Signals and Systems - Electrical Engineering

8.3 Discrete-Time Systems 479

nExample 8.19

A square-root computation system.The input–output relation characterizing a discrete-time sys-

tem is nonlinear if there are nonlinear terms that include the inputx[n], the outputy[n], or both

(e.g., a square root ofx[n], products ofx[n] andy[n], etc.). Consider the development of an iterative

algorithm to compute the square root of a positive real numberα. If the result of the algorithm

isy[n] asn→∞, theny^2 [n]=αand likewisey^2 [n−1]=α, thusy[n]=0.5(y[n−1]+y[n−1]).

Replacingy[n−1]=α/y[n−1] in this equation, the following difference equation, with some

initial conditiony[0], can be used to find the square root ofα:

y[n]=0.5

[

y[n−1]+

α

y[n−1]

]

n> 0

Find recursively the solution of this difference equation. Use the results of finding the square roots

of 4 and 2 to show the system is nonlinear. Solve the difference equation and plot the results for

α=4, 2 with MATLAB.

Solution

The given difference equation is first order, nonlinear (expanding it you get the product ofy[n]

withy[n−1] andy^2 [n−1], which are nonlinear terms) with constant coefficients. This equation

can be solved recursively forn>0 by replacingy[0] to gety[1], and use this to gety[2] and so

on—that is,

y[1]=0.5

[

y[0]+

α

y[0]

]

y[2]=0.5

[

y[1]+

α

y[1]

]

y[3]=0.5

[

y[2]+

α

y[2]

]

..

.

For instance, lety[0]=1 andα=4 (i.e., we wish to find the square root of 4),

y[0]= 1

y[1]=0.5

[

1 +

4

1

]

=2.5

y[2]=0.5

[

2.5+

4

2.5

]

=2.05

..

.