Signals and Systems - Electrical Engineering

3.5 Analysis of LTI Systems 223

withy( 0 )the initial voltage in the capacitor andi(t)the current through the resistor, inductor, and
capacitor. The above two equations are calledintegro-differentialgiven that they are composed of
an integral equation and a differential equation. To obtain a differential equation in terms of the
inputx(t)and the outputy(t), we find the first and second derivative ofy(t), which gives

dy(t)

dt

=

1

C

i(t) ⇒ i(t)=C

dy(t)

dt

d^2 y(t)

dt^2

=

1

C

di(t)

dt

⇒L

di(t)

dt

=LC

d^2 y(t)

dt^2

which when replaced in the KVL equation gives

x(t)=RC

dy(t)

dt

+LC

d^2 y(t)

dt^2

+y(t) (3.40)

which, as expected, is a second-order differential equation with two initial conditions:y( 0 ), the
initial voltage in the capacitor, andi( 0 )=Cdy(t)/dt|t= 0 , the initial current in the inductor. To find
the impulse response of this circuit, we letx(t)=δ(t)and the initial conditions be zero. The Laplace
transform of Equation (3.40) gives

X(s)=[LCs^2 +RCs+1]Y(s)

The impulse response of the system is the inverse Laplace transform of the transfer function

H(s)=

Y(s)

X(s)

=

1 /LC

s^2 +(R/L)s+ 1 /LC

IfLC=1 andR/L=2, then the transfer function is

H(s)=

1

(s+ 1 )^2

which corresponds to the impulse response

h(t)=te−tu(t)

Now that we have the impulse response of the system, suppose then the input is a unit-step signal,
x(t)=u(t). To find its response we consider the convolution integral

y(t)=

∫∞

−∞

x(τ)h(t−τ)dτ

=

∫∞

−∞

u(τ)(t−τ)e−(t−τ)u(t−τ)dτ

=

∫t

0

(t−τ)e−(t−τ)dτ

=[1−e−t( 1 +t)]u(t)