Unformatted text preview: t he final value if it closely approaches the final value. A widely
accepted m easure of closeness is w ithin 2% o f t he final value. T he t ime required
for t he response to reach and s tay w ithin 2% o f t he final value is called t he s ettling
time t s.t In Fig. 6.36c, we find t s ' " 1 second. A good system has a small overshoot,
a small value o f t r a nd t s a nd a small steadystate error.
A large overshoot, as in t he p resent case, may be unacceptable in many applications. Let us t ry t o d etermine K ( the gain) which yields fastest response without
oscillations. C omplex c haracteristic roots lead to oscillations; to avoid oscillations,
t he c haracteristic roots should be real. In t he present case the characteristic polynomial is s2 + 8 s + K . For K > 16, t he c haracteristic roots are complex; for K < 16,
t he roots are real. T he fastest response without oscillations is o btained by choosing
K = 16. We n ow consider this case.
3. K = 16 + 8 s + 16) S(8 1
= ~ 8 G num=[O 0 K J;Gden=conv([O 1 OJ,[O 1 8 ]);
H num=[O 0 I J;Hden=[O 0 I J;
[ NumTF,DenTFJ=feedback(Gnum,Gden,Hnum,Hden);
s tep(NumTF,DenTF) (s + 4)2 This response a lso a ppears in Fig. 6.36c. T he s ystem with K > 16 is s aid t o be
u nderdamped (oscillatory response), whereas t he system with K < 16 is said to
be o verdamped. For K = 16, the system is said to be c ritically d amped.
T here is a tradeoff between undesirable overshoot and rise time. Reducing
overshoots leads t o higher rise time (sluggish system). In practice, a small overshoot
may b e a cceptable, which is still faster t han t he critical damping. Note t hat p ercent
overshoot P O a nd p eak time tp a re meaningless for the overdamped or critically
d amped cases. I n a ddition t o a djusting gain K , we may need t o a ugment t he s ystem
with some t ype o f c ompensator if t he specifications on overshoot a nd t he s peed of
response are t oo s tringent.
Ramp I nput I f t he a ntiaircraft g un in Fig. 6.36a is t racking a n enemy plane moving with a
uniform velocity, t he gunposition angle must increase linearly with t. Hence, t he
i nput in this c ase is a ramp; t hat is, Bi(t) = tu(t). Let us find the response of t he
s ystem t o t his i nput when K = 80. In this case, ei(S) =
a nd f" 80
s2(s2 + 88 + 80) 0.1 1
8 =  + 2 +
s 0.1(8  2) ,~'82
88 80 + + Use of Table 6.1 yields yet) = [ 0.1 C omputer E xample C 6.4 F ind t he s tep r esponse o f t he feedback s ystem in Fig. 6.36b with G (s) ==
1
if
K = 7, 16, a nd 80. F ind t he u nit r amp r esponse o f t his system for t he case K = 8~+B)
I n c omputer e xample C6.3, we h ave o btained t he t ransfer functions of t his feedback
system. Here, we s hall redo this p art a gain in a nother way t o i llustrate t he use of 'conv'
c ommand w hen t he d enominator o f G (s) is m ade u p o f two factors D I(S) a nd D2(S).
T he c ommand ' conv' multiplies D ,(S) w ith D2(S) a nd gives t he coefficients o f t he p roduct
D I(S)D2(S). 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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