I am trying to find an expression for the Fourier Transform of the frequency response of the cascade system seen here: here

Here is my approach:

$(-1)^n = (-1)^{-n}$

$v[n] = x[n]e^{j\pi n}$

$V(e^{jw}) = X(e^{j(w + \pi)})$

$w[n] = v[n] * h_1[n]$

$W(e^{jw}) = X(e^{j(w + \pi)}).H_1(e^{jw})\,\,...... (1)$

$y[n] = w[n]e^{j\pi n}$

$Y(e^{jw}) = W(e^{j(w + \pi)})\,\,...... (2)$

With equations 1 and 2, I can find $\frac{Y(e^{jw})}{X(e^{jw})}$ where I get $H(e^{jw})$ as $H_1(e^{j(w + 2\pi)})$.

Now, looking at the official solution from the book, I see:


I don't see how $Y(e^{jw}) = W(e^{j(w-\pi)})$ here. Have I made a mistake above?


Try writing your equation (2) as $Y(e^{j\omega})=W(e^{j(\omega + \pi)})=X(e^{j\omega})H_1(e^{j(\omega + \pi)})$, and now try solving for $H(e^{j\omega})$. Remember that phase shifting by $\pi$ and $-\pi$ gives the same result.

  • $\begingroup$ I see - that results in H_1(e^({j(w+\pi})), which is a shift of 2*pi. Do you mean a shift of 2*pi since Discrete Fourier Transforms have a periodicity of 2pi?. $\endgroup$ – perfectace Sep 19 '20 at 14:25
  • $\begingroup$ @perfectace Think of the unit circle, $e^{j(\omega + \pi)}=e^{j(\omega - \pi)}$ $\endgroup$ – Engineer Sep 19 '20 at 16:10
  • $\begingroup$ I was thinking about the wrong thing - got it, thanks. $\endgroup$ – perfectace Sep 19 '20 at 17:29

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