JPH0119320B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a sound image enlarging device used to spread the sound when driving a plurality of speakers to obtain sound. By giving a phase difference between the audio signals that drive the speakers, the sound spreads, and by selecting the characteristics of this phase shift circuit, the sound spread does not cause any unnatural hearing. This paper proposes a device that does this.

Since the start of audio multiplex broadcasting in television broadcasting, television receivers have been equipped with two audio signal paths and a left audio signal connected to each of these audio signal paths in order to reproduce the multiplexed audio. -More and more devices are equipped with a pair of right speakers. These are so-called audio multiplex compatible television receivers, but since they are television receivers, they are limited by their external dimensions.
Therefore, the gap between the left and right speakers is also restricted. For this reason, when the audio signal is a monaural signal or a stereo signal, even if the audio is played back using the left and right speakers of such an audio multiplex compatible television receiver, the expected sense of presence and sound spread may not be achieved. will not be obtained. Therefore, research is being conducted on audio reproduction methods that can provide a sense of presence and sound expansion even with relatively narrow speaker spacing, and already,
By creating a phase difference between the audio signals driving the pair of left and right speakers, the auditory sound image is localized outside the area connecting the listener and the left and right speakers, respectively. Devices aimed at amplifying sound have been proposed. An example of the basic configuration of such a device for enlarging a sound image is shown in FIG. 1 is an audio signal input terminal;
A first audio signal path 2 and a second audio signal path 3 are arranged from here. The first audio signal path 2 is a phase shift circuit 4
The output terminal of the phase shift circuit 4 is connected to the amplifier circuit 5.
through the first speaker, for example, left speaker 6
It is connected to the. The second audio signal path 3 is connected to the audio signal input terminal 1 via a changeover switch 7, and is also directly connected to a second speaker, for example a right speaker 9, via an amplifier circuit 8. The changeover switch 7 connects the second audio signal path 3 to the audio signal input terminal 1 or to another audio signal input terminal 10 .

In such a device, when the audio signal is a monaural signal, the changeover switch 7 connects the audio signal input terminal 1 to the second audio signal path 3,
This audio signal is supplied to the audio signal input terminal 1. Then, the signal enters the first audio signal path 2, is phase shifted by the phase shift circuit 4, is amplified by the amplifier circuit 5, and then drives the left speaker 6. Also, audio signal input terminal 1
The audio signal enters the second audio signal path via the changeover switch 7, is directly amplified by the amplifier circuit 8, and then drives the right speaker 9. In this case, the audio signal that drives the left speaker 6 and the audio signal that drives the right speaker 9 have a phase difference corresponding to the phase shift by the phase shift circuit 4, and if this phase difference is appropriate, And the sounds from the right speakers 6 and 9 have a spread based on this phase difference.
When the audio signal is a stereo signal, the changeover switch 7 connects the audio signal input terminal 10 to the second audio signal path 3, so that the left signal is supplied from the audio input terminal 1 to the first audio signal path, and the right signal is supplied to the first audio signal path. A second audio signal path is supplied from the audio signal input terminal 10. At this time, the phase difference corresponding to the phase shift in the phase shift circuit 4 that occurs for the in-phase components in the left and right signals contributes to the spread of sound from the left and right speakers 6 and 9. However, there are problems with the phase shift circuit 4 when actually configuring the device. That is, it can be said that when the mutual phase difference between the audio signals driving the left and right speakers 6 and 9 described above is relatively large for each frequency (close to +180° or -180°), the sound image enlargement effect is greater. However, if the phase difference becomes 180 degrees, the sound from both speakers will give a so-called out-of-phase sensation to the auditory sense, making it extremely unnatural. The circuit forming the phase circuit 4 does not produce a relatively large phase difference over a wide frequency band, or does not produce a feeling of out-of-phase.
It has some frequencies that cause a phase difference of 180 degrees, and it is not possible to obtain a sufficient sound spread without causing unnaturalness. Moreover, the conventional phase shift circuit 4
The circuit forming the left speaker does not have a flat frequency vs. amplitude characteristic, and the amplitude of the audio signal that causes the phase shift changes depending on the frequency. There was also a risk that the amplitude frequency characteristics of the circuit system for the audio signal driving the right speaker would differ.

