Case studies for CT systems - Continuous and Discrete Time Signals and Systems - page 381

Impulse response Since
H ( s ) = 1
s

′
m
s 2 + 2 ξ n ω n s + ω n

k

,

H ′ ( s )

the impulse response of the dc motor equals the integral of the impulse response
h

′ ′ ′

( s ) is similar
to the transfer function of the spring damping system, Eqs. (8.20)-(8.27) are
used to derive the impulse response h ( t ) of the dc motor. Depending upon the
value of ξ n , we consider three different cases.

( t ), the inverse Laplace transform of H ( s ). Since the form of H

Case 1 ( ξ n = 1) As derived in Eq. (8.21), the inverse Laplace transform of
H

′

( s ) for ξ n = 1isgivenby

′
m
s 2 + 2 ξ n ω n s + ω n

k ←→

′
m t e

−ω n t u ( t )

.

k

′

Taking the integral of h ( t ) yields

t
ω n

+ 1
ω n

′ −ω n t d t ′

m

−ω n t

h ( t ) = k

m t e

=− k e + C for t ≥ 0 . (8.42)

′

Case 2 ( ξ n > 1) Equation (8.24) derives the inverse Laplace transform of H ( s )

for ξ n > 1 as follows:
k

e ω n

√ √ ′

m

′
m
s 2 + 2 ξ n ω n s + ω n

k

←→

ξ n − 1 t ξ n − 1 t −ξ n ω n t −ω n

− 1 e

− e u ( t ) .

2 ω n ξ n

The impulse response of the dc motor is given by

d t

√ √

k ′
m

−ξ n ω n t

e ω n

ξ n − 1 t −ω n ξ n − 1 t

h ( t ) = e − e

2 ω n ξ n − 1

√ √

′ −ξ n ω n t −ω n ξ n − 1 t

e ω n

ξ n − 1 t

k

m e

e

= −

ξ n ξ n ξ n

2 ω n − 1 ξ n ω n + ω n − 1 ξ n ω n − ω n − 1

+ C for t ≥ 0 . (8.43)

′

Case 3 ( ξ n < 1) Equation (8.27) derives the inverse Laplace transform of H ( s )

for ξ n < 1 as follows:
k

′
m

′
m
s 2 + 2 ξ n ω n s + ω n

k ←→

−ξ n ω n t sin

e ω n 1 − ξ n t u ( t ) .

ω n 1 − ξ n

The impulse response of the dc motor is given by

′
m

k

−ξ n ω n t sin

h ( t ) = e ω n 1 − ξ n t d t

ω n

1 − ξ n

′ −ξ n ω n t

=− k

m e

1 − ξ n cos ω n 1 − ξ n t

ω n 1 − ξ n

+ ξ n sin ω n 1 − ξ n t + C for t ≥ 0 . (8.44)