Signals and Systems - Electrical Engineering

3.5 Analysis of LTI Systems 219

s= 0 ; if the system input is x 1 (t)=u(t)so that X 1 (s)= 1 /s, then Y 1 (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)=s/(s+ 2 )^2 will give

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

1

s(s+ 1 )

s

(s+ 2 )^2

=

1

(s+ 1 )(s+ 2 )^2

which will give a zero steady state, even though the system is unstable. This is possible because of the
pole-zero cancellation.
n 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−^2 t)u(t)is zero, while for
s(t)=0.5u(t)−e−tu(t)+0.5e−^2 tu(t)it is0.5. The transient responses are then h(t)− 0 =h(t)and
s(t)−0.5u(t)=−e−tu(t)+0.5e−^2 tu(t). These transients eventually disappear.
n 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 isρ(s), the Laplace transform of the ramp response, and soρ(t)=d^2 h(t)/dt^2 =ds(t)/dt.

nExample 3.23

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

d^2 y(t)

dt^2

+ 3

dy(t)

dt

+ 2 y(t)=x(t)

but now with initial conditionsy( 0 )=1 anddy(t)/dt|t= 0 =0, andx(t)=u(t). Find the complete

responsey(t). Could we find the impulse responseh(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)

dt

∣

∣t= 0 ]+3[sY(s)−y( 0 )]+ 2 Y(s)=X(s)

Y(s)(s^2 + 3 s+ 2 )−(s+ 3 )=X(s)

so we have that

Y(s)=

X(s)

(s+ 1 )(s+ 2 )

+

s+ 3

(s+ 1 )(s+ 2 )

=

1 + 3 s+s^2

s(s+ 1 )(s+ 2 )

=

B 1

s

+

B 2

s+ 1

+

B 3

s+ 2