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# Linear System Analysis I ECE 311

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GPA 3.8

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This 42 page Class Notes was uploaded by Sarina Wintheiser on Tuesday September 22, 2015. The Class Notes belongs to ECE 311 at Colorado State University taught by Ali Pezeshki in Fall. Since its upload, it has received 6 views. For similar materials see /class/210273/ece-311-colorado-state-university in ELECTRICAL AND COMPUTER ENGINEERING at Colorado State University.

## Reviews for Linear System Analysis I

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Date Created: 09/22/15

9 Fourier Transform Properties Recommended Problems P91 Determine the Fourier transform of xt equot 2ut and sketch a IXCw b lttXw c ReXw d I mXw P92 Figure P92 shows real and imaginary parts of the Fourier transform of a signal xt ReXco lmXco 1 1 0 w W W W W Figure P92 a Sketch the magnitude and phase of the Fourier transform Xw b In general if a signal xt is real then X co Xw Determine Whether 33t is real for the Fourier transform sketched in Figure P92 P93 Determine which of the Fourier transforms in Figures P931 and P932 correspond to realvalued time functions a Xw 4X30 Figure P931 P9l Signals and Systems P92 P94 b Re Xw XRw 1mXw X1w 2W Figure P932 P95 a By considering the Fourier analysis equation or synthesis equation show the validity in general of each of the following statements i If xt is realvalued then Xw X w ii If xt x t then Xw is realvalued b Using the statements in part a show the validity of each of the following statements i If xt is real and even then Xw is real and even ii If xt is real and odd then Xw is imaginary and odd a In the lecture we derived the transform of xt e ut Using the linearity and scaling properties derive the Fourier transform of e 39 900 x t b Using part a and the duality property determine the Fourier transform of 11 t2 c If 1 T m 5a nd Rw d xt is sketched in Figure P95 If yt ova2 sketch yt Yw and Xw xt A Figure P95 P96 Fourier Transform Properties Problems P93 P97 Show the validity of the following statements 1 00 a 5130 E JmXw dw 00 b X0 Lo xt dt P98 The output of a causal LTI system is related to the input xt by the differential equation dyt d t 290 960 a Determine the frequency response Hw YwXw and sketch the phase and magnitude of Hw b If xt e ut determine Yw the Fourier transform of the output c Find yt for the input given in part b By rst expressing the triangular signal xt in Figure P98 as the convolution of a rectangular pulse with itself determine the Fourier transform of xt XU j Figure P98 Optional Problems P99 Using Figure P991 determine yt and sketch Yw if Xw is given by Figure P992 Assume wc gt wo Signals and Systems P94 P910 P911 xm Q A m V cos wct Figure P99l Xw Tl39 quot JO 0 0 Figure P992 Compute the Fourier transform of each of the following signals a e cos wotut a gt 0 b 9 339 sin 2t c sin wt sin 21d 1rt 7rt P912 Consider the following linear constantcoef cient differential equation LCCDE d t y 2yt A cos wot dt Find the value of 00 such that yt will have a maximum amplitude of A3 Assume that the resulting system is linear and timeinvariant Suppose an LTI system is described by the following