1. Fourier Transform sin 8 = ej8−e−j8 ej8 e–j8 ej8 + e–j8 ej8 + e–j8
1) The complex or exponential form of a Fourier series:
The form used for the Fourier series in last year consisted of cosine and sine terms. However, there is another form that is commonly used—one that directly gives the amplitude terms in the frequency spectrum and relates to phasor notation. This form involves the use of complex numbers. It is called the exponential or complex form of a Fourier series.
By complex notation, we have:
ej8
= cos j sin 8
(1)
(2)
e–j8
and = cos 8 − j sin 8
Adding equations (1) and (2) gives:
ej8 + e–j8
from which,
= 2cos8
ej8 +e—j8
cos 8 =
(3) 2
Similarly, equation (1) – equation (2) gives:
ej8 + e–j8
= 2sin8
sin 8 = ej8−e−j8
from which,
(4) 2j
Thus, the Fourier series f (x) over any range T
Page 1 of 24

2. 2nnx 2nnx )+ b sin ( )] f (x) = a + Σ [a cos ( T T j2nns –j2nns j2nnsœ
2nnx 2nnx
)+ b sin ( )]
f (x) = a
+ Σ [a cos ( n T n T
n=1
may be written as: œ
j2nns
–j2nns
j2nns
–j2nns
− e 2j
+ Σ [an (
n=1 e T
+ e T T T
f (x) = a
) + b ( e )] 2 n
Multiplying top and bottom of the bn term by −j (and remembering that j 2 = −1) gives: œ
j2nns
–j2nns
j2nns
–j2nns T
+ e T e T
− e T
f (x) = a + Σ [a ( e
0 n
) − jbn ( )] 2 2
n=1
Rearranging gives: œ
f (x) = a0 + Σ [(
n=1
an − jbn
j2nns
an + jbn
–j2nns
)e − ( )e ] (5)
T T
2 2
The Fourier coefficients a0, an and bn may be replaced by complex coefficients C0, Cn and C−n such that:
Page 2 of 24

3. f (x) = C0 + Σ Cne + Σ C−ne j2nns –j2nns (9) + Σ C−ne f (x) = Σ Cne
C0 = a0
(6)
an–jbn
Cn =
(7) 2
an+jbn
And; C−n =
(8) 2
where C−n represents the complex conjugate of Cn. Thus, equation (5) may be rewritten as: œ
f (x) = C0 + Σ Cne
n=1 ∞
+ Σ C−ne
n=1
j2nns
–j2nns T T
(9)
Since e0 =1, the C0 term can be absorbed into the summation since it is just another term to be added to the summation of the Cn term when n = 0. Thus, œ ∞
+ Σ C−ne
n=1
f (x) = Σ Cne
n=0
j2nns
–j2nns T T
(10)
The C–n term may be rewritten by changing the limits n = 1 to n = ∞ to n = −1 to n = −∞. Since n has been made
j2unx
negative, the exponential term becomes e and C–n becomes Cn. Thus: T œ
f (x) = Σ Cne
n=0 −∞
+ Σ Cne
n=−1
j2nns
j2nns T T
Since the summations now extend from −∞ to −1 and from 0 to +∞, equation (10) may be written as:
Page 3 of 24

4. 2) The complex coefficients:œ
f (x) = Σ Cne
n=–œ
j2nns T
(11)
Equation (11) is the complex or exponential form of the Fourier series.
2) The complex coefficients:
From equation (7), the complex coefficient Cn was defined as:
an–jbn
C = . n 2
However, an and bn are defined by: 2
T⁄2
2nnx T
an = T ƒ
f(x) cos (
) dx;
–T⁄2
and, 2
T⁄2
2nnx
bn = T ƒ f(x)sin (
)dx T
–T⁄2
Thus,
Page 4 of 24

5. ƒ(s) cos(2unx)ds –j2 ∫T⁄2 = T —T⁄2 T T —T⁄2 T 2nnx T 2nnx T )dx j2nnx
ƒ(s)sin ( )ds
ƒ(s) cos(2unx)ds –j2 ∫T⁄2
= T —T⁄2 T T —T⁄2 T C n 2 1
T⁄2
2nnx T 1
T⁄2
2nnx T
= T ƒ f(x) cos (
) dx − j T ƒ f(x)sin (
)dx
–T⁄2
–T⁄2
From equations (3) and (4),
j2nnx
−j2nnx
j2nnx
−j2nnx 1
T⁄2 e
e − e 2j T
+ e T 1
T⁄2 T T
Cn = T ƒ
f(x) (
) dx − j T ƒ f(x) (
) dx 2
–T⁄2
–T⁄2
From which,
j2nnx
−j2nnx
j2nnx
−j2nnx 1
T⁄2 e T
+ e T 1
T⁄2
e − e T T
Cn = T ƒ f(x) (
) dx − T ƒ f(x) (
) dx 2 2
–T⁄2
–T⁄2
i.e. 1
T⁄2
−j2nnx
Cn = T ƒ f(x) e T dx
(12)
–T⁄2
Page 5 of 24

