Do not worry about your difficulties in mathematics, I assure you that mine are greater.

Transcription

1 C H A P E R HE FOURIER SERIES 6 Do no worry abou your difficulies in mahemaics, I assure you ha mine are greaer. Alber Einsein Hisorical Profiles Jean Bapise Joseph Fourier (768 83), a French mahemaician, firs presened he series and ransform ha bear his name. Fourier s resuls were no enhusiasically received by he scienific world. He could no even ge his work published as a paper. Born in Auxerre, France, Fourier was orphaned a age 8. He aended a local miliary college run by Benedicine monks, where he demonsraed grea proficiency in mahemaics. Like mos of his conemporaries, Fourier was swep ino he poliics of he French Revoluion. He played an imporan role in Napoleon s expediions o Egyp in he laer 79s. Due o his poliical involvemen, he narrowly escaped deah wice. Alexander Graham Bell (847 9) invenor of he elephone, was a Scoish- American scienis. Bell was born in Edinburgh, Scoland, a son of Alexander Melville Bell, a well-known speech eacher. Alexander he younger also became a speech eacher afer graduaing from he Universiy of Edinburgh and he Universiy of London. In 866 he became ineresed in ransmiing speech elecrically. Afer his older broher died of uberculosis, his faher decided o move o Canada. Alexander was asked o come o Boson o work a he School for he Deaf. here he me homas A. Wason, who became his assisan in his elecromagneic ransmier experimen. On March, 876, Alexander sen he famous firs elephone message: Wason, come here I wan you. he bel, he logarihmic uni inroduced in Chaper 4, is named in his honor. 77

2 78 PAR 3 Advanced Circui Analyses 6. INRODUCION We have spen a considerable amoun of ime on he analysis of circuis wih sinusoidal sources. his chaper is concerned wih a means of analyzing circuis wih periodic, nonsinusoidal exciaions. he noion of periodic funcions was inroduced in Chaper 9; i was menioned here ha he sinusoid is he mos simple and useful periodic funcion. his chaper inroduces he Fourier series, a echnique for expressing a periodic funcion in erms of sinusoids. Once he source funcion is expressed in erms of sinusoids, we can apply he phasor mehod o analyze circuis. he Fourier series is named afer Jean Bapise Joseph Fourier (768 83). In 8, Fourier s genius came up wih he insigh ha any pracical periodic funcion can be represened as a sum of sinusoids. Such a represenaion, along wih he superposiion heorem, allows us o find he response of circuis o arbirary periodic inpus using phasor echniques. We begin wih he rigonomeric Fourier series. Laer we consider he exponenial Fourier series. We hen apply Fourier series in circui analysis. Finally, pracical applicaions of Fourier series in specrum analyzers and filers are demonsraed. 6. RIGONOMERIC FOURIER SERIES While sudying hea flow, Fourier discovered ha a nonsinusoidal periodic funcion can be expressed as an infinie sum of sinusoidal funcions. Recall ha a periodic funcion is one ha repeas every seconds. In oher words, a periodic funcion f()saisfies f()= f( + n ) (6.) where n is an ineger and is he period of he funcion. According o he Fourier heorem, any pracical periodic funcion of frequency ω can be expressed as an infinie sum of sine or cosine funcions ha are inegral muliples of ω. hus, f()can be expressed as f()= a + a cos ω + b sin ω + a cos ω (6.) + b sin ω + a 3 cos 3ω + b 3 sin 3ω + or f()= a }{{} + (a n cos nω + b n sin nω ) n= dc }{{} ac (6.3) he harmonic frequency ω n is an inegral muliple of he fundamenal frequency ω, i.e., ω n = nω. where ω = π/ is called he fundamenal frequency in radians per second. he sinusoid sin nω or cos nω is called he nh harmonic of f(); i is an odd harmonic if n is odd and an even harmonic if n is even. Equaion 6.3 is called he rigonomeric Fourier series of f(). he consans a n and b n are he Fourier coefficiens. he coefficien a is he dc componen or he average value of f(). (Recall ha sinusoids

3 CHAPER 6 he Fourier Series 79 have zero average values.) he coefficiens a n and b n (for n ) are he ampliudes of he sinusoids in he ac componen. hus, he Fourier series of a periodic funcion f () is a represenaion ha resolves f () ino a dc componen and an ac componen comprising an infinie series of harmonic sinusoids. A funcion ha can be represened by a Fourier series as in Eq. (6.3) mus mee cerain requiremens, because he infinie series in Eq. (6.3) may or may no converge. hese condiions on f()o yield a convergen Fourier series are as follows:. f()is single-valued everywhere.. f()has a finie number of finie disconinuiies in any one period. 3. f()has a finie number of maxima and minima in any one period he inegral f() d < for any. hese condiions are called Dirichle condiions. Alhough hey are no necessary condiions, hey are sufficien condiions for a Fourier series o exis. A major ask in Fourier series is he deerminaion of he Fourier coefficiens a,a n, and b n. he process of deermining he coefficiens is called Fourier analysis. he following rigonomeric inegrals are very helpful in Fourier analysis. For any inegers m and n, sin nω d= cos nω d= sin nω cos mω d= (6.4a) (6.4b) (6.4c) sin nω sin mω d=, (m n) (6.4d) cos nω cos mω d=, (m n) (6.4e) sin nω d= cos nω d= (6.4f) (6.4g) Hisorical noe: Alhough Fourier published his heorem in 8, i was P. G. L. Dirichle (85 859) who laer supplied an accepable proof of he heorem. A sofware package like Mahcad or Maple can be used o evaluae he Fourier coefficiens. Le us use hese ideniies o evaluae he Fourier coefficiens.

4 7 PAR 3 Advanced Circui Analyses We begin by finding a. We inegrae boh sides of Eq. (6.3) over one period and obain [ f()d = a + (a n cos nω + b n sin nω ) d = a d + n= [ a n cos nω d n= + b n sin nω d d (6.5) Invoking he ideniies of Eqs. (6.4a) and (6.4b), he wo inegrals involving he ac erms vanish. Hence, or f()d = a d = a a = f()d (6.6) showing ha a is he average value of f(). o evaluae a n, we muliply boh sides of Eq. (6.3) by cos mω and inegrae over one period: f()cos mω d [ = a + (a n cos nω + b n sin nω ) cos mω d = + n= a cos mω d+ [ a n cos nω cos mω d n= b n sin nω cos mω d d (6.7) he inegral conaining a is zero in view of Eq. (6.4b), while he inegral conaining b n vanishes according o Eq. (6.4c). he inegral conaining a n will be zero excep when m = n, in which case i is /, according o Eqs. (6.4e) and (6.4g). hus, f()cos mω d= a n, for m = n or a n = f()cos nω d (6.8) In a similar vein, we obain b n by muliplying boh sides of Eq. (6.3) by sin mω and inegraing over he period. he resul is b n = f()sin nω d (6.9)

