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Find output response using Z-transform, y(n)32y(n1)+12y(n2)=2x(n)+32x(n1)
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Solution:

y(n)32y(n1)+12y(n2)=2x(n)+32x(n1)

Taking Z-transform,

Y(Z)32[Z1Y(Z)+y(1)]+12[Z2Y(Z)+Z1y(1)+y(2)]=2X(Z)+32[Z1X(Z)+x(1)]

Y(Z)32[Z1Y(Z)]+12[Z2Y(Z)+1]=X(Z)[2+32Z1]

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} …

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