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
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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)
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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
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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 =− ∞ 

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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

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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 
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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

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© 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 = −∞
∑ 
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© 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
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© 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
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© 2014, John Bird