5 CHAPER 6 he Fourier Series 7 Be aware ha since f()is periodic, i may be more convenien o carry he inegraions above from /o/ or generally from o + insead of o. he resul will be he same. An alernaive form of Eq. (6.3) is he ampliude-phase form f()= a + A n cos(nω + φ n ) (6.) n= We can use Eqs. (9.) and (9.) o relae Eq. (6.3) o Eq. (6.), or we can apply he rigonomeric ideniy cos(α + β) = cos α cos β sin α sin β (6.) o he ac erms in Eq. (6.) so ha a + A n cos(nω + φ n ) = a + (A n cos φ n ) cos nω n= n= (6.) (A n sin φ n ) sin nω Equaing he coefficiens of he series expansions in Eqs. (6.3) and (6.) shows ha or a n = A n cos φ n, b n = A n sin φ n (6.3a) A n = a n + b n, φ n = an b n a n (6.3b) o avoid any confusion in deermining φ n, i may be beer o relae he erms in complex form as A n φ n = a n jb n (6.4) he convenience of his relaionship will become eviden in Secion 6.6. he plo of he ampliude A n of he harmonics versus nω is called he ampliude specrum of f(); he plo of he phase φ n versus nω is he phase specrum of f(). Boh he ampliude and phase specra form he frequency specrum of f(). he frequency specrum of a signal consiss of he plos of he ampliudes and phases of he harmonics versus frequency. he frequency specrum is also known as he line specrum in view of he discree frequency componens. hus, he Fourier analysis is also a mahemaical ool for finding he specrum of a periodic signal. Secion 6.6 will elaborae more on he specrum of a signal. o evaluae he Fourier coefficiens a, a n, and b n, we ofen need o apply he following inegrals: cos a d = sin a a (6.5a) sin a d = cos a a (6.5b)

6 7 PAR 3 Advanced Circui Analyses cos a d = a cos a + sin a a (6.5c) sin a d = a sin a cos a a (6.5d) I is also useful o know he values of he cosine, sine, and exponenial funcions for inegral muliples of π. hese are given in able 6., where n is an ineger. ABLE 6. Values of cosine, sine, and exponenial funcions for inegral muliples of π. Funcion Value cos nπ sin nπ cos nπ ( ) n sin nπ { ( ) n/, n = even cos nπ, n = odd sin nπ e jnπ ( ) n e jnπ { ( ) (n )/, n = odd, n = even { ( ) n/, n = even e jnπ/ j( ) (n )/, n = odd E X A M P L E 6. f() 3 Figure 6. For Example 6.; a square wave. Deermine he Fourier series of he waveform shown in Fig. 6.. Obain he ampliude and phase specra. Soluion: he Fourier series is given by Eq. (6.3), namely, f()= a + (a n cos nω + b n sin nω ) (6..) n= Our goal is o obain he Fourier coefficiens a,a n, and b n using Eqs. (6.6), (6.8), and (6.9). Firs, we describe he waveform as {, << f()= (6..), << and f()= f( + ). Since =, ω = π/ = π. hus, a = f()d = [ d + d = = (6..3)

7 CHAPER 6 he Fourier Series 73 Using Eq. (6.8) along wih Eq. (6.5a), a n = f()cos nω d = [ cos nπ d + = sin nπ nπ = sin nπ = nπ From Eq. (6.9) wih he aid of Eq. (6.5b), b n = f()sin nω d = [ sin nπ d + = cos nπ nπ = (cos nπ ), nπ = nπ [ ( )n = cos nπ d sin nπ d cos nπ = ( )n nπ, n = odd, n = even (6..4) (6..5) Subsiuing he Fourier coefficiens in Eqs. (6..3) o (6..5) ino Eq. (6..) gives he Fourier series as f()= + π sin π + sin 3π + sin 5π + (6..6) 3π 5π Since f()conains only he dc componen and he sine erms wih he fundamenal componen and odd harmonics, i may be wrien as f()= + π k= sin nπ, n n = k (6..7) By summing he erms one by one as demonsraed in Fig. 6., we noice how superposiion of he erms can evolve ino he original square. As more and more Fourier componens are added, he sum ges closer and closer o he square wave. However, i is no possible in pracice o sum he series in Eq. (6..6) or (6..7) o infiniy. Only a parial sum (n =,, 3,...,N, where N is finie) is possible. If we plo he parial sum (or runcaed series) over one period for a large N as in Fig. 6.3, we noice ha he parial sum oscillaes above and below he acual value of f(). A he neighborhood of he poins of disconinuiy (x =,,,...), here is overshoo and damped oscillaion. In fac, an overshoo of abou 9 percen of he peak value is always presen, regardless of he number of erms used o approximae f(). his is called he Gibbs phenomenon. Summing he Fourier erms by hand calculaion may be edious. A compuer is helpful o compueheermsandplohesumlikehoseshown in Fig. 6.. Hisorical noe: Named afer he mahemaical physicis Josiah Willard Gibbs, who firs observed i in 899.

8 74 PAR 3 Advanced Circui Analyses f() dc componen Fundamenal ac componen (a) Figure 6.3 runcaing he Fourier series a N = ; Gibbs phenomenon. Sum of firs wo ac componens Sum of firs hree ac componens Finally, le us obain he ampliude and phase specra for he signal in Fig. 6.. Since a n =, A n = an + b n = b n = nπ, n = odd (6..8), n = even and φ n = an b { n 9, n = odd = a n, n = even (6..9) he plos of A n and φ n for differen values of nω = nπ provide he ampliude and phase specra in Fig Noice ha he ampliudes of he harmonics decay very fas wih frequency. A n p Sum of firs four ac componens.5 3p 5p p p 3p (a) 4p 5p 6p v Sum of firs five ac componens Figure 6. (b) Evoluion of a square wave from is Fourier componens. f 9 Figure 6.4 p p 3p 4p 5p 6p (b) For Example 6.: (a) ampliude and (b) phase specrum of he funcion shown in Fig. 6.. v

9 CHAPER 6 he Fourier Series 75 P R A C I C E P R O B L E M 6. Find he Fourier series of he square wave in Fig Plo he ampliude and phase specra. Answer: f() = 4 sin nπ, n = k. See Fig. 6.6 for he specra. π n k= f() 3 v A n 4 p Figure 6.5 For Pracice Prob. 6.. f 4 3p 4 5p p p 3p 4p 5p 6p v p p 3p 4p 5p 6p (a) v 9 (b) Figure 6.6 For Pracice Prob. 6.: ampliude and phase specra for he funcion shown in Fig E X A M P L E 6. Obain he Fourier series for he periodic funcion in Fig. 6.7 and plo he ampliude and phase specra. Soluion: he funcion is described as f()= Since =, ω = π/ = π. hen a = f()d = [ d+ {, <<, << d = = 4 o evaluae a n and b n, we need he inegrals in Eq. (6.5): a n = f()cos nω d = [ cos nπ d + = cos nπ d [ cos nπ + sin nπ n π nπ = n π (cos nπ ) + = ( )n n π (6..) (6..) f() 3 Figure 6.7 For Example 6..

