If I convolve a periodic signal $x(t)$ with period = 1sec with an aperiodic signal $h(t)$ whose Fourier transform $H(f)$ is exactly equal to 1 at frequencies $f = 0,\pm 1,\pm 2, \ldots \textrm{Hz}$, but has some arbitrary and finite values at non-integers, will the resulting waveform $y(t) = x(t) \star h(t)$ be the same as $x(t)$?

I think it should, but I want to validate my reasoning.

My reasoning is that since $x(t)$ is periodic with period=1sec, its spectrum is discrete and is non-zero at integer values.

So if $H(f)$ equals 1 at integers, it should not distort $x(t)$ regardless of what its values are in between integers.

Note that I am referring to frequency in terms of Hertz and not $\omega$ (angular frequency).

Is this reasoning valid?

  • 1
    $\begingroup$ Yes you are right. Eventhough it's important to remember the amount of idealization being assumed while arriving at that conclusion. $\endgroup$
    – Fat32
    Commented Aug 2, 2016 at 17:20
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    $\begingroup$ The formula for the output signal of an LTI system excited with a periodic input signal given in this answer should help you to answer your question. $\endgroup$
    – Matt L.
    Commented Aug 2, 2016 at 18:41

1 Answer 1


The convolution in the time domain is translated to simple multiplication in the frequency domain. This means that since the values of the transform of x(t) are at the integer locations, the result of the multiplication will be the transform of x(t), as you defined the integer values for h(t) transform to be 1 at integer bins.


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