JPS61214689A - Picture recorder - Google Patents

Abstract

PURPOSE:To decrease a pseudo color and to obtain a more accurate chrominance components by interpolating color difference signal of a scanning line to be interpolated based on a weight means of color difference signals at a position corresponding to a scanning line before and after the color difference signals. CONSTITUTION:The chrominance components and a luminance signal at each scanning line outputted from a reproducing device 41 are subjected to 1H delay sequentially by delay circuits 42 44. In the output state of each delay circuit at a time, the output of a delay circuit 42 is Eby(i+1), the output of the delay circuit 44 is Ery(i) and the output of the delay circuit 43 is Ey(i+1), where Ery(i+2) and Ey(i+2) are outputs of the reproducing device 41. Thus, the output of an adder 45 is the sum between Ery(i) and Ery(i+2) and the output of an operating device 46 is a half of the said output. Since the output of the operating device 46 and the output Eby(i+1) of the delay circuit 42 via a changeover switch SW11 and the output Ey(i+1) of the delay circuit 41 are inputted respectively to a matrix circuit 47, the chrominance components are obtained.

Description

DETAILED DESCRIPTION OF THE INVENTION (Field of Industrial Application) The present invention relates to an image recording apparatus that directly copies a color image displayed by scanning lines as a hard copy.

(Prior art) Recently, instead of ordinary cameras using silver halide film,
ccolJ! A small still image camera that combines an m-image element and a magnetic disk has been developed.

In this type of still image camera, the image processing device for the magnetic disk processes a luminance signal Ey and two color difference signals Eb.
In particular, a line sequential method is used in which the two color difference signals are recorded alternately for each scanning line.

This kind of composite signal is transmitted by S, which is one of the broadcasting systems.
It is similar to the ECAM system, and the luminance signal Ey and color difference signals Eby and Ery are expressed by the following equation.

Ey-0,3Er +0.59Eg ÷○, 11Eb Eby-Eb-Ey (2>E
rV-Er-EV (3) Here, Eb, EO, and Er are blue and green, respectively.

It is a red color signal.

Still images recorded on this magnetic disk are normally reproduced as shown in FIG. That is, first, the luminance signal Ey written on the magnetic disk by the reproducing device 1 and the color difference signals Ery, Eb
y is output, and a color order determination signal for determining whether the color difference signal is Ery or Eby is read out. Color difference signal Ery.

Since Eby is output alternately for each scanning line, the same information is repeatedly output over two scanning lines using a one-line delay circuit 2. The luminance signal Ey and color difference signals Ery and Eby obtained from each scanning line are input to the matrix circuit 3, and the above (1), <2), (
3) The color signals Eb, Ea, and Er are obtained by performing the inverse calculation of the equation.

(Problems to be Solved by the Invention) In this way, by using the one-line delay circuit 2, the color signals Eb and E are
In the case where g and Er are obtained, distortions such as false colors may occur in the boundary line portions and the like. An example in which this distortion occurs will be explained based on FIGS. 6 to 8.

FIG. 6 is a diagram showing scanning lines, where ○ marks indicate pixels in odd fields (scanning Il), and Δ marks indicate pixels in even fields (
(scanning line). In FIG. 6, 1. If the scanning line has information about the color difference signal Eby, the second scanning line has information about the color difference signal Ery, and the third scanning line has the color difference signal Eby again.

Similarly, in even fields, if the color difference signal Eby is present on the 264th scanning line, the color difference signal Ery is present on the 265th scanning line, and the color difference signal Eby is present on the 266th scanning line.

Now, an example in which false colors occur at the boundary will be described in the case where pixels in a portion A, in which information in the vertical direction perpendicular to the scanning line is surrounded by dotted lines in FIG. 6, are sampled.

In this case, FIGS. 7(a) to (f) show that the color belief Eb,
In this case, Ea-0 is assumed, and only the red color signal Er has a value. In the figure, the horizontal axis is the vertical position. This figure 7 (
In c), (e), and (f), the . mark is the initially assumed value, and based on this information, the luminance signal E is calculated from equation (1) above.
A value of y is generated. In equation (1), Eb −Ea −0
Substituting , Ey becomes Ey-o. 3 [Er, which is 0.3 times the magnitude of the red light Er.

