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A#19_Solutions - Assignment#19 — Solutions — p.1...

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Unformatted text preview: Assignment #19 — Solutions — p.1 ECSE—2410 Signals & Systems — Spring 2009 Due Tue 04/07/09 1020)- (a)(5), Find [1(5), 5, can , sketch pole—zero plot. £2. 2 R229 L=2H H53): 2 W K-L-St’i‘éo {2551f SZLC 1-1 W) W) a... .3 _ 3 WM: W}: L 32- :7 + Zififiag+ a}? :Fébg 61+ §U+S+Q gH+g+ ii- ‘i‘t 2,. Jig-£44: (+4);f (b)(5)' Use MATLAB to plot the Bode magnitude and phase diagrams. Bode -Diagram zeta=0 25 Magnitude (dB) : num=[4}; ' den=[1 1 4]; b0de(num,den) ; ; grid . title('Bode Diagram zeta=0.25') Phase (deg) Frequency (rad/sec) Assignment #19 — Solutions — p.2 ECSE—2410 Signals & Systems — Spring 2009 Due Tue 04/07/09 W 1. Continued (C)(5)- Use Laplace transforms to compute the step response. Express your solution in the form, y(t) = Au(t) + By“ cos(Dt + Eng). 3 as): “is“ «st/ML EM» + W: Sfisz+s+fl 5 size-3+? ”’5 ‘94:? 9* ’“‘ + made Jaw... hinckégs—t 4 @ ’%£(l5“) “214‘? 9% 1“ 07‘s“ 1""? K‘E :aszgmcl KL? Assignment #19 - Solutions — p.3 ECSE—2410 Signals & Systems — Spring 2009 Due Tue 04/07/09 1' Continued, (dXS). Use MATLAB step(num,den), to plot the step response. Compare Matlab STEP command with analytical solution in 1(0). step response using MATLAB step command Amplitude 0 2, i 4 I s 10 i ., 12 Time(sec) 1 5 step response using analytical solution in 1(9) Amplitude Time(sec) num=[4]; den=[1 l 4]; subplot(2,l,1) step(num,den) grid tit1€=('Step response using MATLAB step command') t=linspace(0, 12, 100); ySteP=1*(4/Sqft(15)).*exp(—0.5*t).*Cos((0.5*sqrt(15).*t)-atan(1/sqrt(15»); subplot(2, l ,2) plot(t,ystep) grid titleCSmp response using analytical solution in 1 (e)') ylabel('Arnplitude‘) Xlabel('Time(sec)‘) Assignment #19 — Solutions -— p.4 ECSE-2410 Signals & Systems ~ Spring 2009 Due Tue 04/07/09 2 d y(t)+251—y~(—Q+y=(t) x0), alt2 dt 2(20). A second~order system is described by the differential equation K where K is an unknown constant. (3X5)- Find H (s) using the derivative property of Laplace transforms. K32 315’ *2 $333) :B-EDaM) E \ . H5033 : i ,2 g X: 1 339 “2+st 5 §2+jgs+§ (bXS) Find the value of K that will make the system critically damped :3? :"i 5 fi/ 102—“ ... éfia/gd's 5012i :4... :72 K21 ] 3'!“ K =3? #5 (C. {(2 i W (c)(10) If K: 1, what IS the output, y(t), of this system when the input is x(t) = sin(2t). Note. The problem mg 2: Sin [2%) 48%? :9” is very different than 0623‘): Slet-fitb‘} “>2 Lia} i'" 326%) The first is a steady-State condition while the second includes transient terms due to u(t) If we allow the transient terms to die out as t ~> 00 then the ultimate solutions are the same, i.e., x10) = lim x2 (I). [—900 We will solve this problem both wayS- i (A) Steady state solution. QQH‘H Si 3K (Z‘t’i ”7i: H45) P” J) H}, #8) ”g +j§+i Mi Mi WW (name-=7 i t/wa/artvvtfw’} 61a Hd “- W‘ « i- "‘ (Haw/in “it?“ 7" saw?“ _. 56’ £12. Assignment #19 - Solutions — p.5 ECSE-2410 Signals & Systems - Spring 2009 Due Tue 04/07/09 2(c). Continued. £3) %M§§m% (COWHL‘R) géi‘. i2fi%: sfilfif)k§9~;l%yg Pa? ggfifi EQJ‘NE Lg); } . Z 2 g 3 3+5 2" 2' 32+ 2 éfifi" (9g: + Sfl ”f 82%?