LCCDE d2yt 2dyt 4dxt dt2 dt 3ym dt xt Fourier Transform Properties Problems P95 a Show that the lefthand side of the equation has a Fourier transform that can be expressed as AwYw Where Yw 7yt Find Aw b Similarly show that the righthand side of the equation has a Fourier transform that can be expressed as BwXw Where Xw 57xt c Show that Yw can be expressed as Yw HwXw and nd Hw P913 From Figure P913 nd yt Where x smtw0t and Mt Slnw0t xt gt ht L gt yt Figure P913 P914 a Determine the energy in the signal 30t for which the Fourier transform Xw is given by Figure P914 l l I quot l 1 3 Figure P914 b Find the inverse Fourier transform of Xw of part a Signals and Systems P96 P915 Suppose that the system F takes the Fourier transform of the input as shown in Figure P9151 xt gt F gt yt21rX w Figure P9151 What is wt calculated as in Figure P9152 x t gt F gt 1 gt F gt w t Figure P9152 P916 Use properties of the Fourier transform to show by induction that the Fourier trans form of tn l at xt n 1 e ut a gt O is Xw l a Mquot MIT OpenCourseWare httpocw mit edu Resource Signals and Systems Professor Alan V Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare but has been provided by the author as an individual learning resource For information about citing these materials or our Terms of Use visit httpocwmiteduterms 9 Fourier Transform Properties Solutions to Recommended Problems S91 The Fourier transform of xt is co Xwf xte j quot dtf e 2ute j dt 891 1 Since ut 0 for t lt 0 eq 8911 can be rewritten as D XOD I e l2jwtdt 0 2 l ij It is convenient to write Xw in terms of its real and imaginary parts 2 1 j2co 2 340 1ij 1j2w 14 2 2 4w 1 4w2 1 4032 X00 7 2 V1 4w2 Xw tan 2w tan 2w 2 4w I 1 402 mm 1 4002 Magnitude of Xw ReXw 30 IX w F to Figure S911 b Figure S912 S9l Signals and Systems 892 0 ReXw Figure S9l3 d Im Jaw 1 I NIH 1 Figure 8914 892 a The magnitude of X0 is given by Xw VXiw X w Where X 30 is the real part of Xw and Xw is the imaginary part of Xw It follows that 2 M lt w Xw to gt W IXw 7 0 w W Figure 8921 The phase of Xw is given by 1 X10 1 ltZXw tan XRw tan 1 col lt W Fourier Transform Properties Solutions ZQXw 1 i w W W Figure 8922 b 1 39 lt W Xw a M O otherWISe 1 j lwl lt W X w 0 otherw1se 1 j w lt W Xogt 0 otherWISe Hence the signal is not real 893 For xt to be realvalued Xw is conjugate symmetric Xw Xw a Xw lXwe X Xw COSltfXw j Xw I sinltIXw Therefore Xw X wlcosltIX w jXwlsinltIXw XwCOSltIXw J39lXwlsinltIXw Xw Hence xt is realvalued b Xw XRCw J39XKw Xw XIXco jX1w XRw jXw 27r for w gt O Xw XRw jXKw Therefore Xw 95 X w Hence xt is not realvalued 94 00 a i Xw Lo xte39j dt We take the complex conjugate of both sides to get 00 Xw 00 xtej dt 893 Signals and Systems S94 ii b i ii 895 Since xt is realvalued X Xw f ooxtej dt Therefore 00 X w I xte j quotdt X80 1 xt J Xwequot quot do 27139 00 Taking the complex conjugate of both sides we have 1 xt I Xwe quot dw 27139 00 Therefore 1 x t J Xwequotquot do 277 oo Since xt x t we have 1 1 f Xwequot quot do I Xwequotquot dw 27139 00 27F 00 This shows that Xw must be realvalued Since xt is real Xw X w Since xt is real and even it satis es xt x t and therefore Xw is real Hence Xw X w X w It follows that Xw is real and even If xt is real Xw X w Since xt is real and odd xt x t an analysis similar to part aii