As a result, the luminance signal Ey becomes 0 as shown in FIG. 7(a).
The value is indicated by the mark. Also, as shown in Figure 6, the color difference signal Ery is obtained every other scanning line for each of the odd and even fields, so it is represented by a circle at a period of 2 sample points, as shown in Figure 7(b). Since accurate values are obtained and a delay is applied by the 1-line delay circuit 2, the same data is repeated in odd fields and even fields, and
The gap is compensated by the value given by the X mark. The way to compensate is
A method is used to fill in areas where no data exists with data on adjacent scanning lines within the same field (for example, an odd field). For example, in the case of FIG. 7(b), the data at point K1 is supplemented with the data at point K2 adjacent in the same field. The blue color difference signal Eby is (2
), Eby--Ey is obtained, so the interpolation point is at the opposite position from (b) as shown in FIG. 7<d).

When the color signals Er and Eb are reproduced from the color difference signals Ery and Eby obtained in this way, FIG. 7(c) and
The value is represented by the Q mark in (e).

FIG. 7(f) shows the reproduction of the color signal Eo from this information and the W11 signal Ey.

From this result, especially at the edge where the brightness changes,
The difference from the initially assumed luminance value becomes significant, and false colors occur. That is, E in FIG. 7(f).

Taking a signal as an example, a prominent color signal E(+1) causes a false color (green in this case) to be generated.

In addition, the green signal Eo-0 is shown in FIGS. 8(a) to (f),
This shows the case where the color changes from the red color signal Er to the blue color signal Eb. As in the case of Fig. 7, the difference from the initially assumed luminance value is particularly large at the edge where the color changes. It's becoming noticeable.

In the conventional technology, the one-line delay circuit 2 repeatedly outputs the color difference signal Ers+ and the color difference signal Ebyll-, which causes distortion at the edge portion.

The present invention has been made in view of these points, and
The purpose is to realize a high-quality color image recording device that can reduce false colors and obtain more accurate color signals.

(Means for Solving the Problems) The present invention, which solves the above-mentioned problems, is an image recording apparatus that records a color image displayed by scanning lines, in which the color difference signal is a color image signal of a line sequential system. In this case, when obtaining each of the three primary colors from the luminance signal and color difference signal,
The present invention is characterized in that a color difference signal of a scanning line to be interpolated is interpolated based on a weighted average of color difference signals of corresponding positions of scanning lines before and after the interpolation.

(Operating Principle) First, the operating principle of the present invention will be explained. FIG. 1 is a diagram for explaining the operating principle of the present invention, in which the color difference signal Ery
, Eby is shown. Consider the case where the color signals Eb, Er, and Ea regarding the i+1 scanning line in FIG. 1 are determined. For the i+1 scan line, the luminance signal Ey (i
+ 1) and the color difference signal Eby (++1>shi). Therefore, with this amount of information, as is clear from equations (1) to (3), only the Eb signal can be obtained. That is, other Er, The EO signal cannot be determined.

In order to be able to calculate Er and Eo multiplied signals, it is necessary to know the remaining color difference signal 1:ry(i + 1>).

(1) When recording a frame, Ery(i + 1>
It is necessary to generate signals from the scan lines on both sides of the 1 scan line. First, the case of frame recording will be explained. Among the scan lines above the i+1 scan line, the scan lines containing Ery information are i −! -263 scan line skipping one scan line. Among the scanning lines below the i+1 scanning line, the scanning line containing Ery information is the scanning line immediately below the i -1-
There are 2f34 scan lines.

Here, considering the influence of 1 scan line and i + 264 scan line on i + 1 scan line, i scan line is
2 scan lines apart from scan line, i + 264
- The scan line is separated by one scan line from the i+1 scan line.

Therefore, Ery signal Ery of i + 264 scan lines
< ++ 264) is the Ery signal Ery(
It can be considered that the influence is twice as strong as i).

From now on, Ery signal Ery(i +
1) can be expressed as the following equation.

Ery (i + 1) - 1/3XEry (i) + 2/3X Ery (t + 264) The Ery signal Ery (i + 1) of the i+1 scanning line is
When obtained in this way, all the image information has been obtained, so from equations (1) to (3), the color signals Eb, Er
, Eg can be obtained. The Eb, Ee, and Eg multiplied signals calculated by such a weighted average method are each Eb
(i + 1), Er (i + 1).

Assuming that Eo(++1), these can be respectively expressed as follows.

Eb(++1) -EbV(i + 1> +Ey (i + 1>) Next, Er(++1) is obtained from the equation (3) solved for Er and the equation (4), and is expressed by the following equation.