“ {§+Ul Q a A M2 a. ~2sz .3 i A“ $14:th :: %/ «E» 2:: $%+Li)/ 6“ éz’i‘ti); 2-5 $30 xiv-s 92-1 2?- 8 #3 Lég—risa‘ii-«Emmfiefizcw +§k4239+13 22%34- +..... E343 23 23 36 ~ :2 i5: ,... Vi. Assignment #19 —- Solutions — p.6 ECSE—2410 Signals & Systems - Spring 2009 Due Tue 04/07/09 W 2(c). Solution B. Continued. 32H): 42:... tétufl) +- fi, 8;th _ i @Szzéjaw - 3.‘ $iézg‘éfi559 5' as as 25 MW: ‘3‘ iogferH 3 sikf’z‘é) 2:. K 32% (2):) +3); Z$§a€z+)asg§ “i“ gag/263359 @413 Kwsa :2; Biz? R? Z: {‘1‘ Kgifiév— ‘2‘ 3 9,44: 2.53.? 4&0“ CE) (03913 ”‘59 KS! 9‘: 1+ :9 K: S if. 3 <3 golf). is F M? £6; WM ”(3359* 5 sm<z++;;1”)al+) 5 «(My 426$ 5% : 15— $21": (zfi 42m") :2 3W Assignment #19 — Solutions —- p.7 ECSE—2410 Signals & Systems - Spring 2009 Due Tue 04/07/09 W 3(30) A ButterWOI’th lowpass filter of order n ~1 is given by H (s) — .17 wafiBWi \ we 2:: 1 s + (a) Find the transfer function HEP (s) of the corresponding bandpass filter with center frequency 600 =1 and a passband B: 0 1 we a? (b) Use MATLAB to plot the magnitude frequency response of this bandpass filter Indicate on your plot frequencies, (UL 500 , (0H Bandpass fitter with center frequency=1 and 8:0.1 num=[0‘1 0]; den=[10.1 1]; w:[0.5:.01:1.5}; [mag,phase]=bode(num,den,w)g plot(w,mag); grid -............ ................ .............................‘......... ................... ....a..q........... ............................................... >> roots([1 .1 -1]) .................... ' """""":‘""'"“""""""“"""”?' 0...”... ........ ans: 4.0512 0.9512 >>1/O.9512 ans 3 4 . 110513 8.5 0.6 0.7 0.8 935111105“ 1.2 1.3 1.4 1.5 F requeney (rad/sec.) Assignment #19 — Solutions — p.8 ECSE-2410 Signals & Systems - Spring 2009 Due Tue 04/07/09 “MW—W 1 m (a) Derive an exPression for the impulse response 110). Express answer in the form cosQ—g— 1+ 6'), 6 in radians ‘ ESC 4455):. i :W ”£11.? + 5 ans +Zs+x ($+z){§§1rs+i) 93—2 +3}:ij M Cg“? A.) :3. i, {gigs-()4. (SH 3C8$+€J ”521 4:33 + Sari-{1&3} + (jig) sawij wFfiZc/twi‘fifii 5G: fem v g 52‘ (9:: {+3 :7 3:4 a 4(30). Consider a Butterworth Iowpass filter of order n — 3 ,.i e., H (s) = é. ‘. 0 2:: Her-B 0' 7:: =5? 6"" S - l‘t‘C J— _L_ E S+$~L i give. 2. 2,. #19325... ~ 2 z 2 3+: fl 1 2 f(9£i2 'H (sfigzjrg (sir-E)??— 4f 2. 4+ Assignment #19 -— Solutions - p.9 Due Tue 04/07/09 ECSE~24lO Signals & Systems — Spring 2009 “WM—WWW 4. Continued. (b) Use MATLAB to plot h(t) twice as follows: First, using the command ‘Impulse’, and secondly, by Plating your analytical solution in (3). Clearly, you should get the same graph in each case. Impulse response o 'o E '2- < o i ,, , : ., O 1 2 3 4 5 6 7 8 9 10 Time (sec) Analytical Solution 05er } j ,¢,_,,,, , , _, a, ,, 'o 3 3. E < Note that both waveforms are the same! clear all num=[ l ] ; t=[O:0.0l :10]; h:exp(—t)—(2/sqrt( 3 )). *exp(—O.5 *t). *cos(0.5 *sqrt(3)*t+(pi/6)); den=[1 2 2 1 ]; subplot(2, l , l ); impulse(num,den); grid subplot(2, l ,2); grid; plot(t,h); grid; title('Analytical Solution’); ylabel('Amplitude');Xlabel(’Time(sec)'); ...
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