Embodiment) FIG. 1 shows the structure of an image processing apparatus showing an embodiment of the present invention.
FIG. Image data read by reader 1
(For example, 8bit digital data for each of R, G, B)
Luminance (Y) signal used for NTSC signal after signal conversion
And color difference signals (I, Q). Such a conversion is for example
R, G, B data It can be obtained by the following matrix calculation. Here the transformation mat
The Rixx coefficient matches the reader's color separation characteristics, γ characteristics, etc.
And amended accordingly. Such Y, I, Q signals are compressed by a compressor
Disk memo for image data file compressed by 2
It is stored in the file 3. The image data on the disk is
Read and process / edit on IC memory called Mori5
It is. Here, basic processing is performed by hardware to perform high-speed processing.
Performed by the integrated pipeline processor 4.
You. On the other hand, the image data on the image memory 5 is
Seed processing is performed and processing and correction are performed. Editing process is CR
Displayed on color CRT 10 by T controller 9 and edited
You can monitor the situation. The edited result is
Molly 5 returns to original image data through decoder 6
The image data is compatible with the printer by the converter 12.
Color signals (yellow, magenta, cyan, black)
And output to the color printer 7. (Compression encoding) Next, a compression method of image data will be described. Dividing into three color signals of luminance and color difference like Y, I, Q
Better preserves spatial frequency of Y signal as luminance data
If this is done, the spatial frequencies of the I and Q signals
To some extent (cut of high frequency components)
Is known to be small. Therefore, for example, the I and Q signals are m × m blocks (m is an integer).
The color information is represented by the average value of
A data compression method that reduces the amount of data is conceivable. I, Q signal
The block size depends on the required image quality and the allowable memory capacity.
Select a block size such as 2 x 2, 4 x 4, 6 x 6 depending on the amount
It is. For example, if the block size is 4 × 4,
As mentioned above, A4 1page memory capacity of 48MByte is equivalent to Y signal of 16MBy.
te (uncompressed) + I, Q signal 2MByte = 18MByte in total, about 2.7
Compression rate. On the other hand, resolution of Y signal is different from compression of I and Q signals.
A compression method that leaves sufficient data is required. The first method is a block coding method. This method calculates the average value of pixel data x in m × m blocks.
, The standard deviation σ is calculated. Next, the shade information for each pixel
Is represented by about several bits. For example, the calculated value of (x −) / σ
Can be realized by requantizing. This compressed data
The format is as shown in Fig. 2 (a).
Following the quasi-deviation, the density information of each pixel is continued.
The order is made to correspond one-to-one with the pixel position in the block. The second method is a vector quantization method of m × m pixels.
You. This method averages the pixel data in m × m blocks
, The standard deviation σ and the code representing the image pattern
It expresses and measures the compression of data. This compression
The data format is as shown in FIG. In the above, resolution data is taken only for the Y signal, and the
This is an example of taking the average data within m × m.
The IQ signal may be encoded in the same format as the Y signal.
I don't know. Such a compression mode converts one image consisting of N × N pixels into m
Xm small section, and this mxm is fixed to one k-bit
Compressed image data to convert to fixed length code
Is Memory structure consisting of k bits in the depth direction in the address space of
Take the lead. Therefore, the image address information is retained.
Compression method with variable code length such as normal MH, MMH, MR, MMR
Random access such as image search, writing, etc.
It has good access performance and can be said to be optimal for editing work. Scale and rotate image data based on the above compressed data
If you try, you need to rewrite the compressed data itself
Become. FIG. 3 is a view for explaining this, and FIG. 3A shows an image data of m × m pixels.
Indicates the delimiter of compressed data when data is compressed. m × m
The image data contained in each unit is directly converted into one code.
You. (In the future, this m × m compressed pixel will be referred to as
It will be referred to as a “stick pixel”. ). Figure 2 shows the result of rotating the image for each block pixel.
Thus, the letter T is cut in this way.
To avoid this, restore the block pixels as shown in Fig.
To restore the original image data, rotate each pixel
It is necessary to compress again as shown in Fig.
You. At this time, the m × m pixel unit A to be compressed shown in FIG.