10 76 PAR 3 Advanced Circui Analyses since cos nπ = ( ) n ; and b n = f()sin nω d = [ sin nπ d + = sin nπ d [ sin nπ cos nπ n π nπ = cos nπ nπ = ( )n+ nπ Subsiuing he Fourier coefficiens jus found ino Eq. (6.3) yields f()= [ [( ) n 4 + (nπ) n= cos nπ + ( )n+ nπ sin nπ (6..3) o obain he ampliude and phase specra, we noice ha, for even harmonics, a n =,b n = /nπ, so ha A n p p 3p (a) 4p 5p 6p f v Hence, A n φ n = a n jb n = + j nπ A n = b n =, n =, 4,... nπ φ n = 9, n =, 4,... For odd harmonics, a n = /(n π ), b n = /(nπ) so ha ha is, A n φ n = a n jb n = n π j nπ 4 A n = an + b n = n 4 π + 4 n π = 4 + n π n π, n =, 3,... (6..4) (6..5) (6..6) (6..7) From Eq. (6..6), we observe ha φ lies in he hird quadran, so ha p p 3p 4p 5p 6p v (b) Figure 6.8 For Example 6.: (a) ampliude specrum, (b) phase specrum. φ n = 8 + an nπ, n =, 3,... (6..8) From Eqs. (6..5), (6..7), and (6..8), we plo A n and φ n for differen values of nω = nπ o obain he ampliude specrum and phase specrum as shown in Fig P R A C I C E P R O B L E M 6. Deermine he Fourier series of he sawooh waveform in Fig. 6.9.

11 CHAPER 6 he Fourier Series 77 Answer: f()= π n= sin πn. n f() 3 Figure 6.9 For Pracice Prob SYMMERY CONSIDERAIONS We noiced ha he Fourier series of Example 6. consised only of he sine erms. One may wonder if a mehod exiss whereby one can know in advance ha some Fourier coefficiens would be zero and avoid he unnecessary work involved in he edious process of calculaing hem. Such a mehod does exis; i is based on recognizing he exisence of symmery. Here we discuss hree ypes of symmery: () even symmery, () odd symmery, (3) half-wave symmery Even Symmery A funcion f()is even if is plo is symmerical abou he verical axis; ha is, f()= f( ) (6.6) Examples of even funcions are, 4, and cos. Figure 6. shows more examples of periodic even funcions. Noe ha each of hese examples saisfies Eq. (6.6). A main propery of an even funcion f e () is ha: / / f e () d = / f e () d (6.7) because inegraing from / o is he same as inegraing from o /. Uilizing his propery, he Fourier coefficiens for an even funcion become a = a n = 4 b n = / / f()d f()cos nω d (6.8) Since b n =, Eq. (6.3) becomes a Fourier cosine series. his makes sense because he cosine funcion is iself even. I also makes inuiive sense ha an even funcion conains no sine erms since he sine funcion is odd. o confirm Eq. (6.8) quaniaively, we apply he propery of an even funcion in Eq. (6.7) in evaluaing he Fourier coefficiens in Eqs. (6.6), (6.8), and (6.9). I is convenien in each case o inegrae over he inerval / <</, which is symmerical abou he origin. hus, f() A A g() A (a) (b) h() A p p p p (c) Figure 6. ypical examples of even periodic funcions.

12 78 PAR 3 Advanced Circui Analyses a = / / f()d = [ f()d + / / f()d (6.9) We change variables for he inegral over he inerval / <<by leing = x, so ha d = dx, f()= f( ) = f(x), since f()is an even funcion, and when = /,x = /. hen, [ / a = = f(x)( dx) + / [/ f(x)dx+ / showing ha he wo inegrals are idenical. Hence, a = as expeced. Similarly, from Eq. (6.8), a n = [ f()cos nω d+ / f()d (6.) f()d / f()d (6.) / f()cos nω d (6.) We make he same change of variables ha led o Eq. (6.) and noe ha boh f()and cos nω are even funcions, implying ha f( ) = f() and cos( nω ) = cos nω. Equaion (6.) becomes a n = [ / f( x)cos( nω x)( dx) + f()cos nω d / [ / or = = f(x)cos(nω x)( dx) + / [/ f(x)cos(nω x)dx + / / f()cos nω d f()cos nω d (6.3a) a n = 4 f()cos nω d (6.3b) as expeced. For b n, we apply Eq. (6.9), b n = [ / f()sin nω d+ f()sin nω d (6.4) / We make he same change of variables bu keep in mind ha f( ) = f()bu sin( nω ) = sin nω. Equaion (6.4) yields b n = [ / f( x)sin( nω x)( dx) + f()sin nω d / = [ / f(x)sin nω xdx+ f()sin nω d / = [ / f(x)sin(nω x)dx + / f()sin nω d = (6.5) confirming Eq. (6.8).

13 6.3. Odd Symmery A funcion f()is said o be odd if is plo is anisymmerical abou he verical axis: CHAPER 6 he Fourier Series 79 f( ) = f() (6.6) Examples of odd funcions are, 3, and sin. Figure 6. shows more examples of periodic odd funcions. All hese examples saisfy Eq. (6.6). An odd funcion f o () has his major characerisic: / / f o () d = (6.7) because inegraion from / o is he negaive of ha from o /. Wih his propery, he Fourier coefficiens for an odd funcion become b n = 4 a =, a n = / f()sin nω d (6.8) f() A A (a) g() A A which give us a Fourier sine series. Again, his makes sense because he sine funcion is iself an odd funcion. Also, noe ha here is no dc erm for he Fourier series expansion of an odd funcion. he quaniaive proof of Eq. (6.8) follows he same procedure aken o prove Eq. (6.8) excep ha f() is now odd, so ha f() = f(). Wih his fundamenal bu simple difference, i is easy o see ha a = in Eq. (6.), a n = in Eq. (6.3a), and b n in Eq. (6.4) becomes b n = [ [ = = f( x)sin( nω x)( dx) + / / [/ f(x)sin nω xdx+ f(x)sin(nω x)dx + / / / / f()sin nω d f()sin nω d f()sin nω d b n = 4 f()sin nω d (6.9) as expeced. I is ineresing o noe ha any periodic funcion f()wih neiher even nor odd symmery may be decomposed ino even and odd pars. Using he properies of even and odd funcions from Eqs. (6.6) and (6.6), we can wrie f()= [f()+ f( ) + [f() f( ) = f e () + f o () (6.3) } {{}} {{} even odd Noice ha f e () = [f()+ f( ) saisfies he propery of an even funcion in Eq. (6.6), while f o () = [f() f( ) saisfies he propery of an odd funcion in Eq. (6.6). he fac ha f e () conains h() A (b) A Figure 6. (c) ypical examples of odd periodic funcions.

14 7 PAR 3 Advanced Circui Analyses only he dc erm and he cosine erms, while f o () has only he sine erms, can be exploied in grouping he Fourier series expansion of f()as f()= a + a n cos nω + b n sin nω = f e () + f o () (6.3) n= } {{ } even n= } {{ } odd I follows readily from Eq. (6.3) ha when f()is even, b n =, and when f()is odd, a = = a n. Also, noe he following properies of odd and even funcions:. he produc of wo even funcions is also an even funcion.. he produc of wo odd funcions is an even funcion. 3. he produc of an even funcion and an odd funcion is an odd funcion. 4. he sum (or difference) of wo even funcions is also an even funcion. 5. he sum (or difference) of wo odd funcions is an odd funcion. 6. he sum (or difference) of an even funcion and an odd funcion is neiher even nor odd. Each of hese properies can be proved using Eqs. (6.6) and (6.6) Half-Wave Symmery A funcion is half-wave (odd) symmeric if f ( ) = f() (6.3) which means ha each half-cycle is he mirror image of he nex halfcycle. Noice ha funcions cos nω and sin nω saisfy Eq. (6.3) for odd values of n and herefore possess half-wave symmery when n is odd. Figure 6. shows oher examples of half-wave symmeric funcions. he funcions in Figs. 6.(a) and 6.(b) are also half-wave symmeric. Noice ha for each funcion, one half-cycle is he invered f() A g() A A A (a) (b) Figure 6. ypical examples of half-wave odd symmeric funcions.

