# expression for the FT of the frequency response of a system

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

## 1 Answer

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.

• 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?. – perfectace Sep 19 '20 at 14:25
• @perfectace Think of the unit circle, $e^{j(\omega + \pi)}=e^{j(\omega - \pi)}$ – Engineer Sep 19 '20 at 16:10
• I was thinking about the wrong thing - got it, thanks. – perfectace Sep 19 '20 at 17:29