The phenomenon I'm talking about is positive feedback, as known from control engineering: when the microphone is too close to its speaker, it can "hear itself", so the signal will be infinitely amplified. This can be very irritating e.g. during a concert.

But I wonder: why do you only hear a high-pitched tone when this occurs, and not some kind of white noise? Is it because the system acts as some kind of high-pass filter? And if so, why?

  • 4
    $\begingroup$ I take it Wikipedia doesn't help? $\endgroup$ Aug 11 '13 at 18:42
  • $\begingroup$ More like band-pass filter in first approximation. $\endgroup$
    – Ruslan
    Aug 11 '13 at 19:51
  • $\begingroup$ Is it possible to proof more mathematically, e.g. with some realistic transfert-functions, coming with a high pitch resonance frequency? Suppose an amplifying circuit made with a simple opamp circuit (e.g. non-inverting amplifier), and an electrete microphone (which will be a condensator in the circuit?). Why is it always such a high tone, or does it occur with basses too? And maybe even with infra- or ultrasonic sound? $\endgroup$
    Aug 11 '13 at 20:19

There are actually several different things going on. The reason you hear a single tone is because of resonance. The reason that it is usually a higher frequency tone has to do partly with the falloff in amplification with distance and partly (perhaps mainly) with the frequency response of the microphone/amplifier/loudspeaker.

Let's model the situation with $A$ representing the combination of the microphone, amplifier and loudspeaker, and the effect of the sound traveling through the air from the loudspeaker back to the microphone as $B$. In general $A(\omega)$ will be a function of the frequency $\omega$, and will be almost linear until the output signal gets to be large enough that it gets "clipped". That is, $A$ behaves as a complex multiplier where the magnitude of $A$ is the amplification and the angle of $A$ is the phase shift.

$B$ also behaves linearly, and depends mostly on the distance $r$ of the microphone from the loudspeaker. The signal will travel through the air at the speed of sound (~340m/s), and so there will be a delay of $\frac{r}{340\mathrm{m/s}}$, and $B$ will attenuate the signal by some factor, let's say $1/r^2$. Here's the public domain picture from wikimedia commons:

System diagram for feedback

If the input is $x(t)$ and the output is $y(t)$ then we get $ y=A(x+By) $ or $$ y(t) = \frac{A}{1-AB} x(t)\mathrm{.} $$

Oscillations occur when $AB = 1$ (exactly). That means that the attenuation from $B$ is equal to the amplification by $A$ (including clipping) and the phase shift of traveling distance $r$ exactly cancels the phase shift from the amplifier.

Let's consider the simplest possible case where the microphone, amplifier and loudspeaker have no phase shift. Then the phase shift for the feedback is based completely on the distance between the loudspeaker and microphone. The speed of sound is about 340 m/s. That means that a 20Hz sound wave is about 17 meters long, so you'd need to be about 17 meters from the speaker to get 20Hz (and all the harmonics of 20Hz) to be in phase.

So if you put a microphone 17 meters from the speaker will you get 20Hz feedback? Probably not. You'll probably get one of the higher frequency harmonics. Why? Because the microphone, amplifier and loudspeaker are probably much better at amplifying midrange frequencies than they are at amplifying low and high frequencies.

Look at the frequency response for the Shure SM58 microphone (a popular and widely used voice microphone). It looks like this:

Shure SM58 frequency response

Most guitar amps look similar. dB is a logarithmic scale, usually $20 \log_{10}\frac{p}{p_\mathrm{ref}}$ so 5dB is about 1.8x and 10dB is 3.2x. Some harmonic in that range between about 3KHz and 8KHz is probably going to dominate.

Note that the phase shift from the amplifier matters a lot and we didn't incorporate it into our model. For fun I decided to try out some different distances on my home computer to see what would happen. Because of constraints due to wiring the farthest I could get my mike from the speaker was 1.8 meters. In multiple trials I got 383, 379 and 386 Hz. 1.8 meters corresponds to a minimum frequency of 189 Hz in our model. At 10cm I got about 3700 Hz (lowest harmonic at that distance in our model is around 3400) At 20cm I got about 2600 Hz and at 50cm I got about 400 Hz. The lowest harmonic at 50cm would be about 680Hz according to the model.

  • 1
    $\begingroup$ +1. Many audio systems have lower gains outside the mid-range to keep distortion down. Equalizers can be used to make finer adjustments, but they can also be used to shift the feedback tones out of band. $\endgroup$
    – user6972
    Aug 18 '13 at 5:49
  • $\begingroup$ Good answer. Note that the condition is that phase shift is $2n\pi$ - and frequency rolls off more quickly at high frequencies. That's why that is where a combination of phase shift from path plus electronics is likely to hit an integer multiple of $2\pi$. $\endgroup$
    – Floris
    Sep 7 '16 at 14:02
  • $\begingroup$ It's also important if the feedback pushes the amplifier into the non-linear range of amplification (like clipping from high volume). The harmonics will dominate at that point. $\endgroup$
    – EvilSnack
    Oct 30 '20 at 1:50
  • $\begingroup$ A nice answer! Stafford Beer's "Brain of the Firm", chapter 2 ("Concepts and terms") does a nice job of fleshing out similar material. $\endgroup$ May 29 at 19:10

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