(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q2 of Week 10 Quiz Pool == ---- Using the DTFT formula, let assume that <math>H(w)</math> is the frequency response of <math>h[n...) |
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Revision as of 18:42, 26 October 2010
Solution to Q2 of Week 10 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) $.
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