LTI Continuous-Time Systems 155

2.3 LTI Continuous-Time Systems 155

Delay τ 2 × FIGURE 2.17 Echo system with two paths. α 2

where x(t) is the input, and α i ,τ i >

0, for i = 1 and 2, are attenuation factors and delays. Thus, the output is the superposition of attenuated and delayed versions of the input. Typically, the attenuation factors are less than unity. Is this system causal and BIBO stable?

Solution

Since the output depends only on past values of the input, the echo system is causal. To determine if the system is BIBO stable we consider a bounded input signal x(t), and determine if the output is bounded. Suppose x(t) is bounded by a finite value M, or |x(t)| < M < ∞, for all times, which means that the value of x(t) cannot exceed an envelope [−M, M] at all times. This would also hold when we shift x(t) in time, so that

| y(t)| ≤ |α 1 || x(t − τ 1 ) | + |α 2 || x(t − τ 2 ) |< [|α 1 | + |α 2 | ]M

so the corresponding output is bounded. The system is BIBO stable. We can also find the impulse response h(t) of the echo system, and show that it satisfies the abso-

lutely integrable condition of BIBO stability. Indeed, if we let the input of the echo system be x(t) = δ(t) the output is

y(t) = h(t) = α 1 δ( t−τ 1 ) +α 2 δ( t−τ 2 )

and the integral is Z ∞

| h(t)|dt = |α 1 |

■ Example 2.18

Consider a positive feedback system created by a microphone close to a set of speakers that are putting out an amplified acoustic signal (see Figure 2.18). The microphone picks up the input signal x(t) as well as the amplified and delayed signal βy(t − τ ), |β| ≥ 1. Find the equation that connects the input x(t) and the output y(t) and recursively from it obtain an expression for y(t) in terms of past values of the input. Determine if the system is BIBO stable or not—use x(t) = u(t), β =

2, and τ = 1 in doing so.

C H A P T E R 2: Continuous-Time Systems

x (t)

y (t)

FIGURE 2.18 Positive feedback system: the microphone picks

× up input signal

Delay τ

βy x(t) and the amplified and (t − τ )

delayed signal β y(t − τ ), making the system unstable.

Solution

The input–output equation is

y(t) = x(t) + βy(t − τ )

If we use this expression to obtain y(t − τ ), we get that

y(t − τ ) = x(t − τ ) + βy(t − 2τ )

and replacing it in the input–output equation, we get y(t) = x(t) + β[x(t − τ ) + βy(t − 2τ )] = x(t) + βx(t − τ ) + β 2 y(t − 2τ )

Repeating the above scheme, we will obtain the following expression for y(t) in terms of the input y(t) = x(t) + βx(t − τ ) + β 2 x(t − 2τ ) + β 3 x(t − 3τ ) + · · · If we let x(t) = u(t) and β = 2, the corresponding output is y(t) = u(t) + 2u(t − 1) + 4u(t − 2) + 8u(t − 3) + · · · which continuously grows as time increases. The output is clearly not a bounded signal, although

the input is bounded. Thus, the system is unstable, and the screeching sound from the speakers will prove it—you need to separate the speakers and the microphone to avoid it.