proves that Xw must be imaginary Hence Xw X w Xw It follows that Xw is also odd a 7e quot39 939e ut equot u t 1 l ajw ajw 2a a2ogt2 b Duality states that Since 57 90 9 00 Gt lt3 21rg w 2a a2w2 e altl E a Fourier Transform Properties Solutions S95 we have 1 12 lt gt Te lwl 1 7 1 E7 1 w lw3 9 c 1 302 3 we smce xat lal Xltagt d We are given Figure S951 xl Figure S95l T i A Xw A J 6 le dt 39e wT erT jw SinwT wT Sketches of yt Yw and Xw are given in Figure 8952 A 2TA Xw quot 1 V T T Mw yt 4TA 1 lt gt A A w 2r 1 iv V 2T 2T Figure 8952 Signals and Systems S96 Substituting 2T for T in Xco we have sinw2T Yco 22T wZT The zero crossings are at szT War or wz n 96 1 t a xt XweJ do 27F 00 Substituting t 0 in the preceding equation we get 1 00 x0 J Xw dw 27F 00 b Xw J xte 1 dt Substituting 0 0 in the preceding equation we get XO f xt dt 897 a We are given the differential equation d t 3 2W xt 3971 Taking the Fourier transform of eq 8971 we have ijw 2Yw Xw Hence Yw2 I39ij X00 and Yw 1 H w Xw 2 jw 1 1 2 jw 2 jw H w 2jw 2jw2 jwgt 4co2 2 w 4 02 J 4 092 Fourier Transform Properties Solutions S97 4 w2 4 w2 4 w22 4 w22 quot 4 w22 IHGMN2 l Hw W HUGH 1 2 Amw tan12 n r N12 M Figure S97 b We are given xt e ut Taking the Fourier transform we obtain Xw Hw 1 jw 2 jw Hence 1 1 1 Y 0 1jw2jw 1jw 2jw c Taking the inverse transform of Yw we get W e ut ue Signals and Systems S98 S98 A triangular signal can be represented as the convolution of two rectangular pulses as indicated in Figure 898 1 1 1 1 Figure 898 Since each of the rectangular pulses on the right has a Fourier transform given by 2 sin ww the convolution property tells us that the triangular function will have a Fourier transform given by the square of 2 sin ww 4 sin2w X0 2 0 Solutions to Optional Problems 899 We can compute the function xt by taking the inverse Fourier transform of Xw xt 1 mo we dw 21r WC 1 1 wot ont 2 Ce 6 Sin wot t Therefore W sin 000 cos wet t Fourier Transform Properties Solutions From the multiplicative property we have Yw Xw 160 we 1r5w 06 Yw is sketched in Figure 899 i 2 l l l T 0 3900 w39 wL 030 we 030 we C 030 Figure 899 910 a xt 6 cos wotut a gt 0 e utcosw0t Therefore 1 Xw 2 1r a jw 1r5w 00 7r5w 00 12 12 a jw no a jw coo b xt 6 339 sin 2t 6 9 02 e 3ltl lt gt i7 mmt wmwa Mwn 2 Therefore 9 02 6w we so won j3 j3 9m2 9m 2f l X00 sin 1rt sin 21rt 9 x 7 t 1 Xw 392 X1w X2w 1r Where 1 col lt 1r X1 O otherw1se 1 col lt 21r X26 O otherWISe 899 Signals and Systems 8910 Hence Xw is given by the convolution shown in Figure 8910 X1w X2w 1 1 77 77 21r 2n X to 1 i t 39 0 31r 211 1r 1T 21r 37r Figure 910 S9ll We are given the LCCDE d t L 2yt A cos wot dt We can View the LCCDE as d t 2W mo the transfer function of which is given by Hw and xt A cos wot 2 jw We have already seen that for LTI systems 210 Hw0l A cosw0t where as ltIHwo 2A COSw0t p 1 V4w0 Fourier Transform Properties Solutions For the maximum value of yt to be A3 we require 1 4w COIH Therefore coo i 8912 2 a 7 dillth Egg2 3yt w2Yw 2ijw 3Yw wu2 ij 3Yw Aw w2 j2w 3 b 57 xt 4ijw Xw 740 1Xw Bw j4w 1 ACwYw BwXw 393 Yw AOO Xw HwXw Therefore 1 140 Had Aw w2 3 ij l j4w wz 3 j2w S9l3 1 sin Wt I 3 l l I 1139 W W Figure S9l3l X 60 7T 39 t m 111 0 H quotquot0 Q 0 Figure S9132 8911 Signals and Systems 8912 Hw 2 Sln w t TrTQ A L zwo ZwO Figure S9133 Yw 