Signals and Systems - Electrical Engineering

134 C H A P T E R 2: Continuous-Time Systems

FIGURE 2.6

Nonlinear behavior of RL

circuit: (top)I 0 = 1 ,

B= 1 ,

v(t)=u(t),i 1 (t)=u(t),

and (bottom)I 0 = 1 ,

B=2,v(t)= 2 u(t),

i 2 (t)=( 2 −e−t)u(t),

andi 2 (t)6= 2 ii(t).

− 1 0 1 2 3 45

0

0.5

1

v(

t),

i(^1

t)

2 v

(t

),

i(^2

t)

− 1 0 1 2 3 45

0

0.5

1

1.5

2

t (sec)

v(t)

i 1 (t)

2 v(t)

i 2 (t)

Zero initial conditions

Suppose then we perform the above experiments withI 0 =0 whenB=1 and whenB=2. We get

i 1 (t)=( 1 −e−t)u(t)

forB=1, and forB=2 we get

i 2 (t)= 2 ( 1 −e−t)u(t)

= 2 i 1 (t)

which indicates the system is linear. In this case the response only depends on the inputv(t).

Time invariance

Suppose now thatB=1,v(t)=u(t− 1 ), and the initial condition isI 0. The complete response is

i 3 (t)=I 0 e−tu(t)+( 1 −e−(t−^1 ))u(t− 1 )

IfI 0 =0, then the above response isi 3 (t)=( 1 −e−(t−^1 ))u(t− 1 ), which equalsi(t− 1 )(Equa-

tion (2.15) withB=1 andI 0 =0 delayed by 1) indicating the system is time invariant. On the

other hand, whenI 0 =1 the complete response is not equal toi(t− 1 )because the term with the

initial condition is not shifted like the second term. The system in that case is time varying. Thus,

ifI 0 =0 the system is LTI. n