Signals and Systems - Electrical Engineering

480 C H A P T E R 8: Discrete-Time Signals and Systems

FIGURE 8.8

Nonlinear system:

(a) square root of 2,

(b) square root of 4

compared with twice the

square root of 2, and

(c) sum of previous

responses with response

when computing square

root of 2 + 4. Figure (b)

shows scaling does not

hold and (c) shows that

additivity does not hold

either. The system is

nonlinear.

0 1 2 3 4

(a)

(b)

(c)

5 6 7 8 9

0123456789

0

1

2

0

2

4

0 1 2 3 4 5 6 7 8 9

0

2

4

n

n

n

square root of 2

square root of 4

square root of 6

y^1

[n

]

y^2

[n

]

y^3

[n

]

which is converging to 2, the square root of 4 (see Figure 8.8). Thus, as indicated before, when

n→∞, theny[n]=y[n−1]=Y, for some valueY, which according to the difference equation

satisfies the relationY=0.5Y+0.5( 4 /Y)orY=

√

4 =2.

Suppose then that the input isα=2, half of what it was before. If the system is linear, we should

get half the previous output according to the scaling property. That is not the case, however. For

the same initial conditiony[0]=1, we obtain recursively forα[n]= 2 u[n−1]:

y[0]= 1

y[1]=0.5[1+2]=1.5

y[2]=0.5

[

1.5+

2

1.5

]

=1.4167

..

.

This solution is clearly not half of the previous one. Moreover, asn→∞, we expecty[n]=

y[n−1]=Y, forYthat satisfies the relationY=0.5Y+0.5( 2 /Y)orY=

√

2 =1.4142, so that

the solution is tending to

√

2 and not to 2 as it would if the system were linear. Finally, if we add