Solution to Q1 of Week 8 Quiz Pool


$ \begin{align} \text{(a)} \quad & y[n] = 0.6 y[n-1] + 0.4 x[n] \\ & h[n] = 0.6h[n-1] + 0.4\delta[n] \\ \end{align}\,\! $

assume that $ h[n]=0 $ when $ n<0 $.

$ \begin{align} {\color{White}abcde} & h[0]=0.4 \\ & h[1]=0.6h[0]=0.4 \times 0.6 \\ & h[2]=0.6h[1]=0.4 \times (0.6)^2 \\ & \ldots \\ & h[n] = 0.4(0.6)^n u[n] \\ \end{align} $

Quiz8Q1sol 1.jpg


$ \begin{align} \text{(b)} \quad & y[n] = y[n-1] + 0.25 (x[n]-x[n-3]) \\ & h[n] = h[n-1] + 0.25(\delta[n]-\delta[n-3]) \\ \end{align}\,\! $

assume that $ h[n]=0 $ when $ n<0 $.

$ \begin{align} {\color{White}abcde} & h[0]=0.25 \\ & h[1]=h[0]=0.25 \\ & h[2]=h[1]=0.25 \\ & h[3]=h[2]-0.25=0 \\ & h[4]=h[3]=0 \\ & \ldots \\ & h[n] = 0.25(u[n]-u[n-3]) \\ \end{align} $

Quiz8Q1sol 2.jpg


Back to Lab Week 8 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett