Analysis of LTI Systems 219

3.5 Analysis of LTI Systems 219

s) = 1/(s 2 ( s + 1)). There will be no steady state because of the double pole s = 0. On the other hand, X 2

s = 0; if the system input is x 1 ( t) = u(t) so that X 1 ( s) = 1/s, then Y 1 (

2 ( s) = s/(s + 2) will give

Y 2 ( s) = H(s)X 2 ( s) =

2 s + 2) = ( s + 1)(s + 2) 2

s(s + 1) (

which will give a zero steady state, even though the system is unstable. This is possible because of the pole-zero cancellation.

The steady-state response is the response of the system away from t = 0, and it can be found by letting t → ∞ (even though the steady state can be reached at finite times, depending on how fast the transient goes to zero). In Example 3.22, the steady-state response of h(t) = (e −t −e −2t ) u(t) is zero, while for s(t) = 0.5u(t) − e −t u(t) + 0.5e −2t u(t) it is 0.5. The transient responses are then h(t) − 0 = h(t) and s(t) − 0.5u(t) = −e −t u(t) + 0.5e −2t u(t). These transients eventually disappear.

The relation found between the impulse response h(t) and the unit-step response s(t) can be extended to more cases by the definition of the transfer function—that is, H(s) = Y(s)/X(s) so that the response Y(s) is connected with H(s) by Y(s) = H(s)X(s), giving the relation between y(t) and h(t). For instance, if x(t) = δ(t), then Y(s) = H(s) × 1, with inverse the impulse response. If x(t) = u(t), then Y(s) = H(s)/s is S(s), the Laplace transform of the unit-step response, and so s(t) = dh(t)/dt. And if x(t) = r(t), then

Y(s) = H(s)/s 2 is ρ(s), the Laplace transform of the ramp response, and so ρ(t) = d h(t)/dt 2 = ds(t)/dt.

■ Example 3.23

Consider again the second-order differential equation in the previous example,

d 2 y(t)

dy(t) dt 2 +3 dt + 2y(t) = x(t)

but now with initial conditions y(0) = 1 and dy(t)/dt| t=0 = 0, and x(t) = u(t). Find the complete response y(t). Could we find the impulse response h(t) from this response? How could we do it?

Solution

The Laplace transform of the differential equation gives

[s 2 Y(s) − sy(0) − dy(t)

] + 3[sY(s) − y(0)] + 2Y(s) = X(s) dt t=0

Y(s)(s 2 + 3s + 2) − (s + 3) = X(s) so we have that

X(s)

s+3

Y(s) =

s + 1)(s + 2)

( s + 1)(s + 2)

1 + 3s + s 2 B 1 B 2 B 3

s(s + 1)(s + 2)

s+1

s+2

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

after replacing X(s) = 1/s. We find that B 1 = 1/2, B 2 = 1, and B 3 = −1/2, so that the complete response is

(3.36) Again, we can check that this solution satisfies the initial condition y(0) and dy(0)/dt (this

y(t) = [0.5 + e −t − 0.5e −2t ]u(t)

is particularly interesting to see, try it!). The steady-state response is 0.5 and the transient [e −t − 0.5e −2t ]u(t).

According to Equation (3.36), the complete solution y(t) is composed of the zero-state response, due to the input only, and the response due to the initial conditions only or the zero-input response. Thus, the system considers two different inputs: One that is x(t) = u(t) and the other the initial conditions.

If we are able to find the transfer function H(s) = Y(s)/X(s) its inverse Laplace transform would be h(t). However that is not possible when the initial conditions are nonzero. As shown above, in the case of nonzero initial conditions, we get that the Laplace transform is

X(s)

I(s)

Y(s) =

A(s) + A(s)

where in this case A(s) = (s + 1)(s + 2) and I(s) = s + 3, and thus we cannot find the ratio Y(s)/X(s). If we make the second term zero (i.e., I(s) = 0), we then have that Y(s)/X(s) = H(s) = 1/A(s) and h(t) = e −t u(t) − e −2t u(t).

■ Example 3.24

Consider an analog averager represented by

y(t) =

where x(t) is the input and y(t) is the output. The derivative of y(t) gives the first-order differential equation

dy(t)

1 = [x(t) − x(t − T)]

dt

with a finite difference for the input. Let us find the impulse response of this analog averager.