| Line 16: | Line 16: | ||
\end{align}</math> | \end{align}</math> | ||
| − | Frequency Response | + | Frequency Response <math>H(\omega)</math> |
| + | <math>\begin{align} | ||
H[e^{j\omega }] &= \frac{1+e^{-j\omega }}{2} \\ | H[e^{j\omega }] &= \frac{1+e^{-j\omega }}{2} \\ | ||
&= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}+e^{-j\frac{\omega }{2}}}{2} \right) \\ | &= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}+e^{-j\frac{\omega }{2}}}{2} \right) \\ | ||
| Line 26: | Line 27: | ||
b) | b) | ||
| − | <math> | + | <math>y[n]= \frac{x[n]-x[n-1]}{2}</math> |
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function<br> | Applying Z-transform on both sides and grouping terms, we can obtain the transfer function<br> | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | + | Y[z]&= \frac{X[z]-X[z].z^{-1}}{2} \\ | |
\frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\ | \frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\ | ||
| − | + | H[z] &= \frac{1-z^{-1}}{2} \\ | |
\end{align}</math> | \end{align}</math> | ||
| − | Frequency Response | + | Frequency Response <math>H(\omega)</math> |
| − | + | <math>\begin{align} | |
| + | H[e^{j\omega }] &= \frac{1-e^{-j\omega }}{2} \\ | ||
&= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2} \right) \\ | &= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2} \right) \\ | ||
&= je^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2j} \right) \\ | &= je^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2j} \right) \\ | ||
Revision as of 10:14, 29 October 2011
Contents
Homework 6, ECE438, Fall 2011, Prof. Boutin
Question 1
a)
$ y[n]= \frac{x[n]+x[n-1]}{2} $
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function
$ \begin{align} Y[z]&= \frac{X[z]+X[z].z^{-1}}{2} \\ \frac{Y[z]}{X[z]}&= \frac{1+z^{-1}}{2} \\ H[z] &= \frac{1+z^{-1}}{2} \\ \end{align} $
Frequency Response $ H(\omega) $ $ \begin{align} H[e^{j\omega }] &= \frac{1+e^{-j\omega }}{2} \\ &= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}+e^{-j\frac{\omega }{2}}}{2} \right) \\ &= e^{-j\frac{\omega }{2}} cos \left( \frac{\omega }{2} \right) \\ \end{align} $
b)
$ y[n]= \frac{x[n]-x[n-1]}{2} $
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function
$ \begin{align} Y[z]&= \frac{X[z]-X[z].z^{-1}}{2} \\ \frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\ H[z] &= \frac{1-z^{-1}}{2} \\ \end{align} $
Frequency Response $ H(\omega) $ $ \begin{align} H[e^{j\omega }] &= \frac{1-e^{-j\omega }}{2} \\ &= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2} \right) \\ &= je^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2j} \right) \\ &= je^{-j\frac{\omega }{2}} sin \left( \frac{\omega }{2} \right) \\ \end{align} $
Question 2
Question 3
Question 4
Question 5
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