Complex Fourier Series. Adding these identities, and then dividing by 2, or subtracting them, and then dividing by 2i, will show that

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1 Mah 344 May 4, Complex Fourier Series Par I: Inroducion The Fourier series represenaion for a funcion f of period P, f) = a + a k coskω) + b k sinkω), ω = π/p, ) can be expressed more simply using complex exponenials. Moreover, because of he unique properies of he exponenial funcion, Fourier series are ofen easier o manipulae in complex form. We will see one ineresing example in Par II of his handou. More will be discussed in Week. The ransiion from he real form o he complex form of a Fourier series is made using he following ideniies, called Euler Ideniies, e i θ = cosθ) + i sinθ) and e i θ = cosθ) i sinθ). Adding hese ideniies, and hen dividing by, or subracing hem, and hen dividing by i, will show ha cosθ) = eiθ + e iθ and sinθ) = eiθ e iθ i. ) These formulas for he sine and cosine should be compared o he definiions of he hyperbolic cosine and sine, cosh and sinh: coshθ) = eθ + e θ and sinhθ) = eθ e θ. The complex Fourier series is obained from ) by wriing coskω) and sinkω) in heir complex exponenial form and rearranging as follows: f) = a + e ikω + e ikω a k + b k e ikω e ikω i = a + a k ib k )e ikω + a k + ib k )e ikω = c + c k e ikω + c k e ikω. The las line is obained by making he definiions Observe ha c k is he complex conjugae of c k : c k = c k. c = a, c k = a k ib k ), c k = a k + ib k ). 3) The complex sum ha was jus obained can also be wrien in one of he following hree equivalen forms. Alernae Forms for a Complex Fourier Series f) = c + f) = c + f) = Rc k e ikω ) 4a) k c k e ikω c k e ikω The firs form, 4a), is a consequence of he fac ha c k e ikω is he complex conjugae of c k e ikω. This implies ha c k e ikω + c k e ikω = Rc k e ikω ) 4b) 4c)

2 where Rc k e ikω ) denoes he real par of he expression c k e ikω. Some commens:. Formula 4a) makes he ransiion from he complex series o is real form. Is use is recommended when ploing parial sum approximaions in complex form using a compuer algebra sysem like Maple.. An explici complex Fourier series is usually wrien using formula 4b). This is because he consan erm, c, is almos never equal he formula obained by subsiuing k = ino he formula for c k. On he oher hand, i is always he case ha c k can be obained from he formula for c k by replacing k wih k. 3. Commen does no preclude he possibiliy of wriing he complex series in he form given in 4c) and adding separae formulas for c and c k. This is worh emphasizing because 4c) is clearly he nices of hese hree alernaives. Wha makes i even nicer is he fac ha here is jus one inegral formula for he compuaion of all of he complex coefficiens, c k. This will be shown below. 4. A word of cauion: If parial sums for 4b) or 4c) are used o approximae he funcion f, hen he sum mus be in he symmeric form S n x, ) = c k e ikω because his will yield he n h Fourier approximaion. An inegral formula for c k The k h complex coefficien c k can always be obained from he real coefficiens using he formula c k = a k ib k ). This would require wo inegraions. However, c k can also be obained direcly from he funcion f wih jus one inegraion. This is because To see why, observe ha c k = a k ib k ) = P = P = P c k = P f)e ikω d. 5) f) coskω) d i ) P f) sinkω) d P f)coskω) i sinkω)) d f)e ikω d. When he calculaion of c k is made using 5), he inegral can be evaluaed over any periodic inerval. Moreover, once c k has been found, boh real Fourier series coefficiens can hen be obained by solving he defining equaion c k = a k ib k ) for a k and b k as follows. Here, Rc k ) and Ic k ) denoe he real par and he imaginary par of c k. The following example illusraes hese observaions. a k = Rc k ) and b k = Ic k ). 6) Example. Obain he complex Fourier series represenaion for he funcion f of period P = defined as follows for < <. {, < < f) =, < < Skech he approximaion S 3 ) and analyze he error. Use he complex coefficien formula o obain he real Fourier series coefficiens for f. Soluion. Le s begin wih a picure of he waveform.

3 ..5 K3 K K 3 4 The funcion f has period. Since P =, he fundamenal frequency is ω = π/p = π, herefore f) = c ke ikπ. By inspecion of he graph, c = + /)/ = /4 and using inegraion-by-pars wih u = and dv = e ikπ d, c k = f)e ikπ d = e ikπ d = e ikπ ikπ + ) = e ikπ d ikπ = = ) e ikπ ikπ + e ikπ i k π = ) ) k kπ i + )k k π The ransiion o he final formula was aided by he fac ha e ±ikπ = ) k when k is an ineger. The following plo shows he funcion and is hird Fourier approximaion, S 3 )...5 Error Analysis K3 K K 3 4 The funcion f and S 3 ). The family of complex exponenial funcions { e ikω} is orhogonal on [, P ] wih respec o he complex inner produc f, g = f)g) d. Moreover, eikω = P. Because of his, he complex form of he n h mean square error has he following simple form. Compare formula 7) below o he MSE formula for a real Fourier series in he Classical Fourier Series handou. MSE = f) d c k. 7) P The calculaion of MSE for he n h approximaion for his example proceeds as follows. MSE = = f) d d = n = 5 48 c + This is an inegral calculaion. c k = c + c k because c is real and c k = c k. 4 c k ) c k k π + )k ) k 4 π 4 k π + )k ) ) k 4 π 4 3 ) )

