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


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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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