I have a system where $H(w) = \dfrac{1}{(1-\frac{1}{4}e^{-jw})(1-\frac{1}{3}e^{-jw})}$. 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 $x = e^{-jw}$
Thus $H(w) = \dfrac{1}{(1-\frac{1}{4}e^{-jw})(1-\frac{1}{3}e^{-jw})}$ becomes $H(w) = \dfrac{1}{(1-\frac{1}{4}x)(1-\frac{1}{3}x)}$
I then proceeded to solve both of these using Wolfram, expecting the answers to be the same:
I can't figure out how to get from one answer to the other thus confirming both are equal. Any suggestions would be appreciated.
Wolfie's answer is $$-\frac{3/4}{1/x-1/4}+\frac{4/3}{1/x-1/3}+1.$$ Now $$-\frac{3/4}{1/x-1/4}=\frac{3}{1-4/x}=\frac{3x}{x-4} =\frac{12}{x-4}+3.$$ Also $$\frac{4/3}{1/x-1/3}=\frac{4}{3/x-1}=-\frac{4x}{x-3} =\frac{-12}{x-3}-4.$$ Now put these together....