Using partial fraction decomposition to find inverse Fourier transform

Questions
1

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:

wolf1.JPG
wolf1.JPG809×371 23.6 KB

1611wolf2.JPG
1611wolf2.JPG795×352 23.6 KB

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.

2

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)