Er (+ + 1) −Ery(i + 1) +Ey lt +
1)-1/3xEry)+2/3XEry(i-1-264>+EV(i+1
) Furthermore, EQ (i + 1) is obtained by solving equation (1) for EO and adding Er(++1) to Er, Eb, and EV, respectively.
), Eb (i + 1>, Ey (i +
It is obtained by substituting 1) and is expressed by the following equation.

0.59E(J (i + 1> -Ey (i + 1) -0,3Er (i +
1) -O, 11Eb (i + 1) If both sides are divided by 0.59, the above equation can be expressed as the following equation.

E(1(i+1)−1,69Ey(i+1>
-O,51Ey (i + 1) -O,19Eb (i + 1> Now the signal E(++1) for the i+1 scan line.

This means that Er (++ 1) and EQ (++ 1>) have all been found.

(2) Field recording In this case, scanning lines in the same field (odd field or even field) are interpolated using the same weighted average method as described above.

That is, the scanning line numbers t, i+1 . Just think about i+2. Now consider the case of i+1 scan line. As in the case of frame recording, Ery(
Since information regarding i + 1) is not included, it is necessary to create it from the scanning lines + and ++2 on both sides of it.

The distances from the i scan line and the 12 scan line to the 1+1 scan line are both separated by two scan lines and are equal. Therefore, the degree of influence on the i+1 scan line is equal. From now on i
The +1 scanning line Ery signal Ery (1↓ 1) can be expressed as the following equation.

Ery(i+1>11/2xEry(i)+1/2x Ery(i -!-2>1+1+) When the Ery signal Ery(i+1) of 1 scanning line is found in this way, all image information is Since it has been found, from equations (1) to (3), 0 confidence @Eb(i +
1), Er (i + 1), E(+ (i +
1) can be obtained. In this case as well, it can be determined in the same manner as in the case of frame recording. Eb(i
+1) is obtained from the equation (2) solved for Eb, and is expressed by the following equation.

Eb (+ + 1> -Eby(i + 1) -', Ey (i +
1>Er (i=1) is obtained from the equation (3) and the equation (8), and is expressed by the following equation.

Er (i + 1) −[:ry(i + 1> +EV (i = 1)
=1/2xEry(i)+1/2XEry(i+
2) +Ey (i book 1) Ea (i + 1) can be found by solving equation (1) for EO in the same way as equation (7).

Eg (i÷1) −1,69EV (i + 1)−O,51Er
<i + 1)-O19Eb (i ±
1) With the above, the color signal Eb (+ +
1), Er (i+ 1>, EQ (i+ 1>
This means that all of these have been found.

(Example) Hereinafter, an example of the present invention will be described in detail with reference to the drawings.

FIG. 2 is a block diagram showing an embodiment of the present invention. The embodiment shown in the figure shows an interpolation circuit during frame recording. In the figure, reference numeral 11 denotes a reproducing device for reproducing recorded image signals, and the reproducing device 11 outputs a luminance signal @Ey and color difference signals Ery and Eby for each scanning line.

12 is a sampling hold circuit that samples and holds the Ey multiplied signal at a constant cycle, 13 is a sample hold circuit that samples and holds the Ery and Eby signals at a constant cycle, and 14.15 converts the outputs of the sample and hold circuits 12 and 13 into digital data. Convert to A/
It is a D converter.

Reference numerals 16 and 17 indicate line memories that receive output data from the A/D converters 14 and 15 and store them as image data for one frame, and 18 to 20 indicate line memories 1 and 1, respectively.
Latch circuits 21 to 25 sequentially latch each scanning line data outputted from line memory 17, and latch circuits 21 to 25 sequentially latch scanning line data outputted from line memory 17. The scanning line data output from the line memory 16 is the luminance signal E.
y, and the scanning line data output from the line memory 17 are color difference signals Ery and Eby, and these color difference signals are Ery−+Eby→Ery→E for each field.
By is output alternately. The image signals output from each latch circuit are as shown in the figure when shown in correspondence with FIG. 1. That is, the latch circuit 20 outputs the luminance signal Ey(i+1) of the i↓1 scanning line, and the latch circuit 21 outputs the color difference signal Ery(i+2> of the i+2 scanning line).
The latch circuit 22 outputs the i+264 scanning line color difference signal Er.
y(i −1−264> is i+ from the lattice circuit 23
The color difference signal Eby (i + 1) of one scanning line is output from the latch circuit 24 as the color difference signal Eby of i + 263 scanning lines.
(i + 263), and the latch circuit 25 outputs the color difference signal Ery(i) of the i scanning line.

