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A#19_Solutions-1

# A#19_Solutions-1 - Assignment#19 Solutions ECSE-2410...

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Unformatted text preview: Assignment #19 - Solutions ECSE-2410 Signals and Systems — Spring 2007 Due: Fri, 04/20/07 1(45). The following feedback system. X(s) Y (S) For all root locus plots, for K > 0, calculate all that apply: 0 breakaway points (BAP), re-entry points (or Break-in points — BIP), asymptote angle (AA), centroid of the asymptotes (also called asymptote location —AL), value of K and a) when the root locus crosses the imaginary axis (called j a) —axis crossing). __1__ (s+le+3)‘ (a)(5). Sketch the root locus for G(s) = S (b)(10). Sketch the root locus for G(s) = m s s . What is the effect of the zero, compared to (a)? A#l9 - Solutions. p.1 (3+4) 1.(c)(10), Sketch the root locus for G(s) = ( +1)( +3) 5 s . What is the effect of the zero compared to (a)? ' “94¢: , Fe ma gm mm M3 ~:_4_ Aaawﬁtok 3‘? M :L‘EQE ”2% M», ~N<ﬁ~ usajw / ﬂ 7. .1 81? I Ber? @1th Bl? qEAP: K: -_ \ ~ ._.. <31+4+\$+3> (31\$) 31% (11:; ”é (\$+¢i)<2s+%)»ts2+%s+2>§) :0 :7 ga+8\$+1§ a o GLS (SH) 6L (5+Lt3L—‘3 =0 Mote. 14253112 am; W M m tmsﬁw Tdcihﬂ We (>an d 9.2.4 794%; Land“ Wm WM M +29 We mam 4m 8m, 0 03\$ S=~‘1’. Oats tﬁutﬂgtmw w mm mm m. 8% 719,, mt Low; w is. maﬁa am MM was (3‘, We. cam Prom WA; Ohm .423: HMO no g2+c4+ns+té+wzo M gem—:3 3 ”ct; (X43924 mow +CZ+LH<]= o . l : 5% Han/Q emaaem/g and; W311 a ,. Emil: ﬁ—abﬁéﬁgxt—Cawﬂao gulp; {mm M9 2w; +CLtH<>é =0 :7 Ke .. 2x4; W 2232+ (Lt—21M) Q<+ (3—8 we) 2 O 2. 2.. no air >< «ea tax—H; % aft/«Chg wimid Maw m W5 www- A#19 — Solutions. p.2 l s(s +1)(s + 3) What is the location of all three poles when K = 12 ? 1.(d)(10). Sketch the root locus for G(s) = . What is the effect of the pole compared to (a)? PEE': 3 “O gf’légmf‘l-o‘les six .. 1&9: : o 2 3/ AA “ g 7 A L : 2 {Dale lot. ~ "Egan la” P235 ’ :2 - if. 3 W K: ﬂats; :: ”(SC&H)LS+33> ~ ~ (334459 +33) it 0 WC 3£a+85+3> Eat: g: W 1 & Q 3%? « “Lgﬁ —.= «aflgl‘l _wg\> i; l0 ”ﬁe r :3 t 3 :3- 4— K I 131:? ﬁajfém m +9 ESQ/“32’“? 4f. 0 ZIZ. 5 K Mali? [4‘ A#19 - Solutions. p.3 1 s(s +1XS + 3) What is the location of all three poles when K = 12 ‘? l.(d)(10). Sketch the root locus for 0(5) = . What is the effect of the pole compared to (a)? A#19 - Solutions. p.4 (e)(lO). Use MATLAB to ﬁnd the root locus for G(s) = -———§:~%-——— Hand in your plot, (s + 1)(s::§§j,25) with your name in the title and indicate, by hand drawn arrow the direction of travel of the closed loop poles as K increases. 