In view of this, the present invention provides a phase shift circuit as described above to provide a mutual phase difference between the audio signals driving the plurality of speakers. When configuring a device for enlarging a sound image, by making the phase shift circuit have a selected transfer function, problems that exist in conventional devices can be solved, and a sound image that can obtain a natural spread of sound. A magnifying device is provided. Hereinafter, a sound image enlarging device according to the present invention will be explained with reference to FIGS. 2 to 7.

FIG. 2 shows an example of a sound image enlarging device according to the present invention. The basic configuration of this device is the same as that shown in Figure 1, and parts corresponding to those in Figure 1 are given the same numbers as in Figure 1, and their connection relationships and operations are indicated. Although detailed explanation will be omitted, this device is characterized by the circuit forming the phase shift circuit 4. Here, the phase shift circuit 4 is configured as a CR circuit using, for example, capacitance C and resistance R, and its transfer function G(s) is generally G(s)=S ² −ω ₀ /Q・S+ω ² / ₀ /S ² +ω ₀ /Q・
S + ω ² / ₀ ... (1) (However, S = jω, ω is the angular frequency, ω ₀ is the resonance angular frequency, for example, in the case of a CR circuit, it is 1/CR, Q is the goodness of the circuit, and the same applies below), It is formed of a circuit with a biquadratic phase function expressed as . The amplitude characteristic of a circuit having the transfer function G(s) in equation (1) above is |G(s)|=1, and the phase characteristic is argG(s)=-2tan ^-1 ω ₀ ω/Q (ω ² / ₀ − ω ² ). Since this amplitude characteristic is always 1 regardless of ω,
The frequency versus amplitude a characteristic is flat, as shown in FIG. On the other hand, the phase characteristic is ω=0~∞
The phase θ changes from 0 to -2π in the range, 0 at ω=0, -π at ω=ω ₀ , -2π at ω=∞, and
The curve of the change in phase θ varies depending on the value of Q, as shown in FIG. 4. Furthermore, the circuit 11 described above has the frequency ₀ (=ω ₀ /2π) corresponding to ω ₀ in the transfer function of equation (1) in the middle region of the frequency band of the audio signal transmitted through the audio signal path 2. is set to exist.

In this way, the phase shift circuit 4 is formed by the circuit 11 which has a transfer function as shown in equation (1) and where ₀ is set in the middle region of the audio signal frequency band. The audio signal in path 2 undergoes a phase shift over its entire frequency band, and in particular undergoes a relatively large phase shift of around -180° in the mid-range region; , that is, the phase is reversed only at frequency ₀ . Further, since the circuit 11 has a flat frequency versus amplitude characteristic, the amplitude frequency characteristic of the audio signal path 2 for the audio signal transmitted through it does not change. Therefore, only the phase of the audio signal in the audio signal path 2 is changed, and the audio signal that drives the speaker 6 and the audio signal that drives the speaker 9 have a mutual phase difference over the entire frequency band. , which has a significant impact on hearing.

Although there will be a relatively large phase difference in the midrange, which is close to being out of phase, when it becomes a sound, an unnatural sense of out-of-phase will occur.There is only one specific frequency that is in out-of-phase with each other. , that is, there are only ₀ . Moreover, no particular change is caused in the respective amplitudes. Therefore, the sound from the speakers 6 and 9 has a sufficient spread, and the spread has very little chance of causing a sense of inverse phase, so that a natural sound spread can be obtained as a whole. be.

