hw7-solns - D 90 e—eCOSVQ um gm 9*205 s‘ $2 Z L260...

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Unformatted text preview: D 90:): e—eCOSVQ um gm: 9*205 . s‘ : $2+Z L260 00 (Dwai’zéb 503.444) from, Lu‘éll.‘ é’Wm <=> X(s~0Q So: 3(9-— €209 <=> (-1.6% ‘X(5 + W _ SSH? +— 2. G‘CS\ _ (5+1 (M111 4- 4) ’r0 fmoL 4m Wolu we. w?“ (/94. em = HCs\-—‘5- => Ms}: S 61(3) 5° now" = W {ml W9 ><(S S3+551+3rs+5 (53+ 331+ ls +5)Y(s\ = (53 + is?- + 3 3) HS) VkOUU' wt W HAL muMS‘rC kph/U. (13 801%.. 8:09.55 RESUdH/{j M: 15% + 333% + 3,1303 + We {my 2390A +3 72m a2 “K (+3 = 8M (mum 9M z [He‘ 6+ 2.8M (10 — V Cos (19] um a\ Jrcflkmj HAL Laplouu. ”I'msf-flVI/K J5» BOW/L X(s\— 52—31:; 1 4' - _‘/§__ - YCS) 5+ \ 5‘44 5‘4“! ' (5+OLs'-+'~i) 30 now- wt. dLLf/LWVLL M MAPUIUE 215119091953 H03 1 Y((S;_ IO SSH» 3} l0 (st-+43 X S (5+1Xs‘w'): 55+s‘+qs+ ’1 (55+ 31+- 45 4— 0W3) =:o(s‘ H} ><(S) ‘5‘”me 55m +4 9‘0 4; 4 We) = I 0 ( 55%“ + 41““ b) SMCQ. 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CWLL‘L Jet SIZJJ'L—thj mg: .__..___ro» Hm: €37» LSHXS‘t-r—fir') Wm, we, \Aauut 0s. {Db- Ai' 5;: ‘l ) actual/v Whom: M Ssta/vt <3 BIBO SMELL q M IO ”(400$ = 3w + I )Cuyumuj Rapmse fchWVL INCH”. : l0 Maqufiuote 1+ wi- - ¢=émf‘ “'03 WW .\ {m «M wwmotl fwdm, 01mm“ .uc Mum up”); {(35921} H) frefuemcv (up), (WILL Sp}- ?é— F0 26cc. 19/16 eoM. at)!“ MWSUAJL 0x Mam /M«M.d. 5/3/06 3:39 PM C:\Documents and Settin s\Elvis Silva\M Decuments\c...\HW7 2e.m 1 of 1 ===================================i=============g=============================== %Homework 7 problem 2e % Define the numerator and denominator of the system num = 10*[1 0 4]; den = [1 1 4 4]; % Define the system using the tf command sys = tftnum.den): % Now determine the system response due to an impulse and step: T = linspacet0,30,1000); % Define time range (0 < t < 30 ) h = impulse{sys,T); % Determine the impulse response g = step(sys,T): % Determine the step response hold on % Plot the step in impulse together plot(T,h,'r','LineWidth',2) plot(T,g,'—-','Linewidth‘,2} % Legend, Labels, title and axis scaling legend('Impulse response','Step response'); xlabel('Time [sec]');ylabel(‘Magnitude'); title('Response Due to Various Inputs'); L = axis;axis{{L(1) L(2) M3) L(4)*l-2]); Magnitude Response Due to Various Inputs Impulse response 10 15 20 25 Time [sec] 30 Hts): S(S+1\ (5" +33 +1X51+s — 1) A H(\' \(CQ _ SZ+1S S )((S\ .. 51+453+3sZ—<{s-‘/ flufm, (5W 433+3sz—z/s —7)Y(ss = (3‘4 mm) It 'lj”(t]+‘{15“(0+5g(e)-Lllj(fl- 41d (4;): X (0 4— 21(0 ‘63 0mg, mama.) we. cam Mama. 