15 CHAPER 6 he Fourier Series 7 version of he adjacen half-cycle. he Fourier coefficiens become a = 4 / f()cos nω d, for n odd a n =, for n even 4 / f()sin nω d, for n odd b n =, for n even (6.33) showing ha he Fourier series of a half-wave symmeric funcion conains only odd harmonics. o derive Eq. (6.33), we apply he propery of half-wave symmeric funcions in Eq. (6.3) in evaluaing he Fourier coefficiens in Eqs. (6.6), (6.8), and (6.9). hus, a = / / f()d = [ f()d + / / f()d (6.34) We change variables for he inegral over he inerval / << by leing x = + /, so ha dx = d; when = /,x = ; and when =,x = /. Also, we keep Eq. (6.3) in mind; ha is, f(x /) = f(x). hen, a = = [/ [ f / ( x f(x)dx+ ) dx + / / f()d f()d = confirming he expression for a in Eq. (6.33). Similarly, a n = [ / f()cos nω d+ f()cos nω d / (6.35) (6.36) We make he same change of variables ha led o Eq. (6.35) so ha Eq. (6.36) becomes a n = [/ ( f x ) ( cos nω x ) dx / (6.37) + f()cos nω d Since f(x /) = f(x) and ( cos nω x ) = cos(nω nπ) = cos nω cos nπ + sin nω sin nπ = ( ) n cos nω subsiuing hese in Eq. (6.37) leads o (6.38)

16 7 PAR 3 Advanced Circui Analyses a n = [ ( )n / f()cos nω d 4 / (6.39) f()cos nω = d, for n odd, for n even confirming Eq. (6.33). By following a similar procedure, we can derive b n as in Eq. (6.33). able 6. summarizes he effecs of hese symmeries on he Fourier coefficiens. able 6.3 provides he Fourier series of some common periodic funcions. ABLE 6. Effecs of symmery on Fourier coefficiens. Symmery a a n b n Remarks Even a a n b n = Inegrae over / and muliply by o ge he coefficiens. Odd a = a n = b n Inegrae over / and muliply by o ge he coefficiens. Half-wave a = a n = b n = Inegrae over / and muliply a n+ b n+ by o ge he coefficiens. ABLE 6.3 Funcion he Fourier series of common funcions. Fourier series. Square wave f() A f()= 4A π n= n sin(n )ω. Sawooh wave f() A 3. riangular wave f()= A A π n= sin nω n f() A f()= A 4A π n= (n + ) cos(n )ω

17 CHAPER 6 he Fourier Series 73 ABLE 6.3 (coninued) Funcion Fourier series 4. Recangular pulse rain f() A f()= Aτ + A n= nπτ sin n cos nω 5. Half-wave recified sine f() A f()= A π + A sin ω A π n= 4n cos nω 6. Full-wave recified sine f() A f()= A π 4A π n= 4n cos nω E X A M P L E 6. 3 Find he Fourier series expansion of f()given in Fig f() Figure 6.3 For Example 6.3. Soluion: he funcion f()is an odd funcion. Hence a = = a n. he period is = 4, and ω = π/ = π/, so ha

18 74 PAR 3 Advanced Circui Analyses P R A C I C E P R O B L E M 6. 3 Hence, / b n = 4 f()sin nω d = 4 [ sin nπ 4 d+ sin nπ d = nπ f()= π cos nπ n= which is a Fourier sine series. = nπ ( cos nπ n ( cos nπ ) sin nπ Find he Fourier series of he funcion f()in Fig ) f() π π π π 3π Figure 6.4 For Pracice Prob Answer: f()= 4 π k= sin n, n = k. n E X A M P L E 6. 4 Deermine he Fourier series for he half-wave recified cosine funcion shown in Fig f() 5 3 Figure A half-wave recified cosine funcion; for Example 6.4. Soluion: his is an even funcion so ha b n =. Also, = 4,ω = π/ = π/. Over a period,

19 CHAPER 6 he Fourier Series 75, << f()= cos π, <<, << a = / f()d = [ cos π 4 d+ = π sin π = π a n = 4 / f()cos nω d= 4 [ 4 d cos π nπ cos d + Bu cos A cos B = [cos(a + B) + cos(a B). hen a n = [cos π (n + ) + cos π (n ) d For n =, For n>, a = a n = [cos π + d = [ sin π π + = π(n + ) sin π (n + ) + π(n ) sin π (n ) For n = odd (n =, 3, 5,...), (n + ) and (n ) are boh even, so sin π (n + ) = = sin π (n ), n = odd For n = even (n =, 4, 6,...), (n + ) and (n ) are boh odd. Also, sin π (n + ) = sin π (n ) = cos nπ = ( )n/, n = even Hence, hus, a n = ( )n/ π(n + ) + ( )n/ π(n ) = ( )n/ π(n ), f()= π + cos π π n=even n = even ( ) n/ (n ) cos nπ o avoid using n =, 4, 6,... and also o ease compuaion, we can replace n by k, where k =,, 3,...and obain f()= π + cos π ( ) k cos kπ π (4k ) which is a Fourier cosine series. k= P R A C I C E P R O B L E M 6. 4 Find he Fourier series expansion of he funcion in Fig. 6.6.

20 76 PAR 3 Advanced Circui Analyses f() Answer: f()= 4 π k= cos n, n = k. n p p 4p Figure 6.6 For Pracice Prob E X A M P L E 6. 5 f() 3 4 Figure 6.7 For Example 6.5. Calculae he Fourier series for he funcion in Fig Soluion: he funcion in Fig. 6.7 is half-wave odd symmeric, so ha a = = a n. I is described over half he period as f()=, = 4,ω = π/ = π/. Hence, / << b n = 4 f()sin nω d Insead of inegraing f()from o, i is more convenien o inegrae from o. Applying Eq. (6.5d), b n = 4 sin nπ [ sin nπ/ 4 d = cos nπ/ n π /4 nπ/ = 4 [ sin nπ ( n π sin nπ ) [ cos nπ ( nπ + cos nπ ) = 8 n π sin nπ 4 nπ cos nπ since sin( x) = sin x as an odd funcion, while cos( x) = cos x as an even funcion. Using he ideniies for sin nπ/ and cos nπ/ in able 6., 8 n b n = π ( )(n )/, n = odd =, 3, 5,... 4 nπ ( )(n+)/, n = even =, 4, 6,... hus, f()= b n sin nπ where b n is given above. n= P R A C I C E P R O B L E M 6. 5 Deermine he Fourier series of he funcion in Fig. 6.(a). ake A = and = π. Answer: f()= ( π n π cos n + ) sin n,n= k. n k=

