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I f we desire t o change the angular position of the object instantaneously, we
need to apply a step input. We m ay t hen w ant to know how long t he system
takes t o position itself a t t he desired angle, whether it reaches the desired angle,
and whether it reaches the desired position smoothly (monotonically) or oscillates
a bout t he final position. I f t he system oscillates, we m ay want to know how long i t
takes for the oscillations to settle down. All these questions can be readily answered
by finding the o utput ()o(t) when the input ()i(t) = u(t). A s tep input implies the
instantaneous change in the angle. This i nput would be one of the most difficult 6 430 C ontinuousTime S ystem Analysis Using the Laplace Transform
Output
Potentiometer 6.7 Application to Feedback a nd C ontrol S tate E quations 431 t o follow; if t he s ystem c an perform well for this i nput, i t is likely to give a good
account of itself under most o ther e xpected situations. This is t he reason why we.
t est control systems for a s tep i nput,
For the step i nput Bi(t) = u (t), 8 i (s) = l /s a nd
1
K G(s)
8 0 (s) = ; T(s) = s[1 + KG(8)] L et t he m otor (with load) transfer function relating t he load angle Bo(t) t o t he
motor i nput voltage b e G(s) = s(s~8) [see Eq. (1.65)]. This yields (a) K
S (S2 Motor and
load Amplifier i ,. [ " K ,. G {s) I ,. + 8s + K ) L et us investigate the system behavior for t hree different values o f g ain K . 90 1. K = 7 I 7 ( b) 7 8 0 (8) = 8(8 2 + 88 + 7) 8(8 + 1)(8 + 7) 1
1
= _ __I _+_6_
6
s
s +1
s +7 a nd Bo(t) = ( 1 ~et + ~e7t) u(t) T his response, illustrated in Fig. 6.36c, a ppears r ather sluggish. To speed u p t he
response let us increase t he g ain to, say, 80.
2 . K = 80
( c) 80 8 0 (8) = 8(s2 + 88 + 80) 8(S +4  80
j8)(8 + 4 + j 8) ( 4 2 a nd O'l· Bo(t) = [1 ///'~/"':?::: (d) ( F ig. 6 .36 (a) An automatic position control system (b) its block diagram (c) the unit
step response (d) t he unit ramp response. + :.:{e 4t cos (8t + 153°)] u(t) T his response, depicted in Fig. 6.36c, is c ertainly faster t han in t he earlier case
( K = 7), b ut u nfortunately t he i mprovement is achieved a t t he cost of ringing
(oscillations) with high overshoot. In t he p resent case t he p ercent o vershoot P O
is 21%. T he response reaches its peak value a t p eak t ime t p = 0.393 seconds.
T he r ise t ime, defined as t he t ime required for t he response to rise form 10% t o
90% o f i ts s teadystate value, indicates t he s peed o f response. t I n t he p resent case
t r = 0.175 seconds. T he s teadystate value o f t he response is u nity so t hat t he
s teadystate e rror is zero. Theoretically it takes infinite time for t he response to
t Delay t ime t d, defined as t he t ime required for t he r esponse t o r each 50% o f i ts s teadystate
value, is a nother i ndication o f speed. For t he p resent case, t d = 0.141 seconds 432 6 ContinuousTime System Analysis Using t he Laplace Transform reach t he d esired value of unity. In practice, however, we may consider t he response
t o have settled t o...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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