FIG. 5 shows an example of a specific configuration of a circuit 11 having a transfer function as expressed by equation (1) above. In this example, a low-pass filter (LPE) 12 having a third-order Butterworth characteristic and a high-pass filter (HPF) 13 having a third-order Butterworth characteristic are connected in parallel. These low pass filters 1
2 and the high-pass filter 13, respectively, for example.
Configured as a CR circuit, low pass filter 12
Let FL(s) be the transfer function of the high-pass filter 13.
If the transfer function of is FH (s), then FL (s) = ω ³ / ₀ / ω ³ / ₀ +2ω ² / ₀ S + 2ω ₀ S ² + S ³
...(2) FH(s)=S ³ /ω ³ / ₀ +2ω ² / ₀ S+2ω ₀ S ² +S ³ ...(
3) becomes.

If the transfer function of the parallel connected circuit is F(s), then F(s)=FL(s)+FH(s) =ω ³ / ₀ +S ³ /ω ³ / ₀ +2ω ² / ₀ S+2ω ₀ S ² +S ³ = S ² - ω ₀ S + ω ² / ₀ / S ² + ω ₀ S + ω ² / ₀ . This is the case when Q=1 in the transfer function G(s) of equation (1). That is, a low-pass filter 12 having a third-order Butterworth characteristic and a high-pass filter 13 having a third-order Butterworth characteristic.
The parallel connection circuit with the above forms a circuit in which Q=1 among the circuits having the transfer function G(s) of equation (1). An example of a specific circuit configuration of such a low-pass filter 12 and a high-pass filter 13 is shown in FIG. FIG. 6A shows an example of the low-pass filter 12. Here, +B is a power supply, transistor 14 is a buffer with an amplification degree of 1, and resistors 15, 16 and 17 whose resistance value is R and whose capacitance value is 2C, C/2 and C, respectively. Certain capacitors 18, 19 and 20 are the transfer function elements of this circuit. Now, the voltages at the input terminal 21, the intermediate connection points 22 and 23, and the output terminal 24 are
v ₁ , v ₂ , v ₃ and v ₄ respectively, and a resistor 15
If the current flowing through the capacitor 18, the current flowing through the capacitor 18, and the current flowing through the capacitor 19 through the resistor 16 are i ₁ , i ₂ , and i ₃ , respectively, i ₁ = i ₂ + i ₃ ...(4) i ₁ = v ₂ −v ₁ /R ……(5) i ₂ =jω ₂ C・(v ₃ −v ₂ ) ……(6) Here, if we ignore the base current of transistor 14 and the voltage drop between base and emitter, i ₃ = v ₃ − _{v 2} /R ...(7) Substituting (5), (6), and (7) into (4) and rearranging, v ₂・(2+jω ₂ CR) −v ₃・( 1 + jω ₂ CR) = V ₁ ... (8) Also, v ₃ = 2 / jωC / R + 2 / jωC · v ₂ , so v ₂ = (1 + jωCR / 2) · v ₃ ... (9) (8) 9) and rearranging, v ₃ / v ₁ = 1/1 + jωCR−ω ² C ² R ² ...(10) Also, v _{3 by v 4} = 1 / jωC / R + 1 / jωC · _{v 3} =(1+jωCR)・v ₄ ...(11) Substituting (11) into (10) and rearranging, v ₄ /v ₁ =1/1+2jωCR−2ω ² C ² R ² −jω ³ C ³ R ³ … …(1
2) Substituting S=jω and ω ₀ = 1/CR in (12) to find the transfer function FL(s)′, FL(s)′=v ₄ /v ₁ =ω ³ / ₀ /ω ³ / ₀ + ^2ω2 / _0S + _2ω0S + ^S3 , which is the same as the transfer function FL(s) in equation (2).

