732 Appendix E: Previous EEE480 Exams. Rules: One sheet permitted, calculators permitted. GWC 352,

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1 732 Aedix E: Previous EEE0 Exams EEE0 Exam 2, Srig 2008 A.A. Rodriguez Rules: Oe 8. sheet ermitted, calculators ermitted. GWC 32, Problem Aalysis of a Feedback System Cosider the feedback system defied by the oe loo trasfer fuctio L def s + 2 ] 0 = PK = 3 ss s s + 0 ] 2 s ] ] ] s s s a Determie the aroximate closed loo trasfer fuctio T ry ad the associated domiat oles. b What is the associated time costat τ, settlig time t s, udamed atural frequecy ω, damig factor ζ, damed atural frequecy ω d, overshoot M to a ste commad ad associated time to eak t? c Aroximate the uit ste resose y. d Exlai why the overshoot will be much larger tha M ad how it ca be reduced to achieve M? Problem 2 Root Locus Costruct a root locus for the egative feedback system defied by L above. Determie the umber of asymtotes, the agle of each asymtote, the ceter of gravity, ad all imagiary crossovers. Carefully label all imortat features o your lot. Determie the uward gai margi GM ad dowward gai margi GM. Provide a comlete stability summary. Home: Redo this roblem with def s + 2 ] 0 L home = PK = 3 ss s s + 0 ] s ] ] ] s s s Problem 3 Bode Sketch Bode magitude ad hase lots for the oe loo trasfer fuctio L give above. Idetify all hase crossover frequecies o your lot. How do these relate to your root locus lot? Idetify the uward gai margi GM, dowward gai margi GM, ad hase margi PM o your lot. Comute the hase margi PM ad delay margi DM? Home: Redo this roblem with L home give above. Problem 4 Nyquist Sketch a Nyquist lot for the oe loo trasfer fuctio L give above. Idetify all hase crossover frequecies o your lot. How do these relate to your root locus lot? Idetify the uward gai margi GM, dowward gai margi GM, ad hase margi PM o your lot. Provide a comlete stability summary. Home: Redo this roblem with L home give above. Problem Desig for BW ad PM Suose that we have a lat P = s s+ ] e s with omial values = 0 ad = π 00. a Desig a feedback cotrol system that is as simle as ossible such that the closed loo system is stable, 2 exhibits costat steady state error to ram iut disturbaces d i, 3 exhibits a uity gai crossover frequecy of ω g = rad/sec, 4 exhibits a hase margi of PM = 0. Describe your desig rocess. Suort your desig with a root locus lot. b How would you reduce the overshoot due to a ste referece commad? c How small a value for ca be tolerated by your desig structure? How would you modify your desig?

2 Aedix E: Previous EEE0 Exams 733 Problem Aalysis of a Feedback System: Partial Aswer a We ote that L def s + 2 ] = PK 3 ss Notig that there is o re-filer W =, it follows that T ry = PK + PK 3s + 2 ss + 3s + 2 = 3s + 2 s 2 + 2s + 2 The aroximate domiat closed loo oles are s = ± j 3 4. E.30 E.302 b From this, we have τ = Re ole = E.303 t s = τ = E.304 ω = Re ole 2 + Im ole 2 = E.30 4 Re ole θ = ta = ta = 3 E.30 Im ole 3/4 ζ = si θ = 0.8 E.307 ω d = ω ζ2 = Im ole = 3 4 t = π Im ole = 4π 3 E.308 E.309 M = e ζω t = e π 3 = e 4π 3 = 0.0 E.30 It must be emhasized that T ry is NOT i stadard secod order form. Cosequetly, t ad M will NOT be achieved without a commad re-filter. That is, a commad re-filter is eeded so that T ry is i stadard secod order form. See art d below. c To determie the ste resose, we roceed as follows: From this, we obtai where Y = A s + B s + j E.3 y = A + 2 B e t cos0.7t + B E.32 A = T ry 0 = E.33 B = lim s + j0.7y s. s +j0.7 E.34 d It is imortat to ote that the overshoot will be very large because of the umerator term i T ry. The s i the umerator term s+ 2 differetiates the ste. This results i a imulsive-like sigal ad cosiderable outut overshoot. A commad re-filter is required to achieve t ad M. W = 2 s + 2 E.3

3 734 Aedix E: Previous EEE0 Exams Problem 2 Root Locus: Partial Aswer You should begi by costructig the hase of L def s + 2 ] 0 = PK = 3 ss s s + 0 ] 2 s + 0 This ca give isight ito the imagiary crossovers o the root locus. At low frequecies, L 3 s+ 2 ss 0 ] ] ] s s s ]. The hase is 270 at very low frequecies. As ω icreases toward about 0 rad/sec, the hase rises toward 90. Sice the hase crosses 80, we have a hase crossover at a frequecy ω. It ca be foud from the followig aroximate closed loo characteristic equatio: Φ cl s ss + 3k s + 2 From this, we obtai k = 3 ad ω = gai margi is give by GM = k <. 2 Above 0 rad/sec ad below 0 4 rad/sec, we have = s 2 + 3k s + 2k = 0. E.3. Examiatio of the root locus, shows that the dowward L 3 0 ] 2 s s s + 0. Over this rage, the hase dros toward 40. Sice the hase crosses 80, we have aother hase crossover at ω 2. Sice ω 2 < 0 3, it ca be foud as follows: ta 3 ω 2 0 ω2 2 = 80. E.37 This yields ω 2 90 = ta 0 ω2 2 2 = or ω ω 2 0 = 0. E.38 From this, MATLAB yields sad 7. as roots. It thus follows that ω 2 = 7. rad/sec < 0 3 rad/sec. E.39 Over the frequecy rage 0 3, 0 ], we have ] 0 2 ] s L 3 s s 2 0 O this iterval, the hase rises toward 270. Over the frequecy rage 0,, we have L 3 s 0 s 2 ] 2 s 0 ] ] ] s s s O this iterval, the hase dros toward 900. Sice the hase crosses 80, we have aother hase crossover at ω 3 that is associated with a hase agle of 40. To determie ω 3, we must solve: 270 7ta ω = 40. E.320

