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A causal LTI system is described y[n]=34y[n1]18y[n2]+x[n].Where y[n] response of the system and x[n] is excitation to the system.

(i) Determine impulse response of the system.

(ii) Determine step response of the system.

(iii) Plot pole-zero pattern and state whether system is stable.

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

1 Answer
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(i) Impulse response of the system:

By z-transform,

Y(z)=34z1Y(z)18z2Y(z)+X(z)Y(z)34z1Y(z)+18z2Y(z)=X(z)Y(z)(134z1+18z2)=X(z)H(z)=Y(z)X(z)=1134z1+18z2H(z)=z2z234z+18H(z)=z2(z12)(z14)H(z)z=z(z12)(z14)

Solving the above equation using partial fractions

H(z)z=A(z12)+B(z14)

enter image description here

H(z)z=2(z12)1(z14)H(z)=2(112z1)1(114z1)

By Inverse z-transform,

enter image description here

Step response of the system: x(n)=u(n) …………………(step response)

X(z)=11z1 …………………(By z-transform)

we know, H(z)=Y(z)X(z)Y(z)=H(z).X(z)=1(134z1+18z211z1=z2z234z+18zz1Y(z)z=z2z12)(z14)(z1)

By partial fraction,

Y(z)z=A(z12)+B(z14)+C(z1)

enter image description here

Y(z)z=2(z12)+1/3(z14)+8/3(z1)Y(z)=2z(z12)+z/3(z14)+8z/3(z1)

By Inverse z-transform,

y(n)=2(12)nu(n)+13(14)nu(n)+83(1)nu(n)

(iii) The system function is given by,

H(z)=z2(z12)(z14)

Poles at: z=1/2, z=1/4.

Zeros at: z=0, z=0.

Pole zero diagrams:

enter image description here

Since all the poles lie inside the unit circle, the system is stable.

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