Signals and Systems - Electrical Engineering

2.3 LTI Continuous-Time Systems 139

Remarks

n We will see that the impulse response is fundamental in the characterization of linear time-invariant
systems.
n Any system characterized by the convolution integral is linear and time invariant by the above construction.
The convolution integral is a general representation of LTI systems, given that it was obtained from a
generic representation of the input signal.
n We showed before that a system represented by a linear differential equation with constant coefficients and
no initial conditions, or input, before t= 0 is LTI. Thus, one should be able to represent that system by a
convolution integral after finding its impulse response h(t).

nExample 2.9

Obtain the impulse response of a capacitor and use it to find its unit-step response by means of

the convolution integral. LetC=1 F.

Solution

For a capacitor with a initial voltagevc( 0 )=0, we have that

vc(t)=

1

C

∫t

0

i(τ)dτ

The impulse response of a capacitor is found by letting the inputi(t)=δ(t)and the outputvc(t)=

h(t), which according to the above equation becomes

h(t)=

1

C

∫t

0

δ(τ)dτ=

1

C

t> 0

and zero ift<0, orh(t)=( 1 /C)u(t). ForC= 1 F, to compute the unit-step response of the

capacitor we let the inputi(t)=u(t), andvc( 0 )=0. The voltage across the capacitor is

vc(t)=

∫∞

−∞

h(t−τ)i(τ)dτ=

∫∞

−∞

1

C

u(t−τ)u(τ)dτ

and since, as a function ofτ,u(t−τ)u(τ)=1 for 0≤τ≤tand zero otherwise, we have that

vc(t)=

∫t

0

dτ=t

fort≥0 and zero otherwise (as the input is zero fort<0 and there are no initial conditions),

orvc(t)=r(t). The above result makes physical sense since the capacitor is accumulating charge

and the input is providing a constant charge, so that the result is a ramp function. Notice that the

impulse response is the derivative of the unit-step response.