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I convolute two periodic function:

$f(x) = \sin(20 \cdot 2 \pi x)$

$g(x) = \sin(15 \cdot 2 \pi x)$

$h = f \ast g$

The function h consists of frequenzy portions of 5 and 35 HZ. I understand the 5 HZ peak (of the fourier transform), because of the reflection of the signal on the Nyquist frequency of 10 HZ.

Can someone explain why I get a peak at 35 HZ too?

Fourier transform of h: 5 and 35 HZ peak

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1 Answer 1

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How did you calculate that convolution? And exactly what sort of "convolution" do you mean? If you're talking about the default convolution for functions of period $1$, namely $$f*g(x)=\int_0^1f(x-t)g(t)\,dt,$$then the convolution of those two functions is $0$.

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  • $\begingroup$ Thanks for your reply, I used this formula: $f \ast g(t) = 1/T \int_{0}^{T} f(\tau) \cdot g(t-\tau) d\tau$ I can provide my matlab code too, if it could help $\endgroup$
    – dnrhead
    May 31, 2018 at 15:50
  • $\begingroup$ Somewhere you need to set $T=1$ to get the actual convolution. Using matlab for that integral is a little sad - it's easy to work out. (Probably the easiest way to do the integral is to first write the sine in terms of complex exponentials.) $\endgroup$ May 31, 2018 at 16:00
  • $\begingroup$ In any case, that convolution is $0$. $\endgroup$ May 31, 2018 at 16:02

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