772 Yw XwHw J L co w0 w0 Figure S9134 39 t Therefore yt 2 914 1 2 a Energy 2 IXw dw 77 00 mm 4 1 1 w 2 1 l 2 Figure S914l Area 42 211 10 5 Energy 7T Fourier Transform Properties Solutions 8913 b Xw 1 1 2 2 Figure S9l42 sin t sin 21 330 7rt 1rt 915 Given that 31 27rX wwt we have 00 y1t 2 f u xuej du Similarly let y2t be the output due to passing xt through F twice 312 21quot I 21r I xuej du em dv v m u CD 2702 xuJ ej dv du 2102 xu21r6t u du 2103 x t Finally let y3t be the output due to passing xt through F three times 21 27r3x uej du u w y3t wt 21r4 J 00 3quot xu du 27F4Xt 8916 We are given n l t at xt e ut a gt O Signals and Systems 8914 H H Let n xt emu Xogt agt0 ajw ll 3 Let n xt te ut d Xwjl w 1 mmr Assume it is true for n 7 d 39n tact 9 39X a jw SI ce de w n 1 L 1 jwquot 900 e mu Xw We consider the case for n 1 xt file tum X0 ii ml joy mva cwmawrrv 1 a jwnl Therefore it is true for all n MIT OpenCourseWare httpocw mit edu Resource Signals and Systems Professor Alan V Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare but has been provided by the author as an individual learning resource For information about citing these materials or our Terms of Use visit httpocwmiteduterms 8 ContinuousTime Fourier Transform Recommended Problems P8l P82 Consider the signal xt which consists of a single rectangular pulse of unit height is symmetric about the origin and has a total width T1 a Sketch xt b Sketch 50t which is a periodic repetition of xt with period T0 3T12 c Compute Xw the Fourier transform of xt Sketch Xw for le S 61rT1 d Compute ak the Fourier series coef cients of Mt Sketch ak for k O t 1 i 2 i3 e Using your answers to c and d verify that for this example 1 a Firm 0 w 21rkT0 f Write a statement that indicates how the Fourier series for a periodic function can be obtained if the Fourier transform of one period of this periodic function is given P83 Find the Fourier transform of each of the following signals and sketch the magni tude and phase as a function of frequency including both positive and negative frequencies a 5t 5 b 639 ut c e lj2tut a real positive In this problem we explore the de nition of the Fourier transform of a periodic signal a Show that if x3t axlt bx2t then X3w aX1w bX2w b Verify that 1 em E 21r5w w0eJ do 7139 00 From this observation argue that the Fourier transform of 9 is 215w coo c Recall the synthesis equation for the Fourier series 00 cat Z akejk21rTt k oo P8l Signals and Systems P82 By taking the Fourier transform of both sides and using the results to parts a and b show that 00 2 Xw Z 21rak5 w k oo 1 Sketch Xw for your answer to Problem P81d for w S 41rT0 P84 a Consider the oftenused alternative de nition of the Fourier transform which we will call Xa f The forward transform is written as 00 Xaoc L woe 1392 dt where f is the frequency variable in hertz Derive the inverse transform formula for this de nition Sketch XaCf for the signal discussed in Problem P81 b A second alternative de nition is Xb39v xte 3 dt 00 P805 Consider the periodic signal t in Figure P851 which is composed solely of impulses i0 f11a11fnljifm 7 6 5 4 3 2 1 0 1 2 3 Figure P851 a What is the fundamental period To b Find the Fourier series of 520 c Find the Fourier transform of the signals in Figures P852 and P853 i 3511 1 2th l