hwk8_solns - . ECE 300 Signals and Systems Homework 8 Due...

Info iconThis preview shows pages 1–17. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . ECE 300 Signals and Systems Homework 8 Due Date: Matlab/Prelab, Tuesday October 30 at the beginning of class Problems 1—6; Thursday November 1 at the beginning of class Problems 1. ZTF, problem 4—4. /2. Using the duality grogegy, find the corresponding Fourier transform for the following: 3) g(t) = sincz(Bt) b) g(t) = sinc(Wt) C) g(t) = 5(1) d) g(t) = cosmot) Do not just look up the pairs from the table (though you can use any other pairs except the one you are trying to find). . f 3. Consider a linear time invariant system with transfer function given by —j2w S 2 Hm) = {Se ml 0 else 2(t-1) 72' ‘with inputx(t) = —8—sincz[ ; 7r . The output of the system is y(t). a) Determine X (0)). b) Sketch the spectrum of X (0)) (magnitude and phase) accurately labeling the axes and important points. c) Sketch the spectrum of H ((0) (magnitude and phase) accurately labeling the axes and important points. d) Determine y(t) , the output of the system. Answer y(t) = Esme (t — 3)] + Esine2 (t — 3)] 7r 7: 7r 7r Fall 2007 @onsider a linear time invariant system with impulse response given by h(t) = isinc£££j with input x(t) = isine cos(t). The output of the system 27: 27! 7! 72’ is y(t). a) Determine X (co) . b) Sketch the spectrum of X (cu) (magnitude and phase) accurately labeling the axes and important points. 0) Determine the energy in x(t) d) Determine H ((0) . e) Sketch the spectrum of H (w) (magnitude and phase) accurately labeling the axes and important points. f) Determine y(t), the output of the system. 9) Determine the energy in y(t). /5. Find the fraction of the total signal energy (as a percentage) contained between 100 and 300 Hz in the signalx(t)given below: x(t)=55inc( t +5sinc( t ) Answer 56% 0.002 0.001 A. In this problem we’ll look at a real world situation when we have to truncate a signal. This actually happens more with digital signal processing, but we can get the basic idea using our continuous time abilities. a) Find an expression for the Fourier transform of f (t) = cos(4t)+cos(5t). b) Now assume we look at f (t) for a finite time, say T seconds. What we see is actually y(t) = f (t)rect(t/T). Determine an expression for the Fourier transform of y(t), and write your answers in terms of sinc functions. c) Plot, using Matlab, Y(a)) for a) between 0 and 10 when T =1, T =6, T=10, T =20, and T =40. Can you clearly tell there are two cosines present when you are looking at Y (w) for all values of T? What happens as T gets larger (you are looking at more and more data)? Think in terms of the width of the sinc function (the distance between the first nulls). Note: The sinc function exists in Matlab. Fall 2007 . tf command and Bode commands will be really useful here. Note that you can click on the curve on the Bode plot to read it more accurately. Turn in your plot. c) Matlab’s command r = pole(H), where H is the transfer function of the Buttewvorth filter, returns the poles of the transfer function in the array r. Using Matlab’s commands abs and angle, relate the magnitude of the poles to (up, then plot the pole locations in the complex plane on a circle with radius mp. (Note that angle returns angles in radians, and you probably want angle in degrees.) Note that the pole locations are all separated by an angleB. What is this angle? d) For mp =15 rad/sec, ms = 35 rad/sec, and A 2 28 dB, determine