Revision as of 12:26, 8 October 2010 by Han83 (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)



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]=0.2 \\ & h[1]=0.8h[0]=0.2 \times 0.8 \\ & h[2]=0.8h[1]=0.2 \times (0.8)^2 \\ & \ldots \\ & h[n] = 0.2(0.8)^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


Credit: Prof. Charles Bouman

Back to Lab Week 8 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang