LTI Continuous-Time Systems 133

2.3 LTI Continuous-Time Systems 133

The solution of this differential equation is given by

i(t) = [I 0 e − t + B(1 − e − t ) ]u(t)

(2.15) which satisfies the initial condition i(0) = I 0 and the differential equation. In fact, if t = 0+

(slightly larger than 0) we have that the solution gives i(0+) = I 0 , and that for t > 0 when we replace in the differential equation the input voltage by B, i(t), and di(t)/dt (using the above solution), we get

0 e − t + B(1 − e − ) ] + [Be − t − I 0 t e − ] = B |{z} t>0

di(t)/dt

or an identity indicating i(t) in Equation (2.15) is the solution of the differential equation.

Initial Condition Different from Zero

When I 0 = 1 and B = 1, the complete solution given by Equation (2.15) becomes

i 1 ( t) = [e − t +( 1−e − t ) ]u(t)

(2.16) The zero-state response (i.e., the response due to v(t) = u(t) and zero initial condition) is

= u(t)

1zs ( t) = (1 − e − ) u(t)

which is obtained by letting B = 1 and I = 0 in Equation (2.15). The zero-input response, when

v(t) = 0 and the initial condition is I 0 =

1, is

i 1zi ( t) = e − t u(t)

obtained by subtracting the zero-state response from the complete response in Equation (2.16).

1, the complete solution is given by

If we then consider B = 2 (i.e., we double the original input) and keep I 0 =

i 2 ( t) = [e − t + 2(1 − e − t ) ]u(t)

− 2−e t ) u(t)

which is completely different from the expected 2i 1 ( t) = 2u(t) for a linear system. Thus, the system is not linear (see Figure 2.6). In this case we have that the zero-state response due to v(t) = 2u(t) and zero-initial conditions is doubled so that

i 2zs ( t) = 2(1 − e − t ) u(t)

while the zero-input response remains the same, as the initial condition did not change. So,

i 2zi ( t) = e − t u(t)

and we get the complete solution shown above. The output in this case depends on the input v(t) and on the initial condition, and when testing linearity we are only changing v(t).

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

2v(t) Nonlinear behavior of RL

i 2 (t ) circuit: (top) I 0 2 = 1, i

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

and (bottom) I 0 = 1, B = 2, v(t) = 2u(t), 0

i 2 ( t) = (2 − e − t ) u(t),

0 1 2 3 4 5 and i 2 ( t) 6= 2i i ( t).

t (sec)

Zero initial conditions

Suppose then we perform the above experiments with I 0 = 0 when B = 1 and when B = 2. We get

i 1 ( t) = (1 − e − t ) u(t)

for B = 1, and for B = 2 we get

i 2 ( t) = 2(1 − e − t ) u(t) = 2i 1 ( t)

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

Time invariance

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

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

0, then the above response is i 3 ( t) = (1 − e −( t−1) ) u(t − 1), which equals i(t − 1) (Equa- tion (2.15) with B = 1 and I 0 = 0 delayed by 1) indicating the system is time invariant. On the other hand, when I 0 = 1 the complete response is not equal to i(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, if I 0 = 0 the system is LTI.

If I 0 =