After the transfer operation, the image is transformed as shown in A '
Straddle between elementary units. That is, in order to perform high definition Needs to be converted. Now consider the case of rotating the image,
When rotating back to image data of
Memory capacity for storing the original image data
The amount becomes extremely large. 48Mb on par with the worst image data
An intermediate buffer of yte data capacity is required. Extremely simple operation of decoding → rotation → compression encoding
The method of realization is proposed below. First, the method of rotation will be described. Generally any angle
The rotation of θ can be realized by the following affine transformation. here Is the coordinates of the image before rotation, Is the coordinates of the rotated image, and θ is the rotation angle. This two-dimensional
After conversion, the coordinates of the image before rotation Is at the address point of the image data (which can be expressed as an integer)
Is the coordinates of the rotated image Is generally irrational. Therefore, the rotation
For high precision, some non-integer coordinates Address point of data after rotation from data of (Integer) data must be obtained by two-dimensional interpolation
No. Also, the data obtained as a result of rotation as shown in FIG.
A considerable amount of capacity, as it is not trivial to clean in m units
Without the intermediate buffer, recompression becomes difficult. So 2
Equation (1) of the dimensional affine transformation is expressed as follows in the x direction and the y direction.
Into two products. First one-dimensional x-direction transformation T ₁ Then: That is, one-dimensional conversion T ₁ The value of y does not change,
The direction width is reduced to cosθ times. Diagram 4
The rectangle in (a) is a one-dimensional conversion T in the x direction. ₁ And as in (b)
Be transformed. Next, the one-dimensional conversion T in the second y direction _Two So next
Become like That is, one-dimensional conversion T in the y direction _Two Then, Fig. 4-(A)
As shown in (b) to (c), the value of x does not change in the y direction
Image is enlarged by 1 / cos θ times and rotated by angle θ
(C) is obtained. As described above, the one-dimensional conversion T in the x direction ₁ Then, 1 in the x direction
If the line data is read, the coordinates of the converted data will be
Data of the address point of the converted image
Can be obtained by one-dimensional interpolation. That is, in the x direction
Processing on a line-by-line basis is possible. Therefore, one-dimensional change
Exchange T ₁ Collecting m lines of processing results of
Shrinkage is easily possible. Similarly, one-dimensional conversion T in the y direction _Two To
Also, one line of data in the y direction is read out and one line is read.
In-unit processing is possible. However, compressed data is stored
Memory (5 in FIG. 1) in two directions, x-direction and y-direction.
The configuration must be accessible. On the other hand, in the above one-dimensional conversion, FIG.
In the conversion from (a) to (b), image shrinkage occurs in the x direction.
In the conversion from (b) to (c), the image extends in the y direction.
You. For this reason, two decodings in the x and y directions → compression operation
Deterioration of image data is large. Specifically, once you have an image
Image sampling pitch is constant when shrunk
Therefore, image data is lost. Next, enlarge the image data
However, since this missing image data is not restored,
The displayed image has a small amount of information from which high-frequency components have been removed.
Transform into data. Therefore, in the present embodiment, Is converted as follows. Such conversion is performed by (1) one-dimensional conversion operation without image expansion / contraction.
No. (2) Therefore, even if compression and decoding are repeated, the amount of image information is lost.
I can't. There is a feature. That is, T _Three Or T ₁ The operation of Which indicates a mere Shear in the x-direction. The conversion from (a) to (b) in FIG.
In the figure, the shearing in the x direction is performed, and the width β
Does not change. Therefore, it can be realized with simple hardware.
You. In other words, the change of the raster start address of the raster image
It only needs to be changed. Also, when decoding and shearing the compressed data,
Transfer the decoded image data to the line memory of at least m lines.
It is sufficient to shift the head address when touring. Therefore,
There is no loss of image due to the reduction of. Further intermediate line
The size of the buffer is the image width, which is economical. (Enlarge
If there is a reduction, an intermediate buffer larger than the image width is required
Becomes Note that the following equation is used instead of equation (2)-
You can also. Here, the temporary element conversion T ′ ₁ , T ′ _Three Is Shear in the y direction,
T ′ _Two Is a Shear in the x direction, and these transforms
There is no shrinkage, and the same processing as in (2) -expression is possible
That is clear. Next, the conversion of compressed data → decoding → rotation → compression encoding
Details will be described based on the embodiment shown in FIGS.
