JPH0457157B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION Field of the Invention The present invention relates to a hue adjustment circuit that performs hue adjustment using demodulated color signals.

BACKGROUND ART AND PROBLEMS Hue adjustment is generally performed by adjusting the phase of subcarriers supplied to a demodulator. Some methods utilize color signals, such as red and blue color difference signals.

FIG. 1 shows an example. Reference numerals 2 and 3 indicate demodulators, to which a carrier color signal S _C is supplied through a terminal 1 and a subcarrier CW is supplied through a terminal 4. 5 is a 90° phase shifter. A pair of demodulated color difference signals R-Y, B-Y
is supplied to the hue adjustment circuit 10. in this case,
Considering the level, the pair of demodulated color difference signals should be expressed as RY/1.14 and BY/2.03, respectively, but 1.14 and 2.03 are omitted for simplicity. The same applies to all subsequent explanations.

The hue adjustment circuit 10 includes a pair of variable amplifiers 11,
12, and a pair of combiners 15 and 16,
The amplification degrees A ₁ and A ₂ of the pair of variable amplifiers 11 and 12 are made equal and controlled in conjunction with each other. In this case, the second variable amplifier 12 is configured as a variable inverting amplifier so that A ₁ =−A ₂ . _A1 ,
The absolute values of A ₂ are −1≦A ₁ ≦1 and −1≦A ₂ ≦, respectively.
Set to 1.

Then, the blue color difference signal B-Y is transmitted to the variable amplifier 11.
The amplification degree of the signal is controlled by the synthesizer 15, and the red color difference signal RY is synthesized and outputted to the terminal 17a. Similarly, the amplification degree of the red color difference signal R-Y is controlled by the variable amplifier 12, and then combined into the blue color difference signal B-Y by the synthesizer 16, and the signal is synthesized at the terminal 17.
It is output to b.

In this circuit, a pair of demodulated color difference signals R-
Let the levels of Y and B-Y be v _RY and v _BY , respectively, and let the degree of saturation (level) at that time be v. In Figure 2, when = v _BY = v _RY ... (1) , the level of these color difference signals v _RY ,
If v _BY is expressed using v and phase difference θ, it becomes = v _BY = v cos θ (2) = v _RY = vsin θ (3). In this state, when the variable amplifiers 11 and 12 are varied so that |A ₁ |=|-A ₂ |=A, the outputs v' _RY and v' _BY of the combiners 15 and 16 are line segments, In this case, v′ _RY ==+=vsinθ+Avcosθ……(4) v′ _BY ==+=vcosθ−Avsinθ……(5) By changing the amplification degree, the input color difference signals R−Y, B−Y Each level (hereinafter simply referred to as a color difference signal) changes to v′ _RY and v′ _BY . Therefore,
By controlling the amplification degrees of the pair of variable amplifiers 11 and 12 and changing the synthesis ratio of the blue and red input color difference signals v _{BY and} v _RY with respect to the red and blue input color difference signals v RY and v _BY , can be adjusted.

By the way, if we eliminate A from equations (4) and (5), we get v′ _RY・sinθ+v′ _BY・cosθ=v (6), and since equation (6) always holds true regardless of the value of A, The composite vector locus of the color difference signals v _RY and v _BY expressed by equations (4) and (5) is orthogonal to the line segment l.

On the other hand, when the amplification degree is A, the output color difference signal
The composite vector of v′ _RY and v′ _BY is located on the line segment l. In other words, the line segment l has the B-Y axis as the x-axis, and the R
If the −Y axis is the y axis, then it can be expressed as y=+( oa ) ² /ob− oa /ob・x...(7), and by substituting equations (2) to (4) into equation (7), we get This is clear from the fact that equation (5) is obtained. Therefore, by varying the amplification degree, the vector locus representing the degree of saturation moves on the line segment l, and when the hue is adjusted, the degree of saturation changes. In order for the degree of saturation to be constant, there must be a vector locus on the circumference l′ whose radius is the line segment.

Purpose of the Invention Therefore, in this invention, the degree of saturation remains constant even if the hue is adjusted.

SUMMARY OF THE INVENTION Therefore, in this invention, in addition to the variable amplifier for hue adjustment, a pair of variable amplifiers for saturation adjustment are provided, and when adjusting the hue, the saturation can be adjusted by simultaneously controlling the amplification degrees of these pair of variable amplifiers. It is designed not to change.

Embodiment Next, an example of the present invention will be described in detail with reference to FIG. 3 and subsequent figures.