6. Note:
Care needs to be taken when determining C0. If n appears in the denominator of an expression the expansion can be invalid when n = 0. In such circumstances it is usually simpler to evaluate C0 by using the relationship: 1
T⁄2
C0 = a0 = T ƒ
f(x) dx
(13)
–T⁄2
Example 1:.
Determine the complex Fourier series for the function defined by: 0,
f (x) = { 5,
wℎen − 2 ≤ x ≤ −1 wℎen − 2 ≤ x ≤ −1 wℎen ≤ x ≤ 2
The function is periodic outside this range of period 4.
Solution:
The function f (x) is shown in Figure (1), where the period, T = 4. From equation (11), the complex Fourier series is given by:
Page 6 of 24

7. j2nns f (x) = Σ Cne −j2nnx dx −j2nnx dx + ƒ 0 dx } Fig.(1)
The function f (x) is shown in Figure (1), where the period, T = 4. From equation (11), the complex Fourier series is given by: œ
j2nns
f (x) = Σ Cne T
n=–œ
where Cn is given by:
T⁄2 1
−j2nnx
Cn = T ƒ f(x) e T dx
–T⁄2
With reference to Figure (1), when T = 4, 1
–1 1
−j2nnx 2
dx + ƒ 0 dx } 1
Cn = 4 {ƒ 0 dx + ƒ 5 e 4
–2 –1
Page 7 of 24

8. = 1 ∫1 5 e–j [ e–j –jun ej 2 – e—j 2 –jun C dx = 5 [ e ] = ]
—junx 1
= 1 ∫1
5 e–j
unx 2 2 –5
[ e–j
unx 2 1 C
dx = 5 [ e
] = ] n
–jun 4 –1 4
j2nn –1 2 –1
un un
–jun –5 un
2 j 2 5
nn 2j
ej 2 – e—j 2
j2nn
Cn = nn sin ( 2 ) from equation (4)
Hence, from equation (11), the complex form of the Fourier series is given by: 5 nn œ
f (x) = Σ Cne
n=–œ ∞
j2nns
= Σ nn sin 2 e
n=−∞ 5 nn
jnns T 2
(14)
From equation (13), 1
T⁄2 1
1 5 5 2
C0 = a0 = ƒ f(x) dx = ƒ 5 dx = [x]1 = T 4 4 −1
–T⁄2 –1
Since, 5 nn
C =
sin nn , tℎen n 2
C = 5 sin n = 5
n 2 n 5 2n 1
C =
sin 2n = 0 2 2
Page 8 of 24

9. jnx j3nx j5nx j7nx −jnx −j3nx −j5nx −j7nx e + e − e + ⋯ + e e e e
(in fact, all even terms will be zero since sin nn = 0) 5 3n
C =
sin 3n = − 5 3 2 3n
By similar substitution,
C = 5 , C = − 5 and so on. 5 5n 7 7n
Similarly,
C = − 5 sin −n = 5 –1
n 2 n
C = − 5 sin −2n = 0 = C
2n 5
−3n
= C and so on. –2 2 –4 –6
C =
sin −3n = − 5 –3 2 3n
C = 5 , C = − 5 and so on. –5
5n 7n –7
Hence, the extended complex form of the Fourier series shown in equation (14) becomes:
5 5 5 3n 5 5n 5 7n
jnx
j3nx
j5nx
j7nx 5 5 3n 5 5n 5 7n
−jnx
−j3nx
−j5nx
−j7nx
f(x) = + e
2 −
e + 2
e − 2
e + ⋯ + e 2
2 − e 2 + e 2 − e 2
+ ⋯
2 n n
Page 9 of 24

10. 1 cos ( jnx −jnx j3nx −j3nx j5nx −j5nx + e 2 ) + ( e + e 2 ) − ⋯
5 5
jnx −jnx 5
j3nx
−j3nx 5
j5nx −j5nx
f(x) = + ( e 2 + e 2 ) − ( e 2
+ e 2 ) + ( e + e 2 ) − ⋯ 2
2 n 3n 5n
jnx −jnx
j3nx −j3nx
j5nx
−j5nx
5 5
e e 2 5
e + e
2 2 5
e + e 2 2
f(x) = + (2) (
) − (2) (
) + (2) (
) − ⋯
2 n 2 3n 2 5n 2
From equation (3):
f(x) = 5 10
cos (
) − cos ( ) + cos 5n ( x 2 n 3n 5n
i.e.:
[cos nx
) − 1
3nx) +
1 cos (
5nx)
nx 10
3nx 10 +
) − ⋯ + (
− ⋯ ]
2 n 2 3 2 5 2
Hence, ∞ 5 nn
jnns
Σ nn sin 2 e 2
is equivalent to:
n=–∞ 5 10
[cos ( nx
) − 1
cos (
3nx
) +
5nx 2 n 3 +
) − ⋯ ]
Page 10 of 24

