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) $, which put some constraints on the magnitude and phase reponse of $ H(w) $.

That is, the magnitude response must be even $ |H(w)|=|H(-w)|\,\! $,

and the phase reponse must be odd $ \angle H(w) = - \angle H(-w) $.


Back to Lab Week 9 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood