Solution to Q2 of Week 10 Quiz Pool
Using the DTFT formula, let assume that $ X(w) $ is the frequency response of $ x[n] $ such that $ X(w) = \sum_{n=-\infty}^{\infty} x[n] e^{-jwn} $.
Then, what is the DTFT of $ x^{\ast}[n] $ ?
Start with $ X(w) = \sum_{n=-\infty}^{\infty} x[n] e^{-jwn} $.
If we apply conjucation to both sides,
then, $ \begin{align} X^{\ast}(w) & = \sum_{n=-\infty}^{\infty} x^{\ast}[n] (e^{-jwn})^{\ast} \\ & = \sum_{n=-\infty}^{\infty} x^{\ast}[n] e^{jwn} \\ \end{align} $.
Changing the variable ($ w'=-w $) to make the right-side as DTFT formula of $ x^{\ast}[n] $,
then $ X^{\ast}(-w') = \sum_{n=-\infty}^{\infty} x^{\ast}[n] e^{-jw'n} $.
This implies that the frequency response of $ x^{\ast}[n] $ is $ X^{\ast}(-w) $.
Since $ x[n]=x^{\ast}[n] $, thus $ X(w)=X^{\ast}(-w) $, which put some constraints on the magnitude and phase reponse of $ X(w) $.
That is, the magnitude response must be even $ |X(w)|=|X(-w)|\,\! $,
and the phase reponse must be odd $ \angle X(w) = - \angle X(-w) $.
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