A#20_Solutions-1 - Assignment #20 — Solutions ECSE-2410...

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Unformatted text preview: Assignment #20 — Solutions ECSE-2410 Signals & Systems — Spring 2007 Tue 04/24/07 105). For the system X(s) Y(s) (A)(15). Calculate the step response when __ _ (3+9) * (5—9) (a) Hm” (3+10)(s+1)) (b) Hm— (s+10)(s+1)) (C) Hm“ (s+10)(s+1))' \ i (Q)? (.9): ‘\ =L+£+§Dzémkfi ’ Sig? 5(9+1)&+!o) 3 3+1 5 *5 9+1 54,10 PM Jt ~01: @ {3mg} 7;. -—L Lia-2 ~77L8 ab? +5715 611 MIA (o —- 52/ go , 0 .... ~ M [Q \r 9” :— fi+£+ ——C-- s: W“! 9H S—HO ’33; ’ soy—IVSHO) 5 SH 9H0 5 _f 401: K L+)=%Mfi~%c at” “4'15 2% (c5 ... . V ‘5? maize“ smbém w F S(S+D(s+lo) $ 91 31-10 s 9+! “t \[01’2 fgqufi); “um—+7136 szg‘ 6 Lia—)1 \ IO A#20 ~ Solutions. p.1 (B)(8). Use MATLAB to plot the three step responses above. Step Response num=[1]; den=[111 10]; step(num,den) gnd num=[1 9]; den=[1 11 10]; step(num,den) hold on num=[1 ~9]; den=[1 11 10]; step(num,den) Amplitude . holdoff O 1 2 T’mefsec) 4 5 6 (C) C 4— o «e W _ RM 0 Sand“ 51“ flaw HA9 " é+Fi>€Sh+ FL) I 9+1“ s+ 4“ 5+5 [ “‘73-‘25; figeéez éWJHP-Q m + SLS F2) 6’ “HH‘ sCS-H’n) 8 S30 sg‘fx 5““- 2.. ~ Lawn (7‘3"?) {imam +(‘3‘w 5% v ' L "RL‘FfiFJ ('FJ"PL+F‘) Em? saM‘cvx ) ta: 6i) a: “1 cs“— mmel/ Aw W. sew wfamwbvzfl was We m PM}, HOW W (MEAth 04 {Hide Mfmwfivls me ,dg+mmc‘m J m POW, aw . A#20 — Solutions. p.2 253 w 35 + 4 s4 +s3 +Ks2 +2s+3' (a)(1 5). Use the Routh-Hurwitz criterion to calculate the range of K that makes the system stable. Mes—sec; as; Sl-tg’fl‘gatzgi‘gw K 3 2(20). A negative feedback system has a closed-loop transfer function, T (s) : Si 1 33 l '2. $1 14¢ 3 Fax glabiltlfl i<—2>o:§l 2LK*Zl-3>o S! ZUL‘Q"; k6“ K"; K72 fig) :hq so 5 10.121! WK rill 2. 3:5 (WWWELQ in W 7’ 3% (b)(5). Plot the root locus using MATLAB and use the cursor to find the K that makes the system stable. 7> 2. amt 24¢ sin 334-14 $24 2%; = o 2 gm +2s+ 3+ Ks. a o 2.. Z. Ci: H. “SEW... no 3 e 1‘»> 61$) =- SLL’ 9 fis’ig sflsazgg 2) Root Locus 8 I i z ‘1 num=il 0 0]; 6 _ """""""""""" . System: sys """"""""""""""" _________ U" denzu 1 0 2 3]; 5 Gain: 3,57%,“ {4 flab rlocus(num.deni 4 _ ________________________ _A Pole: -o.0315 +1.43i ______________ Damping: 0.022 : Overshoot (°/o): 93.8 . 2 _ __________________ Frequency (rad/sec): 1.43 ___________________ ___________________ ,_ \lr—-———___.;.<__* i i i . i . . i i Imaginary Axis Real Axis A#20 ~ Solutions. 133 __._£___ 3(5 +2Xs +8). (a)(15). Sketch the root locus, calculating asymptotes, asymptote locations and angles, break-away points and K at break—away points, and the value of a) and K at the j a) ~ax1's crossing. 3(55). A unity-feedback system has an open-loop transfer function, 0(5) = FEE—:3 \ \9 mots =7 5:.- «Whats ~ 53370 cm)»; at m\ Nd”... a+ a. 39m 04 k:_£_g(g+zxsfeflz 7.04 l $3 "05?qu jwmls OXOQSl‘Mé chm . gm 3%. loga+les+t<= O $3 I M; 32 IO K ‘ . _ fl \ 5‘ [32:5 ca Wager Jul-@795 antic gfiM K460 O (b)(10). When K = 160 , the closed-loop system has a pole at S 2 ~10. Calculate the location of the other two closed~loop poles. W K Pita—M jwmowtslvj 01mg Pelt/i QM ails: A#20 — Solutions. p.4 (c)(15). When K = 80 , sketch the straight-line Bode plot (magnitude and phase). GCQ :: “EX-«W 7. 83/5 5 . M 2: C =1 * Stealer) swag ? 