FIG. 3 shows an example of the hue adjustment circuit 10 according to the present invention, in which third and fourth variable amplifiers 13 and 14 for saturation adjustment are provided downstream of combiners 15 and 16, respectively. The amplification degrees A ₃ and A 4 of the third and fourth variable amplifiers 13 and ₁₄ are selected to be equal, and the amplification degrees of the first and second variable amplifiers 11 and 1 are selected to be equal.
2, the degree of amplification is varied.

In this configuration, the output color difference signals v″ _RY and v″ _BY obtained at the output terminals 17a and 17b, respectively, are as follows: v″ _RY = A ₃ v′ _RY (8) v″ _BY = A ₄ v′ _BY = A ₃ v _BY ......(9) In order for the saturation level before and after the weight adjustment to be constant, (v″ _RY ) ² + (v″ _BY ) ² = A ₃ {v′ _RY ² + (v ′ _BY ) ² }=v ₂ ...(10) It must be. By substituting equations (4) and (5) into equation (10), we get Therefore, the amplification degree A ₃ and
If you adjust A ₄ at the same time, the saturation will not change even if you adjust Hue.

In equations (4) and (5), the phase change when -1≦A ₁ ≦1 is: when A ₁ = 1, v' _RY = v (sin θ + cos θ) v' _BY = v (-sin θ + cos θ) ...(12) Therefore, the phase angle α ₁ (see Figure 2) is α ₁ = θ
It becomes +45°.

Also, when A ₁ = -1, v' _RY = v (sin θ - cos θ) v' _BY = v (sin θ + cos θ) ... (13), so the phase angle α ₂ is α ₂ = 45° - θ. Become.

Therefore, the hue adjustment range is ±45°.

FIG. 4 and subsequent figures show other examples of the present invention. In the case of FIG. 4, the output signal of the synthesizer 16 is used as the input signal of the first variable amplifier 11, and the output signal of the synthesizer 15 is used as the input signal of the second variable amplifier 12 to perform the hue adjustment. This is the case if you choose to do so.

In this example, the third and fourth variable amplifiers 13,
The amplification degree of 14 is selected as A ₃ =A ₄ =√1+ ² , and the hue adjustment range is ±45°.

In the example of FIG. 5, the third and fourth variable amplifiers 1
In the case where each output signal of 3 and 14 is used as an input signal of the first and second variable amplifiers 11 and 12,
In this example, the amplification degrees A ₃ and A ₄ are A ₃ = A ₄ = 1/√1
+A ² is selected, and the hue adjustment range at this time is ±90°.

The examples from FIG. 6 to FIG. 8 all include synthesizers 15, 16 and third and fourth variable amplifiers 13, 1.
This is a modification example when the connection relationship of 4 is reversed,
In the example shown in FIG. 6, the input signal of the fourth variable amplifier 14 is controlled by the first variable amplifier 11, and the input signal of the third variable amplifier 13 is controlled by the second variable amplifier 12. This is a case where the hue can be adjusted while the value is constant. In this example,
The amplification degrees A ₃ and A ₄ are selected as A ₃ =A ₄ =√1+ ² , and the hue adjustment range at that time is ±90°.

In the example shown in FIG. 7, the output signal of the fourth variable amplifier 14 is controlled by the first variable amplifier 11, and the output signal of the third variable amplifier 13 is controlled by the second variable amplifier 12. This is the case when the following is done.

In this example, A ₃ =A ₄ =1/√1+ ² is chosen, giving a hue adjustment range of ±45°.

In the example of FIG. 8, the output signals of the combiners 15 and 16 are used as input signals to the first and second variable amplifiers 11 and 12, and A ₃ =
A ₄ =√1+ ² is selected, and the hue adjustment range at that time is ±45°.

Next, a specific example of FIG. 3 will be shown. However, in this example, as shown in FIG. 9 due to the circuit configuration, the third and fourth variable amplifiers 13 and 14 are each divided into two parts and connected to each input stage of the combiners 15 and 16. , variable amplifiers 13A and 13B are used as the first variable amplifier 13, and variable amplifiers 14A and 14B are used as the second variable amplifier 14. 18,
19 is a fixed amplifier for output. A specific example shown in FIG. 9 is shown in FIG. 10.

The variable amplifiers 13A to 14B are all composed of differential amplifiers, and a red input color difference signal is input to a pair of transistors Q ₁ and Q ₂ that constitute the variable amplifier 13A.
v _RY is supplied. Furthermore, a blue input color difference signal v _BY is supplied to a pair of transistors Q ₈ and Q ₉ that constitute the variable amplifier 11 . The variable amplifier 11 is composed of transistors Q ₈ , Q ₉ , Q ₁₅ , and Q ₁₆ and also serves as a multiplier for the variable amplifier 13A, and a differential amplifier 2 is connected to the collector side of each of the transistors Q ₈ and Q ₉ .
0 and 21 are connected, and by controlling the respective base currents, the output level of the variable amplifier 11 is controlled.

