6.003 Homework #9 Solutions

Transcription

1 6.00 Homework #9 Soluions Problems. Fourier varieies a. Deermine he Fourier series coefficiens of he following signal, which is periodic in 0. x () 0 0 a 0 5 a k sin πk 5 sin πk 5 πk for k 0 a k 0 πk j e 0 e j πk 0 jπk πk j e 0 d πk j e 0 d 0 πk πk j e j 0 e 0 jπk πk j e 0 j πk 0 πk sin 5 sin πk 5 πk + 0 πk j e 0 j πk 0 b. Deermine he Fourier ransform of he following signal, which is zero ouside he indicaed range. x () X () sin sin X () e d + e j e e d e + e e j + e sin sin

2 6.00 Homework #9 Soluions / Fall 0 c. Wha is he relaion beween he answers o pars a and b? In paricular, derive an expression for a k (he soluion o par a) in erms of X () (he soluion o par b). a k πk j 0 X 0 he Fourier series coefficiens a k are a k 0 X () πk 0 d. Deermine he ime waveform ha corresponds o he following Fourier ransform, which is zero ouside he indicaed range. X () x () sin sin π x () π e j e j jπ e d + π + e j e j jπ e d e jπ sin sin π + e jπ e. Wha is he relaion beween he answers o pars b and d? In paricular, derive an expression for x () (he soluion o par d) in erms of X () (he soluion o par b). x () X j( ) or X j() π π he relaion beween his answer and ha of he previous par is dualiy. x () π X () Since X () is real and even, π X (j) would also work.

3 6.00 Homework #9 Soluions / Fall 0. Fourier ransform properies Le X() represen he Fourier ransform of e 0 < < x(). 0 oherwise Express he Fourier ransforms of each of he following signals in erms of X(). x () X () X() + X( ) x () x() + x( ) x( ) x( )e d X () X() + X( ) x () x()e d X( ) X () X() X( ) x () x() x( ) X () X() X( ) x () X () ( + e )X() x () x() + x( + ) x( + ) x( + )e d X () ( + e )X() x()e ( ) d e X()

4 6.00 Homework #9 Soluions / Fall 0. Fourier ransforms Find he Fourier ransforms of he following signals. 4 a. x () e cos() X () + ( ) + + ( + ) e u() e cos() πδ( ) + πδ( + ) herefore, by he muliplicaion propery, e cos() + ( ) + + ( + ) sin(π) b. x () π( ) X () e (u( + π) u( π)) sin(π) π u( + π) u( π) sin(π) sin(π π) sin(π( )) e (u( + π) u( π)) π( ) π( ) π( )

5 6.00 Homework #9 Soluions / Fall 0 { c. x 0 < < () 0 oherwise j X () e + e + j ( e ) 5 x () (u() u( )) u() + πδ() u() u( ) + πδ() e e πδ() e d f() j F () d d f() F () d ( ) d ( e ) u() u( ) d j j j x( ) e + e ( + e ) d. x 4 () ( ) u( + )u( ) X 4 () ( cos ) Le p() u( + 0.5) u( 0.5). hen x 4 () p() p() and X 4 () P (). e P ( ) e d 4 sin ( cos ) X 4 ( j ) sin You could also solve his by differeniaing wice in he ime domain o ge a sequence of dela funcions, compuing he ransform, and muliplying wice by [he dela funcions in he inegraion propery have zero weigh].

