Signals and Systems - Electrical Engineering

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

IfSis the transformation corresponding to an LTI system, so that the response of the system is

y(t)=S[x(t)]for an inputx(t)

then we have that

S





∑

k

Akx(t−τk)



=

∑

k

AkS[x(t−τk)]=

∑

k

Aky(t−τk)

S

[∫

g(τ)x(t−τ)dτ

]

=

∫

g(τ)S[x(t−τ)]dτ=

∫

g(τ)y(t−τ)dτ

In the next section we will see that this property allows us to find the response of a linear time-invariant

system due to any signal, if we know the response of the system to an impulse signal.

nExample 2.7

The response of an RL circuit to a unit-step sourcev(t)=u(t)is

i(t)=( 1 −e−t)u(t)

Find the response to a sourcev(t)=u(t)−u(t− 2 ).

Solution

Using superposition and time invariance, the output current due to the pulsev(t)=u(t)−u(t− 2 )

volts is

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

Figure 2.8 shows the responses tou(t)and u(t− 2 )and the overall response tov(t)=u(t)

−u(t− 2 ). n

nExample 2.8

Suppose we know that the response to a rectangular pulsev 1 (t)is the currenti 1 (t)shown in

Figure 2.9. If the input voltage is a train of two pulses,v(t), find the corresponding currenti(t).

Solution

Graphically the response tov(t)of the LTI system is given byi(t)as shown in Figure 2.9. n

2.3.5 Convolution Integral.................................................................

In this section we consider the computation of the output of a continuous-time linear time-invariant

(LTI) system due to any continuous-time input signal.