Analysis of LTI Systems 223

3.5 Analysis of LTI Systems 223

with y(0) the initial voltage in the capacitor and i(t) the current through the resistor, inductor, and capacitor. The above two equations are called integro-differential given that they are composed of an integral equation and a differential equation. To obtain a differential equation in terms of the input x(t) and the output y(t), we find the first and second derivative of y(t), which gives

dy(t)

1 dy(t)

= i(t) ⇒ i(t) = C

dt

C dt

d 2 y(t)

1 di(t)

di(t)

d 2 y(t)

which when replaced in the KVL equation gives

dy(t)

d 2 y(t)

(3.40) which, as expected, is a second-order differential equation with two initial conditions: y(0), the

x(t) = RC

dt + LC dt 2 + y(t)

initial voltage in the capacitor, and i(0) = Cdy(t)/dt| t=0 , the initial current in the inductor. To find the impulse response of this circuit, we let x(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

Y(s)

1/LC

H(s) =

X(s) = s 2 + (R/L)s + 1/LC

If LC = 1 and R/L = 2, then the transfer function is

1 H(s) =

( s + 1) 2

which corresponds to the impulse response

h(t) = te −t u(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

Z t = ( t − τ )e −(t−τ ) dτ

= [1 − e −t ( 1 + t)]u(t)

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

which satisfies the initial conditions and attempts to follow the input signal. This is the unit-step response.

In the Laplace domain, the above can be easily computed as follows. From the transfer function, we have that

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

s + 1) 2 s

where we replaced the transfer function and the Laplace transform of x(t) = u(t). The partial fraction expansion of Y(s) is then

Y(s) =

s+1

( s + 1) 2

and after obtaining that A = 1, C = −1, and B = −1, we get

y(t) = s(t) = u(t) − e −t u(t) − te −t u(t)

which coincides with the solution of the convolution integral. It has been obtained, however, in a much easier way.

■ Example 3.27

Consider the positive feedback system created by a microphone close to a set of speakers that are putting out an amplified acoustic signal (see Figure 3.18), which we considered in Example 2.18 in Chapter 2. Find the impulse response of the system using the Laplace transform, and use it to express the output in terms of a convolution. Determine the transfer function and show that the system is not BIBO stable. For simplicity, let β = 1, τ = 1, and x(t) = u(t). Connect the location of the poles of the transfer function with the unstable behavior of the system.

Solution

As we indicated in Example 2.18 in Chapter 2, the impulse response of a feedback system cannot

be explicitly obtained in the time domain, but it can be done using the Laplace transform. The input–output equation for the positive feedback is

y(t) = x(t) + βy(t − τ )

Positive feedback created by closeness of a microphone to a set of speakers.