I do. Fig. 5 shows compressed data → decoding → rotation → compression encoding.
FIG. 2 is a block diagram of the entire circuit configuration. The compressed data is
The data area to be input data and the processed data
It is divided into two areas 511 and 512
You. Before executing the processing, the CPU 8 outputs data to be input data.
The data area is selected by instructing the selector 521. Selection
The selector 521 receives the compressed data memo by the signal 502 from the CPU 8.
One of the address lines 1 and 2 (511 and 512)
Input data side memory address 591 and data write
Connected to the input side compressed data line 503 and the other
Address line to output data side memory address 592,
Then, connect the data line to the output side compressed data line 501.
And the address bus and data for both 511 and 512
Disconnect all tabus (switch to high impedance state)
3) different states. Thereafter, 521 stores the compressed data memory 1 in the input data memo.
And the compressed data memory 2 as the output data memory.
The description will proceed assuming that they are connected. Address calculation times
The path 57 connects the main-scanning synchronous clock and the sub-scanning synchronous clock.
The input data address 59 of the input compressed data memory
1 and output compressed data to which the processed data should be output
Output data address 592 to memory and during processing
Outputs the necessary addresses 593, 594 in the intermediate buffer.
You. The compression memory 511 corresponds to the input data address 591
The data is output to the input side compressed data line 503, and the
3 decodes this and passes through selector 522 to intermediate memory 1 or
2 (541 or 542). Selector 522 sets the value of 593
Intermediate memories 1 and 2 are toggled each time the number becomes m multiples.
Change Selector 523 operates similarly to 522, except that 522
When the intermediate memory 1 is connected, the intermediate memory 2 is connected,
When 522 connects intermediate memory 2, connect intermediate memory 1.
522 and 523 make the data bus 541 and 542
They are alternately connected. Here, m = 4
Let's proceed with the following case and example. 591 is the address
Although it was a bus, the main scanning address and sub-scanning address
Each address is the third bit counted from the lower bit
(Ie the bit in the 22nd place)
It is. Also, the input side intermediate buffer address 593 is
The sub-scanning address is the third bit 1 counted from the lower bit.
Only the bits are input to selectors 522 and 523,
Only main scan addresses are stored in memories 1 and 2 (541 and 542)
Is entered. Intermediate memories 1 and 2 each have 4
It consists of line buffers for rasters,
Output to one of the intermediate memories through selector 522
Is done. On the opposite side of the intermediate memory, the same input intermediate buffer
Address is input, and the address of one of the intermediate memories is
The decoded data is written to the address. On the other hand,
Four lines from the same main scanning address in one intermediate memory
The decrypted data one block line before each buffer is
The signal is output to the interpolation processing circuit 55 through the selector 523. interpolation
The processing circuit 55 is provided with an intermediate buffer main scanning address on the output side of the 594.
Enter the fractional part (composed of integer and fractional parts)
For each of the four line buffer outputs, two consecutive
Create interpolated data from the two data and 594 integers
Intermediate memory 3 or the address corresponding to the address given by the section
Is output to 4 through the selector 524. Selector 524,525
The intermediate memory 3 and the selectors 522 and 523 described above
4 is a selector pair for selecting 4 by a toggle. At this time, 524
And 525 are the sub-scan addresses of the output side intermediate buffer address 594.
One bit of the third bit counting from the lower bits of the dress
A mark is input as a selection signal. Intermediate memory 3 and
4 is a main scanning address of the same output side intermediate buffer address.
Address is input, and the interpolation process is
The data from the circuit is input and the other memory
, The interpolated data one block before is selected by the selector 52.
Output to the compressor 56 in parallel for 4 rasters through 5
I have. The compressor 56 receives the data and inputs the data to each of the four rasters.
Compressed data is output when data for 4 pixels is input from
Output to the selector 521 via the force compression data bus 501. Selection
The selector 521 converts the data on the data bus 501 into a compressed data memo.
Designated by the output compressed data address 592 of file 2 (512)
Output to the address. Next, the address operation circuit 57 will be described. Fig. 6
As shown in the figure, 57 is a clock switching unit 61, an address calculation unit.
62, comprising output switching units 63 and 66 and counters 64 and 65. Above
Rotate the input side address (x, y) and output side
(X, Y), the relationship of equation (1) holds.
