(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...) |
|||
| Line 5: | Line 5: | ||
---- | ---- | ||
| − | Using the DTFT formula, let assume that <math> | + | Using the DTFT formula, let assume that <math>X(w)</math> is the frequency response of <math>x[n]</math> such that |
| − | <math> | + | <math> X(w) = \sum_{n=-\infty}^{\infty} x[n] e^{-jwn} </math>. |
| − | Then, what is the DTFT of <math> | + | Then, what is the DTFT of <math>x^{\ast}[n]</math> ? |
---- | ---- | ||
| − | Start with <math> | + | Start with <math>X(w) = \sum_{n=-\infty}^{\infty} x[n] e^{-jwn} </math>. |
If we apply conjucation to both sides, | If we apply conjucation to both sides, | ||
then, <math>\begin{align} | then, <math>\begin{align} | ||
| − | + | X^{\ast}(w) & = \sum_{n=-\infty}^{\infty} x^{\ast}[n] (e^{-jwn})^{\ast} \\ | |
| − | & = \sum_{n=-\infty}^{\infty} | + | & = \sum_{n=-\infty}^{\infty} x^{\ast}[n] e^{jwn} \\ |
\end{align}</math>. | \end{align}</math>. | ||
| − | Changing the variable (<math>w'=-w</math>) to make the right-side as DTFT formula of <math> | + | Changing the variable (<math>w'=-w</math>) to make the right-side as DTFT formula of <math>x^{\ast}[n]</math>, |
| − | then <math> | + | then <math> X^{\ast}(-w') = \sum_{n=-\infty}^{\infty} x^{\ast}[n] e^{-jw'n} </math>. |
| − | This implies that the frequency response of <math> | + | This implies that the frequency response of <math>x^{\ast}[n]</math> is <math>X^{\ast}(-w)</math>. |
| − | Since <math> | + | Since <math>x[n]=x^{\ast}[n]</math>, thus <math>X(w)=X^{\ast}(-w)</math>, which put some constraints on the magnitude and phase reponse of <math>X(w)</math>. |
| − | That is, the magnitude response must be even <math>| | + | That is, the magnitude response must be even <math>|X(w)|=|X(-w)|\,\!</math>, |
| − | and the phase reponse must be odd <math>\angle | + | and the phase reponse must be odd <math>\angle X(w) = - \angle X(-w)</math>. |
---- | ---- | ||
Latest revision as of 18:04, 26 October 2010
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) $.
Back to Lab Week 10 Quiz Pool
Back to ECE 438 Fall 2010 Lab Wiki Page
Back to ECE 438 Fall 2010
