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A#23_Revised_Solutions - Assignment#23 — Solutions p.1...

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Unformatted text preview: Assignment #23 — Solutions - p.1 ECSE—2410 Signals & Systems - Spring 2009 Due Fri 04/24/09 W 1(10). The root locus shown represents a system with the . ~ Root Locus closed—100p transfer functlon in ~ ,. , ,, , ,, ,, ,, ,, Y(s)_ KG(S) . The X(s) I + K G(s) breakaway point is = 4.44. Note: All the poles of G(s) have real and imaginary parts that are integers (no complicated fractions 0r decimals!). 10 (I1 Imaginary Axis K 3) as :. L“? ”mg sea—am) tfl. x :- 45 e t : , , : ; 3&1“? 31+ (20 *lgng 20 ~5 V4.5 ,4 “3,5 3. "2‘5 V2 4.5 x: .35 Real Axis (a) (5) Find the gain K 80 that one of the closedvloop poles is at S1 = -— 2. g in ”L? ”a scsw‘yésts‘) ._ t- z)(z_X§) W1 '4‘ 614 (3+3) (2) ’ M3 b 2 F d h$W h 1 3"“:- t rtw (W m ”0 e “POSS Frt 1472. (QM e’gw s. g—t‘is Hflsasttg 0 $3 + ‘75 +18 M 3+2 MM? +3386 \52+7$+18£;(3ng+ 3:3 W 791% 323613 WW 7?: :9 <3; ..,_f_7_ 9333 S {as-ts: S 2*” 2- } £1232“ zest-36 (C) (3) Find the range of K that will make the closed-loop system overdamped. Zyfi 57K PM m g site-it 533%)?“ ~ (—5”: year: at M at? he?” ” 6:9: 4% 0% 5:4??? 3: ~¢ffi€+ ( 3 ' meww Assignment #23 —- Solutions - p.2 ECSE-2410 Signals & Systems - Spring 2009 Due Fri 04/24/09 Root Locus 2(10). The root locus below 8 represents a system with the closed—loop transfer function 33(31_ KG(s) X(s) " 1+KG(s)' Find K as the root locus crosses the ja) axis. Note: All the poles of G(s) have real and imaginary parts that are integers (no complicated fractions , _ or decimalsl). a» I -5 WWwWWWWWWWWWW (54:3)LC5H37‘i-fl “8 -1 G «Mm‘.m:émflfifiwm ,, 7" "WW W.;é.,., , V W “4 , , , , , >_ , , , , ,, , , . ,7 ,, , , ._ ,, , Real Axis Imaginary Axis 6(9) 2 K (3+ 3) ($895} ‘3 33+ 332+ it s 1&5 4—K) -= O 4mg? i+£<é©= 1+ Assignment #23 — Solutions - p.3 ECSE—2410 Signals & Systems - Spring 2009 Due Fri 04/24/09 K(s+2) _ K(s+ 2) en loo transfer function is G s H s =-———-—~—-——- 3(20>. Text 1127- OP 1’ r) U 52+2s+4 (my-+3 Y(s) (a) Sketch root locus of the Closed loop transfer function X(s) Paar :Lasdm a :83" 31? g 2-,. ~...L. : __ saggy-l 35 (3* 9+1. W § d4:- _. (Srzflzs +2) .— Qs"+a9r‘r)o) a o '38 A w (9+2? 2.32.4- ?Ag+‘i$+%—%f~(2<—}(= (3 5(9—6‘63 =0 Bi? @ $=atf (b) Find smallest K such that the closed—loop impulse response does not exhibit any oscillatory behavior. U0 osdllafims at Pale; mg 57.1 m “341%. {Max}; P0125 cal“ 3.3-31. Assignment #23 - Solutions - p.4 ECSE—2410 Signals & Systems - Spring 2009 D F ' 04/24/ ue I1 09 4(30). Using Nyquist and Bode plots and Bode a ' - . _ . _ . pprox1mat10ns, flnd galn constant K at h' h systems shown are Just On the verge of 1nstab11ity. Sketch the Nyquist and Bode (Straighvtvlihce thiegfliffgjck and smooth phase) diagrams. 1 (a) CG): =—~«—-—————— (8+1)3 53 +3SZ+3S+1 Nyquist plot for general K: Assignment #23 — Solutions - p.5 ECSE-2410 Signals 8: S ystems - Spring 2009 Due Fri 04/24/09 W W 4(a). Continued. Use Matlab. The Nyquist plot for K :1 is: num=[1]; den=[1 3 3 1]; nyquist(num,den) To find the frequency at Which the Nyquist plot asses throu h — ‘ ' - - Gm tells us how many times we must increase tlije gain beforge thgeplhtpgorel: tiii”1(:111thli3 tialfjlmafgln (Gm), Executing the code, g e pomt. num=[1]; den=[1 3 3 1]; [Gm, Pm,ch ch]=margin(num,den) gives the values: Gm :2 8.0011 Pm = —180 ch 21.7322 ch :0 In this case, we can increase the gain by 80011 (1-6., K = 8.0011) before the system is on the verge of instability. Compare this to our approximate (Bode) gain constant of (K 2 6.96) — not too bad Assignment #23 — Solutions - p.6 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 04/24/09 4(a). Continued. Now let’s put this exact gain constant (K = 8.0011) into MATLAB to see if the lKG(ja))I :1 when AGUw) = —7z: Bode Diagram Gm = 3.449005 dB (at 1.73 rad/sec) , F’m = -0.00238 deg (at 1.73 rad/sec) 20 ”“_ .. l L J "‘ l 1 num=[8.0011]; a 0 ‘r‘ut'l- In. In v+o 1" ---------- [r k v d r--. IIIIII 'I 1 - 1n- u- -u n a - - dgn:{1 3 3 1]; 