Signals and Systems - Electrical Engineering

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 conditioni( 0 )=I 0 and the differential equation. In fact, ift= 0 +
(slightly larger than 0) we have that the solution givesi( 0 +)=I 0 , and that fort>0 when we
replace in the differential equation the input voltage byB,i(t), anddi(t)/dt(using the above
solution), we get

︸︷︷︸B

v(t)

=[I 0 e−t+B( 1 −e−t)]

︸ ︷︷ ︸

i(t)

+[Be−t−I 0 e−t]

︸ ︷︷ ︸

di(t)/dt

=B t> 0

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

Initial Condition Different from Zero
WhenI 0 =1 andB=1, the complete solution given by Equation (2.15) becomes

i 1 (t)=[e−t+( 1 −e−t)]u(t)

=u(t) (2.16)

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

i 1 zs(t)=( 1 −e−t)u(t)

which is obtained by lettingB=1 andI=0 in Equation (2.15). The zero-input response, when
v(t)=0 and the initial condition isI 0 =1, is

i 1 zi(t)=e−tu(t)

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

If we then considerB=2 (i.e., we double the original input) and keepI 0 =1, the complete
solution is given by

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)= 2 u(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 tov(t)= 2 u(t)
and zero-initial conditions is doubled so that

i 2 zs(t)= 2 ( 1 −e−t)u(t)

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

i 2 zi(t)=e−tu(t)

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