Case studies for CT systems - Continuous and Discrete Time Signals and Systems - page 376

Case 1 ( ξ n = 1) For ξ n = 1, Eq. (8.17) reduces to

1 / M
s 2 + 2 ω n s + ω n

1 / M
( s + ω n ) 2 ,

H ( s ) =

=

(8.20)

=−ω n , −ω n .Taking the inverse transform, the impulse

with repeated roots at s

response is given by

1
M t e

−ω n t u ( t ) .

h ( t ) = (8.21)

Case 2 ( ξ n > 1) For ξ n > 1, the poles p 1 and p 2 of the spring damping system
are real-valued and given by

p 1 =−ξ n ω n + ω n ξ n − 1 and p 2 =−ξ n ω n − ω n ξ n − 1 . (8.22)

The transfer function of the spring damping system can be expressed as follows:

1 / M
s 2 + 2 ξ n ω n s + ω n

1 / M

H ( s ) =

=

p 2 ) ,

(8.23)

( s − p 1 )( s −

which leads to the impulse response

1
M

1
( p 1 −

1

p 2 ) [e p 1 t

− e p 2 t ] u ( t ) =

−ξ n ω n t

h ( t ) =

− 1 e

2 ω n M ξ n

e ω n

√ √

ξ n − 1 t − e −ω n ξ n − 1 t

u ( t ) . (8.24)

Case 3 ( ξ n < 1 )For ξ n > 1, the poles p 1 and p 2 of the spring damping system
are complex and are given by

1 − ξ n .

p 1 =−ξ n ω n + j ω n 1 − ξ n and p 2 =−ξ n ω n − j ω n (8.25)

By repeating the procedure for Case 2, the impulse response of the spring
damping system is given by

1 / M
s 2 + 2 ξ n ω n s + ω n

1 / M

H ( s ) = =

p 2 ) ,

(8.26)

( s − p 1 )( s −

which leads to the impulse response

1

−ξ n ω n t sin

h ( t ) = e ω n 1 − ξ n t u ( t ) . (8.27)

ω n M

1 − ξ n

Figure 8.7 shows the impulse response of the spring damping system for the
three cases considered earlier. We set M

= 10 and ω n = 0 . 3 radians/s in each

case. For Case 1 with ξ n = 1, the impulse response decreases monotonically

approaching the steady state value of zero. Such systems are referred to as
critically damped systems.
For Case 2 with ξ n

= 4, the impulse response of the spring damping system
again approaches the steady state value of zero. Initially, the deviation from the
steady state value is smaller than that of the critically damped system, but the