21 CHAPER 6 he Fourier Series CIRCUI APPLICAIONS We find ha in pracice, many circuis are driven by nonsinusoidal periodic funcions. o find he seady-sae response of a circui o a nonsinusoidal periodic exciaion requires he applicaion of a Fourier series, ac phasor analysis, and he superposiion principle. he procedure usually involves hree seps. Seps for Applying Fourier Series:. Express he exciaion as a Fourier series.. Find he response of each erm in he Fourier series. 3. Add he individual responses using he superposiion principle. he firs sep is o deermine he Fourier series expansion of he exciaion. For he periodic volage source shown in Fig. 6.8(a), for example, he Fourier series is expressed as v() = V + V n cos(nω + θ n ) (6.4) n= (he same could be done for a periodic curren source.) Equaion (6.4) shows ha v() consiss of wo pars: he dc componen V and he ac componen V n = V n θ n wih several harmonics. his Fourier series represenaion may be regarded as a se of series-conneced sinusoidal sources, wih each source having is own ampliude and frequency, as shown in Fig. 6.8(b). i() (a) (b) V V u + + I o + I Z(v = ) Z(v ) V + v() Periodic Source i() + Linear nework V cos(v + u ) V cos(v + u ) V n cos(nv + u n ) + + Linear nework + V u + + I Z(v ) Figure 6.8 (a) (a) Linear nework excied by a periodic volage source, (b) Fourier series represenaion (ime-domain). (b) + he second sep is finding he response o each erm in he Fourier series. he response o he dc componen can be deermined in he frequency domain by seing n = orω = as in Fig. 6.9(a), or in he ime domain by replacing all inducors wih shor circuis and all capaciors wih open circuis. he response o he ac componen is obained by he phasor echniques covered in Chaper 9, as shown in Fig. 6.9(b). he nework is represened by is impedance Z(nω ) or admiance Y(nω ). Z(nω ) is he inpu impedance a he source when ω is everywhere replaced by nω, and Y(nω ) is he reciprocal of Z(nω ). V n Figure 6.9 u n + I n Z(nv ) Seady-sae responses: (a) dc componen, (b) ac componen (frequency domain).

22 78 PAR 3 Advanced Circui Analyses E X A M P L E 6. 6 Finally, following he principle of superposiion, we add all he individual responses. For he case shown in Fig. 6.9, i() = i () + i () + i () + = I + I n cos(nω + ψ n ) n= (6.4) where each componen I n wih frequency nω has been ransformed o he ime domain o ge i n (), and ψ n is he argumen of I n. 5 Ω + v s () + H v o () Figure 6. For Example 6.6. Le he funcion f()in Example 6. be he volage source v s () in he circui of Fig. 6.. Find he response v o () of he circui. Soluion: From Example 6., v s () = + π k= sin nπ, n = k n where ω n = nω = nπ rad/s. Using phasors, we obain he response V o in he circui of Fig. 6. by volage division: V o = jω nl R + jω n L V s = jnπ 5 + jnπ V s For he dc componen (ω n = orn = ) V s = V o = V o.5 p Figure 6. p. 3p 4p.3 5p 6p. 7p v For Example 6.6: Ampliude specrum of he oupu volage. his is expeced, since he inducor is a shor circui o dc. For he nh harmonic, V s = nπ 9 (6.6.) and he corresponding response is nπ 9 V o = 5 + 4n π an nπ/5 nπ 9 = 4 an nπ/ n π In he ime domain, ( 4 v o () = 5 + 4n π cos nπ nπ an 5 k= (6.6.) ), n = k he firs hree erms (k =,, 3orn =, 3, 5) of he odd harmonics in he summaion give us v o () =.498 cos(π 5.49 ) +.5 cos(3π 75.4 ) +.57 cos(5π 8.96 ) + V Figure 6. shows he ampliude specrum for oupu volage v o (), while ha of he inpu volage v s () is in Fig. 6.4(a). Noice ha he

23 CHAPER 6 he Fourier Series 79 wo specra are close. Why? We observe ha he circui in Fig. 6. is a highpass filer wih he corner frequency ω c = R/L =.5 rad/s, which is less han he fundamenal frequency ω = π rad/s. he dc componen is no passed and he firs harmonic is slighly aenuaed, bu higher harmonics are passed. In fac, from Eqs. (6.6.) and (6.6.), V o is idenical o V s for large n, which is characerisic of a highpass filer. P R A C I C E P R O B L E M 6. 6 If he sawooh waveform in Fig. 6.9 (see Pracice Prob. 6.) is he volage source v s () in he circui of Fig. 6., find he response v o (). Answer: v o () = sin(πn an 4nπ) π n V. + 6n π n= v s () Ω + + F v o () Figure 6. For Pracice Prob E X A M P L E 6. 7 Find he response i o () in he circui in Fig. 6.3 if he inpu volage v() i() has he Fourier series expansion ( ) n v() = + + n (cos n n sin n) n= v() + Soluion: Using Eq. (6.3), we can express he inpu volage as ( ) n v() = + cos(n + + n an n) n= =.44 cos( + 45 ) cos( ).6345 cos( ).485 cos( ) + We noice ha ω =,ω n = n rad/s. he impedance a he source is Z = 4 + jω n 4 = 4 + jω n8 4 + jω n = 8 + jω n8 + jω n he inpu curren is 4 Ω Ω i o () H Figure 6.3 For Example 6.7. Ω I = V Z = + jω n 8 + jω n 8 V where V is he phasor form of he source volage v(). By curren division, 4 I o = 4 + jω n I = V 4 + jω n 4 Since ω n = n, I o can be expressed as V I o = 4 + n an n For he dc componen (ω n = orn = ) V = I o = V 4 = 4

24 73 PAR 3 Advanced Circui Analyses For he nh harmonic, V = ( )n + n an n P R A C I C E P R O B L E M 6. 7 so ha I o = 4 + n an n In he ime domain, i o () = 4 + ( ) n + n n= an n = ( ) n cos n A ( + n ) ( )n ( + n ) Ω i o () v() + F Ω Figure 6.4 For Pracice Prob If he inpu volage in he circui of Fig. 6.4 is v() = 3 + ( π n cos n π ) sin n V n deermine he response i o (). Answer: 9 + n= n= + n π n π 9 + 4n cos ( n an n ) 3 + an nπ A. 6.5 AVERAGE POWER AND RMS VALUES + v() i() Figure 6.5 Linear circui he volage polariy reference and curren reference direcion. Recall he conceps of average power and rms value of a periodic signal ha we discussed in Chaper. o find he average power absorbed by a circui due o a periodic exciaion, we wrie he volage and curren in ampliude-phase form [see Eq. (6.) as v() = V dc + V n cos(nω θ n ) (6.4) i() = I dc + n= I m cos(mω φ m ) (6.43) m= Following he passive sign convenion (Fig. 6.5), he average power is P = vi d (6.44) Subsiuing Eqs. (6.4) and (6.43) ino Eq. (6.44) gives P = I m V dc V dc I dc d + cos(mω φ m )d m= V n I dc + cos(nω θ n )d (6.45) n= V n I m + cos(nω θ n ) cos(mω φ m )d m= n=