O 1 Figure P852 ContinuousTime Fourier Transform Problems P83 ii x2 I Nlr 1ll 0 l 2 3 4 5 Figure P853 d 50t can be expressed as either x1t periodically repeated or x2t periodically repeated ie 00 out Z 9010 km or P85 1 k oo I 56t Z 0620 kn P85 2 k oo Determine T1 and T2 and demonstrate graphically that eqs P851 and P85 2 are valid e Verify that the Fourier series of 56t is composed of scaled samples of either X1w or X2w P86 Find the signal corresponding to the following Fourier transforms a Xaoo 7 Ma 390 Xbltw 2 w 7 7 Figure P861 c Xena 9 wz See Example 48 in the text page 191 d Xdw XawXbw Where Xaw and Xbw are given in parts a and b respec tively Try to simplify as much as possible Signals and Systems P84 e IXew A X e w 1 o co 1 1 3w Figure P862 Optional Problems P87 In earlier lectures the response of an LTI system to an input xt was shown to be yt xt ht where ht is the system impulse response a Using the fact that 20 96t ht f x7ht 7 d7 show that Yogt f f xrht e39j d1 dt b By appropriate change of variables show that Yw XwHw where Xw is the Fourier transform of xt and Hw is the Fourier transform of ht P88 Consider the impulse train pa Z 6t kT k oo shown in Figure P881 ContinuousTime Fourier Transform Problems P85 W I I 1 I 1 t 2T T 0 T 2T Figure P88l a Find the Fourier series of pt b Find the Fourier transform of pt c Consider the signal xt shown in Figure P882 where Tl lt T x t I t T 11 i T 2 2 Figure P882 Show that the periodic signal 500 formed by periodically repeating xt satis es 560 x0 pt 1 Using the result to Problem P87 and parts b and c of this problem nd the Fourier transform of 9t in terms of the Fourier transform of xt MIT OpenCourseWare httpocw mit edu Resource Signals and Systems Professor Alan V Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare but has been provided by the author as an individual learning resource For information about citing these materials or our Terms of Use visit httpocwmiteduterms 8 ContinuousTime Fourier Transform Solutions to Recommended Problems 81 a xt T1 T1 2 2 Figure 8811 Note that the total Width is T1 00 t 1 L a l amp 4119 2 1 3 T1 3T1T0 2 2 2 2 2 2 Figure 1312 c Using the de nition of the Fourier transform we have m T12 Xw I xte j quotdt I le jw dt since xt 0 for ltl gt T2 le m2 1 2 sm 7 ij12 eijlZ e m jw See Figure 8813 TlZ JO 0 S8l Signals and Systems 882 sin T 7 mam w T1 6Tr 21 1 3 T1 T1 T1 T1 T1 T1 Figure 8813 d Using the analysis formula we have 1 ak I Ete 1quot 0 dt T0 T0 where we integrate over any period T02 1 T12 ak f 2006 jk21rTot dt f e jk21rTotdt To To2 To T12 ak i 1 e jkrT1T0 ejkrTlTo Sin k77T1T0 Sin 7r2153 To a Wk Wk J To sin 1r 21 Iakl 3 rrk Z 3 0 m z I 1139 I 1 J TIL v T y k 3 2 1 0 1 2 3 Figure 8814 Note that ak 0 Whenever 21103 1rm for m a nonzero integer e Substituting 27rkT0 for 0 we obtain 1 2 sin7rkT1T0 sin 1rkT1T0 a w21rkT0 T0 27rkT0 7716 k 1 F0 Xw f From the result of part e we sample the Fourier transform of xt Xw at w 27rlcT0 and then scale by 1 To to get ak ContinuousTime Fourier Transform Solutions S83 882 a Xw I xte 1 dt J 6t 5e 1 dt e j5 cos 5w j sin 5w by the sifting property of the unit impulse Xw lej5 1 for all 0 1 ImXw 1 sin 5w 4mquot ta ReXwgt quot ta cossw39 5w IXw AXQ J 1 w u Figure ss21 J 6 ute 139 dt J 9 e 1 quot dt 0 00 b X0 00 ajwt 00 f e