22 is a differential amplifier for current conversion to which the hue adjustment voltage V _HUE is supplied, and transistors Q ₁₅ and Q ₁₆
The collector current controlled by V _HUE is supplied to multiplication differential amplifiers 20 and 21 for gain control of the variable amplifier 11. The voltages v ₁₇ and v ₁₈ related to V _HUE converted into voltage by a pair of resistors R ₁₇ and R ₁₈ connected to the collector sides of transistors Q ₁₅ and Q ₁₆ are obtained from a pair of transistors Q ₁₉ and Q ₂₀ . The selected voltage _V17 or _V18 is selected by the switching circuit 23.
is converted into a current by a resistor R ₂₃ and then supplied to a current mirror circuit 25 .

This is transistor Q ₂₁ ~ Q ₂₃ and diode D ₂₃
The output current i ₂₂ is supplied to the square characteristic circuit 26. This square characteristic circuit 26 is composed of a pair of diodes D ₂₄ and D ₂₅ connected in series and a current source 27.

The squared output is supplied to a conversion circuit 28 for converting it into √1+ ² . this is a transistor
Consisting of Q ₂₄ to _{Q 27} and a plurality of current sources 29 to 31, each current source 32 and 33 of variable amplifiers 13A and 13B is controlled by the output thereof, and (11) A level-controlled red output color difference signal v″ _RY as shown in the equation is obtained.

The circuit system for obtaining the blue output color difference signal v″ _BY is constructed in the same way, and the corresponding parts are marked with “′”.
Shown with . The components from the differential amplifier 22 for adjusting the hue to the conversion circuit 28 for controlling the current source provided in the variable amplifiers 14A and 14B are used in common.

Next, let's explain its operation.

First, if the current related to the input color difference signal v _RY is i _s1 and the currents flowing from the variable amplifier 11 side to the collector sides of transistors Q ₁ and Q ₂ are i _a and i _b , then
The emitter currents i ₃ and i ₄ of transistors Q ₃ and Q ₄ are as follows: i ₃ = 1/2 I 1 − i _s1 − _{i a} ……(15) i ₄ = 1/2 I ₁ + i _s1 − _{i b} ……( 16) Therefore, if the output current flowing through the output differential amplifier 18 is i _p , the emitter currents i ₆ and i ₇ of the transistors Q ₆ and Q ₇ are: i ₆ = 1/2I ₂ + i _p ...(17 ) i ₇ = 1/2I ₂ −i _p ...(18) Also, if the voltage between the base and emitter of transistor Q _i is V _BEi , then according to this circuit configuration, V _BE3 + V _BE6 = V _BE4 + V _BE7 ... ...(18) From the diode characteristics (static characteristics) of these transistors, equations (19) can be obtained from equations (18).

i ₃ i ₆ = i ₄ i ₇ ...(19) Substituting equations (15) to (18) into equation (19) and rearranging, i _p = 2i _s1 + (i _a − i _b )/I ₁ −(i _a +i _b )・I ₂ /2…
...(20) In the variable amplifier 13B, each differential amplifier 20,
If the control current related to the input color difference signal v _BY flowing through 21 is i' _s1 , then i _a = -1/2I ₁ + i' _s1 ... (21) i _b = -1/2I ₁ -i _s1 ... (22) From equations (20) to (22), i _p = 1/2I ₁ (i _s1 + i' _s1 ) I ₂ ... (23) Therefore, the output color difference signal obtained at terminal 17a
v″ _RY is v″ _RY = R ₇・i _p = i _s1 + i′ _s1 /2I ₁ I ₂ R ₇ ……(24) In this way, the output color difference signal v″ _RY is the input color difference signal v _RY , v It is the vector sum of the currents i _s1 and i′ _s1 related to _BY .

Now, the relationship between the input color difference signal v _RY and the signal current i _s1 is i _s1 = v _RY / R ₁ + R ₂ ... (25), so they are proportional, and when the hue is not adjusted (hue center) Then, since i′ _s1 = 0, the equation (24) is v″ _RY = R ₇ I ₂ /2I ₁・i _s1 = R ₇ I ₂ /2I ₁ (R ₁ +R ₂ )・v _RY … ...(26) From this, an output color difference signal v″ _RY proportional to the input color difference signal _vRY is obtained.

