# Using partial fraction decomposition to find inverse Fourier transform

Posted 4 months ago
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 I've reduced my problem to $H(w) = \dfrac{1}{(1-\frac{1}{4}e^{-jw})(1-\frac{1}{3}e^{-jw})}$. I need its inverse discrete Fourier transform.My thinking is that I could use partial fraction decomposition to break this into two fractions of the form $\dfrac{1}{(1-ae^{-jw})}$ which I can then inverse Fourier transform to get a result of the form $h[n] = a^nu[n]$To simplify, I let $S = e^{-jw}$Thus $H(w) = \dfrac{1}{(1-\frac{1}{4}e^{-jw})(1-\frac{1}{3}e^{-jw})}$ becomes $H(w) = \dfrac{1}{(1-\frac{1}{4}S)(1-\frac{1}{3}S)}$I tried to work these and both answer are different:SHouldn't these be equal? If so, I cannot figure out the necessary steps to get from one to the other. Any help would be appreciated.
 With Mathematica I have:  f[w_] := 1/((1 - E^(-I w)/3)*(1 - E^(-I w)/4)); InverseFourierSequenceTransform[f[w], w, n] (* Piecewise[{{4/3^n - 3/4^n, n >= 0}}, 0] *) g[w] = f[w] // Apart (*1 + 4/(-1 + 3 E^(I w)) - 3/(-1 + 4 E^(I w)) *) InverseFourierSequenceTransform[g[w], w, n] (* Piecewise[{{4/3^n - 3/4^n, n >= 0}}, 0] *) Results are the same. You can check this code in here.(Only you must sign in)