Fill in the following table for the functions shown below.

Transcription

1 By: Carl H. Durney and Neil E. Coer Example 1 EX: Fill in he following able for he funcions shown below. he funcion is odd he funcion is even he funcion has shif-flip symmery he funcion has quarer-wave symmery a v = 0 (DC offse) All he a k are zero All b k are zero for even-numbered subscrips g() h() rue False rue False g() h()

2 By: Carl H. Durney and Neil E. Coer Example 1 (con.) ANS: g() h() rue False rue False he funcion is odd 1 8 he funcion is even 2 9 he funcion has shif-flip symmery 3 10 he funcion has quarer-wave symmery 4 11 a v = 0 (DC offse) 5 12 All he a k are zero 6 13 All b k are zero for even-numbered subscrips 7 14 SOL'N: Answers are explained per superscrip number. 1 g() is no odd because i is no equal o he original g() afer being flipped around he verical and horizonal axes. 2 g() is even because i is symmerical (and periodic, of course) around he verical axis. Cosines are also even funcions and so consiue he erms of Fourier series for even funcions. 3 Shif-flip symmery means he funcion is equal o a copy of iself ha is shifed one-half cycle o he righ and flipped upside down. A cycle for g() is he widh of one hump. If we shif g() o he righ by half he widh of a hump and hen flip i upside down, hen we obain a funcion ha is always negaive. hus, i is clearly no equal o he original g(). 4 Quarer-wave symmery means he funcion has shif-flip symmery and is symmeric o he lef and righ around he poin a /4 as well as o he lef and righ around he poin a 3/4. (Imagine placing a verical axis a /4 or 3/4 and looking for mirror-image symmery around ha axis.) he period of g()

3 By: Carl H. Durney and Neil E. Coer Example 1 (con.) is one lobe, and we do no have mirror-image symmery around /4 and/or 3/4: We have a zero DC offse only if g() has equal area above and below he horizonal axis. We may hink of g() as being made of buer ha we smooh ou unil i is perfecly fla. If he fla heigh is nonzero, hen he DC offse is no zero. 6 he a k coefficiens are for cosine erms. Since g() is even (and nonzero) and cosine erms are even funcions, we mus have some cosine erms. (he sum of even funcions is an even funcion.) hus, he a k are no all zero. 7 he b k coefficiens for even-numbered subscrips are for sine erms wih an even number of cycles per period of g(). Since g() is even, we have only cosine erms, and all a k are zero wheher k is even or odd. 8 h() is odd because is equal o he original g() afer being flipped around he verical and horizonal axes. 9 h() is no symmerical around he verical axis. hus, i is no even.

4 By: Carl H. Durney and Neil E. Coer Example 1 (con.) 10 Shif-flip symmery means he funcion is equal o a copy of iself ha is shifed one-half cycle o he righ and flipped upside down. he following figures show h() shifed righ by 1/2 cycle and hen flipped upside down: h() one cycle h() shifed righ 1/2 cycle 1/2 cycle one cycle h() shifed righ 1/2 cycle and flipped verically 1/2 cycle one cycle he las funcion is no he same as he original h(). hus h() does no have shif flip symmery.

5 By: Carl H. Durney and Neil E. Coer Example 1 (con.) 11 Clearly, h() is no symmerical abou he quarer-wave poins: We have a zero DC offse if h() has equal area above and below he horizonal axis. We may hink of h() as being made of buer ha we smooh ou unil i is perfecly fla. Clearly, he area under h() above he horizonal will exacly fill he area carved ou by h() below he horizonal axis. hus, he heigh afer flaening is zero, and he DC offse is zero. 13 he a k coefficiens are for cosine erms. Since h() is an odd funcion and sines are odd funcions, we will have only sine erms. hus, he a k are all zero. 14 he b k coefficiens for even-numbered subscrips are for sine erms wih an even number of cycles per period of h(). he figures below show h(), sin(2 2π/) where is he period, and h()sin(2 2π/). he coefficien, b 2, is equal o 2/ imes he area under (i.e., inegral of) one period of h()sin(2 2π/). his area appears o be zero, bu he srange shapes preven a definie conclusion from visual inspecion alone.

6 By: Carl H. Durney and Neil E. Coer Example 1 (con.) h() one cycle sin(2 2π/) h() sin(2 2π/) hus, we ackle he problem mahemaically. b k = 2 0 h()sin k2π d Exploiing symmery around /2 for k even, we have 2 b k = 2 /4 0 4 sin k2π d + 2 /2 /4 1 2 sin k2π d. From inegral ables or a calculaor, we have

7 By: Carl H. Durney and Neil E. Coer Example 1 (con.) x sin(ax)dx = 1 a 2 sin(ax) x a cos(ax). We may also assume ha = 1 wihou affecing our answer. 1/4 b k = 4 4 sin( k2π) d /2 1/4 ( ) sin k2π d 1 b k = 16 (k2π) sin ( k2π ) 2 cos k2π k2π ( ) 1/ /2 cos k2π k2π ( ) 1/4 For k even and = 0, k2π = 0 and sin(k2π) = 0. For k even and = 1/4, k2π = ineger π and sin(k2π) = 0. For = 0, cos(k2π) = 0. hus, we have b k = 16 cos k2π k2π ( ) 1/ /2 cos k2π k2π ( ) 1/4 b k = 4 cos( k2π /4) 2 1 cos( k2π /2) cos( k2π /4) k2π k2π k2π b k = 2 cos( k2π /4) 2 1 cos( k2π /2) k2π k2π For k = 2, we have b 2 = 1 2π cos ( π ) 1 cos( 2π) = 1 ( 1+ 1) = 0 2π 2π

8 By: Carl H. Durney and Neil E. Coer Example 1 (con.) So i seems ha b k = 0 migh be rue for all k even. Considering k = 4, however, we have b 4 = 1 cos( 2π) 1 cos( 4π) = 1 1 (1+ 1) = 4π 4π 4π 2π. I follows ha b 4 is no zero, and he b k for k even are no all zero. Noe ha his problem exhibied an unusual symmery ha made he firs even numbered b coefficien zero. Someimes, he mah is necessary. he sandard ypes of symmery and he picures ha go wih hem, however, end o give resuls ha are more obvious. Noe: In his problem, we could acually deermine ha b 2 is zero by observing ha he iniial sloped par of h() is symmeric around is cener poin locaed a /8. he firs hump in sin(2 2π/) is also symmeric around /8. he heigh of h() a /8 is equal o he heigh of he fla segmen ha follows. As we move o he lef and righ of /8, we will muliply sin(2 2π/) by values ha are equally above and below he heigh of h() a /8. hus, when we compue he inegral of (i.e., area under) h() sin(2 2π/) from 0 o /4, we will have exacly he same value as we have for he inegral of h(/8) sin(2 2π/) from 0 o /4. In oher words, we ge he same answer as we would ge if we used a consan value for h(), and ha consan value is exacly he heigh of he fla segmen ha follows he iniial sloped par of h(). By symmery, we conclude ha we may replace he enire h() by he consan value h(/8). Now b 2 is seen o be proporional o he inegral of a consan imes sin(2 2π/). he inegral of any sinusoid of an ineger number of cycles is zero, however, so we conclude ha b 2 is zero.