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Find output response using Z-transform, y(n)−32y(n−1)+12y(n−2)=2x(n)+32x(n−1)
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written 2.3 years ago by |
Solution:
y(n)−32y(n−1)+12y(n−2)=2x(n)+32x(n−1)
Taking Z-transform,
Y(Z)−32[Z−1Y(Z)+y(−1)]+12[Z−2Y(Z)+Z−1y(−1)+y(−2)]=2X(Z)+32[Z−1X(Z)+x(−1)]
Y(Z)−32[Z−1Y(Z)]+12[Z−2Y(Z)+1]=X(Z)[2+32Z−1]
Since x(n) is causal signal x(−1)=0
$ \begin{aligned} & x(n)=\left(\frac{1}{4}\right)^n u(n) \Rightarrow X(Z)=\frac{1}{1-\frac{1}{4} Z^{-1}}=\frac{Z}{Z-\frac{1}{4}} \\\\ \therefore & Y(Z)\left[1-\frac{3}{2} Z^{-1}+\frac{1}{2} Z^{-2}\right]=\frac{Z}{Z-\frac{1}{4}}\left[2+\frac{3}{2} \\Z^{-1}\right]-\frac{1}{2}\\ \end{aligned} …