6 6.00 Homework #9 Soluions / Fall 0 Engineering Design Problem 4. Parseval s heorem Parseval s heorem relaes ime- and frequency-domain mehods for calculaing he average energy of a signal as follows: 0 x() d a k where a k represens he Fourier series coefficiens of he periodic signal x() wih period. a. We can derive Parseval s heorem from he properies of C Fourier series.. Le y() x(). Find he Fourier series coefficiens b k of y(). [Hin: x() x()x ().] b. Le x () represen he inpu o an LI sysem, where x () x( ) 0 α k e jk π 4 for 0 < α <. he frequency response of he sysem is H() x () b k a k e k a e j < W 0 oherwise. π k y()e j k d x()x ()e π π j k π j le π a j l π a me m j k e l m a a e j π ( k l+m) l m d l m a l a mδ[k l + m] l m k m a +m a m. Use he resul from he previous par o derive Parseval s heorem. j π k x() d b 0 a k a k a k d d 6

7 6.00 Homework #9 Soluions / Fall 0 Wha is he minimum value of W so ha he average energy in he oupu signal will be a leas 90% of ha in he inpu signal. he signal x () is periodic wih period 8 and has Fourier series coefficiens a k α k. he average energy in he inpu signal is α x() d a k k + α + α α α he lowpass filer passes some number K of he harmonic componens of x() so ha he average energy in he oupu signals is K+ + K+ K K y() d α k α + α + α α α α α k K o make he energy in he oupu a leas 90% of ha in he inpu + α α K+ ( + α ) 0.9 α α + α α K+ 0.9( + α ) α K+ 0. ( + α ) + (K + ) log α α log 0 log + α 0 K > (inequaliy swiches because he logs are negaive) log α Because he harmonics are spaced a π ( ) log + α 0 W > log α π 8 inervals in frequency 7

8 6.00 Homework #9 Soluions / Fall 0 5. Filering he poin of his quesion is o undersand how he magniude of a filer affecs he oupu and how he angle of a filer affecs he oupu. Consider he following RC circui as a filer. R + v + i v o C 8 Assume ha he inpu v i () is he following square wave. v i () 0 If he fundamenal frequency of he square wave ( π ) is equal o he cuoff frequency of he RC circui ( RC ) hen he oupu v o() will have he following form. v o () 0 We can hink of he RC circui as filering he square wave as shown below. H() H() π /RC /RC he RC filer has wo effecs: () he ampliudes of he Fourier componens of he inpu (verical red lines in upper panel) are muliplied by he magniude of he frequency response ( H() ). () he phase of he Fourier componens (red dos in lower panel) are shifed by he phase of he frequency response ( H()). a. Deermine (using whaever mehod you find convenien) he oupu ha would resul if v i () were passed hrough a filer whose magniude is H() (as above) bu whose phase funcion is 0 for all frequencies. Compare he resul wih v o () above.

9 6.00 Homework #9 Soluions / Fall 0 he following plo shows he sum of he firs 46 erms of he series expansion for he square wave, wih each erm filered by he magniude (bu no he phase) of he RC lowpass filer. he original oupu is also shown (dashed green) for reference. v o () 0 he asymmery in he charging and discharging porions of he RC response is gone. he effec of he filer is o reduce he magniudes of he higher harmonics wihou adding phase delay. b. Deermine (using whaever mehod you find convenien) he oupu ha would resul if v i () were passed hrough a filer whose phase funcion is H() (as above) bu whose magniude funcion is for all frequencies. Compare he resul wih v o () above. he following plo shows he sum of he firs 46 erms of he series expansion for he square wave, wih each erm filered by he phase (bu no he magniude) of he RC lowpass filer. 9 v o () 0 he asymmery in he charging and discharging porions of he RC response is even more pronounced han before. Because he magniudes of he higher harmonics are no aenuaed, hey now accumulae o make a subsanial peak ha was no seen in he RC response. We can undersand he large overshoo as follows. Each harmonic componen in he square wave is a sinusoid: v i () ( ) πk sin k πk. k odd Excep for he fundamenal, he phase shifs are nearly a quarer cycle. Delaying each of he harmonic componens by a quarer cycle aligns he peaks of each componen a as shown below

10 6.00 Homework #9 Soluions / Fall 0 0 v x () 0 where v x () ( ) πk sin πk π/ k k odd.

11 MI OpenCourseWare hp://ocw.mi.edu 6.00 Signals and Sysems Fall 0 For informaion abou ciing hese maerials or our erms of Use, visi: hp://ocw.mi.edu/erms.