awt e 0 O a jw Since Rea gt 0 e goes to zero as t goes to in nity Therefore 1 XwajwO 1 ajw X X X 12 1 1 gt121 w w w jaw awn Xw Xw a 2 a2 wg Xw Xw w 2 quot a2 wz 1 12 ltIXw tan R6Xw tan a The magnitude and angle of Xw are shown in Figure 882 2 Signals and Systems 884 X w l a I 4L J AXw I quot 4 I a L I w a n I 2 i Figure 8822 oo 00 J e1j2tute jwt I e lj2te jwt 0 CXJ 0 X0 00 1 el lj2 wlt 1 2 m Since Re 1 j2 0 lt O limb0 e j2 quot 0 Therefore 0 X m 1X0 XltwXwgtr2 Remw X Em 1 C 2y Immw X X w1 22 ltIXw tan tan w 2 The magnitude and angle of Xw are shown in Figure 8823 ContinuousTime Fourier Transform Solutions Xw A X to 1 4 T F w 1 2 3 Z 1 Figure 8823 Note that there is no symmetry about 0 0 since xt is not real S83 a X3w f x3te 139 dt Substituting for 0030 we obtain 00 X3w le ax1t bx2te 1 quot dt I ax1te j39 dt J bx2te 1 quot dt aj xte M dt b J x2te W dt aX1w bX2w b Recall the sifting property of the unit impulse function I ht6t to dt ht0 Therefore I 21500 w0ej quot dw 21rej 0t 885 Signals and Systems 886 Thus 1 t t 2150 w0eJ dw 6quot 0 277 00 Note that the integral relating 27r6w 00 and ejwo is exactly of the form 1 xt f Xwequotquot dw 27F 00 Where xt ej and Xw 2160 wo Thus we can think of em as the inverse Fourier transform of 216m 00 Therefore 21r5w no is the Fourier transform of em c Using the result of part a we have X00 34590 57 Z akejk 239T Z ak i7 ejk27quot k oo oo From part b 21k 3 k21rTt 2 6 e 7r w T gt Therefore 2 Xw Z 27rak5 w k oo T d X 2 i1 2 sin 2 quot 3 2 sin 2 quot 3 3 A 22 T 411 4n Slfl 3 111 3 Figure 83 884 a We see that the new transform is XaCf X00 w21rf We know that 1 0 xt i f Xwequot quot do ContinuousTime Fourier Transform Solutions 887 Let w 21rf Then do 21rdf and l oo x0 quot2 7 I X21rfe12 27r df Jw Xafe121rft df oo Thus there is no factor of 27r in the inverse relation IXfl T1 quot3F il 1 T1 Figure 884 b Comparing Xbv xte m dt and Xw J xte 139 dt 1 or Xw 27rXbw 1 Xb v 3X00 f The inverse transform relation for Xw is 011 1 1 0 xt J Xwequot dw J 27r Xbwequot quot dw 27139 oo 21F 00 1 I m X velv dv W a Where we have substituted 1 for 0 Thus the factor of 1 27r has been distributed among the forward and inverse transforms 885 a By inspection T0 6 l b ak I m e Jquot2 T dt T0 T0 We integrate from 3 to 3 1 3 1 1 k 2 6t ak 6t16t 5t 1 9 quot dt 6 3 2 2 1 1 1 1 27k J2 n k6 J21rk6 626 129 6lt1cos 6 Signals and Systems S88 so 00 1 M jk27r6z xt kw6ltl cos 6 6 c 1 X100 f x1te j dt 0 6t 1 6t lad 1e jw dt ejw1 e j 1cosw ii X200 J x2te M dt J 600 6t 1 gm 5e w dt 1 e j e d We see that by periodically repeating x1t with period T1 6 we get 500 as shown in Figure S851 3611 1 l 2 l T l O l x1t 6 U1 a Oi NH H IL N x1t6 1 l 2 I g A 7 6 5 0 5ct l l ij i 2i J JJJ 7 6 5 l 0 1 5 6 7 Figure S851 Similarly we can periodically repeat 3320 to get 5t Thus T2 6 See Figure 8852 ContinuousTime Fourier Transform Solutions 889 xzt 39 6 5 1 32t l L I t t f f f 6 5 1 0 l 5 6 7 11 Figure 8852 e Since 7 ct is a periodic repetition of x1t or 7620 the Fourier series coef cients of 7 ct should be expressible as scaled samples of X1w Evaluate X1w at w 27rk6 Then 2 k 1 2 X1w 1 cos quot 6akgtak XI w21rk6 6 6 6 Similarly we can get ak as a scaled sample of X2w Consider X221rk6 21rk 1 1 121rk6 110rk6 X2 6 1 2 e 2 e e lewkS e j101rk621rk ej21rk6 2 k 21rlc X2 1 cos 