Signals and Systems - Electrical Engineering

222 C H A P T E R 3: The Laplace Transform

Its inverse is

y(t)=r(t)−r(t− 1 )

wherer(t)is the ramp signal. This result coincides with the one obtained graphically in

Example 2.12 in Chapter 2.

n In the second case,X(s)=H(s)=L[u(t)−u(t− 1 )]=( 1 −e−s)/s, so that

Y(s)=H(s)X(s)=

( 1 −e−s)^2

s^2

=

1 − 2 e−s+e−^2 s

s^2

which corresponds to

y(t)=r(t)− 2 r(t− 1 )+r(t− 2 )

or a triangular pulse as we obtained graphically in Example 2.13 in Chapter 2. n

nExample 3.26

To illustrate the significance of the Laplace approach in computing the output of an LTI system by

means of the convolution integral, consider an RLC circuit in series with input a voltage source

x(t)and as output the voltagey(t)across the capacitor (see Figure 3.17). Find its impulse response

h(t)and its unit-step responses(t). LetLC=1 andR/L=2.

Solution

The RLC circuit is represented by a second-order differential equation given that the inductor and

the capacitor are capable of storing energy and their initial conditions are not dependent on each

other. To obtain the differential equation we apply Kirchhoff’s voltage law (KVL)

x(t)=Ri(t)+L

di(t)

dt

+y(t)

wherei(t)is the current through the resistor, inductor and capacitor and where the voltage across

the capacitor is given by

y(t)=

1

C

∫t

0

i(σ)dσ+y( 0 )

FIGURE 3.17

RLC circuit with input a voltage source

x(t)and output the voltage across the

capacitory(t).

−

+

x(t)

RL

C y(t)

+

−