7‘”) waterway Using this value of gain margin, what is the largest value of K just before instability occurs? L‘g MWOI Lat A#20 — Solutions. p.5 Remember this is an approximation, since we used Bode magnitude approximations. The error is due to the two poles being relatively close together. Let’ 5 use MATLAB to calculate the exact value: Bode Diagram Gm = 6.02 dB/(at 4 rad/sec) , Pm = 16.8 deg (at 2.77 rad/sec) ...................... . t...i.._,.......WWMHWHT.m...”...V,,W...V.IN,WNW.,,,,.....,_,V,,,.V.NWT...“ «WWW... _,. was” num=[80]; den=[1 1016 0]; l x i l i margin(num,den) Magnitude (dB) '3 400 r 450 .r \ hasten“ 'LAJMV m ‘ _.:J_i..ls_‘ rut,“ ~90 435 *- _130l. -------------------------------------- -- ' ------------------------------------------------------------- ~- Phase (deg) ~225 .— -270L. _, .4, L_.x 1.4.1.4.} I 4.4.4.1 1.4.14 _..i l 1.4 w 3 10" 10° 101 1o2 10 Frequency (rad/sec) Our frequency calculation on the previous page was exact because we used symmetry to find a) = 4 . Since the exact GM above is 6.02 dB, we need to increase the gain by this factor to reach the point of instability. Thus 6.02 = 2010g10 (Abs GM) yields Absolute GM =2. Thus the gain for instability is 80-2 = 160 , just as we found using root locus above. A#20 ~ Solutions. p.6 (d)(15). Find the phase margin when (You can use MATLAB here) K , s(s+2)(s + a) "' 33 +10s2 +163 G(S) '= (it) PM for K=20. What is the damping ratio of the complex roots? . t num=[20];. den=[1 10160]; _ i , [GM,PM,ch,ch]=marg1n(num,den) MATLAB gives answer: GM = 8.0000, PM =’ 53.7086, [Gm,Pm,ch,ch] = MARGIN(SYS) computes the gain margin Gm, the phase margin Pm, and the associated frequencies ch and Wop, for the SISO open-loop model SYS (continuous or discrete). The gain margin Gm is defined as l/G where G is the gain at the —180 phase crossing. The phase margin Pm is in degrees. - System sys j 1 ‘ ; Gain: 20 Pole: -0.803 + 1.3“ 0 5 . p - i Dawning: 0.522 _ I . Overth (%)1 14.6 ' Frequency (rad/sec): 1.54 Imaginary Axis V . __-..'1 .__._ '2 “WNW "11.5 » -1 -0.5 0 - 0.5 A#20 —— Solutions. p.7 (d)(1 5). ’Find the phase margin when (Y on can use MATLAB here) continued. (ii) K =40. What is the damping ratio in this case? num=[40]; “dew—"[1 10160]; _ ' [GM,PM,ch,ch]marginmumfien) MATLAB gives: GM = 4.0000, 'PM = 35.1376,... And the root locus for K=40: . Root Locus 3 num=[l]; _ ' V d6n=[1 .10 16 O]; 2 ' ', , rlocus(num,den) ' System sys axis _3 0'5 _3 3 Gain: 40. . g Hale: -0.643 + 2.04i r . 1 ‘ - . Dan'pingz'OS ‘ r r g i Overshoot(%): 37.2 _ A. I '2 Frequency (rad/sec): 2114 s o ‘ .5; g E -1 -2 -3 , ~3 ’2 5 —2 -1.5 -1 -o 5 o o 5 Real Axis (iii)If the phase margin decreases, what happens to perCent overshoot? Summarizing the results: K = 20,PM = 53.72; = 0.522 K = 40,PM = 35.12; = 0.3 As PM decreases, 5 decreases, which meansthe overshoot increases. A#20 —— Solutions. p.8 ...
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This homework help was uploaded on 04/10/2008 for the course ECSE 2410 taught by Professor Wozny during the Spring '07 term at Rensselaer Polytechnic Institute.

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A#20_Solutions-1 - Assignment #20 — Solutions ECSE-2410...

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