25 CHAPER 6 he Fourier Series 73 he second and hird inegrals vanish, since we are inegraing he cosine over is period. According o Eq. (6.4e), all erms in he fourh inegral are zero when m n. By evaluaing he firs inegral and applying Eq. (6.4g) o he fourh inegral for he case m = n, we obain P = V dc I dc + V n I n cos(θ n φ n ) (6.46) n= his shows ha in average-power calculaion involving periodic volage and curren, he oal average power is he sum of he average powers in each harmonically relaed volage and curren. Given a periodic funcion f(), is rms value (or he effecive value) is given by F rms = f () d (6.47) Subsiuing f() in Eq. (6.) ino Eq. (6.47) and noing ha (a + b) = a + ab + b, we obain Frms = [ a + a A n cos(nω + φ n ) n= + A n A m cos(nω + φ n ) cos(mω + φ m ) d n= m= = a d + a A n cos(nω + φ n )d n= + A n A m cos(nω + φ n ) cos(mω + φ m )d n= m= (6.48) Disinc inegers n and m have been inroduced o handle he produc of he wo series summaions. Using he same reasoning as above, we ge Frms = a + A n or n= F rms = a + A n (6.49) In erms of Fourier coefficiens a n and b n, Eq. (6.49) may be wrien as F rms = a + (an + b n ) (6.5) If f() is he curren hrough a resisor R, hen he power dissipaed in he resisor is n= n= P = RF rms (6.5)

26 73 PAR 3 Advanced Circui Analyses Or if f()is he volage across a resisor R, he power dissipaed in he resisor is P = F rms (6.5) R One can avoid specifying he naure of he signal by choosing a - resisance. he power dissipaed by he - resisance is P = F rms = a + (an + b n ) (6.53) n= Hisorical noe: Named afer he French mahemaician Marc-Anoine Parseval Deschemes ( ). his resul is known as Parseval s heorem. Noice ha a is he power in he dc componen, while /(an + b n ) is he ac power in he nh harmonic. hus, Parseval s heorem saes ha he average power in a periodic signal is he sum of he average power in is dc componen and he average powers in is harmonics. E X A M P L E i() v() Ω F Figure 6.6 For Example 6.8. Deermine he average power supplied o he circui in Fig. 6.6 if i() = + cos( + ) + 6 cos( ) A. Soluion: he inpu impedance of he nework is Hence, Z = For he dc componen, ω =, jω = (/jω) + /jω = + jω I V = IZ = + 4ω an ω I = A V = () = V his is expeced, because he capacior is an open circui o dc and he enire -A curren flows hrough he resisor. For ω = rad/s, I = ( ) V = + 4 an For ω = 3 rad/s, I = 6 45 V = hus, in he ime domain, = (6 45 ) + 36 an 6 = 44.5 v() = + 5 cos( 77.4 ) + cos( ) V

27 CHAPER 6 he Fourier Series 733 We obain he average power supplied o he circui by applying Eq. (6.46), as P = V dc I dc + V n I n cos(θ n φ n ) n= o ge he proper signs of θ n and φ n, we have o compare v and i in his example wih Eqs. (6.4) and (6.43). hus, P = () + (5)() cos[77.4 ( ) + ()(6) cos[44.5 ( 35 ) = = 4.5 W Alernaively, we can find he average power absorbed by he resisor as P = V dc R + V n R = n= = = 4.5W which is he same as he power supplied, since he capacior absorbs no average power. P R A C I C E P R O B L E M 6. 8 he volage and curren a he erminals of a circui are v() = 8 + cos π + 6 cos(36π 3 ) i() = 5 cos(π ) + cos(36π 6 ) Find he average power absorbed by he circui. Answer: W. E X A M P L E 6. 9 Find an esimae for he rms value of he volage in Example 6.7. Soluion: From Example 6.7, v() is expressed as v() =.44 cos( + 45 ) cos( ).6345 cos( ).485 cos( ) + V Using Eq. (6.49), V rms = a + A n n= = + [ (.44) + (.8944) + (.6345) + (.485) + =.786 =.649 V

28 734 PAR 3 Advanced Circui Analyses P R A C I C E P R O B L E M 6. 9 his is only an esimae, as we have no aken enough erms of he series. he acual funcion represened by he Fourier series is v() = πe sinh π, π <<π wih v() = v( + ).he exac rms value of his is.776 V. Find he rms value of he periodic curren i() = cos sin + 5 cos 4 sin 4 A Answer: 9.6 A. 6.6 EXPONENIAL FOURIER SERIES A compac way of expressing he Fourier series in Eq. (6.3) is o pu i in exponenial form. his requires ha we represen he sine and cosine funcions in he exponenial form using Euler s ideniy: cos nω = [ejnω + e jnω (6.54a) sin nω = j [ejnω e jnω (6.54b) Subsiuing Eq. (6.54) ino Eq. (6.3) and collecing erms, we obain f()= a + [(a n jb n )e jnω + (a n + jb n )e jnω (6.55) n= If we define a new coefficien c n so ha c = a, c n = (a n jb n ), c n = cn = (a n + jb n ) (6.56) hen f()becomes f()= c + (c n e jnω + c n e jnω ) (6.57) or n= f()= c n e jnω n= (6.58) his is he complex or exponenial Fourier series represenaion of f(). Noe ha his exponenial form is more compac han he sine-cosine form in Eq. (6.3). Alhough he exponenial Fourier series coefficiens c n can also be obained from a n and b n using Eq. (6.56), hey can also be obained direcly from f()as c n = f()e jnω d (6.59)

29 CHAPER 6 he Fourier Series 735 where ω = π/, as usual. he plos of he magniude and phase of c n versus nω are called he complex ampliude specrum and complex phase specrum of f(), respecively. he wo specra form he complex frequency specrum of f(). he exponenial Fourier series of a periodic funcion f () describes he specrum of f () in erms of he ampliude and phase angle of ac componens a posiive and negaive harmonic frequencies. he coefficiens of he hree forms of Fourier series (sine-cosine form, ampliude-phase form, and exponenial form) are relaed by or c n = c n θ n = A n φ n = a n jb n = c n (6.6) a n + b n an b n /a n (6.6) if only a n >. Noe ha he phase θ n of c n is equal o φ n. In erms of he Fourier complex coefficiens c n, he rms value of a periodic signal f()can be found as Frms = f () d = [ f() c n e jnω d n= [ = c n f()e jnω (6.6) n= = c n cn = c n or n= n= Equaion (6.6) can be wrien as F rms = c n (6.63) n= F rms = c + c n (6.64) n= Again, he power dissipaed by a - resisance is P = F rms = n= c n (6.65) which is a resaemen of Parseval s heorem. he power specrum of he signal f()is he plo of c n versus nω.iff()is he volage across a resisor R, he average power absorbed by he resisor is Frms /R; iff() is he curren hrough R, he power is Frms R.

