Solution of Difference Equation using ZT

Updated on 2017/08/06 22:43

Solution of Difference equation using ZT

The response of any discrete time system can be decomposed as:

Total Response = zero state response + zero input response

* Zero state response / Forced Response: Response of the system due to input alone when the initial state of the system (condition) is zero.

* Zero input response / Natural response: Response of the system depends only on the initial state of the system i.e. the input is zero.

Expected Questions

  1. Calculate impulse response of the system: take input x (n) as impulse  sequence.
  2. Calculate step response of the system: take input x (n) as unit step sequence.
  3. Calculate impulse response of the system: i.e. calculate h(n).
  4. Calculate system / transfer function i.e. H(z). H(z)=Y(z)/X(z).
  5. Calculate zero state and input response of the system.

Calculate impulse response of the system

Questions.
Compute the impulse response of the system given by difference equation: 

y (n)=0.7 y (n-1)-0.12 y(n-2)+x(n-1)+x(n-2)

Solution:

As we need to calculate the impulse response of the system that means input is impulse function.

x(n)=\delta (n) , x(n-1)=\delta (n-1) , x(n-2) =\delta (n-2)

y(n) = 0.7 y(n-1)-0.12 y(n-2)+\delta (n-1)+\delta (n-2)

Take ZT on both sides,

Z[y(n)]=Y(z) ,

According to time shifting property,

Z[y(n-1)] = Z^{^{-1}} Y(z)

Z[y(n-2)] = Z^{^{-2}} Y(z)

Z[\delta (n-1)] = Z^{^{-1}}

Z[\delta (n-2)] = Z^{^{-2}}

\therefore Y(Z) = 0.7 Z^{^{-1}} Y(Z) -0.12 Z^{-2}Y(Z)+Z^{-1}+Z^{-2}

\therefore Y(Z) - 0.7 Z^{^{-1}} Y(Z) +0.12 Z^{-2}Y(Z)=Z^{-1}+Z^{-2}

\therefore Y(Z)(1- 0.7 Z^{^{-1}} +0.12 Z^{-2})=Z^{-1}+Z^{-2}

\therefore Y(Z) = \frac{Z^{-1}+Z^{-2}}{(1- 0.7 Z^{^{-1}} +0.12 Z^{-2})}

Convert it into positive powers of z, multiply numerator and denominator by Z^2,

\therefore Y(Z) = \frac{(Z^{-1}+Z^{-2})Z^{2}}{(1- 0.7 Z^{^{-1}} +0.12 Z^{-2})Z^{2}}

\therefore Y(Z) = \frac{(Z^{1}+1)}{(Z^{2}- 0.7 Z^{^{1}} +0.12 )}

\therefore Y(Z) = \frac{(Z+1)}{(Z-0.4)(Z-0.3)}

Take IZT using partial fraction expansion method, 

\therefore\frac{ Y(Z)}{Z} = \frac{(Z+1)}{Z(Z-0.4)(Z-0.3)}

Write equation in partial fraction expansion form,

\frac{Y(Z)}{Z}=\frac{A}{Z}+\frac{B}{Z-0.4}+\frac{C}{Z-0.3}

A=\left |Z\frac{Z+1}{Z(Z-0.4)(Z-0.3)} \right |_{Z=0}    = 8.33

B=\left |(Z-0.4)\frac{Z+1}{Z(Z-0.4)(Z-0.3)} \right |_{Z=0.4} = 35

C=\left |(Z-0.3)\frac{Z+1}{Z(Z-0.4)(Z-0.3)} \right |_{Z=0.3} = -44.3

\frac{Y(Z)}{Z}=\frac{8.33}{Z}+\frac{35}{Z-0.4}+\frac{-44.3}{Z-0.3}

Taking IZT,

Y(Z)=8.33+ 35 \frac{Z}{Z-0.4}-44.3 \frac{Z}{Z-0.3}

h(n)= 8.33 \delta (n)+35(0.4)^{n} u(n)-44.3 (0.3)^{n}u(n)

References

  • Notes by Prof. Priyanka Bhosale
  • WikiNote Foundation
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Created by PRIYANKA BHOSALE on 2017/08/03 19:02