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
x(n)=(14)nu(n)⇒X(Z)=11−14Z−1=ZZ−14∴Y(Z)[1−32Z−1+12Z−2]=ZZ−14[2+32Z−1]−12
Y(Z)=ZZ−14(2+32Z−1)(1−32Z−1+12Z−2)−121−32Z−1+12Z−2Y(Z)=ZZ−14(2Z2+32ZZ2−32Z+12)−12Z2Z2−32Z+12
Y(Z)Z=(2Z2+32Z)(Z−14)(Z−1)(Z−12)−12Z(Z−1)(Z−12)=(2Z2+32Z)−12Z(Z−14)(Z−14)(Z−1)(Z−12)=2Z2+32Z−12Z2+18Z(Z−14)(Z−1)(Z−12)=32Z2+138Z(Z−14)(Z−1)(Z−12) =A(Z−14)+B(Z−1)+C(Z−12)
32Z2+138Z=A(Z−1)(Z−12)+B(Z−14)(Z−12)+C(Z−14)(Z−1)
AtZ=14:A=83
At Z=1:B=253
At Z=12:C=−192
∴Y(Z)Z=83Z−14+253Z−1−192Z−12Y(Z)=83ZZ−14+253ZZ−1−192ZZ−12
Applying inverse Z transform,
y(n)=83(14)nu(n)+253u(n)−192(12)nu(n)