the required Butterworth filter order for this filter. (Remember it must be an integer). Using the Table at the end of this problem, plot the Bode plot of your Butterworth filter and verify that all frequencies w>ws have magnitude (power) less than Am. Matlab’s tf command and Bode commands will be really useful here. Turn in your plot. e) Matlab’s command r = pole(H), where H is the transfer function of the Butten/vorth filter, returns the poles of the transfer function in the array r. Using Matlab's commands abs and angle, relate the magnitude of the poles to cop, . then plot the pole locations in the complex plane on a circle with radius cop. (Note that angle returns angles in radians, and you probably want angle in degrees.) Note that the pole locations are all separated by an angle6l. What is this angle? n denominator(s) S 1 +1 (up 2 2 {3—) +1.414[i]+1 mp mp 3 2 3 —S— +2 1 +2 1 +1 mp (OP (0P ' 4 3 2 4 i +2.6131 —s— +3.4142 i +2.6131 :3— +1 (Up (Up mp (0P 5 4 3 2 5 i +3.2361—i— +5.2361i +5.2361—i— +3.2361—S- +1 (up 60’, (Up (OF (UP Table 1: Denominators of Buten/vorth filter for filter orders 1—5. The numerator for the Butten/vorth filter is 1. Fall 2007 7. (Matlab/Prelab Problem) A Butteiworth filter has the property that it is maximally flat in the passband. An nth order Butterworth filter has the magnitude squared response 1 —_—2n “[2] (017 where (up is the passband frequency. At this frequency the power has been reduced by one half or 3 dB, |H(w)lz= 1 1 orlolog10 lH(a)p) |2= 101og10 [a] = —3dB |H(60,,)lz=———-= 2n { F] To determine the required order of a filter we often look at the desired stopband frequency, ms. Usually we want to indicate the minimum required power difference between the passband and the stopband, A. A is the rejection.Hence we have i 2 A = zologm |H(0) l —2010g,0 |H(ws)l or A =—1010g10 a) . . . . 5 Is called the transmon rat/o. A 1:1[1010 -1) n = ——————— mp Note that n must be an integer, so we always round up (to the next larger integer). The ratio (up a) Show that we can write b) For wp =10 rad/sec, ms 2 20 rad/sec, and A = 18 dB, determine the required Butterworth filter order for this filter. (Remember it must be an integer). Using the Table at the end of this problem, plot the Bode plot of your Butterworth filter and verify that all frequencies a2>w5 have magnitude (power) less than Amax .Matlab’s Fall 2007 @ 27F paw/97W 9) 4212(an 77g) ,5 aid/4,. my) 1" f ma aim/é -OU 6Q : fir/de/Mfl/é -/ f 1%“) 3/)? £154? -00 - 03 f p)\ mm“ a data” 7» 0 $0 31+er “kg 3““0 : j/rmuoqwfifl ” o be ' 9mg) ws(wt\l£ + 5?,“er 003 (w E‘Léi- ~03 0 WW: 4: . V 0hr : ’0“: i o m f r S «(-03 00$(w(—r\>[—JF) + 54rt>w§lwt\°“' GO /\\ l? ¢ J1 0 Quinn JAN“ {NW hfln‘é‘S 7W mwani m 0Q : 5 «(wow F‘\a€¢+ gomwwmvlc 0 EM : :2 5 wiboongJi— N m (N 24/.» :~ :2 5 «(a w§(—w‘t\p/1§~ 2 2 5 6‘ow w S/w°c\df~ c) 0 Jam ‘5) Wam 147:) a; «xv/an a/aol/r cflraflrwn , $0 mfi'h7qu5 zero ‘ m x 90> 3:395)» (wt>0[tJ/' 1 -j E: S we) gfmw‘Ba-(E 1-; 7995/6 {wflo/a -3g (3 (a! o“: ’ O A Jag ’v/LL I fix) 5m [40 14)) Fair) /;>v : «VHfl’Smw “aim/0”] (>0 : 50 7cmka flue") J”- _ p ’ _ 00 yo 37w) / 2/ ‘g 61793 A; fidfia/ZL ‘- .- ‘ M 1/ W) ’ -1) IO d(t>5}0[—wf>JC/' = ’30“) ‘ IO X/M)J‘o rMégf/vgaf/ fl/yw/fl/j ‘3, o/ w in”) :2 21733180.? (gt é—B 62,va : 2W4 (33> % G//W> : gyS/ficz E ’ = WShCz %f>é~> 6260) :2fl‘g'{_w> :21]. 