6.003 Homework #8 Solutions

Transcription

1 6.003 Homework #8 Soluions Problems. Fourier Series Deermine he Fourier series coefficiens a k for x () shown below. x ()= x ( + 0) 0 a 0 = 0 a k = e /0 sin(/0) for k 0 a k = π x()e k d = 0 0 π e 0 k d = 0 e π 5 k π 5 k 0 = ( e /5) j Noice ha his expression is badly formed a k = 0. We could use l Hôpial s rule o evaluae his expression, bu an easier mehod (which is also more robus agains errors) is o simply evaluae he average value of x () o find ha a 0 = /0. his soluion could also be wrien in erms of sinusoids as a k = { 0 k = 0 e/0 sin(/0) k 0.

2 6.003 Homework #8 Soluions / Fall 0 Deermine he Fourier series coefficiens b k for x () shown below. x ()= x ( + 0) 0 b 0 = 5 b k = e /5 sin(/5) for k 0 b k = π x()e k d = 0 0 π e 0 k d = 0 e π 5 k π 5 k 0 = ( e /5) j As wih he previous par, his expression is badly formed for k = 0. We herefore obain b 0 = /5 by calculaing he average value of x (). his soluion could also be wrien in erms of sinusoids as { b k = 5 k = 0 e/5 sin(/5) k 0.

3 6.003 Homework #8 Soluions / Fall 0 3 Deermine he Fourier series coefficiens c k for x 3 () shown below. x 3 ()= x 3 ( + 0) 3 0 c 0 = 0 c k = j e3/0 sin(/5) sin(/0) for k 0 x 3 () = x () x ( ) π x ( )e k d = π c k = a k e k a k = π x ()e k(+) π d = e k π x ()e k π d = e k a k ( e /5) ( e /5) j he average value of x 3 () is zero, so c 0 = 0. his soluion could also be wrien in erms of sinusoids as { 0 k = 0 c k = j e3/0 sin(/5) sin(/0) k 0.

4 6.003 Homework #8 Soluions / Fall 0 Deermine he Fourier series coefficiens d k for x () shown below. x ()= x ( + 0) 3 0 d 0 = 5 d k = 0 π k e 3/0 sin(/5) sin(/0) for k 0 x 3 () = dx () d c k = j π kd k { 5 k = 0 d k = c j π k k = 5 ( π k e /5 ) ( e /5) k 0 his soluion could also be wrien in erms of sinusoids as { d k = 5 k = 0 0 e 3/0 sin(/5) sin(/0) k 0. π k

5 6.003 Homework #8 Soluions / Fall 0 5. Inverse Fourier series Deermine he C signals wih he following Fourier series coefficiens. Assume ha he signals are periodic in =. Ener an expression ha is valid for 0 < (oher values can be found by periodic exension). { a. a k = jk; k < 3 0 oherwise x() = sin(π/) sin(π) for 0 <. x() = a k e j π k = je π je π + je j π + je j π = sin(π/) sin(π) b. b k = { ; k odd 0; k even x() = δ() δ( ) for 0 <. x() = b k e j π k = k = k odd e j/ Unforunaely, his sum is no easy o close. However, i is closely relaed o he synhesis formula for an impulse rain, x δ () = δ( k ) = a k e j/ = e j/. If here were wo impulses per period insead of one, hen x δ () = δ( k ) + δ( k ) = e j/ + e j( )/ = e j/ ( + e j) = k = k even e j/ I follows ha x() = x δ () x δ() = x δ () x δ () so ha x() is an alernaing sequence of impulses x() = δ( l) δ( l) l=

6 6.003 Homework #8 Soluions / Fall Maching Consider he following Fourier series coefficiens. a k b k = { 3 jk k = ±, ±, ±, ±5, ±7,... 0 k = 0, ±3, ±6, k k c k d k k k e k f k k k a. Which coefficiens (if any) corresponds o he following periodic signal? ( ) π x () = cos 3 a k, b k, c k, d k, e k, f k, or None: e k From he consan, i is clear ha he zero coefficien is. Since cos θ = e jθ + eθ, he coefficiens for k = ± are. herefore he answer is e k.

7 6.003 Homework #8 Soluions / Fall 0 7 b. Which coefficiens (if any) corresponds o he following periodic signal wih period = 3? x () a k, b k, c k, d k, e k, f k, or None: f k his signal is real and even, so is FS coefficiens mus be real and even. Also, he signal has no DC value, so he zero coefficien mus be zero. his leaves a k or c k or f k. he Fourier series for an impulse rain mus have infinie exen. herefore he bes candidae is f k. Solving a k = π x ()e k d = (δ() δ( ) δ( + )) e π 3 k d = 3 ( cos(/3)) = { 0 if k is evenly divisible by 3 oherwise So he answer is f k. c. Which (if any) se corresponds o he following periodic signal wih period = 3? x 3 () π 0 3 π a k, b k, c k, d k, e k, f k, or None: b k he signal is real and odd, so is FS coefficiens mus be purely imaginary and odd. hus he only candidae is b k. Solving a k = π x 3 ()e k d = 0 πe π 3 k d + πe π 3 k d = ( e /3 0 ) jk e/3 = 0 jk ( e j/3 e /3 ) = { 0 if k is evenly divisible by 3 ( cos(/3)) = jk 3/jk oherwise So he answer is b k.

