CHAPTER 106 THE COMPLEX OR EXPONENTIAL FORM OF A FOURIER SERIES EXERCISE 370 Page 1098 0, when − π ≤ t ≤ 0 1. Determine the complex Fourier series for the function defined by: f(t) = 2, when 0 ≤ t ≤ π The function is periodic outside of this range of period 2π. The periodic function is shown in the diagram below f (t ) = The complex Fourier series is given by: ∞ ∑ cn e j 2π nt L n = −∞ 2π nt 1 L /2 −j L dt ( ) e f t L ∫ − L /2 cn = where π 2π nt π 1 0 1 −j 1 π − j nt 1 e − j nt cn = e = − [ e − j n π − e0 ] ∫ −π 0 d t + ∫ 0 2 e 2π d t == ∫ 0 2π π − jn 0 jπ n π i.e. =− = j jπn 2 j nπ − 1 [e− = ] j − j nπ ( e − 1) πn j j nπ − 1] [cos nπ − j sin= [cos nπ − 1] πn πn for all integer values of n f (t ) = Hence, ∞ 2π nt cn e L = ∑ j n =− ∞ c= a= mean value = 0 0 c3 = − j π 2 , 3π ∑ n =− ∞ j nπ ( cos nπ − 1) e j nt 2×π = 1 2π j 2 c1 = (−1 − 1) =− j , π ∞ c5 = − j c= 2 j 2π (1 − 1) = 0 and all even terms will be zero 2 , and so on 5π 1558 © 2014, John Bird c−1= 2 j (−2)= j , π −π 2 Thus, f(t) = 1 − j i.e. 1 j f (t ) =− 2 j , (−2) = j 3π −3π c= −3 π e jt − j c5 = j 2 , and so on 5π 2 j 3t 2 j 5t 2 2 − j 3t 2 − j 5t e −j e − .... + j e − j t + j e +j e 3π 5π 3π 5π π 2 j t 1 j 3t 1 j 5 t 2 − j t 1 − j 3t 1 − j 5 t e + e + e + ... + j e + e + e + ... 3 5 3 5 π π 2. Show that the complex Fourier series for the waveform shown below, that has period 2, may be ∞ represented by: f(t) = 2 + j2 ∑ π n ( cos nπ − 1) e π j nt n = −∞ ( n ≠ 0) ∞ ∑c f (t ) = The complex Fourier series is given by: n e j 2π nt L n =− ∞ cn = where 2π nt 1 L /2 −j L dt ( ) e f t L ∫ − L /2 1 i.e. 1 e − j π nt 1 1 − j 2π nt 2 2 ∫ e − j π nt = 2 = − cn = [ e − j π n − e0 ] ∫0 4 e 2 = 0 2 jπ n − jπ n 0 =− = 2 jπ n nπ − 1 [e− j= ] j 2 − j nπ ( e − 1) πn j2 j2 nπ − 1] [cos nπ − 1] [cos nπ − j sin= πn πn for all integer values of n c= a= = mean value 0 0 Hence, 1 1 1 1 t 4 d= 4t ]= 2 [ ∫ 0 2 0 2 ∞ j2 2+ ∑ f (t ) = ( cos nπ − 1) e j π nt n π n = −∞ ( n ≠ 0) 1559 © 2014, John Bird 3. Show that the complex Fourier series of Problem 2 is equivalent to: f(t) = 2 + j2 4 (−1 − 1) =− j , π (1) π c1 = c3 =j c−1= Thus, 2 4 , (−2) = −j 3π 3π j2 4 (−2)= j , −π π 8 1 1 sin π t + sin 3π t + sin 5π t + ... 3 5 π c= 2 j2 (1 − 1) = 0 and all even terms will be zero 2π c5 = − j c= −3 4 , and so on 5π j2 4 , (−2) = j −3π 3π c5 = j 4 , and so on 5π ∞ j2 2+ ∑ f (t ) = ( cos nπ − 1) e j π nt n π n =− ∞ ( n ≠ 0) 4 i.e. f(t) = 2 − j i.e. f (t ) =2 − j π = 2− j 4 j 3π t 4 j 5π t 4 4 − j 3π t 4 − j 5π t e −j e − ... + j e − jπ t + j e e +j 3π 5π 3π 5π π 4 j π t 1 j 3π t 1 j 5π t 4 − j π t 1 − j 3π t 1 − j 5π t + e + ... e + e + e + ... + j e + e 3 5 3 5 π π 4 j π t − j π t 1 j 3π t − j 3π t 1 j 5π t − j 5π t ( e − e ) + ( e − e ) + ( e − e ) + ... 3 5 π = 2 − j2 i.e. e jπ t − j 8 e j π t − e − j π t 2j π 1 e j 3π t − e − j 3π t + 2j 3 1 e j 5π t − e − j 5π t + 2j 5 + ... 