1:), QOL l ‘ margin(nurn,den) E3 l I“ i h grid 1 T III E -60 l i II E 80' I | H H I“ ‘ III -100 J I ll. ' i i. 0 II II W E II § —90 L ll * “I : J E U) o. -180 IN"... ... .. +1 ll ............. {. . .l> .l, u-.. . m .. . . . ....... ... ., . . I -270 J IIII 10’2 10‘1 100 1o1 102 Frequency (rad/sec) Yes! The gain is just about zero dB (absolute magnitude = 1) when phase is —180 degrees. Note above that ~180 degrees of phase occurs at a gain crossover frequency of 1.73 rad/sec, while our Bode approximation gives us an approximate frequency of 1.91 rad/sec. Again , not bad! Now check to see if the Nyquist plot passes through the —1 point. Yes! num=[8.0011]; den:[1 3 3 1]; Nyquist Diagram nyquist(num,den) axis([—1.5 1.5 —1.5 15]) Assignment #23 - Solutions - p.7 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 04/24/09 1 4. Continued. (b) G(S) = —9 s(s +1)‘ I will not do the MATLAB plots for this problem, just the Bode approximations. Nyquist plot for general K: A“ Assignment #23 - Solutions - p.’8 ECSE—2410 Signals & S ystems — Spring 2009 Due Fri 04/24/09 ~58 4. Continued. (0) 0(5) : . Let’s do MATLAB plots for this problem, because it shows how to use s(s +1) time delays. Nyquist plot for general K: K665} flake. Assignment #23 — Solutions - p.9 ECSE—2410 Signals & Systems ~ Spring 2009 Due Fri 04/24/09 e —5 .1' s(s+1). 4. Continued. Matlab. (c) G(s): num=[1]; den=[1 1 0]; G=tf(num,den,'ioDelay', 5) Nyquist plot for K = 1 : nyquist(nurn,den) axis([—10 10 —10 10]) To find the frequency at which the Nyquist plot passes through the —1 point, let’s plot the Bode plot and use the cursor to find the frequency at which the phase is —180 degrees; Bode Diagram 40 E I H num=[1]; g 30?, , den-:{l 1 0]; f; System: G ‘ G=tf(num,den,'ioDelay', 5); -o goj Frequency (rad/sec): 0.262 _ _ r2 1 Magnitude (dB): 11.3 w—logspace( 2’0) g ‘ «I Bode(G,w) 2 10 'd gm 0 f -107 -90\ § -180 3 System: G g _270 3 Frequency (rad/sec): 0.262 g Phase (deg): «180 -360 \ j—L! E 41505 5 10' 10‘1 Frequency (rad/sec) Compare this exact (0%, = 0.262 rad/sec to the approximation (Bode) 606g = {52— = 0.2618. Not bad! The exact (MATLAB) magnitude at 6ch = 0.262 is 11.3 dB, while the Bode approximation is 11.6 dB. Assignment #23 - Solutions - p.10 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 04/24/09 _ 4(c). Continued. Another method of getting the magnitude at the gain crossover frequency: num=[1]; den=[1 1 0]; G=tf(num,den,'ioDelay‘, 5); w=.262; [mag, phase]=Bode(G,W) magdB=20*log10(rnag) The alphanumeric output is: mag 2: 3.6922 (absolute value, not dB) phase = ~179.739O magdB = 11.3456 Choosing K so that the lKG(a)Cp) l 1 K:———-———_ mag — 3.6922 = 1 results in, = 0.2708 Recall the Bode approximation for the gain constant was K = f:— = 0.2618. Plotting the Bode plot With K = 0.2708 so the feedback system is on the verge of instability shows that the gain is zero dB (close!) when the phase is —180° : Bode Diagram 10 WW ___________________ “Imam...“ W ..... W W 3 WT num=[.2708]; A 5; «l den=[l 1 0]; “35‘, G=tf(num,den,'ioDelay', 5); 1% 0: I :ystem G w=logspace(— l ,0) E) _5i f Frequency (rad/sec): 0.26 B0de(G’W) Magnitude (dB): 0.0616 :1: . t H r ~90 Wkt Hi 1 grid ED: System: G E L Frequency (rad/sec): 0.262 E '270 Phase (deg): 480 m l ’ —360 l -450 = I 0.1 .262 Frequency (rad/sec) Assignment #23 - Solutions - p.11 4(0). Continued. Plotting the Nyquist plot with K = 0.2708 so the feedback 3 t ' . , , result in the plot passing through the _1 p oint: ys em Is on the verge of 1nstab111ty should For visualization, we limit the range of frequencies to 10”1 g a) £100: num=[.2708]; den=[l 1 0]; G:tf(num,den,'ioDelay' , 5); w=logspace(~ 1 ,0) Nyquist(G,w) Nyquist Diagram Assignment #23 — Solutions - p.12 ECSE—2410 Signals & Systems - Spring 2009 Due Fri 04/24/09 400 (a) Sketch both the Bode Straight line magnitude and the smooth phase diagrams, (b) Use Bode approximations to show that this system is unstable. (C) Sketch the Nyquist plot. { KGKM‘IMPJ 5(30). Using the feedback block diagram in problem 4 above where G(s) = ...
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