I do. At this time, the rotation calculation is performed as shown in equations (2)-(1).
3 T ₁ , T _Two , T _Three Think separately. T ₁ = T _Three Is now T ₁ And T _Two Describe the conversion of
You. (2)-from equation The one-dimensional transformation is Is the main scanning of the y-th raster
The inspection clock takes the x-th pixel as the (x, y) pixel
When handling Y '= y (4) -X' (0,0) = 0 ... (4) -X '(0, j + 1) = X' (0, j) + α ... (4) -X '(I + 1, j + 1) = X' (i, j + 1) (4)- Therefore, the operation of equation (3), Y ', is
It can be calculated only by counting the time
No matter how many main scan synchronization clocks are inserted between clocks, it is unchanged.
Therefore, the address calculation unit 62 is configured as shown in FIG.
Then, the sub-scan synchronous clock is counted up by the counter 76.
Y ′. Also, the CPU 711
The value of α is set, and 731 is set to 0 as an initial value.
Has been reset. The adder 721 has a sub-scanning clock.
Adds the value stored in 731 and the value stored in 711 each time is entered
And outputs the sum. 731 latches the output value,
Hold as a new value and execute calculation of (4) -expression
You. Next, 732 is set to 0 by the CPU as an initial value.
Adder 722 determines the value held in 732 and the value held in 712.
Is added each time the main scanning clock is input.
Perform calculations. The selector 74 indicates that the sub-scanning clock has entered.
731 output is selected and output only, and 722 output is selected
Power. 732 is the main scanning clock or sub-scanning
When at least one of the clocks is input, the output of the selector 74 is output.
Latch the force and calculate the result of (4)-or (4)-
It is taken as X 'as appropriate. Then T _Two In exactly the same way, Y ″ (0,0) = 0 (6) −Y ″ (i + 1,0) = Y ″ (i + 0) + sin θ (6) −Y ″ (i + 1, j + 1) = Y ″ (i + 1.j) +1 ... (6)-In this case, T _Two Is calculated by the main scan synchronization clock described above.
And the sub-scanning synchronous clock were interchanged.
The same holds true. Therefore, in this case, CPU 8
Switch 61, 63, 66 beforehand, and set
The values to be set are sin θ and 1, respectively. Seventh
This is shown in FIG. Also, the compressed data address is assigned to the input and output sides.
And intermediate buffer addresses are provided
The buffer is the main run of the compressed data memory in the case of formula (3).
The scanning address and sub-scanning address are assigned to the main scanning
And as the same address on the sub-scanning side,
In the case of execution of the expression (5), the main scan of the compressed data memory
The side address direction is the address of the intermediate buffer in the sub-scanning direction.
In addition, the address direction of the sub-scanning side of the compressed
It is used to correspond to the main scanning direction address of the software.
is there. This allows the intermediate buffer to be a raster buffer.
However, in the case of the equations (3) and (5),
Is also available. Next, the interpolation processing circuit 55 will be described. Interpolation times
The path 55 has the decrypted data and the output intermediate buffer address.
Fraction of 594 (2 ^-1 Second place, 2 ^-2 Place and 2 ^-3 Each rank
Tut) and 2 of the integer part ⁰ 4 bits in total
Is entered. The interpolation processing circuit 55 is configured as shown in FIG.
4 sets of 801 parts and other parts 1 of the circuit configured in
Each of the 801 parts, each of the four rasters
Each raster corresponds to one raster and operates in parallel. Fig. 8
Circuit holds two successive data of the decrypted data
811,812 for output and intermediate buffer address on output side
802 part that detects whether the integer part of the source has changed
And the interpolation coefficient from the decimal part of the output side intermediate buffer address
803 that outputs the 802 is the output side intermediate buffer
The least significant bit of the integer part indicates whether the integer part of the dress has changed.
Tut (20th bit)
Outputs by detecting whether it has changed. 803
Is the output intermediate buffer address of the two consecutive clocks.
From the decimal part, (1−α ′) V (n) + α′V (n−1) (7) where V (i) is the decoded data of the i-th clock.