4 Aedix E: Previous EEE0 Exams 73 Doig so yields 270 ω 3 = 0 7 ta = E.32 7 With this frequecy, we have a associated gai k 3 = Ljω ω 3 ω ω From the above, we see that we have 3 imagiary crossovers o the root locus. The umber of asymtotes is give by:. umber of asymtotes = umber of fiite oles umber of fiite zeros = 2 4 = 8. E.322 The ceter of gravity c.g. is give by fiite oles fiite zeros c.g. = umber of asymtotes The agles of the asymtotes are foud by solvig: = E.323 s 8 = β ke j0,e ±j30,e ±j720,e ±j080,e j440 E.324 From this, it follows that the asymtote agles are give by: s = 0, ±4, ±90, ±3, 80. E.32 Sice β < 0 high frequecy gai of L, it follows that oe must use the eve real-axis rule; i.e. the real-axis ortio of the root locus lies to the left of a eve umber of fiite real-axis oles ad/or zeros.

5 73 Aedix E: Previous EEE0 Exams For def s + 2 ] 0 L home = PK = 3 ss s s + 0 we have the followig: ] s ] ] ] s s s The hase is 270 at low frequecies. It rises toward 90. Sice the hase crosses 80, we have a hase crossover at a frequecy ω. It ca be foud from Φ cl s ss + 3k s + 2 = s 2 + 3k s + 2k = 0. E.32 From this, we obtai k = 3 ad ω 2 =. Examiatio of the root locus, shows that the dowward gai margi is give by GM = k. The hase the dros toward 270. Sice the hase crosses 80, we have aother hase crossover at ω 2 = 0 3. With this frequecy, we have a associated gai k 2 = Ljω = Examiatio of the root locus, shows that the uward gai margi is give by GM = k 2. The hase the rises toward 90. Sice the hase crosses 80, we have aother hase crossover at ω 3 0. With this frequecy, we have a associated gai k 3 = Ljω The hase the dros toward 720. Sice the hase crosses 80, we have aother hase crossover at ω 4 that is associated with a hase agle of 80. Sice the hase crosses 40, we have aother hase crossover at ω that is associated with a hase agle of 40. To determie ω 4, we must solve 90 7ta ω = 80. E.327 Doig so yields With this frequecy, we have a associated gai k 4 = To determie ω, we must solve Doig so yields With this frequecy, we have a associated gai k = 90 ω 4 = 0 7 ta = E Ljω ω 4 ω ω ta ω 0 7 = 40. E ω = 0 7 ta = E Ljω 3 0 ω ω ω From the above, we see that we have imagiary crossovers o the root locus. It will have asymtotes: umber of asymtotes = umber of fiite oles umber of fiite zeros = 0 4 =. The ceter of gravity c.g. is give by fiite oles fiite zeros c.g. = umber of asymtotes The agles of the asymtotes are give by 0, ±0, ±20, 80. =.. E E.332

6 Aedix E: Previous EEE0 Exams 737 Problem 3 Bode: Partial Aswer ] s+ 2 The uity gai crossover is ω g 3 rad/sec. The hase margi PM is foud usig L 3 ss. It is give by The delay margi is give by: PM = 80 + Ljω g E.333 = 80 + ta ωg ta ω ] g E.334 2/ = ta ωg 90 + ta ω g E.33 2/ 3 3 = ta 90 + ta E.33 2/ = E.337 =.72. E.338 DM = PM ω g =.72 π 80 rad = 39. msec E rad/sec

7 738 Aedix E: Previous EEE0 Exams Problem Desig for BW ad PM: Partial Aswer a Ks = gs + z s gs + z Ls = PsKs = ss z = 90 ω g PM+90 ta ta ωg + ta ωg PM + 90 ta ω g E.340 ] e s E.34 s + + ta +ω g ωg E ω g ta + ta 0 + π = π E.343 E.344 E.34 This requires =.273. We therefore choose = 2. b To reduce overshoot, we eed a refilter ] z Ws =. E.34 s + z c To determie the miimum tolerable, we roceed as follows 90 PM + 90 ta ω g + ta ωg + ω g ta + ta 0 + π ta π E.347 E.3 E.349 From this, we obtai ta 2 E.30 or = 99.9 ta. E.3 From this, we obtai mi def = Whe dros below mi, should be icreased to 3. ta 99.9 =.7. E.32