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Elmo 41’ equal moL oppogk (m 553% lamina/LS 31A Mu, dfikg'mv—{h m W94 fit“), HLSUQ: it}: jrcquL1 mslwwiz'wu‘ww if» out, Ltd“- hw‘i Wfirqx‘uob. J]: $21143. {inlawrx 23 ‘:f we [mama 0L Solmwm‘m.’ of folks.“ Md ‘Ewos CADOU" “A 1 M W3 W7 MES in“; All pom kw: M 3 YOU: M GAG-um «have; thuwc/tj RFSfW‘W Macini‘vcll fizz-W, WC w”! \(LMC Mamba-f? from Pubs we} Cow/16d 001‘- “)3be Wfifikudfig from gems : M1. M31 “I; \H (3013‘ F“ Mp‘ MP1- ”f3 3w). ‘UALW Masai-bell: m 2:50:11: MP| -’ ”1‘ , Mfr. ‘ M“- uh ‘ M355 U02, Mk We M MaflML-vde ka f‘rLQL/Wbbf Mesa/L39. Whom 01:»qu equal +v one, 5‘5 HtflzHe’ <=>Y(s\ 35+ ' =_L_).31$+| X.(+\= um 2:» x15) = as l __ Q8 '1 HZCfU=ZQoS(+\'SM(Q<=> {30:}! Effl— - 31H - {St-Fl 3‘ +1 ( +3 H(5\' Y (S) \4 (S)_ 213+: %5 X203) X (S) 5;“ .. é— SMF%»MLJ, - 2 . . 43145-1) —_ -5~\';l;‘=+7- HEX ' (snAG’u—SH) 5 +51... 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PM") + @6an [7103* + (“0% + qfiofi 87‘ 4» [1y 0‘) 615 + [2+ 5((0’3 “*CKD') 4- lflo-il -— 0 [filwhqm [1+ ita3+g(0'3+§(0']] = 0 My; aw. ’wo earumfiths Oil/1d 3. Mbfiwwg (715035, qfO‘),g(0‘33 we, mou‘A m4 monwL Q?voJ\/UV'L ‘1) Solue fin 4L1 _' mknowws. ’To cg} WON/LIA equail‘upm we web/d afiso mud cquu-izk ra-«(smL—oul'pd (dobmsbupfl'fflus 4‘0 39h», u£¢3mWJM, m oLo moi I’mwe. ELFIOUSLL, Mformduom 5/3/06 3:41 PM C:\Documents and Settin s\Elvis Silva\M Documents\C...\HW7 5c.m l of 1 =====================================g==============3L================================= %Homework 7 problem 5c % Define the numerator and denominator of the system num = [—1 4 2]; den = [1 1 *1 2]: % Define the system using the tf command sys = tf(num,den); % Determine the system response due to a input with no I.C.: T = linspace(0,50,1000); % Define time range (0 < t < 50 ) u = sin(T): % Define the input to the system y = lsim<sys,u,T); % Determines the output (Y) % Determine output of system duet to same input with I.C.: y2 = 2*costT)—sin(T); % Determines the output (Y2) subplot(2,1,1) plotIT,y2,'--','Linewidth',2) % Plot the system response titlei'Response Due to Various Inputs'); xlabelt‘Time [sec]');ylabel('Magnitude'): legend('Response with I.C.'}; subplott2,1,2) plot(T,y,'r',’LineWidth',2) % Plot the system.response L = axis;axis{[L(1} L(2) —100 100]); xlabel('Time [sec]');ylabel('Magnitude'): legend('Response without I.C.'); Response Due to Various Inputs M e s n o D. s e R — - - muBEmmS. 0' C. Response with Time [sec] 93.2%: Time [sec] c3 4&1 dbfluuww @Lwatm 33m. hue gawk/'11.? stun-mm, M “M. PainuS rife!“ @0ng Jam Did W M W 3345?me [MS mm‘flrai McLuHu—mj‘, Wig/g, am OHM/L mseomfirfl does not ...
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