(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q1 of Week 11 Quiz Pool == ---- The transfer function of the first and second systems are :<math>H_1(z)=1-z^{-1}\,\!</math> :<ma...) |
|||
| Line 10: | Line 10: | ||
Then, the transfer function of the combined system, <math>(T_1+T_2)[x[n]]</math> is | Then, the transfer function of the combined system, <math>(T_1+T_2)[x[n]]</math> is | ||
| − | :<math>\begin{align}H(z)=H_1(z)+H_2(z)&=1 | + | :<math>\begin{align}H(z)=H_1(z)+H_2(z)&=1-z^{-1}+\frac{1}{1-\frac{1}{2}z^{-1}} \\ &=\frac{2-\frac{3}{2}z^{-1}+\frac{1}{2}z^{-2}}{1-\frac{1}{2}z^{-1}}\end{align}</math> |
Thus, the impulse response <math>h[n]</math> of the combined system is (if we assume 'casual'), | Thus, the impulse response <math>h[n]</math> of the combined system is (if we assume 'casual'), | ||
Latest revision as of 08:52, 4 November 2010
Solution to Q1 of Week 11 Quiz Pool
The transfer function of the first and second systems are
- $ H_1(z)=1-z^{-1}\,\! $
- $ H_2(z)=\frac{1}{1-\frac{1}{2}z^{-1}} $
Then, the transfer function of the combined system, $ (T_1+T_2)[x[n]] $ is
- $ \begin{align}H(z)=H_1(z)+H_2(z)&=1-z^{-1}+\frac{1}{1-\frac{1}{2}z^{-1}} \\ &=\frac{2-\frac{3}{2}z^{-1}+\frac{1}{2}z^{-2}}{1-\frac{1}{2}z^{-1}}\end{align} $
Thus, the impulse response $ h[n] $ of the combined system is (if we assume 'casual'),
- $ h[n]=\delta[n]-\delta[n-1]+(0.5)^n u[n]\,\! $
And the difference equation for the combined system is
- $ \begin{align}&H(z)=\frac{Y(z)}{X(z)}=\frac{2-\frac{3}{2}z^{-1}+\frac{1}{2}z^{-2}}{1-\frac{1}{2}z^{-1}} \\ &\Rightarrow Y(z)(1-\frac{1}{2}z^{-1})=X(z)(2-\frac{3}{2}z^{-1}+\frac{1}{2}z^{-2}) \\ &\Rightarrow y[n]-\frac{1}{2}y[n-1]=2x[n]-\frac{3}{2}x[n-1]+\frac{1}{2}x[n-2] \end{align} $
Back to Lab Week 11 Quiz Pool
Back to ECE 438 Fall 2010 Lab Wiki Page
Back to ECE 438 Fall 2010