S W t is Eby (i + 263) and Ery (+
+264) is a selector switch that selects either one, SWz is a selector switch that selects either one of ErV(+ +2> and Ery(i)), and 26 is a switch that shifts the output of selector switch SWl by 1 bit. This is a shift circuit that doubles the input data by shifting the input data one bit to the left.This means that the input data is doubled.This changes the weight.27 is the output of the shift circuit 26 and the selector switch S. 28 is an arithmetic unit that multiplies the output of the adder 27 by 1/3. 29 is an adder that adds the output of the arithmetic unit 28 and the luminance signal Ey (i + 1). , 3o is an adder that adds the color difference signal Eby (i + 1) and the luminance signal Ey (i + 1).
Ws is added to the adder 29 by a command from the playback 8111.
．． A changeover switch for changing the color order that changes 30 outputs,
31 is the output of the changeover switch S W 3 and the brightness signal E
y (i + 1), each color signal Er (i
+ 1), E(1(i+1), Eb (i +
This is a matrix circuit that outputs 1>. The operation of the circuit configured as described above will be explained as follows.

Each image signal Ey reproduced by the reproducer 11.

Ery and Eby are sampled and held at regular intervals by sample and hold circuits 12 and 13. Of these, E
The ry signal and the Eby signal are alternately sampled and held. The sampled and held analog signal is
After being converted into digital data by converters 14 and 15, it is stored in line memories 16 and 17 as image data for one frame. The data stored in the line memories 16 and 17 are sequentially output, and each latch circuit 1
8 to 24 are latched in sequence. As a result, each latch circuit outputs a temporally shifted color difference signal D as shown in the figure.
1 to D5 and a luminance signal Y3 are output.

The color difference signals output from each latch circuit enter changeover switches SWI and SW2. These changeover switches SWz,
SWz selects and outputs Dl and D4 or Dl and D5 bare from among the input data D1 to D4. The selection method differs depending on whether D3 and 02 are the same type of color difference signal. D if it is the same type of color difference signal
Bears l and D4 are selected, and when they are of different types, Dl
and D5 bear is selected. Now, D3 is Ebb(i+1
>, Dl is Ery(i + 264>, and the types of color difference signals are different. Therefore, bare Dl and D5 are selected, and Ery(+ + 264> is input to shift circuit 2.
6 and is doubled, and the changeover switch SW 2 is set to Ery.
Select (i).

Output 2E of shift circuit 26 and changeover switch SW 2
ry(i + 264) and Ery(i > are each added by adder II 27. The addition output becomes Ery(i >, +2Ery(i +-264)). This addition output is added by the following arithmetic unit 28. It will be multiplied by 1/3.

As a result, the output of the adder 28 is 1/3xEry(i)
+2/3XEry(i + 264). Adder 2
9 is the output of the adder 28 and the luminance signal Ey (i +
1> is added. Therefore, the output of the adder 28 is 1/3xEry(i)+2/3xEry(i+2
64>+Ey (i + 1). As is clear from equation (6), this is the color signal E
r(i+1). On the other hand, the adder 30 calculates Eby(i
+ 1) and the luminance signal Ey (i + 1). Therefore, the output of the adder 30 becomes E by (i+1) +EV (i + 1>. This means that the color signal Eb(i
+1). The outputs of these adders 29 and 30 are input to the matrix circuit 31 after the color order is changed by the changeover switch S W s. The matrix circuit 31 is
Signal Er (i + 1).

Receiving Eb (i + 1) and Ey (i + 1>), color signals Er (i + 1>, Eb (i + 1)
, EIJ (i + 1〉), the green signal EO
(+ + 1) is Er (i+ 1), Eb (i
+ 1) and Ey (i + 1>), it can be obtained from equation (7).

FIG. 3 is a block diagram showing another embodiment of the present invention. The illustrated embodiment shows an interpolation circuit during field recording. In the figure, 41 is a reproducing device that reproduces the recorded image signal, and 42 is a delay circuit that delays the color difference signal Fry or Eby output from the reproducing device 41 by a time H corresponding to one horizontal scanning frequency (one horizontal scanning time). , 43 is a delay circuit that similarly delays the luminance signal Ey output from the reproducing device 41 by one horizontal scanning time H. As these delay circuits 42 and 43, for example, analog delay lines are used.

