Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm not sure if this is on topic or not. The tag may also not actually fit.

If you add together two sinusoidal waves of different frequencies, how do you calculate the frequency of the resulting function as perceived by a human?

share|improve this question
    
What does "as perceived by a human" mean? I'm at least half serious. –  Zev Chonoles Jun 28 '12 at 21:40
    
@ZevChonoles: Maybe it would be clearer to talk about the tone that a human hears. Maybe you could rephrase this as "What is the frequency of the sinusoidal wave that has the same tone as this sound?" –  yakiv Jun 28 '12 at 21:44
    
Yes, but there would in fact be a range of frequencies within which any human would perceive indistinguishable tones. There is no exact answer to any question about the real world. I would recommend you remove that aspect of your question. –  Zev Chonoles Jun 28 '12 at 21:46
    
Is there a good way to calculate something like the center of that range of frequencies? –  yakiv Jun 28 '12 at 21:47
    
You wouldn't know what the exact range of frequencies is, either, and you'd only find out the (approximation of the) range by doing an experiment, not with math. –  Zev Chonoles Jun 28 '12 at 21:51
show 5 more comments

2 Answers

When two sine wave audio signals are added, and the frequencies are sufficiently different, we hear both frequencies, not some sort of average of the two.

For example, listen to 250 hz and 600 hz sine wave audio signals added together.

You can hear both signals. It does not sound like a single sine wave signal, of, say 425 hz.

If the signals have frequencies close together, then we hear beats caused by the interference of the two signals.

For example, listen to 400 hz and 410 hz sine waves added together. We hear the 10 hz difference in the the signal.

Adding two sinusoidal audio signals does not result in a signal perceived as a single sinusoidal signal.

share|improve this answer
5  
The human hearing apparatus is very interesting. The information that goes up the auditory nerve is encoded in the frequency domain, not in the time domain. Many phenomena result from this. One is that you can get a ringing in the ears, a perception of a hallucinated constant pure frequency overlaid over the other, real stimuli. Another is that you can tell the difference between a single pure tone and two superimposed pure tones. –  MJD Jun 28 '12 at 23:32
1  
The URL of your last link got truncated somehow, so I took the liberty of fixing it. I hope it is as you intended. –  Rahul Jun 29 '12 at 3:12
    
Thanks, Rahul. I appreciate it. –  Matthew Conroy Jun 29 '12 at 5:13
add comment

Identical Amplitudes

When two sinusoidal waves of close frequency are played together, we get $$ \begin{align} \sin(\omega_1t)+\sin(\omega_2t) &=2\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\\ &=\pm\sqrt{2+2\cos((\omega_1-\omega_2)t)}\;\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\tag{1} \end{align} $$ Unless played together, two tones of equal frequency, but different phase sound just the same, so the "$\pm$" goes undetected (the sign flips only when the amplitude is $0$), and what is heard is the average of the two frequencies with an amplitude modulation which has a frequency equal to the difference of the frequencies.

$\hspace{1.5cm}$enter image description here

The green curve is the sum of two sinusoids with $\omega_1=21$ and $\omega_2=20$; its frequency is $\omega=20.5$. The red curve is the amplitude as given in $(1)$, which has frequency $\omega=|\omega_1-\omega_2|=1$.

Differing Amplitudes

A similar, but more complex and less pronounced, effect occurs if the amplitudes are not the same; let $\alpha_1< \alpha_2$. To simplify the math, consider the wave as a complex character: $$ \begin{align} \alpha_1e^{i\omega_1 t}+\alpha_2e^{i\omega_2 t} &=e^{i\omega_2t}\left(\alpha_1e^{i(\omega_1-\omega_2)t}+\alpha_2\right)\tag{2} \end{align} $$ The average frequency, $\omega_2$, is given by $e^{i\omega_2 t}$ (the frequency of the higher amplitude component), and the amplitude and a phase shift is provided by $\alpha_1e^{i(\omega_1-\omega_2)t}+\alpha_2$:

$\hspace{3.5cm}$enter image description here

The amplitude (the length of the blue line) is $$ \left|\alpha_1e^{i(\omega_1-\omega_2)t}+\alpha_2\right|=\sqrt{\alpha_1^2+\alpha_2^2+2\alpha_1\alpha_2\cos((\omega_1-\omega_2)t)}\tag{3} $$ The phase shift (the angle of the blue line) is $$ \tan^{-1}\left(\frac{\alpha_1\sin((\omega_1-\omega_2)t)}{\alpha_1\cos((\omega_1-\omega_2)t)+\alpha_2}\right)\tag{4} $$ The maximum phase shift (the angle of the green lines) to either side is $$ \sin^{-1}\left(\frac{\alpha_1}{\alpha_2}\right)\tag{5} $$ This phase modulation has the effect of varying the frequency of the resulting sound from $$ \omega_2+\frac{\alpha_1(\omega_1-\omega_2)}{\alpha_2+\alpha_1} =\frac{\alpha_2\omega_2+\alpha_1\omega_1}{\alpha_2+\alpha_1}\tag{6} $$ (between $\omega_2$ and $\omega_1$) at peak amplitude to $$ \omega_2-\frac{\alpha_1(\omega_1-\omega_2)}{\alpha_2-\alpha_1} =\frac{\alpha_2\omega_2-\alpha_1\omega_1}{\alpha_2-\alpha_1}\tag{7} $$ (on the other side of $\omega_2$ from $\omega_1$) at minimum amplitude.

Equation $(3)$ says that the amplitude varies between $|\alpha_1+\alpha_2|$ and $|\alpha_1-\alpha_2|$ with frequency $|\omega_1-\omega_2|$.

$\hspace{1.5cm}$enter image description here

The green curve is the sum of two sinusoids with $\alpha_1=1$, $\omega_1=21$ and $\alpha_2=3$, $\omega_2=20$; its frequency varies between $\omega=20.25$ at peak amplitude to $\omega=19.5$ at minimum amplitude. The red curve is the amplitude as given in $(3)$, which has frequency $\omega=|\omega_1-\omega_2|=1$.

Conclusion

When two sinusoidal waves of close frequency are played together, the resulting sound has an average frequency of the higher amplitude component, but with a modulation of the amplitude and phase (beating) that has the frequency of the difference of the frequencies of the component waves. The amplitude of the beat varies between the sum and the difference of those of the component waves, and the phase modulation causes the frequency of the resulting sound to oscillate around the frequency of the higher amplitude component (between the frequencies of the components at peak amplitude, and outside at minimum amplitude).

If the waves have the same amplitude, the phase modulation has the effect of changing the frequency of the resulting sound to be the average of the component frequencies with an instantaneous phase shift of half a wave when the amplitude is $0$.

share|improve this answer
    
This assumes that the two tones have the same amplitude. Things are somewhat more complicated when the amplitudes differ. You probably have to do the usual division by the square root of the sum of the squares of the amplitudes, but, since the frequencies differ, things will get interesting. –  marty cohen Jun 29 '12 at 3:18
1  
@martycohen: finally finished it. :-) –  robjohn Jun 29 '12 at 17:45
    
@martycohen: thanks for suggesting different amplitudes. I've learned a lot writing this answer. I never knew that the frequency varies with the amplitude. –  robjohn Jun 30 '12 at 2:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.