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==ECE438 Week9 Quiz Question5 Solution== | ==ECE438 Week9 Quiz Question5 Solution== | ||
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a. System impulse response is the system output when input is impulse signal. | a. System impulse response is the system output when input is impulse signal. | ||
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | H(e^{j\omega})&=\sum_{k=-\infty}^{infty}h[k]e^{-j\omega k} \\ | + | H(e^{j\omega})&=\sum_{k=-\infty}^{\infty}h[k]e^{-j\omega k} \\ |
&=\frac{1}{M_1+M_2+1}\sum_{k=-M_1}^{M_2}e^{-j\omega k} \\ | &=\frac{1}{M_1+M_2+1}\sum_{k=-M_1}^{M_2}e^{-j\omega k} \\ | ||
&=\frac{1}{M_1+M_2+1}\frac{e^{j\omega M_1}-e^{-j(M_2+1)}}{1-e^{-j\omega}} \\ | &=\frac{1}{M_1+M_2+1}\frac{e^{j\omega M_1}-e^{-j(M_2+1)}}{1-e^{-j\omega}} \\ | ||
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</math> | </math> | ||
| − | c. | + | c. Hint: The magnitude response looks like a sinc function with cut off frequency of <math>\pm \frac{2\pi}{5}</math> |
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| + | ------------------------------ | ||
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Latest revision as of 17:50, 20 October 2010
ECE438 Week9 Quiz Question5 Solution
a. System impulse response is the system output when input is impulse signal.
Let
$ x[n]=\delta [n] $
Then
$ h[n]=\frac{1}{M_1+M_2+1}\sum_{k=-M_1}^{M_2}\delta[n-k] $
$ h[n]=\left\{\begin{array}{ll}\frac{1}{M_1+M_2+1}, & -M_1\le n\le M_2 \\ 0, & otherwise \end{array}\right. $
b.
$ \begin{align} H(e^{j\omega})&=\sum_{k=-\infty}^{\infty}h[k]e^{-j\omega k} \\ &=\frac{1}{M_1+M_2+1}\sum_{k=-M_1}^{M_2}e^{-j\omega k} \\ &=\frac{1}{M_1+M_2+1}\frac{e^{j\omega M_1}-e^{-j(M_2+1)}}{1-e^{-j\omega}} \\ &=\frac{1}{M_1+M_2+1}e^{-j\omega (M_2-M_1+1)/2}\frac{e^{j\omega (M_1+M_2+1)/2}-e^{-j(M_1+M_2+1)/2}}{1-e^{-j\omega}} \\ &=\frac{1}{M_1+M_2+1}e^{-j\omega (M_2-M_1)/2}\frac{e^{j\omega (M_1+M_2+1)/2}-e^{-j(M_1+M_2+1)/2}}{e^{j\omega /2}-e^{-j\omega /2}} \\ &=\frac{1}{M_1+M_2+1}e^{-j\omega (M_2-M_1)/2}\frac{sin[\omega(M_1+M_2+1)/2]}{sin(\omega /2)} \end{align} $
c. Hint: The magnitude response looks like a sinc function with cut off frequency of $ \pm \frac{2\pi}{5} $
