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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

x(n)=(14)nu(n)X(Z)=1114Z1=ZZ14Y(Z)[132Z1+12Z2]=ZZ14[2+32Z1]12

Y(Z)=ZZ14(2+32Z1)(132Z1+12Z2)12132Z1+12Z2Y(Z)=ZZ14(2Z2+32ZZ232Z+12)12Z2Z232Z+12

Y(Z)Z=(2Z2+32Z)(Z14)(Z1)(Z12)12Z(Z1)(Z12)=(2Z2+32Z)12Z(Z14)(Z14)(Z1)(Z12)=2Z2+32Z12Z2+18Z(Z14)(Z1)(Z12)=32Z2+138Z(Z14)(Z1)(Z12) =A(Z14)+B(Z1)+C(Z12)

32Z2+138Z=A(Z1)(Z12)+B(Z14)(Z12)+C(Z14)(Z1)

AtZ=14:A=83

At Z=1:B=253

At Z=12:C=192

Y(Z)Z=83Z14+253Z1192Z12Y(Z)=83ZZ14+253ZZ1192ZZ12

Applying inverse Z transform,

y(n)=83(14)nu(n)+253u(n)192(12)nu(n)

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