On the other hand, each transistor of the differential amplifiers 20 and 21
The emitter currents i ₁₀ to i ₁₃ flowing through Q ₁₀ to _{Q 13} are expressed as i ₁₀ = 1/2, where the currents flowing through the differential pair of transistors Q ₁₀ , Q ₁₁ and Q ₁₂ , Q ₁₃ are i _s21 and i _s22 . (I ₁ /2 - i _s2 ) - i _s21 ... (27) i ₁₁ = 1/2 (I ₁ /2 - i _s2 ) + i _s21 ... (28) i ₁₂ = 1/2 (I ₁ /2 + i _s2 ) + i _s22 ... (29) i ₁₃ = 1/2 (I ₁ /2 + i _s2 ) - i _s22 ... (30) Let i _s3 be the current due to the hue adjustment voltage V _HUE , and each of the transistors Q ₁₅ and Q ₁₆ The emitter current of transistors Q ₁₇ and Q ₁₈ connected in series on the collector side is
Assuming i ₁₇ and i ₁₈ , i ₁₇ = I ₃ /2 − i _s3 ... (31) i ₁₈ = I ₃ /2 + i _s3 ... (32) Diodes of transistors Q ₁₀ - Q ₁₃ , Q ₁₇ , Q ₁₈ From the characteristics, we get i ₁₇ i ₁₀ = i ₁₈ i ₁₁ ... (33) i ₁₇ i ₁₃ = i ₁₈ i ₁₂ ... (34), so from this and equations (27) to (32), i _s21 = −1/2I ₃ (I ₁ /2−i _s2 ) i _s3 ……(35) i _s22 = −1/2I ₃ (I ₁ /2+i _s2 ) i _s3 …(36) ∴i′ _s1 = i _s21 −i _s22 = i _s2・i _s3 /I ₃ ...(37) i _s2 and i _s3 are respectively , and since V _d is a fixed bias applied to transistor Q ₁₅ , from equations (37) and (38), i' _s1 = v _BY / (R ₈ + R ₉ ) (R ₁₅ + R ₁₆ ) I ₃ (V _HUE −V _d ) ...(39) Therefore, this current i′ _s1 that flows into the collector side of transistors Q ₁ and Q ₂ and is combined with the signal current i _s1 is:
Input color difference signal v _BY proportional to hue adjustment voltage V _HUE
A part of the current is converted into current.

From equations (24), (25), and (39), the output color difference signal
v″ _RY is v″ _RY = R ₇ I ₂ /2I ₁ {v _RY /R ₁ +R ₂ +v _BY / (R ₈ + R ₉ ) (R ₁₅ + R ₁₆ ) I ₃ (V _HUE −V _d )} ... (40) Here, R≡R ₁ = R ₂ = R ₈ = R ₉ = R ₁₅ = R ₁₆ ... (41) Then, equation (40) becomes v″ _RY = R ₇ I ₂ /4I ₁ R{v _RY +v _BY /2RI ₃ (V _HUE −V _d )} (42) Similarly, the output color difference signal v″ _BY obtained at the output terminal 17b is also obtained. Although a detailed explanation thereof will be omitted, if R≡R ₁ ′=R ₂ ′=R ₈ ′=R ₉ ′ (43), then the output color difference signal v″ _BY will be as follows.

v″ _BY =R ₇ ′I ₃ /4I ₁ R{v _BY −v _RY /2RI ₃ (V _HUE −V _d )} ...(44) By the way, the emitter potential v ₁₉ of transistors Q ₁₉ and Q ₂₀ is From the higher potential of either v ₁₇ or v ₁₈
The potential is lower by V _BE19 (= V _BE20 ). That is,
When V _HUE < V _d , the potential is (v ₁₈ − V _BE20 ),
When V _HUE > V _d , the potential is (v ₁₇ −V _BE19 ). This change characteristic is shown in FIG. The reason for such a change characteristic is to match the phase characteristic in the positive and negative directions from the hue center during hue adjustment.

The diode current i ₂₃ when the potential is v ₁₉ is i ₂₃ = v ₁₉ −V _BE23 /R ₂₃ (44) Because of the current mirror circuit 25, i ₂₂ = i ₂₃ , so i ₂₂ = v ₁₉ −V _BE23 /R ₂₃ ...(45) As is clear from Fig. 11, the potential v ₁₉ also includes the bias voltage V _B , so the current source 2
When the current I ₄ corresponding to the bias voltage is subtracted from this current i ₂₂ by 7, only the current i _s3 related to V _HUE is obtained. And this current i _s3 flows through a pair of diodes
A square characteristic is imparted by D ₂₄ and D ₂₅ and the resulting signal is input to the conversion circuit 28 at the subsequent stage.