11. j2nnt f (t) = Σ Cne Example 2:.
Determine the complex Fourier series for the periodic function defined by:
† (t ) = t in the range t = 0 to t = 1.
Solution:
The saw tooth waveform is shown in Figure (2): T
Fig. (2)
From equation (11), the complex Fourier series is given by: œ
f (t) = Σ Cne
n=–œ
j2nnt T
and when the period, T =1, then:

12. e−j2nnt dt = −j2nn e–j2nnt 1 Cnej2nnt = ƒ t e−j2nntdt f (t) = Σ −j2nntœ
f (t) = Σ
n=–œ
Cnej2nnt
where, from equation (12),
T⁄2 1
−j2nnt 1 T
−j2nnt
Cn = T ƒ
f(t) e T
dt = T ƒ f(t) e T dt
–T⁄2
and when T =1 and f (t )=t , then: 1
= ƒ t e−j2nntdt Cn
Using integration by parts let u = t , from which, du = 1, and dt = du, and let dr = e–j2unt , from which, dt
e−j2nnt
r = ƒ e
−j2nnt
dt = −j2nn 1
Cn = ƒ t e−j2nntdt = ur − ƒ rdu
e–j2nnt 1
1 e–j2nnt e–j2nnt e–j2nnt 1
= t −j2nn ] − ƒ
−j2nn dt = [t −j2nn − (−j2nn)2]

13. = (−j2nn − (−j2nn)2) − (0 − (−j2nn)2)
e–j2nn e–j2nn e0
= (−j2nn − (−j2nn)2) − (0 − (−j2nn)2)
From equation (2),
cos2nn − jsin2nn − cos2nn − jsin2nn) + ( 1
C = ( ) n
−j2nn (−j2nn)2 (−j2nn)2
However, cos2un = 1 and sin2un = 0 for all positive and negative integer values of n.
1 1 1
C = ( −
) + ( ) n
−j2nn (−j2nn)2
(−j2nn)2 1
1(j)
= −j2nn = −j2nn(j) j
2nn
i.e. Cn
From equation (13), 1
T⁄2 T 1 1 5 2
C0 = a0 = ƒ f(t) dt = ƒ f(t) dt = [x]1 = T T 4 −1
–T⁄2 0
t2 1 1 1 1 2
= ƒ t dt = [ ] =
1 2

14. ej2nnt ej2nnt Odd – Even property: j2nnt f (t) = Σ Cne
Hence, the complex Fourier series is given by: œ
f (t) = Σ Cne
n=–œ
j2nnt T
from equation (11)
i.e. ∞
ej2nnt
f(t) = 1 + Σ 2
n=−∞ œ Σ
n=–œ
ej2nnt
1 j
= +
2 2n n
Odd – Even property:
If even or odd symmetry is noted in a function, then time can be saved in determining coefficients. The Fourier coefficients present in the complex Fourier series form are affected by symmetry. Summarizing from previous lecturer notes:
An even function is symmetrical about the vertical axis and contains no sine terms, i.e. bn = 0. For even symmetry, 1 T
a0 = T ƒ f(x) dx , and

15. n = 0 and, 2nnx T 2nnx T 2nnx T 2nnx T 2nnx T (15) 2
an = T ƒ f(x) cos (
) dx 4
T⁄2
2nnx T
an = T ƒ f(x) cos (
) dx
An odd function is symmetrical about the origin and contains no cosine terms, a0 = an = 0. For odd symmetry, 2 T
2nnx T
bn = T ƒ f(x) sin (
) dx 4
T⁄2
2nnx T
bn = T ƒ f(x) sin (
) dx
an – j bn
From equation (7), n 2
Thus, for even symmetry, bn = 0 and,
T⁄2
2nnx T n 2 2
Cn =
= T ƒ f(x) cos (
) dx
(15)
For odd symmetry,
n = 0 and,

16. (nn) 2nnx T 2nnx T f(x) cos ( 2nnx T nnx 2 sin (nnx) 1 2 2 2 2 −2j
ƒ f(x) sin (
2nnx T
Cn = =
) dx
(16) T
For example 1 the function f (x) is even, since the waveform is symmetrical about the y axis. Thus equation (15) could have been used, giving:
T⁄2 2
2nnx T
Cn = T ƒ
f(x) cos (
) dx 2 2
2nnx T
= ƒ f(x) cos (
) dx 4 1 2 2
[ƒ 5 cos ( ) dx + ƒ 0 dx]
0 1 1
nnx = 2 5
= [
sin (nnx) 1 2
5 sin nn
] =
(nn) 2 nn 2 2
5 nn
Cn = nn sin 2
which is the same answer as in example 1; however, a knowledge of even functions has produced the coefficient more quickly.