4 When n = 3, MSE =.44 yielding a relaive RSE error of MSE/ f =.94, or abou 3%. This is raher large, reflecing he error a he jump disconinuiy. However, in erms of he average power in one cycle, he relaive MSE error associaed wih S 3 ) is less han 9%: / MSE f) d =.866 = 8.66%. The large difference in he wo relaive errors is also clearly eviden from he ampliude and power specra of f. The Ampliude and Power Specra for he Complex Series When he complex Fourier series is used o represen a periodic funcion, hen he ampliude specrum, skeched below, is wo-sided. I consiss of he poins kω, c k ) for k =, ±, ±, ±3, K8 p K6 p K4 p K p p 4 p 6 p 8 p The wo-sided ampliude specrum of he funcion f in Example. The power specrum for f is also wo-sided, consising of he poins kω, c k ) for k =, ±, ±, ±3, K8 p K6 p K4 p K p p 4 p 6 p 8 p The wo-sided power specrum of he funcion f in Example. Calculaion of he real coefficiens. Repeaing he formula for c k, bu in sandard form for a complex number, shows ha c k = )k k π + )k kπ i, a k = Rc k ) = )k k π and b k = Ic k ) = )k kπ. The following example is relaed o Example 5 in he Classical Fourier Seried handou. Example. Obain he complex Fourier series represenaion for he funcion of period P = π defined as f) = sin) for < < π and for π < < π. See is graph below. Plo f and S 3 ), make an error analysis, including he ampliude and power specra, and obain he real Fourier series coefficiens...5 Kp p p 3 p The funcion f is neiher even nor odd. 4

5 Soluion. Since P = π, he fundamenal frequency is ω = π/p =. Therefore, f) = c k e ik. The k h coefficien, c k, will be calculaed using he complex formula for sin). This simplifies he inegraion considerably. c k = π Some commens on his calculaion: π π f)e ik d = π i ei e i )e ik d = π e i k) e i+k) d line ) 4πi = ) =π e i k) 4πi k)i + e i+k) + k)i = = e iπ k) ) 4πi + e iπ+k) line 4) k + k = + )k 4π k + ) line 5) + k = + )k π k ). The ransiion from line 4 o line 5 used he fac ha, because k is an ineger, e iπ k) = e iπ+k) = ) k.. The final formula for c k shows ha c = /π. 3. The value of c mus be obained using a separae calculaion. Working from line, subsiue k = o obain c = 4πi π e i d = 4πi ) + e i π i = π 4πi = i 4. Puing his all ogeher, we ge he following complex Fourier series represenaion. Noe ha c = c = i/4. f) = π i 4 ei + i 4 e i + π k,± + ) k k e ik. The plo of f along wih S, S, and S 3 is displayed below. S 3 is he same as S, and is hardly visible...5 Kp p p 3 p The funcion f and is approximaions S doed), S dashed), and S 3 same as S ). Error Analysis 5

6 The calculaion of MSE for he n h approximaion proceeds as follows. MSE = π f) d c k π = ) π sin ) d c + c k π ) = 4 π ) k ) 4π k ) = 8 π + π k= k= + ) k ) k ) When n = 3, MSE =.6 yielding a relaive RSE error of π MSE/ f =.68, or abou 7%. This is small, reflecing he excellen qualiy of he approximaion. See he ampliude specrum for f, skeched below..3.. K K9 K8 K7 K6 K5 K4 K3 K K The wo-sided ampliude specrum of he funcion f. In erms of he average power in one cycle, he relaive MSE error associaed wih S 3 ) is less han.5%: / MSE π π f) d =.465 =.465%. The power specrum for f, skeched below, shows ha S ) carries essenially all of he power in his waveform..8.4 Calculaion of he real coefficiens. K K9 K8 K7 K6 K5 K4 K3 K K The wo-sided power specrum of he funcion f. Since c = i/4, i has no real par. Therefore, a = and b = Ic ) = /. On he oher hand, when k, c k is real so b k =, and a k = c k = + ) k )/π k ). Using his informaion, he real Fourier series for f is f) = π + sin) + π = π + sin) + π k= + ) k k cosk) cosk) 4k. If his looks familiar, i should. In Example 5 of he Classical Fourier Series handou we derived he Fourier series for he similar waveform skeched below...5 Kp p p 3 p The funcion g) = sin), < x < π, period P = π. 6

7 There i was shown ha g has he following Fourier series represenaion. g) = π + 4 π cosk) 4k These wo waveforms are paricularly useful in engineering applicaions. 7

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