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1、 Solutions# SolutionsSolutionsP3.1 Tlie unit step response of a ceitaiii system is given by c(f)= i+k-严，tn。(a) Detenniiie the impulse response of the system(b) Detenniiie the transfer fimction C(s)/R(s) oftlie systemSolution: Tlie impulse response is the differential of coirespondiiig step response,
2、 i.e.g =豊2 =灭）-戶+茲fAs we know that the transfer fimction is the L叩lace transfbmi of conespondiiig impulse response, i.e.# Solutions# SolutionsP3.2 Consider the system desci ibed by the block diagram shown ill Fig. P3.2(a). Detenniiie tlie polarities of two feedbacks for each of the following step re
3、sponses shown in Fig. P3.2(b), where “(T indicates tliat tlie feedback is open.(a) Block diagramW)Parabolic(b) Unit-step resp onsesFigure P32Solution: In general we haveC(s) _2R(s) s2 土 k=s 土 k出o 2 o 1 2Note that tlie chai acteristic polynomial isA(s) = s2 kskxk0.0.wiiere the sign of k2s is depended
4、 on the outer feedback and the sign of kxk2 is depended on the inter feedback.Case (1). Tlie response presents a sinusoidal. It means that tlie system has a pair of pure iinagiiiaiy roots, i.e. the chaiacteristic polynomial is in tlie fonn of 力(s) = s2 + kxk2 Obviously, the outlet feedback is - and
5、tlie iimer feedback is 0：Case (2). Hie response presents a diveied oscillation. Tlie system has a pair of complex coiijugate roots witli positive real parts, i.e. tlie characteristic polynomial is in the fonn of (s) = s? - s + 上他.Obviously, the outlet feedback isand tlie inner feedback is 一二Case (3)
6、. Hie response presents a converged oscillation. It means that tlie system has a pair of complex coiijugate roots witli negative real parts, i.e. the characteristic polynomial is in the fonn of 4(s) = s2 + k2s + Obviously, both the outlet and inner feedbacks are “ 一二Case (4). In fact tliis is a ramp
7、 response of a fnst-order system Hence、the outlet feedback is CT to prochice a ramp signal and tlie iimer feedback is “ -二Case (5) Consideiiiig that a parabolic fimction is the integral of a ramp fimction, both the outlet and inner feedbacks are (F.P3.3 Consider each of tlie following closed-loop tr
8、ansfer fimction. By consideiiiig the location of tlie poles on tlie complex plane, sketch tlie unit step response, explaining tlie results obtained.20金 2 +12金 + 206s + +lls + 6(d)(s) =12.53 +2s + 5)(s + 5)Solution: (a) (s)= =+12S + 20(s + 2)(s + 10)By inspection, the characteristic roots ai e -2, -1
9、0. Tliis is an overdamped second-order system. Tlierefbre, consideruig tliat the closed-loop gain is = , its unit step response can be sketched as shown.(b) (s)= ；=八 +6“ +11S + 6 (s + l)(s + 2)(s + 3)By inspection, the characteristic roots aie 一1 , 一2 , -3 Obviously, all tliree transient components
10、are decayed exponential tenns. Hierefore, its unit step response, witli a closed-loop gain J = 1, is sketched as shown.4(S + l)2 +1Tliis is an underdamped second-order system, because its characteristic roots are -1 j Hence, baiisient component is adecayed sinusoid. Noting that tlie closed-loop gain
11、 is % = 2、tlie unit step response can be sketched as shown12.5_12.53 +2s + 5)(s + 5)(s+12 +2?(s + 5)By inspection, the characteristic roots are -1 j2 , 一5 Since(d) 0(s)|- 0.1 5|, tliere is a pair of dominant poles, -1 j2 , for this system. The unit step response, with a closed-loop gain = 0.5、 is sk
12、etdied as shown.P3.4 Tlie open-loop transfer fimction of a unity negative feedback system isG(s) =s(s +1)Detenuiiie tlie rise time, peak time, percent overshoot and setting time (using a 5% setting criterion).Solution: Writing he closed-loop transfer fimctions? +2阿$+此we get = 1, (; = 0.5 . Since thi
13、s is an underdainped second-order system with $ = 0.5 , the system peiionnaiice can be estimated as followsRising timen 一 arccosq 叫Jl*龙 一 aiccos05 2.42 sec.# Solutions# SolutionsPeak time tp = =-= 3.62 sec (Dn Ji* 1 a/1-0.52Percent oveishoot ap =严 W xlOO% = e0-51-05 xlOO%16.3%Setting time t = 6 sec.
14、 (using a 5% setting aiterion)ga)n 0.5x1P35 A second-order system gives a unit step response shown ill Fig. P35 Find the open-loop transfer fimction if the system is a tuiit negative-feedback system.Solution: By inspection we have5 = 131x100% = 30%p 1Figure P3.5Solving tlie fbnnula for calculating t
15、lie overshoot,=e 陷r = 0.3 , we have 0362+ lir 6Since tp = lsec., solving the fonnula for calculating the peak time, tp =-=, we geta)n = 33.7 rad / secHence, the open-loop transfer fiinction is塚 1135.7G(s)=s(s + 2阿)s(s + 24.4)P36 A feedback system is shown in Fig. P3.6(a and its unit step response cuive is sliowii in Fig. P36(b). Determine the values of kx, 込，and a.Solution: Tlie transfer fiinction between tlie input and output is given by %)% k2R(s) s? + +匕