44 is a delay circuit that further delays the output of the delay circuit 42 by one horizontal scanning time H; 45 is a color difference signal Ery (or Eb
y) and the output of the delay circuit 43, and 46 is an arithmetic unit that multiplies the output of the adder 44 by 1/2. SW+
+ is a changeover switch that receives the output of the arithmetic unit 46 and the delay circuit 42 and changes the color order according to a command from the playback device 41; 47 receives the output of the changeover switch SW+t and changes each color difference signal Er, Ea, Eb. This is a matrix circuit that outputs. In this way, the circuit of the present invention is entirely composed of analog circuits. The operation of the circuit configured as described above will be explained as follows.

The color signal and luminance signal for each scanning line output from the reproducing device 41 are sequentially delayed by 1H in delay circuits 42 to 44. Therefore, the output state of each delay circuit at a certain time is
The output of playback 141 is Ery (i + 2), Ey
<i + 2), the results are as follows. That is, the output of the delay circuit 42 is Eby(i + 7)
, the output of the delay circuit 44 is Ery(i), and the output of the delay circuit 43 is Ey (i + 1>. Therefore, the output of the adder 45 is Ery(i) + Ery(i + 2). Therefore, The output of the arithmetic unit 46 which multiplies this output by 1/2 is 1/2XEry(i >+1/2xEry(i + 2
> becomes.

The matrix circuit 47 has the output of the arithmetic unit 46 shown in equation (12) and the output Eby(i + 1
Since the output Ey (i + 1>) of the delay circuit 43 is input through the changeover switch SW+t, all the color signals Er (
i + 1), Eg(i + 1> and Eb(i+1
) can be obtained. These obtained color signals are
It is output from the matrix circuit 47. In this way, when recording in the field, it can be configured with analog circuits,
The circuit is simplified.

FIG. 4 is a diagram showing the results of comparing the effect of interpolation according to the present invention with that of the conventional method. All numerical values shown in the figure indicate square errors. In the figure, a test pattern for a television signal was used as image 1, and a portrait of an adult woman was used as image 2. Er.

E(+, Eb indicates the color signal in each case. In both images 1 and 2, the square error is smaller in the case of the method of the present invention than in the case of the conventional method. Therefore, the square error of the method of the present invention is smaller than that of the conventional method. It can be seen that the interpolation method according to the invention is superior to the conventional interpolation method. It can also be seen that among the methods of the invention, the frame interpolation method is superior to the field interpolation method.

(Effects of the Invention) As described in detail above, according to the present invention, when the color difference signal is a line sequential color image signal, when obtaining each of the three primary color signals from the luminance signal and the color difference signal, , by configuring the color difference signal of the scanning line to be interpolated to be interpolated based on the weighted average of the color difference signals of the corresponding positions of the scanning lines before and after the interpolation, thereby obtaining accurate color signals Er, Eg,
Eb can be obtained. Therefore, according to the present invention, false colors can be reduced and more accurate color signals can be obtained.

[Brief explanation of the drawing]

FIG. 1 is an explanatory diagram of a scanning line signal to explain the operating principle of the present invention, FIG. 2 is a block diagram showing an embodiment of the invention, and FIG. 3 is a diagram showing another embodiment of the invention. 4 is a diagram showing the effect of the interpolation method according to the present invention, FIG. 5 is a diagram showing an example of a conventional image reproducing circuit, FIG. 6 is a diagram showing a scanning state, FIGS. 7 and 8. is an explanatory diagram of a conventional interpolation method. 1.11.41...Player 2...1 line delay circuit 3.31.47...Matrix circuit 12.13...Sample bold circuit 14.15...
・A/D converter 16.17...Line memory 18-25...Latch circuit 26...Shift circuit 27.29.30.45...Adder 28.46...Arithmetic unit 42- 44...Delay circuit S
W t~SW3.8Wt t... Changeover switch Patent applicant Konishi Roku Photo Industry Co., Ltd. Representative Patent attorney Fuji Ijima 1 person from outside Japan 4 illustrations 7 figures 7 M8 illustrations 8

Claims (1)

[Claims] In an image recording device that records a color image displayed by scanning lines, when the color difference signal is a line-sequential color image signal, interpolation is required when obtaining each of the three primary colors from the luminance signal and the color difference signal. An image recording apparatus characterized in that a color difference signal of a scanning line is interpolated based on a weighted average of color difference signals of corresponding positions of scanning lines before and after the scanning line.