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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) $.


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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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