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From a practical point of view, when you connect two audio signals to one speaker the result is that both audio signals can be heard. That means the two voltage waves were added, correct?

However, when you connect two batteries in parallel, the resulting voltage is not the sum of both batteries, it's more like the average.

This is all from real-world observasions. I don't know if this two cases are supposed to be conceptually the same.

Why does the voltage bahave differently in this two cases?

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  • \$\begingroup\$ It's all about the frequency. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 25 '15 at 1:22
  • \$\begingroup\$ @IgnacioVazquez-Abrams, can you elaborate? \$\endgroup\$ – GetFree Jan 25 '15 at 1:39
  • \$\begingroup\$ They are not added, but rather interfered \$\endgroup\$ – Jason Pyeron Jan 25 '15 at 13:00
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Output is always the weighted average (superposition) of inputs.

Consider the following circuit in which two voltage sources are connected to the same load.

schematic

simulate this circuit – Schematic created using CircuitLab

Let V1 and V2 be DC sources then by superposition theorem, $$V_O = (V_1 + V_2)/3 \tag{1}$$

Assumed that R1,R2 = RL.

If V1 and V2 be monotonic ac signal sources then we can write, \$V1 = A_1 \cos (2\pi f_1 t)\$ and \$V2=A_2\cos (2\pi f_2 t)\$. Then,

case1: \$f_1 = f_2 = f\$ then, $$V_O = \frac{(A_1+A_2)}{3}\cos(2 \pi f t)\tag2$$

Here both sources generates same tone when connected to a speaker (load). Then output is same tone with output amplitude = superposition (weighted average) of the two input signals.

case2: \$f_1 \ne f_2 \$ then, $$V_O = \frac{A_1 cos(2 \pi f_1 t) +A_2 cos(2 \pi f_2 t)}{3} \tag3$$

Here both sources generates different tone when connected to a speaker (load). When both connected together, output signal will have both the tones (\$f_1\$ and \$f_2\$) as given by expression (3). Hence both the tones can be heard at the output.

If V1 and V2 are audio signals then they will have lot of frequency components in it (given by the Fourier transform). When connected to same speaker, output will also contain all these frequencies and hence one can hear both the audio signals.

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  • \$\begingroup\$ I think I get it. The voltage is always the average, but since the average is just the sum divided by a constant, the human ear doesn't notice the difference. \$\endgroup\$ – GetFree Jan 25 '15 at 10:32
  • \$\begingroup\$ @GetFree Yes. the amplitude of the audio signals are also reduced in the same way. \$\endgroup\$ – nidhin Jan 25 '15 at 18:35
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Connecting batteries in parallel is sometimes done to boost current output, but is not a good idea if you have two different voltage batteries. Unless they are exactly the same voltage--and often even those of the same kind aren't, due to discharging--one will be charging the other. Not only can this be dangerous if the batteries aren't designed to be recharged, but the voltage you measure might not be either of the two batteries' rated voltages.

If you connect two signal lines to the same speaker, you're right: you do hear both signals. However, the same principle applies; anytime one of your audio signals is at a higher voltage (i.e. is 'louder') than the other (for example, you could have vocals on one channel, and guitar+drums on the other), the louder will drive current through the other. This 'cross-talk' leads to distortion. Mixers--which are designed to add audio signals in the way you imagine, linearly--resolve this issue by using op-amps: they use a virtual ground at the point where the signals meet, so that no current wants to flow through the other signal line-- thus distortion is minimized.

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  • \$\begingroup\$ You mention the problems of doing one or the other, but my question is why voltage seems to behave differently in each case. In one case there seems to be an addition (audio signals) in the other there is not (batteries) \$\endgroup\$ – GetFree Jan 25 '15 at 2:35
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First the bad news. There is absolutely no difference between the math performed when joining two batteries together as when mixing two audio signals. Both are a weighted average based on the impedances in the circuit.

So what about the audio signals? Well, as in all things audio, we'll start with pure sine waves.

