# Solution of Difference Equation using ZT

## 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 =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]=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 Vishal E on 2019/01/11 08:54

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