The impulse response of an LTI system is aperiodic and I pass in a periodic signal. Now I need to find the Fourier series representation of the output.

Since Convolution in time domain is multiplication in frequency domain, I think that converting both signals in their respective frequency domain and multiplying them would work.

Therefore, finding the Fourier series of first and transform for second. Is it a correct way?

My question being is it legal to multiply fourier series and transform the way I'm doing?


My question being is it legal to multiply fourier series and transform the way I'm doing?

The Fourier series is still a time domain thing, with coefficients being multiplied with time-domain harmonic functions (and then summed up).

So, no. Not directly. You need to multiply frequency domain things.

However, a periodic function's Fourier series is tightly linked to the Fourier transform of that function:

A periodic function's Fourier transform is discrete, i.e. it's a functional that only consists of \$\delta\$ dirac impulses at the frequencies of its periodicities (which are exactly the frequencies of the harmonic functions you see in the Fourier series!), weighted with a coefficient (which happens to be the matching coefficient from the Fourier series).

So, using the Fourier series, you can trivially calculate the Fourier transform.

Now, multiplying any function with a "comb" of dirac impulses just sets the result to 0 everywhere but at the locations of the dirac impulses; that aligns nicely with reality, since an LTI system can never introduce new frequencies to a signal, and those were the only frequencies present in the original signal.

So, multiply the coefficients with the values of the Fourier transform at the exact frequencies of your periodic signal. Transform back: Obviously, the result will be periodic again, have the same harmonic frequencies as your original signal, but with other Fourier series coefficients.

Signals and Systems basics – this is really fun to learn about, and once you've been through your SaS textbook and exercised your skills a bit, this will be all very natural to you :)

  • \$\begingroup\$ Wonderful Explanation! \$\endgroup\$ – Adarsh Kumar Oct 13 '19 at 13:30
  • \$\begingroup\$ there is a tool that does this, with 8--10 useful examples included: Signal Wave Explorer. Check out the JB filter, to examine why active low-pass filters are not useful for far-out-of-band rejection. \$\endgroup\$ – analogsystemsrf Oct 13 '19 at 21:04

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