If we take two sine waves with the same amplitude, frequency, and phase and mix them, we get an identical sine wave as the output. This is... not that exciting a revelation, nor should it be.

If we take two sine waves with the same frequency and phase but different amplitudes, the result is a weighted average of the two input waves. So we end up with a sine wave with a different amplitude (or no signal at all, if the amplitudes were the same magnitude but opposite polarities).

Now we take two sine waves with different frequencies and mix them. The result... is no longer a sine wave. But one thing remains true: the voltage of the resultant wave at any given point is the weighted average of the voltages of the input waves at the same point. This remains true if we mix three sine waves together. Ten sine waves. A hundred. A thousand. A hundred thousand.

So then the big question is why does it sound like the two original audio waves mixed together instead of one big mass of sound? That's a question for Biology.SE, but the spoiler is that the cochlea performs a Fourier transform of incoming pressure waves, and the primary auditory cortex turns the results of the transform into "hearing". TL;DR: We hear frequencies, not impulses.

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  • \$\begingroup\$ Long story short, it's just an ilusion. Two audio signals don't add to each other, they are averaged exactly like two batteries. That's it? \$\endgroup\$ – GetFree Jan 25 '15 at 10:20
  • \$\begingroup\$ That is the gist of it, yes. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 25 '15 at 14:56
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schematic

simulate this circuit – Schematic created using CircuitLab

$$ V_{load} = (V_1+V_2)\frac{R_{load}}{R_{out}+2R_{load}} $$ For a simple comparison, assume Rout << Rload, as in the power supply case $$ V_{load} \approx \frac{(V_1 + V_2)}{2} $$ Even though, this is the same for the audio case as the amplitudes of the individual V1 or V2 have been halved, both sources (which are AC signals) appear at the load and you can hear both of them, so the perception is that they add together.

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You have to distinguish different cases when adding voltages.

  • 1) The signals are DC: The sum is the algebraic sum of all voltages (Eg. U₁ = 5V, U₂ = 3V: Usum = 5V + 3V = 8V. (Eventually averaged when applicable i.e. 8/2 = 4)

  • 2) The signals are pure sine waves with the exact same frequencies and no phase shift exists between them. The sum is the algebraic sum of all voltages (See 1, DC) (Eventually averaged when applicable)

  • 3) The signal are sine waves with the exact same frequency, but a phase shift of φ degrees exists between them. In that case the angle-adjusted cosine rule applies to both voltages. (Eg. U₁ = 5V, u₂ = 3V, φ=30⁰: Usum = √(U1² + U2² + 2*U1*U2*cosφ) = 25 + 9 + 2*5*3.cos(30) = 7.7V (Eventually averaged when applicable)

  • 4) The signals are sine waves with different frequencies or are non-sinusoidal in nature: The sum is the square root of the squared sums (RMS value): Usum = √(U1² + U2²) = √(5² + 3²) = 5.83V

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Strictly speaking, it is incorrect to connect voltage sources in parallel and current sources in series since a conflict between them will appear (each of them will try to impose its "desired value" on the common output). However, these incorrect connections are widely used in electronics. For example, in a differential stage with dynamic load, both the cases can be seen - a voltage conflict and a current conflict.

Circuit 'conflicts'

In practice, voltage sources always have internal resistances, or we deliberately add external resistors in series with them. Thus we obtain the extremely useful circuit of a resistive summer with weighted inputs... that wonderfully illustrates the principle of superposition. I have created a lot of stories about this humble passive circuit and its op-amp implementations that can be useful for you:

Walking along the resistive film - a fancy interpretation of the 18th century Ohm's experiment

Parallel voltage summer - a Wikibooks story about the resistive summer

Voltage-to-voltage summer - an animated Flash story about the ubiquitous circuit

My favorite way to present this circuit is by the voltage diagram (the distribution of the local voltages along the resistors).

The resistive summer illustrated by a voltage diagram

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