6 6 Although X10 a X209 they are equal for w 21rk6 Signals and Systems 8810 86 a By inspection 57 e ut lt gt 1 30 Thus 7t 57 e ut lt gt 7 jw Direct inversion using the inverse Fourier transform formula is very dif cult b Xbw 25w 7 26w 7 x0 1 1 f Xbwequot do I 25w 7 5w 7e1w dw 27 oo 21F 00 l 1 2 e J7 e cos 7t 7r 1r 1r c From Example 48 of the text page 191 we see that 37 2a lt gt a2 2 at 6 However note that 7 let lt gt since I axte 1 dt a I xte 1 dt ozXw Thus 1 i7 1 1 7 alt 9 gt 3ltl 2a 6 a2 w2 9 2 e d XawXbw Xaw26w 7 253 7 2Xa75w 7 2Xa76w 7 Xdw 50 7 5 3 3750 7 2 7 j7 1 2 2 5 7 jwt d xda 21rfm 7 j7 50 7 7 37 to e w j7t 7 j7e 7j7 l wdt 1 Note that e j7t 1r4 ej7t vr4 cos 7t Z 71r 4 ContinuousTime Fourier Transform Solutions SSll toe 3 05005 1 e Xew we j3 l lt w 5 0 0 elsewhere 1 1 l 0 0 quot J Xwe quot dw we 1393quot dw I we quotj3wej quot do 21r oo 21r o 1 Note that fwe dxi7ax l Substituting a jt 3 into the integrals we obtain ejt 3w jt 3w WU 3w 1 which can be simpli ed to yield cost 3 1 sin t 3 t 32 t 3 0 y 1 2 t 3w 1 1 l x ii 0 quot an 3 xt Solutions to Optional Problems 887 a Yw Jm yte 391 dt LOW 100 xrht 1 d1 e j dt 00 00 x7ht Te 3 d1 dt b Let r t 1 and integrate for all 139 and 739 Then Yw lo w fw xrhre M M dr d7 I00 xre 139 d7 00 hre 139 dr XEZSZm S88 a Using the analysis equation we obtain T2 1 a 5 t e jquot2T dt k T T2 T Thus all the Fourier series coef cients are equal to 1 T b For periodic signals the Fourier transform can be calculated from ak as Xw 21r Z ak5 w k oo Signals and Systems 8812 In this case k 00 T P w 2 7r F 7 Q 31 21 0 21 3 T T T T Figure S88 c We are required to show that 5w Mt pt Substituting for pt we have xtpt xt Z 6t m k 00 Using the associative property of convolution we obtain mo pa Z W so km k 00 From the sifting property of 6t it follows that 00 mo pa Z w M out k 00 Thus xt pt is a periodic repetition of xt with period T d From Problem P87 we have X o XwPw Xw i gg ltw 2 Egt k 00 0 27r 27rk X 6 kZ oo T w w T gt Since each summation term is nonzero only at w 2716 T 0 21r 27rk 27146 X X 5 T w T From this expression we see that the Fourier series coef cients of 50t are l 21rk X ak T lt T gt which is consistent with our previous discussions MIT OpenCourseWare httpocw mit edu Resource Signals and Systems Professor Alan V Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare but has been provided by the author as an individual learning resource For information about citing these materials or our Terms of Use visit httpocwmiteduterms ECE311 FS Problems Fall Semester 20147 CSU Ft Collins DTFS Problems 1 Find the DTFS coef cients of the following signals a 1 sin cos gtw4 i Mni namp 2 Use the de nition of DTFS to determine the time signals represented by the following DTFS coef cients a X kl J39sin 42 608 k b Xk 6k 7 2m 7 2m 3m 3 Prove the Parseval property for DTFS 1 N Z lrlnllZ 2 Wk 2 nltNgt kltNgt Given 1M zMi E Xw s1n En evaluate with the following DTFS coef cients using only the DTFS properties a Yk Xk75Xk5 b Yk Xk Xk 4 Problem 351 from textbook 5 Problem 337 from textbook CTFS Problems 1 Problem 322c from textbook 2 Problem 324 from textbook 3 Problem 334 from textbook 4 Problem 341 from textbook 5 Problem 362 from textbook 6 Problem 363 from textbook

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