Solution to Q2 of Week 9 Quiz Pool
Using the DTFT formula, let assume that $ H(w) $ is the frequency response of $ h[n] $ such that $ H(w) = \sum_{n=-\infty}^{\infty} h[n] e^{-jwn} $.
Then, what is the DTFT of $ h^{\ast}[n] $ ?
Start with $ H(w) = \sum_{n=-\infty}^{\infty} h[n] e^{-jwn} $.
If we apply conjucation to both sides,
then, $ \begin{align} H^{\ast}(w) & = \sum_{n=-\infty}^{\infty} h^{\ast}[n] (e^{-jwn})^{\ast} \\ & = \sum_{n=-\infty}^{\infty} h^{\ast}[n] e^{jwn} \\ \end{align} $.
Changing the variable ($ w'=-w $) to make the right-side as DTFT formula of $ h^{\ast}[n] $,
then $ H^{\ast}(-w') = \sum_{n=-\infty}^{\infty} h^{\ast}[n] e^{-jw'n} $.
This implies that the frequency response of $ h^{\ast}[n] $ is $ H^{\ast}(-w) $.
Since $ h[n]=h^{\ast}[n] $, thus $ H(w)=H^{\ast}(-w) $.
Back to Lab Week 9 Quiz Pool
Back to ECE 438 Fall 2010 Lab Wiki Page
Back to ECE 438 Fall 2010