082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

21.11 Transfer functions 667
u a (t)
K = 5
K = 0.5
K = 0.375
t
Figure21.17
Time response forvariousvalues ofK.
This is shown in Figure 21.17 and is termed acritically damped response. It corresponds
tothe fastestrisetime of the systemwithoutovershooting.
ForK=5
a
(s)
d
(s) = 2.5
s 2 +s+2.5
2.5
a
(s) =
s(s 2 +s +2.5)
Rearranging toenable standard formstobe inverted gives
a
(s) = 1 s −
s+0.5
(s +0.5) 2 +1.5 − 0.5
2 (s +0.5 2 ) +1.5 2
θ a
(t) = 1 −e −0.5t cos1.5t − 1 3 e−0.5t sin1.5t
The trigonometric terms can be expressed as a single sinusoid using the techniques
given inSection 3.7.Thus,
θ a
(t) = 1 −1.054e −0.5t sin(1.5t +1.249) for t 0
ThisisshowninFigure21.17andistermedanunderdampedresponse.Thesystem
overshoots its final value.
Inapracticalsystemitiscommontodesignforsomeovershoot,provideditisnot
excessive,asthisenablesthedesiredvaluetobereachedmorequickly.Itisinteresting
tocomparethesystemresponseforthethreecaseswiththenatureoftheirrespective
transfer function poles. For the overdamped case the poles are real and unequal, for
the critically damped case the poles are real and equal, and for the underdamped
casethepolesarecomplex.Engineersrelyheavilyonpolepositionswhendesigning
a system to have a particular response. By varying the value of K it is possible to
obtain a range of system responses and corresponding pole positions.
EXERCISES21.11
1 Find the transfer functionforeachofthe following
equationsassuming zeroinitialconditions:
(a) x ′′ +x=f(t)
(b) 2 d2 x
dt 2 + dx
dt −x=f(t)
(d)
(c) 2 d2 y
dt 2 +3dy +6y =p(t)
dt
3 d3 y y
dt 3 +6d2 dt 2 +8dy +4y =g(t)
dt
(e) 6 d2 y
dt 2 +8dy dt +4y=3df dt +4f,
f isthe systeminput
(f) 6x ′′′ +2x ′′ −x ′ +4x = 4f ′′ +2f ′ +6f,fisthe
systeminput