17. Example 4:
Obtain the Fourier series, in complex form, for the square wave shown in Figure (4).
Fig.(4)
Solution:
The square wave shown in Figure (4) is an odd function since it is symmetrical about the origin. The period of the waveform, T = 2u.
Thus, using equation (16):
−2j T
T⁄2
ƒ f(x) sin (
2nnx T
Cn =
) dx
−2j n
2nnx =
ƒ 2 sin (
) dx 2n 2n 2 n
= −j n ƒ sinnx dx

18. f (t) = Σ −j 2 (1 − cosnπ) ejns
2 −cosnx n
= −j n [ ] n 2
= −j nn (−cosnπ + cos0)
= −j nn (1 − cosnπ)
It is clear that this method is by far the shorter than the method without even-odd property. From equation (11), the complex Fourier series is given by: 2 œ
f (t) = Σ Cne
n=–œ
j2nns T œ
f (t) = Σ −j 2 (1 − cosnπ) ejns nn
n=–œ

19. 4) The frequency spectrum:
In the Fourier analysis of periodic waveforms seen in previous lecturer notes, although waveforms physically exist in the time domain, they can be regarded as comprising components with a variety of frequencies. The amplitude and phase of these components are obtained from the Fourier coefficients an and bn; this is known as a frequency domain. Plots of amplitude/frequency and phase/frequency are together known as the spectrum of a waveform.
Example 5:
A pulse of height 20 and width 2 has a period of 10. Sketch the spectrum of the waveform. The pulse is shown in Figure (5).
Fig.(5)
Solution:
The complex coefficient is given by equation (12):
T⁄2 1
−j2nnt
Cn = T ƒ f(t) e T dt
–T⁄2

20. = nn [ ] = 2[1 − (− −jnnt −jnnt 1 −jnn jnn 5 jnn −jnn 2j 1)] = 4 1
Cn = 10 ƒ 20 e 5 dt –1
−jnnt 1
20 e 5
20 5
−jnn
jnn
Cn = 10 [ −j nn]
= [ e 5 − e
10 −jnn
5 ] 5 –1
jnn
−jnn
20 e 5 − e 5 ]
= nn [ 2j
i.e.
20 nn
Cn = nn sin 5 from equation (4),
From equation (13), 1
T⁄2 1
1 20
C0 = T ƒ f(t)dt = 10 ƒ 20dt = [t
] = 2[1 − (− 1
1)] = 4 10 –1
–T⁄2
C1 = n sin 5 = 3.74 and –1
20 n
20 −n 20 n
C–1 = −n sin
= n sin 5 = 3.74 5
Further values of cn and c−n, up to n = 10, are calculated and are shown in the following table:
Page 20 of 24

21. n cn
c−n 4 1
3.74 2
3.03 3
2.02
0.94 5 6
-0.62 7
-0.86 8
-0.76 9
-0.42 10
A graph of |cn | plotted against the number of the harmonic, n, is shown in Figure (6).
Figure (7) shows the corresponding plot of cn against n.
Since cn is real (i.e. no j terms) then the phase must be either 0o or ±180 shown in Figure (8).
, depending on the sign of the sine, as
When cn is positive, i.e. between n = −4 and n = +4, angle αn = 0o .
When cn is negative, then αn = ±180
; between n = +6 and n = +9, αn is taken as +180◦, and between
n = −6 and n = −9, αn is taken as −180◦.
Figures (6) to (8) together form the spectrum of the waveform shown in Figure (5).

22. Fig.(6)
Fig.(7)

23. Fig.(8)
The coefficients Cn for some functions are real numbers whereas for others are complex numbers> In general, the Cn are complex numbers and can be written as;
A sin nnaT
|Cn| = T | |
nnT
And ∅n = − nna T

24. These complex coefficients constitute a discrete complex spectrum where Cn represents the spectral coefficient of the nth harmonic. Each spectral coefficient couples an amplitude spectrum value |Cn| and a phase spectrum value ∅n. The amplitude spectrum tells us the magnitude of each of the harmonic components and has graphs shown on figures (6-8).