8 1 1 f (t ) = 2 + sin π t + sin 3π t + sin 5π t + ... π 3 5 4. Determine the exponential form of the Fourier series for the function defined by: f(t) = e 2t when –1 < t < 1 and has period 2 The function is shown in the diagram below 1560 © 2014, John Bird The complex Fourier series is given by: f (t ) = ∞ ∑ cn e j 2π nt L n = −∞ cn = where 2π nt 1 L /2 −j L dt ( ) e f t L ∫ − L /2 1 1 1 2t − j 2π nt 1 1 2t − j π nt 1 et ( 2 − j π n ) 1 e( 2 − j π n ) − e −( 2 − j π n ) i.e. e 2 dt = e d cn t = = ∫ −1 e= ∫ 2 2 2 − jπ n −1 2 2 − jπ n 2 −1 ∞ Thus, f(t) = ∑c n n =− ∞ e j 2π nt L = 1 ∞ e(2− j π n ) − e − (2− j π n ) j π nt ∑ 2 − jπ n e 2 n =− ∞ 1561 © 2014, John Bird EXERCISE 371 Page 1101 1. Determine the exponential form of the Fourier series for the periodic function defined by: π −2, when − π ≤ x ≤ − 2 π π f(x) = 2, when − ≤ x ≤ + 2 2 π −2, when + 2 ≤ x ≤ + π and has a period of 2π. The periodic waveform is shown below. It is an even function and contains no sine terms, hence bn = 0 and between –π and +π, the mean value is zero, hence a0 = 0 cn 2 L /2 2 2π nx = f ( x) cos d x ∫ 0 L 2π L = ∫ π {∫ 1 π 0 2π nx f ( x) cos d x 2π π /2 0 2 cos nx d x + ∫ π π /2 since L = 2π } −2 cos nx d x π /2 π 1 2sin nx 2sin nx = − π n 0 n π /2 = ∞ Hence, f(x) = ∑ cn e n =− ∞ j 2π n x L 1 nπ nπ 4 nπ − 0 − 2sin nπ − 2sin 2sin = sin π n 2 2 π n 2 ∞ = 4 nπ 2 ∑ π n sin n =− ∞ j n x e 2. Show that the exponential form of the Fourier series in Problem 1 is equivalent to: f ( x) = 8 1 1 1 cos x − cos 3 x + cos 5 x − cos 7 x + ... π 3 5 7 1562 © 2014, John Bird Since from Problem 1, cn = c0 == 0 , c1 c3 = 4 nπ , then sin 2 πn 4 2π 4 π 4 , c2= sin = 0= c4= c6= c−2= c−4 and so on , sin = 2π 2 2 π π 4 3π 4 , sin = − 3π 2 3π c5 = 4 4 , c7 = − and so on 7π 5π 4 4 −π 4 4 4 −3π , c−3 = , c−5 = and so on c−1 = − sin = − sin = − 5π 2 π π 3π 2 3π ∞ f(x) = ∑ cn e j 2π n x L ∞ nπ 2 4 ∑ π n sin = n =− ∞ n =− ∞ = 4 π e jx − j n x e 4 j3x 4 j5x 4 4 − j3x 4 − j5x e + e + ... + e − jx − e + e − ... 3π 5π 3π 5π π 4 4 4 j3x 4 − j3x 4 j5x 4 − j5x = e jx + e − jx − e + e e + e + − ... π 3π 5π π 3π 5π = = 8 π cos x − 8 e j 3 x + e− j 3 x − 2 3π 8 e j 5 x + e− j 5 x + 2 5π − ... 8 8 cos 3 x + cos 5 x − ... 3π 5π 8 1 1 1 cos x − cos 3 x + cos 5 x − cos 7 x + ... 3 5 7 π f ( x) = i.e. 8 e jx + e − jx π 2 3. Determine the complex Fourier series to represent the function f(t) = 2t in the range –π to +π. The triangular waveform shown below is an odd function since it is symmetrical about the origin The period of the waveform, L = 2π Thus, cn = − j = −j 2 L2 2π nt f (t ) sin dt ∫ L 0 L 2 2π ∫ π 0 2 π 2π nt − j ∫ t sin nt d t 2t sin dt = π 0 2π 1563 © 2014, John Bird π 2 −t cos nt sin nt 2 −π cos nπ sin nπ = −j + = − j + 2 n 0 n n2 π n π i.e. − ( 0 + 0 ) by parts 2 cn = j cos nπ n Hence, the complex Fourier series is given by: ∞ ∑ f(t) = cn e j 2π nt L ∞ = n =− ∞ j2 cos nπ e jnt n n =− ∞ ∑ 4. Show that the complex Fourier series in Problem 3 is equivalent to: 1 1 1 f (t ) = 4 sin t − sin 2t + sin 3t − sin 4t + ... 2 3 4 2 From Problem 3, cn = j cos nπ n When n = 1, c1 =j 2 2 j2 cos π =j ( −1) = − (1) (1) 1 2 2 When n= = 2, c2 j= cos 2π j 2 2 2 2 j2 When n = 3, c3 =j cos 3π =j ( −1) = − 3 3 3π By similar reasoning, c4 = When n = –1, c−1 =j j2 , 4 c5 = − j2 , and so on 5 2 2 j2 cos(−π ) = +j ( −1) = (−1) (−1) 1 2 2 j2 j cos(−2π ) = j − When n = –2, c−2 = (1) = (−2) (−2) 2 2 j2 By similar reasoning, c−3 = j , c−4 = − , and so on 3 4 Since the waveform is odd, c= a= 0 0 0 ∞ From Problem 3, f(t) = ∑ n = −∞ Hence, f(t) = − cn e j 2π nt L ∞ = j2 cos nπ e j nt n n = −∞ ∑ j 2 − j t j 2 − j 2 t j 2 − j 3t j 2 − j 4 t j 2 j t j 2 j 2 t j 2 j 3t j 2 j 4 t e − e + e − e + ... e + e − e + e − ... + 1 2 3 4 1 2 3 4 j 2 − j t j 2 j 2 t j 2 − j 2 t j 2 j 3 t j 2 − j 3t j2 e + e − e +− e + e + ... = − e jt + 1 2 3 1 2 3 1564 © 2014, John Bird e j t − e − jt = − j4 2 j 4 e j 2t − e − j 2 t + 2 2 j 4 e j 3t − e − j 3t − 2 3 + ... e j t − e − jt j 2 4 e j 2t − e − j 2 t j 2 4 e j 3t − e − j 3t = − j2 4 + − + ... by multiplying top 2j 2j 2j 2 3 and bottom by j 4 4 = 4sin t − sin 2t + sin 3t + ... 2 3 i.e. 1 1 1 f(t) = 4 sin t − sin 2t + sin 3t − sin 4t + ... 2 3 4 Hence, f(t) = ∞ 1 1 1 j2 cos nπ e jnt ≡ 4 sin t − sin 2t + sin 3t − sin 4t + ... n 2 3 4 n = −∞ ∑ 1565 © 2014, John Bird EXERCISE 372 Page 1105 1. Determine the pair of phasors that can be used to represent the following voltages: (a) v = 4 cos 4t (b) v = 4 cos ( 4t + π / 2 ) (a) From equation (3), page 1094, cos = θ 1 jθ ( e + e− jθ ) 2 1 v = 4 cos 4t = 4 ( e j 4t + e − j 4t ) = 2e j 4t + 2e − j 4t 2 Hence, This represents a phasor of length 2 rotating anticlockwise (i.e. in the positive direction) with an angular velocity ω of 4 rad/s, and another phasor of length 2 and rotating clockwise (i.e. in the negative direction) with an angular velocity of 4 rad/s. Both phasors have zero phase angle Hence, one phasor is 2∠0° anticlockwise, and the other phasor is 2∠0° clockwise (b) From equation (3), page 1094, cos θ = Hence, 1 jθ ( e + e− jθ ) 2 1 v 4 cos ( 4t + π / 2 ) = 4 ( e j ( 4t + π /2) + e − j ( 4t + π /2) ) = 2e j ( 4t +π 2) + 2e − j ( 4t + π /2) = 2 v = 2e j 4t e j π /2 + 2e − j 4t e − j π /2 i.e. This represents a phasor of length 2 and phase angle π/2 radians rotating anticlockwise (i.e. in the positive direction) with an angular velocity ω of 4 rad/s, and another phasor of length 2 and phase angle π/2 radians and rotating clockwise (i.e. in the negative direction) with an angular velocity of 4 rad/s. Hence, one phasor is anticlockwise, and the other phasor is 2∠ − π / 2 clockwise 2. Determine the pair of phasors that can represent the harmonic given by: v = 10 cos 2t – 12 sin 2t v = 10 cos 2t – 12 sin 2t 1 1 = 10 ( e j 2 t + e − j 2 t ) − 12 ( e j 2 t − e − j 2t ) 2 2 j 6 6 = 5e j 2 t + 5e − j 2 t − e j 2 t + e − j 2 t j j 1566 © 2014, John Bird j 1 −j 1 = or = 5e j 2 t + 5e − j 2 t + 6 je j 2 t − 6 je − j 2 t = (note: ) −j 1 j 1 i.e. v = ( 5 + j 6 ) e j 2 t + ( 5 − j 6 ) e − j 2 t Hence, v = 7.81∠0.88 rad, rotating anticlockwise with an angular velocity, ω = 2 rad/s and v = 7.81∠–0.88 rad, rotating clockwise with an angular velocity, ω = 2 rad/s, as shown in the diagram below 3. Find the pair of phasors that can represent the fundamental current: i = 6 sin t + 4 cos t. i = 6 sin t + 4 cos t 1 1 3 = 6 ( e jt − e − jt ) + 4 ( e jt + e − jt ) = ( e jt − e − jt ) + 2 ( e jt + e − jt ) 2 j 2 j = −3 j ( e jt − e − jt ) + 2 ( e jt + e − jt ) i.e. i = ( 2 − j 3) e jt + ( 2 + j 3) e − jt Hence, i = 3.61∠–0.98 rad, rotating anticlockwise with an angular velocity, ω = 1 rad/s and i = 3.61∠0.98 rad, rotating clockwise with an angular velocity, ω = 1 rad/s, as shown in the diagram below 1567 © 2014, John Bird 1568 © 2014, John Bird
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