aa168 Chapter 5 / Transient and Steady-State Response Analyses(3)Overdamped case(z>1): In this case, the two poles of C(s)/R(s)are negative
real and unequal. For a unit-step input,R(s)=1/sandC(s)can be written
(5–16)
The inverse Laplace transform of Equation (5–16) is
fort 0 (5–17)
where and Thus, the response c(t)
includes two decaying exponential terms.
When zis appreciably greater than unity, one of the two decaying exponentials
decreases much faster than the other, so the faster-decaying exponential term (which
corresponds to a smaller time constant) may be neglected. That is, if –s 2 is located very
much closer to the jvaxis than –s 1 Awhich means @s 2 @@s 1 @B, then for an approximate
solution we may neglect –s 1. This is permissible because the effect of –s 1 on the response
is much smaller than that of –s 2 , since the term involving s 1 in Equation (5–17) decays
much faster than the term involving s 2. Once the faster-decaying exponential term has
disappeared, the response is similar to that of a first-order system, and C(s)/R(s)may
be approximated by
This approximate form is a direct consequence of the fact that the initial values and
final values of both the original C(s)/R(s)and the approximate one agree with each
other.
With the approximate transfer function C(s)/R(s), the unit-step response can be
obtained as
The time response c(t)is then
fort 0
This gives an approximate unit-step response when one of the poles of C(s)/R(s)can
be neglected.
c(t)= 1 - e-Az-^2 z
(^2) - 1 Bvn t
,
C(s)=
zvn-vn 2 z^2 - 1
As+zvn-vn 2 z^2 - 1 Bs
C(s)
R(s)
=
zvn-vn 2 z^2 - 1
s+zvn-vn 2 z^2 - 1
=
s 2
s+s 2
s 1 =Az+ 2 z^2 - 1 Bvn s 2 =Az- 2 z^2 - 1 Bvn.
= 1 +
vn
22 z^2 - 1
ae-s^1 t
s 1
-
e-s^2 t
s 2
b,
-
1
22 z^2 - 1 Az- 2 z^2 - 1 B
e-Az-^2 z
(^2) - 1 Bvnt
c(t)= 1 +
1
22 z^2 - 1 Az+ 2 z^2 - 1 B
e-Az+^2 z
(^2) - 1 Bvnt
C(s)=
v^2 n
As+zvn+vn 2 z^2 - 1 BAs+zvn-vn 2 z^2 - 1 Bs
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