50 qurrg 5036789 (—9 i g WW ‘ J . @ Calf»: gm» ¥VW a,& =\ 65 Gm} 32? ((2le L3 ocmofillk (31%: Gut) :zirgttB @6203 : 239440} :29‘ « 1 'uJT- ,‘LQT’ 0 L60 . r é—> 67F) 3 izg(t+T) réng) f §o ng : é[§(£+7>+ got: 2014—361”) :: (/03 (Tu)\ % 00mm? 32175 :Cu (£3 3' cog (TL->6"; 611.03 7 ivg‘C—VA :1 41044;); I?) T: v30 wqwotv é? TX Sfipwob-fi; gQQWJQ €(\J;Cm—Qv~—va ‘fgmo (M « @ 19/6) 5 i 607:: ‘2 27%) t 1 5727c(2}‘) LOSGf) : 27 217/ 77 77 50 SHEETS 224 42 100 SHEETS 22-144 200 SHEETS 22141 "‘37 Lgfifl/IPAU” E h ,,,,, i z ,; ....... ,, i oE""3OC’-> ,_ r _, _ - ,,,,,,,,,,,,,,,,,,,,,,,, g/Xf‘t’):5$mc \+§Snag LL. 3 0-002 0:00! (mi? W70 $34045? 1194Cch hoodvxd 300 Ht {of SlfiC (Wt\ <'—-> J. ffic‘f‘ \ \V 7mm J— .5 veg/‘1 W: 3:02.? {00 .01 W: 5‘0“ “£000 7 S fed“ W 4. E— md‘ 2.3.. E3; QTMYOV 1000 'K\\Ooo — . 29.. S 4» .29.. B , a 01 Wed." Qfiflfi) Jr 000 reg, LQRI’DC/o “20300 1 SUV 2n'1f0 O _. ~ 1 v 2 Ufifiu‘ — ~2L~ S \th3\ JR) ii“ 2 86,009}; 1 g [o as) pi“) “Dismay ‘IW‘S‘OO ’IV‘ZYC’ y A. names? F 2“ Maw) (a ms)‘ 96 W 7, 7,\zso« (owogjL P 2‘230‘ (Drag)z;(0°\1$ am -23‘130 “213400 [:8an : éfilzig (OvUSZJc/J + 28 {00531645} *2TT‘700 AZT‘ZC 0 ' -. A. an 026%) (000911“ lbfiflsa {o'oflzl 9/6 MHZ ,,,,,,,,, c z A ' __~ _ : zréoyo’“) klmay‘w") ' i faf/b’ may? 3 O'Séa ar/Zg‘ ,LéJi‘ H ‘ 1W5) j ,- 9% +00 V t @MM 7 MW 1. We) 9‘ 7 {£6 2‘ MWWA T=20 10/16/06 1:27 PM C:\Documents and Settin s\throne\M Document.. .\homework5 sinc.m 1 of 1 % . plot for problem 5 of homework 5 % w: T: m Y1 T: a: Y4O linspace(0,lO,lOOO); T/(2*pi); = @(w) (T/2)*(sinc(a*(w+4))+sinc(a*(w—4))+sinc(a*(w+5))+sinc(a*(w—5))) 6; T/(2*pi); = @(w) (T/2)*(sinc(a*(w+4))+sinc(a*(w~4))+sinc(a*(w+5))+sinc(a*(w—5))) 10; T/(2*pi); = @(w) (T/2)*(sinc(a*(w+4))+sinc(a*(w—4))+sinc(a*(w+5))+sinc(a*(w-5))) 20; T/(2*pi); = @(w) (T/2)*(sinc(a*(w+4))+sinc(a*(w—4))+sinc(a*(w+5))+sinc(a*(w—5))) 40; T/(2*pi); = @(w) (T/2)*(sinc(a*(w+4))+sinc(a*(w—4))+sinc(a*(w+5))+sinc(a*(w—5))) orient tall subplot(5,l,l); plot(w,Yl(w)); grid; title('T = 1'); subplot(5,l,2); plot(w,Y6(w)); grid; title('T = 6'); subplot(5,l,3); plot(w,YlO(w)); grid; title('T = 10'); subplot(5,l,4); plot(w,Y20(w)); grid; title('T = 20'); subplot(5,l,5); plot(w,Y40(w)); grid; title('T = 40'); xlabel('\omega '); RC! QHEETS 100 SHEETS 22—144 200 SHEETS 2244? 22-149 1’57 Lama/14mm w? :}Ofa,./Q/ggc W3 3 10 Waco/5,2; 4'3 +(C/Jg , n ‘: ,Qn inju‘g’) q f 2 I M410) I (2%? WJ@ 3w PW” ‘ r ,0 © 1C°r ‘téu'S mPOKQS‘afi—QJ ’(8) “Sidg'é7ao i TKO MK?W{1«CIQS ¥of PacA (‘3 (O raj/gee = (,3? % TL? vl/xaLé’S crime, 455’on -t10 {100 l <3 Z (hog/QC LUS 7 Kg rape/Q0 d 2 +9.? {I ‘5 QW/QD2;:”‘ : 33 WJ@ C29 50t0+ 2 "‘ K 1 Q fibrfizs RHEMLA FOLK am paé 4. W031? 338322) .. .g,gsg/m:53.mo%f i m mw2m1£ucgfi on} W Bmfl/Lec :uflp q; 'NJMg; 4, 0 ) _, , Bode Diagram —2o— - 3 i E 3 533,} e r 3- K;.',,_,_, " '3 SystemzH ; ; 3 g ; . 3 :3 _ EFrequency(rad/sec): 20 ; '40— ' ' j' 'j ‘j 'j : 1 j: ' j’ Magnitude MEX-18.2E 1' :‘ Magnitude (dB) 6: O I : ~ fPrOblem‘7b‘,f3r‘d fordeeriufte'MOrth filter ~100r' 420 i LiiigLLJ igigiiLL 4 iL___iiiil.1 (b 0 Phase (deg) 480 *“ me; ,4; 10' 1o 10‘ 10 10 Frequency (rad/sec) Magnitude (dB) Phase (deg) 50 in o I 400‘” f Problem 2 d, 4th order BufieMorthifilterf Bode Diagram r Tr‘llir‘ r MVTIV‘F‘ v 1 System: H . Frequency (rad/sec): 35 § Magnitude (dB): -29.5 ‘ 480 -270 - ~360 «.— 10' 1o 101 Frequency (rad/sec) 1O ...
View Full Document

This note was uploaded on 11/04/2008 for the course ECE 300 taught by Professor Throne during the Fall '07 term at Rose-Hulman.

Page1 / 17

hwk8_solns - . ECE 300 Signals and Systems Homework 8 Due...

This preview shows document pages 1 - 17. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online