On the other hand, FIG. 6B shows an example of the high-pass filter 13. Here, +B is the power supply, and the transistor 25
is a buffer with an amplification degree of 1, and capacitors 26, 27, and 28 whose capacitance values are C/2, C/2, and C, respectively, and whose resistance values are, respectively.
resistors 29, 30, which are R, 8R, 8R and R;
31 and 32 are transfer function elements of this circuit.
Now, the voltages at the input terminal 33, the intermediate connections 34 and 35, and the output terminal 36 are v ₅ , v ₆ ,
V ₇ and v ₈ , and the current flowing through capacitor 26, the current flowing through resistor 29, and capacitor 2
If the currents flowing in parallel through resistors 30 and 31 through 7 are i ₄ , i ₅ and i ₆ , respectively, then i ₄ = i ₅ + i ₆ ... (13) i ₄ = (v ₆ − v ₅ )・jω・C/2 ... (14) i ₅ = v ₇ - v ₆ /R ... (15) Here, if the base current and base-emitter voltage drop of the transistor 25 are ignored, i ₆ = (v ₇ −v ₆ )・jω・C/2 ...(16) Substituting (14), (15), and (16) into (13) and rearranging, we get v ₆・(2+jω ₂ CR) −v ₇・( 2 + jωCR) = v ₅ · jωCR ... (17) Also, v ₇ = 4R / 4R + 2 / jωC · v ₆ Therefore, v ₆ = (1 + 1 / jω ₂ CR) · v ₇ ... (18) (17) Substituting (18) and rearranging, v ₇ / v ₅ = −ω ² C ² R ² /1 + jωCR−ω ² C ² R ² ... (19) Also, v ₈ = R / R + 1 / jωC・v ₇ Therefore, v ₇ = 1 + jωcR / jωCR · v ₈ ... (20) Substituting (20) into (19) and rearranging, v ₈ /v ₅ = -jω ³ C ³ R ³ /1 + 2jωCR - 2ω ² C ² R ² - jω ³ C ³ R ³ ... (21) Substituting S = jω and ω ₀ = 1/CR in (21) to find the transfer function FH (s)', FH (s)' = v ₈ /v ₅ =S ³ /ω ³ / ₀ +2ω ² / ₀ S+2ω ₀ S ² +S ³ , which is the same as the transfer function FH(s) in equation (3).

FIG. 7 shows another example of the specific configuration of the circuit 11 having the transfer function G(s) as expressed by equation (1). In the circuit shown in FIG. 7, +B is a power supply, and 3
7 is an input terminal and 38 is an output terminal. Buffer circuits 39 and 4 are connected between the input terminal 37 and the output terminal 38.
Two CR phase shift circuits 42 and 43 are connected in cascade via 0 and 41. The CR phase shift circuit 42 includes a transistor 44, a capacitor 45 with a capacitance value of C, and a resistor 46 with a resistance value of R. One end of the capacitor 45 is connected to the collector of the transistor 44, and one end of the resistor 46 is connected to the collector of the transistor 44. are connected to the emitters of the transistor 44, and the other ends of the capacitor 45 and the resistor 46 are connected to each other. The CR phase shift circuit 43 is composed of a transistor 47, a capacitor 48 with a capacitance value of C, and a resistor 49 with a resistance value of R. 46
It is said that the connection relationship is similar to that of . 50 is a resistor forming a feedback path from the buffer circuit 41 to the buffer circuit 39, and 51 is a current source.
In such a configuration, the voltage at input terminal 37, intermediate nodes 52 and 53, and output terminal 38 is
Let them be e ₁ , e ₂ , e ₃ and e ₄ , respectively. Also, resistor 5
Let the feedback current flowing through 0 be 2i. At this time, the feedback current 2i is formed by the current i flowing through the transistors 54 and 55 of the buffer circuit 41 and the transistors 56 and 57 of the buffer circuit 39, as shown. Furthermore, transistors 54,5
Let the internal resistances of the emitters of transistors 5, 56, and 57 be re, the resistance value of the resistor connected between the emitters of transistors 54 and 55 and between the emitters of transistors 56 and 57, respectively, be Re, and the resistance value of resistor 50 be Rf. . Furthermore, the above CR phase shift circuits 42 and 4
The transfer function of each of 3 is 1−S′/1+S′ where S′=jωCR
, so if the transfer function between the intermediate connection points 52 and 53 is T(s'), then the CR phase shift circuits 42 and 43
are connected in cascade, T(s')=
1-S'/1+S'・1-S'/1+S'=(1-S'/1
+S′) becomes ² .