Diodes D ₂₄ , D ₂₅ , transistors Q ₂₄ to _{Q 27}
and a transistor Q ₅ provided in the constant current source 32,
The potential relationship of diode D ₅ is V _D24 + V _D25 = V _BE24 + V _BE25 V _BE5 + V _D5 = V _BE26 + V _BE27 (46) where V _Di is the forward drop voltage of diode D _i . Due to these diode characteristics, (46)
The following current equation is obtained from the equation: However, I ₅ to I ₇ are constant currents of the current sources 29 to 31.

i _s3 ² = I ₅ i ₂₅ I ₁ ² = I ₆ (I ₇ + i ₂₅ ) ……(47) Here, if I ₅ = I ₆ = I ₇ = I ₃ , I ₁ = I ₃ √1 + ( _{s3 3} ) ² ... (49) Also, in equations (42) and (44), if R ₇ = R ₇ ′ = r I ₁ = I ₃ = I ₀ ... (51), then from equations (42) and (44), the following equation can be obtained. can get.

(v″ _RY ) ² + (v″ _BY ) ² = (rI ₀ /4I ₁ R) ² {1 + (V _HUE −V _d /2RI ₀ ) ² } (v ² _RY + v ² _BY )……(52
) (50) Substituting and rearranging the formula, (v″ _RY ) ² + (v″ _BY ) ² = (r/4R) ² (v ² _RY + v ² _BY ) = (r/4R) ² v ² … ...(53) By adjusting the hue adjustment voltage V _HUE from equation (50), the current I ₁ changes, and as is clear from equations (42) and (44), due to this current change,
Blue input color difference signal for red input color difference signal v _RY
The combination ratio of v _BY and the combination ratio of red input color difference signal v _RY to blue input color difference signal v _BY are adjusted, and the red and blue output color difference signals v″ _RY and v″ _BY change.
This allows the hue to be adjusted. As is clear from equation (53), the vector sum of the red and blue color difference signals v″ _RY and v″ _BY remains constant regardless of the hue adjustment voltage V _HUE .

Therefore, by configuring as shown in FIG. 10, it is possible to realize a circuit in which the degree of saturation does not change even if the hue is adjusted.

Application Examples In all of the above embodiments, red and blue color difference signals RY and BY demodulated on the RY axis and the BY axis, respectively, are used as input color signals.

This invention is not limited to these embodiments, but R-Y
The input color signal may be a pair of color signals demodulated on axes shifted by 45 degrees from the B-Y axis, or the I signal and Q signal demodulated on the I and Q axes may be used as the input color signal. It can also be used as a signal.

The former method, in which the demodulation axis is tilted by 45 degrees, can be achieved extremely easily in, for example, a digital television system, and is therefore suitable for use in such systems. In this case, the hue adjustment is performed by first converting it into an analog color signal, so
This has the advantage that the influence on the color killer detector due to changes in hue is removed and the color killer point does not fluctuate. Furthermore, since these demodulation axes are close to the I-axis and Q-axis, they can be handled in the same way as the I-axis and Q-axis.

In the latter case, it is possible to widen the color signal band using the IQ demodulation method, so it is possible to reproduce colors with fine details.

Effects of the Invention As explained above, according to the present invention, even if the hue is adjusted, the degree of saturation can always be kept constant. Therefore, it is possible to obtain a desired hue while keeping the saturation level, that is, the color density unchanged.

[Brief explanation of the drawing]

Figure 1 is a block diagram of a conventional hue adjustment circuit.
FIG. 2 is an explanatory diagram of its operation, FIG. 3 is a block diagram showing an example of the hue adjustment circuit according to the present invention, and FIG.
Figures 1 to 9 are block diagrams showing other examples.
0 is a concrete connection diagram of FIG. 9, and FIG. 11 is an explanatory diagram of its operation. 10 is a hue adjustment circuit, 11 to 14 are first to fourth variable amplifiers, and 15 and 16 are synthesizers.

Claims (1)

[Claims] 1 Comprising first and second variable amplifiers that control the amplification degree of the demodulated first color signal, and third and fourth variable amplifiers that control the amplification degree of the demodulated second color signal. However, the amplification degrees of these first to fourth variable amplifiers are controlled in conjunction with each other, and the first and second variable amplifiers are used as variable amplifiers for hue adjustment, and their respective amplification degrees are equal, The second variable amplifier is configured as an inverting amplifier, and the third and fourth variable amplifiers are used as variable amplifiers for adjusting the degree of saturation. A hue adjustment circuit that allows the hue adjustment to be made in a constant state.