8 6.003 Homework #8 Soluions / Fall 0 8. Inpu/Oupu Pairs he following signals are periodic wih period =. x ()= x ( + ) 0 x ()= x ( + ) 0 x 3 ()= x 3 ( + ) 0 π 0 Deermine if he following sysems could or could no be linear and ime-invarian (LI). We can use he filer idea as follows. Firs calculae he Fourier series coefficiens. hen ask if each Fourier series coefficien in he oupu is a scaled version of he corresponding coefficien in he inpu. x () a k = e π k d = sin k = 0 = k =, 5, 9, 3,... k = 3, 7,, 5,... 0 k =,, 6, 8,... x () = y() y() where y() is he following signal: y()= y( + ) 0 8 y() d k = 8 8 x () b k = d k x 3 () c k = π 0 e π k d = sin sin = π k = π 0 e π k d = sin π k 0 k = 0 k =, 3, 5, 7, 9,, 3,... π k k =, 6, 0,,... π k 0 k =, 8,, 6,...

9 6.003 Homework #8 Soluions / Fall 0 9 x () Sysem A x () he Fourier series coefficiens a k =, 6, 0,... are zero in x bu hese are no zero in x. herefore he sysem could no be LI. x () Sysem B x 3 () he Fourier series coefficiens a k =,, 6, 8, 0,... are zero in x bu hese are no zero in x 3. herefore he sysem could no be LI. x () Sysem C x () All of he nonzero Fourier coefficiens in x are also presen in x. herefore he sysem could be LI. x () Sysem D x 3 () he Fourier series coefficiens a k =, 8,, 6,... are zero in x bu hese are no zero in x 3. herefore he sysem could no be LI. x 3 () Sysem E x () All of he nonzero Fourier coefficiens in x are also presen in x 3. herefore he sysem could be LI. x 3 () Sysem F x () All of he nonzero Fourier coefficiens in x are also presen in x 3. herefore he sysem could be LI. Ener a lis of he sysems ha could NO be LI. If your lis is empy, ener none. answer = A, B, D

10 6.003 Homework #8 Soluions / Fall 0 0 Engineering Design Problems 5. Overshoo a. Wha funcion f() has he Fourier series sin n n? n= You can evaluae he sum analyically or numerically. Eiher way, guess a closed form for f() and hen skech i. Here is a plo of he sum of he firs 30 erms of f(). π I shows ha f() is a sawooh waveform wih period = π and f() = π for 0 < < π. b. Confirm your conjecure for f() by finding he Fourier series coefficiens f n for f(). Compare your resul o he expression in he previous par. Wha happens o he cosine erms? Le g() = df() d. hen g() = +πδ() for π < < π. he Fourier series coefficiens of g() are hen { 0 k = 0 a k = k 0 For k 0, we can find he Fourier series coefficiens for f() by dividing hose for g() by jk. he funcion f() has no average value, so he Fourier coefficien for k = 0 is zero. hus he Fourier series coefficiens for f() are { 0 k = 0 b k = jk k 0 hus he Fourier series is ( f() = e jk e k) = jk k= k= k sin(k) he cosine erms are all zero because he funcion is odd. c. Define he parial sum f N () = N n= sin n n,

11 6.003 Homework #8 Soluions / Fall 0 Plo some f N () s. By wha fracion does f N () overshoo f() a wors? Does ha fracion end o zero or o a finie value as N? If i is a finie value, esimae i. he following plos show he parials sums of 0, 80, and 60 erms. I appears ha he maximum value is approximaely.09 imes he value of f(), i.e., he overshoo is approximaely 9%, as in he Gibb s overshoo o a square wave..09f() I appears ha he maximum value is approximaely.09 imes he value of f(), i.e., he overshoo is approximaely 9%, as in he Gibb s overshoo o a square wave. d. Now define he average of he parial sums: F N () = f () + f () + f 3 () + + f N () N Plo some F N () s. Compare your plos wih hose of f N () ha you made in he previous par, and qualiaively explain any differences. he following plos show F 0, F 80, and F 60.

12 6.003 Homework #8 Soluions / Fall 0.09f() Convergence is much smooher han i was for f 0, f 80, and f 60. he overshoo is gone. here is a sligh undershoo.