Computation of the Convolution Integral

3.5.2 Computation of the Convolution Integral

From the point of view of signal processing, the convolution property is the most important application of the Laplace transform to systems. The computation of the convolution integral is difficult even for simple signals. In Chapter 2 we showed how to obtain the convolution integral analytically as well as graphically. As we will see in this section, it is not only that the convolution property of the Laplace transform gives an efficient solution to the computation of the convolution integral, but that it introduces an important representation of LTI systems, namely the transfer func- tion of the system. A system, like signals, is thus represented by the poles and zeros of the transfer function. But it is not only the pole-zero characterization of the system that can be obtained from the transfer function. The system’s impulse response is uniquely obtained from the poles and zeros of the transfer function and the corresponding region of convergence. The way the system responds to different frequencies will be also given by the transfer function. Stability and causality of the system can be equally related to the transfer function. Design of filters depends on the transfer function.

The Laplace transform of the convolution y(t) = [x ∗ h](t) is given by the product

(3.38) where X(s) = L[x(t)] and H(s) = L[h(t)]. The transfer function of the system H(s) is defined as

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

Y(s)

H(s) = L[h(t)] =

X(s)

H(s) transfers the Laplace transform X(s) of the input into the Laplace transform of the output Y(s). Once Y(s) is found, y(t) is computed by means of the inverse Laplace transform.

■ Example 3.25

Use the Laplace transform to find the convolution y(t) = [x ∗ h](t) when (1) the input is x(t) = u(t) and the impulse response is a pulse h(t) = u(t) − u(t − 1), and

(2) the input and the impulse response of the system are x(t) = h(t) = u(t) − u(t − 1).

Solution

The Laplace transforms are X(s) = L[u(t)] = 1/s and H(s) = L[h(t)] = (1 − e −s )/ s, so that

1−e −s Y(s) = H(s)X(s) = s 2

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

Its inverse is

y(t) = r(t) − r(t − 1)

where r(t) is the ramp signal. This result coincides with the one obtained graphically in Example 2.12 in Chapter 2.

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

−s ) 1−e 2

1 − 2e −s +e −2s

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

which corresponds to

y(t) = r(t) − 2r(t − 1) + r(t − 2)

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

■ Example 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 voltage y(t) across the capacitor (see Figure 3.17). Find its impulse response h(t) and its unit-step response s(t). Let LC = 1 and R/L = 2.