30 736 PAR 3 Advanced Circui Analyses 9 9 Figure 6.7 f() he periodic pulse rain. he sinc funcion is called he sampling funcion in communicaion heory, where i is very useful. As an illusraion, consider he periodic pulse rain of Fig Our goal is o obain is ampliude and phase specra. he period of he pulse rain is =, so ha ω = π/ = π/5. Using Eq. (6.59), and c n = = / / jnω e jnω f()e jnω d = = = e jnω e jnω nω j sin nπ/5 = nπ/5 f()= n= e jnω d jnω (e jnω e jnω ) = sin nω nω, ω = π 5 (6.66) sin nπ/5 nπ/5 ejnπ/5 (6.67) Noice from Eq. (6.66) ha c n is he produc of and a funcion of he form sin x/x. his funcion is known as he sinc funcion; we wrie i as sinc(x) = sin x x (6.68) Some properies of he sinc funcion are imporan here. For zero argumen, he value of he sinc funcion is uniy, sinc() = (6.69) his is obained applying L Hopial s rule o Eq. (6.68). For an inegral muliple of π, he value of he sinc funcion is zero, sinc(nπ) =, n =,, 3,... (6.7) Also, he sinc funcion shows even symmery. Wih all his in mind, we can obain he ampliude and phase specra of f(). From Eq. (6.66), he magniude is while he phase is c n = sin nπ/5 nπ/5 (6.7), sin nπ 5 > θ n = 8, sin nπ (6.7) 5 < Examining he inpu and oupu specra allows visualizaion of he effec of a circui on a periodic signal. Figure 6.8 shows he plo of c n versus n for n varying from o, where n = ω/ω is he normalized frequency. Figure 6.9 shows he plo of θ n versus n. Boh he ampliude specrum and phase specrum are called line specra, because he value of c n and θ n occur only a discree values of frequencies. he spacing beween he lines is ω. he power specrum, which is he plo of c n versus nω, can also be ploed. Noice ha he sinc funcion forms he envelope of he ampliude specrum.

31 CHAPER 6 he Fourier Series 737 c n Figure 6.8 he ampliude of a periodic pulse rain. u n 8 8 Figure E X A M P L E he phase specrum of a periodic pulse rain. Find he exponenial Fourier series expansion of he periodic funcion f()= e, <<π wih f( + π) = f(). Soluion: Since = π, ω = π/ =. Hence, c n = f()e jnω d = π e e jn d π = π jn e( jn) Bu by Euler s ideniy, π = π( jn) [eπ e jπn e jπn = cos πn j sin πn = j = hus, c n = π( jn) [eπ = 85 jn he complex Fourier series is 85 f()= jn ejn n= n

32 738 PAR 3 Advanced Circui Analyses We may wan o plo he complex frequency specrum of f(). Ifwele c n = c n θ n, hen c n = 85 + n, θ n = an n By insering in negaive and posiive values of n, we obain he ampliude and he phase plos of c n versus nω = n, as in Fig c n 85 u n nv nv (a) 9 (b) Figure 6.3 he complex frequency specrum of he funcion in Example 6.: (a) ampliude specrum, (b) phase specrum. P R A C I C E P R O B L E M 6. Obain he complex Fourier series of he funcion in Fig. 6.. Answer: f()= j nπ ejnπ. n = n n = odd E X A M P L E 6. Find he complex Fourier series of he sawooh wave in Fig Plo he ampliude and he phase specra. Soluion: From Fig. 6.9, f() =, <<, = so ha ω = π/ = π. Hence, c n = f()e jnω d = e jnπ d (6..)

33 CHAPER 6 he Fourier Series 739 Bu e a d = ea (ax ) + C a Applying his o Eq. (6..) gives c n = e jnπ ( jnπ ) ( jnπ) Again, = e jnπ ( jnπ ) + 4n π e jπn = cos πn j sin πn = j = so ha Eq. (6..) becomes c n = jnπ 4n π = j nπ (6..) (6..3) his does no include he case when n =. When n =, c = f()d = d= =.5 (6..4) Hence, and f()=.5 + n = n j nπ ejnπ (6..5) c n = n π, n, θ n = 9, n (6..6).5, n = By ploing c n and θ n for differen n, we obain he ampliude specrum and he phase specrum shown in Fig c n.5 u n v 4v 3v v v v (a) v 3v 4v 5v v 5v 4v 3v v v v v 3v 4v 5v v (b) Figure 6.3 For Example 6.: (a) ampliude specrum, (b) phase specrum.

34 74 PAR 3 Advanced Circui Analyses P R A C I C E P R O B L E M 6. Obain he complex Fourier series expansion of f()in Fig Show he ampliude and phase specra. j( ) n Answer: f()= nπ ejnπ. See Fig. 6.3 for he specra. n = n c n.3.3 u n n (a) n 9 (b) Figure 6.3 For Pracice Prob. 6.: (a) ampliude specrum, (b) phase specrum. 6.7 FOURIER ANALYSIS WIH PSPICE i s (a) Figure 6.33 Ω + v o v s + + Ω v o (b) Fourier analysis wih PSpice using: (a) a curren source, (b) a volage source. Fourier analysis is usually performed wih PSpice in conjuncion wih ransien analysis. herefore, we mus do a ransien analysis in order o perform a Fourier analysis. o perform he Fourier analysis of a waveform, we need a circui whose inpu is he waveform and whose oupu is he Fourier decomposiion. A suiable circui is a curren (or volage) source in series wih a - resisor as shown in Fig he waveform is inpued as v s () using VPULSE for a pulse or VSIN for a sinusoid, and he aribues of he waveform are se over is period. he oupu V() from node is he dc level (a ) and he firs nine harmonics (A n ) wih heir corresponding phases ψ n ; ha is, where A n = v o () = a + 9 A n sin(nω + ψ n ) (6.73) n= a n + b n, ψ n = φ n π, φ n = an b n a n (6.74) Noice in Eq. (6.74) ha he PSpice oupu is in he sine and angle form raher han he cosine and angle form in Eq. (6.). he PSpice oupu also includes he normalized Fourier coefficiens. Each coefficien a n is normalized by dividing i by he magniude of he fundamenal a so ha he normalized componen is a n /a. he corresponding phase ψ n is normalized by subracing from i he phase ψ of he fundamenal, so ha he normalized phase is ψ n ψ.

35 CHAPER 6 he Fourier Series 74 here are wo ypes of Fourier analyses offered by PSpice for Windows: Discree Fourier ransform (DF) performed by he PSpice program and Fas Fourier ransform (FF) performed by he Probe program. While DF is an approximaion of he exponenial Fourier series, F is an algorihm for rapid efficien numerical compuaion of DF. A full discussion of DF and F is beyond he scope of his book Discree Fourier ransform A discree Fourier ransform (DF) is performed by he PSpice program, which abulaes he harmonics in an oupu file. o enable a Fourier analysis, we selec Analysis/Seup/ransien and bring up he ransien dialog box, shown in Fig he Prin Sep should be a small fracion of he period, while he Final ime could be 6. he Cener Frequency is he fundamenal frequency f = /. he paricular variable whose DF is desired, V() in Fig. 6.34, is enered in he Oupu Vars command box. In addiion o filling in he ransien dialog box, DCLICK Enable Fourier. Wih he Fourier analysis enabled and he schemaic saved, run PSpice by selecing Analysis/Simulae as usual. he program execues a harmonic decomposiion ino Fourier componens of he resul of he ransien analysis. he resuls are sen o an oupu file which you can rerieve by selecing Analysis/Examine Oupu. he oupu file includes he dc value and he firs nine harmonics by defaul, alhough you can specify more in he Number of harmonics box (see Fig. 6.34) Fas Fourier ransform A fas Fourier ransform (FF) is performed by he Probe program and displays as a Probe plo he complee specrum of a ransien expression. As explained above, we firs consruc he schemaic in Fig. 6.33(b) and ener he aribues of he waveform. We also need o ener he Prin Sep and he Final ime in he ransien dialog box. Once his is done, we can obain he FF of he waveform in wo ways. One way is o inser a volage marker a node in he schemaic of he circui in Fig. 6.33(b). Afer saving he schemaic and selecing Analysis/Simulae, he waveform V() will be displayed in he Probe window. Double clicking he FF icon in he Probe menu will auomaically replace he waveform wih is FF. From he FF-generaed graph, we can obain he harmonics. In case he FF-generaed graph is crowded, we can use he User Defined daa range (see Fig. 6.35) o specify a smaller range. Anoher way of obaining he FF of V() is o no inser a volage marker a node in he schemaic. Afer selecing Analysis/Simulae, he Probe window will come up wih no graph on i. We selec race/add and ype V() in he race Command box and DCLICKL OK. We now selec Plo/X-Axis Seings o bring up he X Axis Seing dialog box shown in Fig and hen selec Fourier/OK. his will cause he FF of he seleced race (or races) o be displayed. his second approach is useful for obaining he FF of any race associaed wih he circui. A major advanage of he FF mehod is ha i provides graphical oupu. Bu is major disadvanage is ha some of he harmonics may be oo small o see. Figure 6.34 ransien dialog box.

