Signals and Systems - Electrical Engineering

552 C H A P T E R 9: The Z-Transform

Remarks

n If the signal is causal, so that{x[i],−N≤i≤− 1 }are all zero, we then have thatZ(x[n−N])=

z−NX(z), indicating that the operator z−^1 is a delay operator. Thus, x[n−N]has been delayed N samples

and its Z-transform is then simply X(z)multiplied by z−N.

n The shifting in time property is useful in the solution of difference equations, especially when it has

nonzero initial conditions as we will see next. On the other hand, if the initial conditions are zero, either

the one-sided or the two-sided Z-transforms could be used.

The analog of differential equations are difference equations, which result directly from the modeling

of a discrete system or from discretizing differential equations. The numerical solution of differen-

tial equations requires that these equations be converted into difference equations since computers

cannot perform integration. Many methods are used to solve differential equations with different

degrees of accuracy and sophistication. This is a topic of numerical analysis, outside the scope of this

text, and thus only simple methods are illustrated here.

nExample 9.16

A discrete-time IIR system is represented by a first-order difference equation

y[n]=ay[n−1]+x[n] n≥ 0 (9.36)

wherex[n] is the input of the system andy[n] is the output. Discuss how to solve it using recursive

methods and the Z-transform. Obtain a general form for the complete solutiony[n] in terms of

the impulse responseh[n] of the system.

For inputx[n]=u[n]−u[n−11], zero initial conditions, anda=0.8, use the MATLAB function

filterto findy[n]. Plot the input and the output.

Solution

In the time domain a unique solution is obtained by using the recursion given by the difference

equation. We would need an initial condition to computey[0], indeed

y[0]=ay[−1]+x[0]

and asx[0] is given, we needy[−1] as the initial condition. Giveny[0], recursively we find the rest

of the solution:

y[1]=ay[0]+x[1]

y[2]=ay[1]+x[2]

y[3]=ay[2]+x[3]

···

where at each step the needed output values are given by the previous step of the recursion.

However, this solution is not in closed form.