Convolution of two sine waves (or tones as called in audio) is theoretically not defined as the integral is infinite. Taking finite duration windowed sine waves and doing there convolution computationally always contains a fundamental frequency equal to that of the lower frequency sine wave. I am not getting an intuitive understanding of this. Anybody who can share more understanding in this?
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$\begingroup$ Could you share your calculations and the corresponding result? $\endgroup$– Matt L.Commented Oct 4, 2014 at 9:03
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$\begingroup$ When I do a windowed computation I don't get the smaller of the two, I get the lowest common multiple. (I.e., 9Hz and 10Hz results in an output of 1Hz, not 9Hz.) This looks similar, in time and frequency domain, to the addition of the two sinusoids. $\endgroup$– user19247Commented Jan 25, 2016 at 20:05
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4 Answers
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Convolution in the time domain is equivalent to multiplication in the frequency domain.
If you window two sinusoids in the time domain to get finite length waveforms, and the two sinusoids are exactly integer periodic in the window width, then the DFT will be impulses. If the frequencies are different, the impulses in the two DFT will be disjoint, and the multiplication of the two spectrum will result in zero output.
If one or both of the two sinusoids are not exactly integer periodic in your window width, then the FT will result in a Sinc function. Then the convolution will be equivalent to multiplying a Sinc function against an impulse or another Sinc function, which will produce a non-zero result.
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1$\begingroup$ Another way to look at that problem would be to interpret the convolution as some modified cross-correlation (time inversion of one of the signals to be precise). Thus for two orthogonal sinusoids correlation is equal to zero. $\endgroup$– jojeck ♦Commented Oct 5, 2014 at 9:49
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The convolution of two windowed sinusoidal signals of different frequencies $f_1$ and $f_2$ can be viewed as the result of passing a windowed sinusoid at frequency $f_1$ through a hypothetical linear filter whose impulse response is a windowed sinusoid at frequency $f_2$. When a linear filter is excited by a tone at frequency $f_1$, the steady-state output is a tone at frequency $f_1$. But, there is also an artifact in the output due to the transient response of the system which wants to put out a signal at frequency $f_2$ if it can.
If the input signal (tone at frequency $f_1$) began at $t = -\infty$ and is continuing unabated into the future eternity, the transient response (which began at $t = -\infty$) will have died away a long time ago and we observe only the steady-state response at ordinary times (say between $t = 0$ and $t = T$). With the input being a windowed signal that is beginning at $t=0$, say, both the steady-state response (tone at frequency $f_1$) as well as the transient response (decaying tone at frequency $f_2$) are observed for $t \geq 0$. Thus, both frequencies should be observable in the output when convolving two windowed tones, and in particular, the smaller frequency should be observable.
answered Oct 5, 2014 at 3:11
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Maybe try utilizing the properties of fourier transform. Off the top of my head maybe utilizing the duality property would help you along.
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I think it's worth looking at this without windowing. The OP didn't mention windowing and there should be an answer without it.
Let's stay in the analog domain for now. We can try to solve this in the frequency domain: The FT of a sine wave is $k\cdot(\delta(\omega_1)+\delta(-\omega_1))$ . Convolution in the time domain is equivalent to multiplication in the frequency domain so the answer would be "0 unless the frequencies are the same". The operation still is a little awkward since you need to multiply zero with something infinite but while the $\delta()$ function has infinite value it has finite energy and so the energy of the multiplication is zero.
The time domain looks a little more awkward. For each output we need to multiply the two sine waves at some time lag and then integrate from $-\infty$ to $+\infty$. The product of two sine waves are two sine waves at the sum and the difference frequencies. However the integral is awkward and at first look doesn't seem to converge.
There may be a trick. First we can split it into two integrals, one over the sum and one over the difference and then evaluate the integrals like this $$\lim_{T \to \infty}\int_{\varphi -T}^{\varphi +T}sin(\omega \cdot t-\varphi )\cdot dt$$
This integral is ALWAYS zero, for any value of T including T = $\infty$. The trick here is to chose the integration interval so that the it's always symmetric around a zero of the sine wave. So the integral over the lower half of the interval is always the negative of the integral over the upper half and both halves will cancel each other out.
I'm not entirely sure though whether this is a "legal" trick, i.e. whether it's okay to force convergence by intelligent selection of the integration interval.
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$\begingroup$ The OP did mention windowing: "Taking finite duration windowed sine waves ...". Also, you probably meant $\delta(\omega\pm\omega_1)$ instead of $\delta(\pm\omega_1)$. Isn't the 'trick' you're mentioning just the Cauchy principal value of the integral? $\endgroup$– Matt L.Commented Oct 5, 2014 at 13:33