Interpolation coefficients α 'and 1 for performing the interpolated interpolation represented
−α ′ is output. 821 has two successive clocks
Enter the decimal part of the input side intermediate buffer address, and
And a 1-α 'output lookup table. (1-α ') V (n) in the equation (7) is converted to 822
In the table, α'V (n-1) is increased to 823
Output each in the table, and use the adder of 83
Is obtained. The result obtained is 802 output 8
In synchronization with 4, intermediate memory 3 or through selector 524
4 is output. The series of operations has been described above.
Inter-memory 3 and 4 are all smaller than 4 raster raster buffers.
522, 523 and 524, 525
Are switched every four main scanning clocks
In this configuration, the memory for four pixels is made up of four buffers.
Of course, it can be handled. The compressed data memories 1 and 2 are physically the same.
The input compressed data is read in time-division
Used as data memory, and when writing, output compressed data
It can be used as a memory. Input compressed data
Is read into the intermediate memory at the time of reading. When writing
Contains the input data area other than the data area
Is not output from Equations (3) and (5).
Is. <Effects> As described above, according to the present invention, an image is an m × m image.
For images represented by compressed data compressed in elementary units
The rotation process to the first image without scaling the image.
First image tilting process to tilt in the direction of, and scaling of the image
Image tilting unit for tilting an image in a second direction without accompanying
And tilt the image in the first direction without scaling the image.
And the third image tilting process is performed separately.
Stored in the compressed data memory for image tilting
Compressed data in the direction of each image tilting process.
, And decode the data sequentially for each m lines.
It is not necessary to decode all compressed data
Therefore, a large capacity to hold all the decoded image data
There is no need to provide an amount of memory, and each image tilts
Missing image information because there is no image scaling due to processing
Image skew processing is performed without image rotation.
Simple configuration, no loss of image information, and high speed
Can be executable.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a block diagram of an image processing apparatus according to the present embodiment, FIG. 2 is a diagram showing a block coding and vector quantization data format, and FIG. 4 (A) and 4 (B) are diagrams for explaining a method of rotating an image by one-dimensional conversion, FIG. 5 is an image processing block diagram, and FIG. 6 is an address calculation. Circuit diagram, FIG. 7 (A),
(B) is a circuit diagram of the address calculation unit, and FIG. 8 is an interpolation processing circuit diagram.

Continuation of front page    (72) Inventor Yuji Nishigaki               3-30-2 Shimomaruko, Ota-ku, Tokyo               Inside Canon Inc.                (56) References JP-A-61-199175 (JP, A)                 JP-A-59-81962 (JP, A)                 JP-A-60-97474 (JP, A)                 JP-A-60-97473 (JP, A)

Claims (1)

(57) [Claims] A compressed data memory storing compressed data obtained by compressing an image in units of m × m pixels; reading means for reading the compressed data from the compressed data memory; decoding the compressed data read by the reading means;
Decoding means for outputting image data for each line; processing means for performing a conversion process of inclining the image data of m lines decoded by the decoding means without enlarging or reducing the image; Compression means for compressing m lines of image data converted by the means in units of m × m pixels, and storage means for storing the compressed data compressed by the compression means in the compressed data memory; The compressed data is read from the compressed data memory in the first direction by the reading means, decoded by the decoding means, and the decoded image data is processed by the processing means into the first image without enlarging or reducing the image. Performs a first conversion process in which the image data is subjected to the first conversion process, is compressed by the compression unit, and is compressed by the storage unit. A first image tilting process to be stored in a memory, and the reading means reads compressed data from the compressed data memory in a second direction orthogonal to the first direction from the compressed data memory, and decodes and decodes the compressed data. The image data is subjected to a second conversion process of inclining the image in the second direction without enlarging or reducing the image by the processing unit, and the image data subjected to the second conversion process is compressed by the compression unit. A second image tilting process for storing the compressed data in the compressed data memory by the storage means, and reading the compressed data in the first direction from the compressed data memory by the reading means, decoding the decoded data by the decoding means, and decoding A first conversion process for inclining the image in the first direction without enlarging or reducing the image by the processing means; A third image tilting process of compressing the processed image data by the compression unit and storing the compressed image data in the compressed data memory by the storage unit allows an image represented by the compressed data stored in the compressed data memory to be processed. An image processing apparatus for performing rotation processing by using