In the settings as described above, the following equations hold true.

i=e ₄ −e ₁ /2(Rf+re) ...(22) e ₂ =e ₁ −i・Re ...(23) e ₃ =e ₂・T(s')=e ₂・(1−S '/1+S') ² ...
(24) e ₄ = e ₃ −i・Re ...(25) Here, the transfer function of the circuit in Figure 7 is expressed as H(s')
Then, H(s')=e ₄ /e ₁ , and by finding e ₁ and e ₄ from the above equations (22), (23), (24), and (25) and substituting them, we get H(s') )=1-2/1+Re/Rf+re・S′+ ^S′2 /1+
2/1+Re/Rf+re・S′+ ^S′2 . Here, if we set m=2/1+Re/Rf+re, H(s′)=1−mS′+S′ ² /1+mS′+S′ ² ……(2
6) becomes. If this equation (26) is rewritten using S=jω and ω ₀ =1/CR, it becomes H(s)=S ² −mω ₀ S+ω ² / ₀ /S ² +mω ₀ S+ω ² / ₀ . This means that in the transfer function G(s) of equation (1),
This is the case when Q=1/m. That is, the circuit shown in FIG. 7 forms a circuit having the transfer function G(s) of equation (1) where Q=1/m.

As described above, in the sound image enlarging device according to the present invention, the audio signal phase shifting circuit is formed of a circuit whose transfer function is selected as expressed by equation (1). For example, without changing the amplitude frequency characteristics of the circuit system forming the audio signal path for audio signals, it is possible to very effectively obtain a natural spread of sound with very little sense of out-of-phase feeling.

[Brief explanation of drawings]

FIG. 1 is a block connection diagram showing an example of the basic configuration of a conventionally known sound image enlarging device, FIG. 2 is a block connection diagram showing an example of the sound image enlarging device according to the present invention, and FIGS. 3 and 4 are A characteristic diagram for explaining the characteristics of the phase shift circuit used in the sound image enlargement device according to the present invention, and FIG. 5 is a block connection diagram showing an example of a specific configuration of the phase shift circuit used in the sound image enlargement device according to the present invention. , FIGS. 6A and 6B are circuit connection diagrams showing an example of the internal configuration of the block shown in FIG.
The figure is a circuit connection diagram showing another example of the specific configuration of the phase shift circuit used in the sound image enlarging device according to the present invention. In the figure, 2 and 3 are audio signal paths, 4 is a phase shift circuit,
5 and 8 are amplifier circuits, 6 and 9 are speakers, 12
13 is a low-pass filter having a third-order Butterworth characteristic; 13 is a high-pass filter having a third-order Butterworth characteristic; 15, 16, 17, 29, and 3.
2 is a resistor with a resistance value of R, 18 is a capacitor with a capacitance value of 2C, 19, 26 and 27 are capacitors with a capacitance value of C/2, 20 and 28 are capacitors with a capacitance value of C, 30 and 31 are resistors with a resistance value of 8R, 39, 40 and 41 are buffer circuits, 42 and 43 are CR phase shift circuits, and 50 is a resistor with a resistance value of Rf.

Claims (1)

[Claims] 1. A first signal path for supplying an audio signal to the first speaker and a second signal path for supplying an audio signal to the second speaker are arranged, A phase shift circuit is inserted in the signal path, and the transfer function of the phase shift circuit, G(s), becomes G(s)=S ² −ω ₀ /Q・S+ω ² / ₀ /S ² +ω ₀ /Q・
S + ω ² / ₀ , (where S = jω, ω is the angular frequency, ω ₀ is the resonance angular frequency, and Q is the quality of the circuit), and the first signal is A sound image enlarging device that causes sound from the first and second speakers to spread by shifting the phase of an audio signal on the path.