Signal Processing and Linear Systems-B.P.Lathi copy

Hence once a signal is applied no m atter how small

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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 steady-state 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 trade-off 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 gun-position 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.

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