Signals and Systems - Electrical Engineering

3.5 Analysis of LTI Systems 225

If we letx(t)=δ(t), the output isy(t)=h(t)or

h(t)=δ(t)+βh(t−τ)

and ifH(s)=L[h(t)], then the Laplace transform of the above equation isH(s)= 1 +βH(s)e−sτor
solving forH(s):

H(s)=

1

1 −βe−sτ

=

1

1 −e−s

=

∑∞

k= 0

e−sk= 1 +e−s+e−^2 s+e−^3 s+···

after replacing the given values forβandτ. The impulse responseh(t)is the inverse Laplace
transform ofH(s)or

h(t)=δ(t)+δ(t− 1 )+δ(t− 2 )+···=

∑∞

k= 0

δ(t−k)

Ifx(t)is the input, the output is given by the convolution integral

y(t)=

∫∞

−∞

x(t−τ)h(τ)dτ=

∫∞

−∞

∑∞

k= 0

δ(τ−k)x(t−τ)dτ

=

∑∞

k= 0

∫∞

−∞

δ(τ−k)x(t−τ)dτ=

∑∞

k= 0

x(t−k)

and replacingx(t)=u(t), we get

y(t)=

∑∞

k= 0

u(t−k)

which tends to infinity astincreases.

For this system to be BIBO stable, the impulse responseh(t)must be absolutely integrable, which
is not the case for this system. Indeed,

∫∞

−∞

|h(t)|dt=

∫∞

−∞

∑∞

k= 0

δ(t−k)dt

=

∑∞

k= 0

∫∞

−∞

δ(t−k)dt=

∑∞

k= 0

1 →∞

The poles ofH(s)are the roots of 1−e−s=0, which are the values ofssuch thate−sk= 1 =ej^2 πkor
sk=±j 2 πk. That is, there is an infinite number of poles on thejaxis, indicating that the system
is not BIBO stable. n