ECE438 Week11 Quiz Q1sol - Rhea

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}