36 74 PAR 3 Advanced Circui Analyses Figure 6.35 X axis seings dialog box. In boh DF and FF, we should le he simulaion run for a large number of cycles and use a small value of Sep Ceiling (in he ransien dialog box) o ensure accurae resuls. he Final ime in he ransien dialog box should be a leas five imes he period of he signal o allow he simulaion o reach seady sae. E X A M P L E 6. V= V= D= F=u R=u PW= PER= + Figure 6.36 V3 V R Schemaic for Example 6.. Use PSpice o deermine he Fourier coefficiens of he signal in Fig. 6.. Soluion: Figure 6.36 shows he schemaic for obaining he Fourier coefficiens. Wih he signal in Fig. 6. in mind, we ener he aribues of he volage source VPULSE as shown in Fig We will solve his example using boh he DF and FF approaches. MEHOD DF Approach: (he volage marker in Fig is no needed for his mehod.) From Fig. 6., i is eviden ha = s, f = = =.5 Hz So, in he ransien dialog box, we selec he Final ime as 6 = s, he Prin Sep as. s, he Sep Ceiling as ms, he Cener Frequency as.5 Hz, and he oupu variable as V(). (In fac, Fig is for his paricular example.) When PSpice is run, he oupu file conains he following resul. FOURIER COEFFICIENS OF RANSIEN RESPONSE V() DC COMPONEN = E- HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG) 5.E E-.E+ -.89E-.E+.E+.E-3 3.6E-3-9.6E+ -9.8E+ 3.5E+.E E E E- (coninued)

37 CHAPER 6 he Fourier Series 743 (coninued) 4.E+.6E E E E+ 5.5E+.73E-.999E E E- 6 3.E+.4E-3 3.8E E E E+ 9.88E-.47E- -.67E+ -.86E+ 8 4.E+.35E E E E E+ 7.65E-.E- -.63E E+ Comparing he resul wih ha in Eq. (6..7) (see Example 6.) or wih he specra in Fig. 6.4 shows a close agreemen. From Eq. (6..7), he dc componen is.5 while PSpice gives Also, he signal has only odd harmonics wih phase ψ n = 9, whereas PSpice seems o indicae ha he signal has even harmonics alhough he magniudes of he even harmonics are small. MEHOD FF Approach: Wih volage marker in Fig in place, we run PSpice and obain he waveform V() shown in Fig. 6.37(a) on he Probe window. By double clicking he FF icon in he Probe menu and changing he X-axis seing o o Hz, we obain he FF of V() as shown in Fig. 6.37(b). he FF-generaed graph conains he dc and harmonic componens wihin he seleced frequency range. Noice ha he magniudes and frequencies of he harmonics agree wih he DFgeneraed abulaed values.. V V s s 4 s 6 s V() ime (a) 8 s s s. V V Hz V() Hz 4 Hz 6 Hz 8 Hz Hz Frequency (b) Figure 6.37 (a) Original waveform of Fig. 6., (b) FF of he waveform. P R A C I C E P R O B L E M 6. Obain he Fourier coefficiens of he funcion in Fig. 6.7 using PSpice.

38 744 PAR 3 Advanced Circui Analyses Answer: FOURIER COEFFICIENS OF RANSIEN RESPONSE V() DC COMPONEN = 4.95E- HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG).E+ 3.84E-.E+ -.78E+.E+.E+.593E- 5.E E+.8E+ 3 3.E+.63E E E+ 3.6E+ 4 4.E E-.56E E+ 5.4E+ 5 5.E+ 6.39E-.8E- -.7E+ 7.E+ 6 6.E E-.676E E+ 9.E+ 7 7.E E-.44E E+.8E+ 8 8.E+ 4.E-.63E E+.6E+ 9 9.E E-.6E E+.44E+ E X A M P L E 6. 3 Ω v s + Ω i() H Figure 6.38 For Example 6.3. If v s in he circui of Fig is a sinusoidal volage source of ampliude V and frequency Hz, find curren i(). Soluion: he schemaic is shown in Fig We may use he DF approach o obain he Fourier coefficens of i(). Since he period of he inpu waveform is = / = ms, in he ransien dialog box we selec Prin Sep:. ms, Final ime: ms, Cener Frequency: Hz, Number of harmonics: 4, and Oupu Vars: I(L). When he circui is simulaed, he oupu file includes he following. FOURIER COEFFICIENS OF RANSIEN RESPONSE I(VD) DC COMPONEN = E-3 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG).E+ 8.73E-3.E E+.E+.E+.7E-4.65E E E+ 3 3.E+ 6.8E-5 7.8E E+ 7.49E+ 4 4.E+ 4.43E E E+ 4.54E+ Wih he Fourier coefficiens, he Fourier series describing he curren i() can be obained using Eq. (6.73); ha is, i() = sin(π ) +.7 sin(π 83.6 ) +.68 sin(π ) + ma

39 CHAPER 6 he Fourier Series 745 We can also use he FF approach o cross-check our resul. he curren marker is insered a pin of he inducor as shown in Fig Running PSpice will auomaically produce he plo of I(L) in he Probe window, as shown in Fig. 6.4(a). By double clicking he FF icon and seing he range of he X-axis from o Hz, we generae he FF of I(L) shown in Fig. 6.4(b). I is clear from he FF-generaed plo ha only he dc componen and he firs harmonic are visible. Higher harmonics are negligibly small. ma VAMPL= FREQ= VOFF= Figure R V R H Schemaic of he circui in Fig I L ma s ms 4 ms 6 ms 8 ms ms I (L) ime ma (a) A Hz 4 Hz 8 Hz Hz 6 Hz Hz I (L) Frequency (b) Figure 6.4 For Example 6.3: (a) plo of i(), (b) he FF of i(). P R A C I C E P R O B L E M 6. 3 A sinusoidal curren source of ampliude 4 A and frequency khz is applied o he circui in Fig Use PSpice o find v(). i s () + v() Ω F Answer: v() = sin(4π ) + µv. he Fourier componens are shown below. Figure 6.4 For Pracice Prob FOURIER COEFFICIENS OF RANSIEN RESPONSE V(R:) DC COMPONEN = E-4 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG).E+3.455E-4.E+ 9.6E+.E+ 4.E+3.85E-6.73E E+ 5.9E+ 3 6.E+3.46E E E+ 3.67E+ 4 8.E+3.E E- 8.77E+ -9.9E+