8+7— . as; 5 II s: ~45: ; 52—93% \$33145 W J ALJW- ’I l «ﬂﬁﬁ— a 5 5 Root Locus 1 5 _~T” I I l I T‘ I I I _ I _: 10 g i I , I l : 5 w 1 wt .9 l ' g; I I? I (V g 0 .. .............. J .................................. W ....................... a (U | E I I . I z -5 l w: I I I I 1 -10 r I “3 I I I I _15 I k I I I l I I -4 -3 5 -3 —2 5 -2 -1.5 -1 ~05 0 Real Axis num=[1 2]; den=[l 9 33 25]; rlocus(num,den) A#l9 - Solutions. p.5 2(20). Consider the feedback system, X(s) Y(s) where [11(3): K +Kls, and ____1____ (s+1)(s+2)‘ (a)(l 0). Use the ﬁnal value theorem to ﬁnd the steady-state error for this system, an = lim e(t), where t~>oo H2(S): e(t) = x(t) — y(t) , when the input is a unit ramp, x(t) = tu(t) . WW * Mia * \$90 i 1“ (“3) Hal K (b)(10). Suppose the controller H1(s) is modiﬁed to include an integrator H1(s) = K + K13 + ——2—. Find 3 the range of values for K 2 so that e” S 0.1 when the input is again a unit ramp.. €53: Lﬁéi) I‘ = «L4? XS‘HXSR) 5‘? K Sevo { + K +14!“ ’33" Sat) 5(S—HXSta) i—ﬂk‘lHSJSi‘KI-j (914192») .: M —.-. 2.: g (0.1 (<2. F1— ﬁéeiéL § \gzOéKZj A#l9 — Solutions. p.6 3(35). A unity feedback system has the following form (a)( 15). Sketch the root locus as K varies from K = O to K -> oo. Calculate all that apply: 0 breakaway points (BAP), re-entry points (or Break~in points - BIP), asymptote angle (AA), centroid of the asymptotes (also called asymptote location ~AL), values of K and a) when the root locus crosses the imaginary axis (called j a) -axis crossing). .—., l ——-‘ %'\$l+l (3+2; O Si sa+s+a+ K s(\$+a)=~ O ~— .— CSHJ S(S+F9 A#l9 - Solutions. p.7 3(a) Continued. ( \$3+ \$2} 3 + L 2”" ="‘ “:23” a2. C2W+23t22+22+l <222++2221223§2 013 (9 +3) meet ~(tsaorj 0.2306: “(956% 0? sﬁzsﬁ o ”223.2 2 2 O 2-? V2045; m (W l W l (b)(10). Using Routh-Hurwitz, ﬁnd the values of both K and j a) at the j a) -axis crossing point. For this value of K, ﬁnd the location of the third root that lies on the negative real axis. Ema “232% «2—2—2222 dam (>213 §+O+K)s"~+a+a2+z=o 53 l 1444 s g‘ ﬁlli- whecgmwleheguoiemlce (1-2143-222— 227k+u<4=o so 2 W} C: K: “*1 if?” . IP84: (L: ~l+JZ irww an 6%.”: {4m 82‘ WW“! \ 20mg Q+g)gl+z=o :7 {232222- =7 \$22—42. c; Fm amt—2a: ﬁfe ad 2222* 2 2* S'FJE— SL-t-J‘LlS3 +ﬁf> Eat/3%?» 3‘7 Mi A#19 — Solutions. p.8 ad .3032an - 03m?» \$54 .mmm \$0 6 m0: V. m, Y. J‘ .N. 4...; z+ «Lie Svd “Auwmim: 3:33..”— o ”E 68890 A, s mmé ”835mb Gamacmi a 0 Mg SEQSO e nmEaEwD . w ”wEQEmD «:8. Hman. mm. W Eon. mm.w MENU owmood ”:60 m “Emﬂgm w “Emﬁﬁm ............................kf O I k , ’ mwmd ”AummESv 5539.1. mfﬂv— )wv A? \ uh I Noo0,o§:oozew>o n0 .5... L‘HbT \$90 ”Aowmﬁm: 3539:... 5ng beugﬁew: 33 5:358 \$3 + 59 ”you 33 Hg .8555 V . a, 3o ”95st 8m .58 4 3‘5 1. a ‘ . . m €me 55 o + who a. ban m: ”58 mmé gooﬁng xocmsami m m “E996 m2 Hg 62290 mid. ”9383 N... + P220 5.0m o EEG w ”Emugm x :2 .55.? um. _ _ _ +6 NW& 3 (65 I it! _ 303 sex iiiiiii z m; .92qu 33 .5ng«.. So» @53 A o H 3 wumﬁwwg 638% zzmoﬂﬂo n onSmEows: n womﬁmvkgo 8m 83%? 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A#19_Solutions-1 - Assignment#19 Solutions ECSE-2410...

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