4. Fourier Series
4.1. Introduction
In this section we will study periodic signals in terms of their frequency
content. Remind that a signal f (t ) is said to be periodic if
f (t + T ) = f (t )
(4.1)
for every t, where T is a number. From this definition it follows that
f (t + nT ) = f (t )
for every integer n, positive or negative. For example if n = 2 then
f (t + 2T ) = f ((t + T ) + T ) = f (t + T ) = f (t ) .
Similarly, for n = −1 the equation
f (t − T ) = f (t' ) = f (t' +T ) = f (t )
holds.
The smallest positive number T for which (4.1) holds is called a period.
Let us consider the well known sinusoidal function
Asin (ω0t + φ )
where A is the amplitude, ω0 is the angular frequency, and φ is the phase.
Since
Asin (ω0 (t + T ) + φ ) = Asin (ω0t + φ + ω0T )
(4.2)
116
and 2π is the smallest angle that satisfies the equation
sinα = sin (α + 2π )
then
T=
2π
.
ω0
(4.3)
Let us consider two periodic signals f 1 (t ) and f 2 (t ) with periods T1 and T2 ,
respectively. The common period of f 1 (t ) and f 2 (t ) , if it exists, is the smallest
positive number T satisfying
f 1 (t + T ) = f 1 (t )
f 2 (t + T ) = f 2 (t )
and
for every t.
Observe that any T, such that f 1 (t + T ) = f 1 (t ) for all t, can be expressed in the
form T = n1T1 where n1 is an integer. Likewise any T, such that f 2 (t + T ) = f 2 (t )
for all t, can be expressed in the form T = n 2T2 , where n2 is an integer. Hence,
the condition under which the functions f 1 (t ) and f 2 (t ) have a common period
is
n1T1 = n2T2 .
The above equation states that the ratio
T1 n2
=
T2 n1
must be a rational number.
The common period is the lowest common multiple (over the positive integers
field) of T1 and T2 . If the ratio T1 T 2 is an irrational number, then f 1 (t ) and
f 2 (t ) do not have a common period. In such a case integers n1 and n2 such that
n1T1 = n2T2 do not exist.
Let h(x ) be an arbitrary function. If f (t ) is a periodic signal with period T
then the function defined by the equation
g (t ) = h( f (t ))
117
is periodic because
g (t + T ) = h( f (t + T ) ) = h( f (t ) ) = g (t ) .
Hence, the period of g (t ) is T or less than T (see Example 2).
Example 4.1
Let us consider the sinusoidal signals
f1 (t ) = 5cos(ω0t + 60°)
f 2 (t ) = 10sin 3ω0t .
The periods of these signals are:
T1 =
2π
ω0
T2 =
,
2π
.
3ω0
Consequently, we obtain
T1
=3
T2
thus, the ratio T1 T 2 is a rational number. Therefore, the two signals have a
common period given by
T = T1 = 3T2 =
2π
ω0
.
Now we consider the signals:
g1 (t ) = 5cos(ω0t + 45°)
g 2 (t ) = 10sin 3 ω0t .
The periods of these signals are:
T1 =
and the ratio
2π
ω0
,
T2 =
2π
3 ω0
118
T1
= 3
T2
is an irrational number. Hence, the signals do not have any common period.
Example 4.2
Let us consider the periodic signal
f (t ) = cost
with period T1 = 2π and let h(x ) = x .
The function g (t ) = h( f (t ) ) is
g (t ) = cost .
It is a periodic function with period T2 =
T1
= π (see Fig.4.1)
2
f(t)
(a)
0
(b)
π
2π
π
2π
t
g(t)
0
t
Fig. 4.1. Plots of signals f (t ) = cos t and g (t ) = cos t
Thus, the period of g (t ) is only half that of f (t ) .
119
4.2. Fourier series. Definition
We consider a periodic signal f (t ) with the period T. If some conditions,
which will be discussed later, are satisfied, we can express the function f (t ) as a
series of sine and cosine functions
f (t ) = a0 + a1cos ω0t + a2cos 2ω0t +
+
+ b1sinω0t + b2sin 2ω0t +
or
f (t ) = a0 +
∞
∑ (a cos nω t + b sin nω t )
n
0
n
0
(4.4)
n =1
2π
. Expression on the right hand side of (4.4) is called the
T
trigonometric Fourier series. The sum is called Fourier series and its terms are
called the harmonics. The n-th harmonic is the term
where ω0 =
wn (t ) = an cos nω0t + bnsin nω0t =
⎛
⎞
an
bn
= an2 + bn2 ⎜
cos nω0t +
sin nω0t ⎟ .
⎜ a 2 + b2
⎟
an2 + bn2
n
⎝ n
⎠
We label
c n = a n2 + bn2
cos θ n =
an
an2
+ bn2
sin θ n =
(4.5)
− bn
an2 + bn2
(4.6)
obtaining
wn (t ) = cn (cos θ n cos ω0t − sin θ nsin nω0t ) = cn cos (nω0t + θ n ) .
(4.7)
Thus, the n-th harmonic is a sinusoidal function with the amplitude c n , the phase
θ n and the angular frequency nω0 .
The first harmonic
120
w1 (t ) = a1cos ω0t + b1sin ω0t = c1cos (ω0t + θ1 )
is called the fundamental and its frequency ω0 is called the fundamental
frequency.
The constant term a 0 is the 0-th harmonic. Hence, an equivalent form of the
Fourier series
f (t ) = a0 + c1cos (ω0t + θ1 ) + c2cos (2ω0t + θ 2 ) +
= c0 +
=
∞
∑ c cos (nω t + θ )
n
0
(4.8)
n
n =1
with c0 = a0 follows.
The plot of c n as a function of n is called the amplitude spectrum and the plot
of θ n as a function of n is called the phase spectrum of the signal f (t ) . Together
they are called the frequency spectra.
Example 4.3
Let us consider a periodic signal f (t ) including the following harmonics:
c0 = 2
w1 (t ) = 1.5cos (ω0t + 30°)
w3 (t ) = cos (3ω0t − 45°)
w5 (t ) = 0.5cos 5ω0t
rad
.
s
Figure 4.2 shows these harmonics and illustrates how the signal f (t ) is built up
from its harmonics.
where ω0 = 1000π
(a)
c0
2
t
121
(b)
w1(t)
t
(c)
w3(t)
t
w5(t)
(d)
t
(e)
f (t)
t
Fig. 4.2. Construction of the signal f (t ) consisting of four harmonics
122
4.3. Evaluation of Fourier series coefficients
Our objective is to evaluate the a n and bn coefficients in the Fourier series
(4.4). The coefficients are called the Fourier coefficients of the signal f (t ) .
In order to determine a 0 we integrate both sides of equation (4.4) from 0 to T
T
∫
0
T
⎛ T
⎞
⎜
f (t ) dt = a0dt +
an cos nω0t dt + bn sin nω0t dt ⎟ .
⎜
⎟
n =1 ⎝
0
0
0
⎠
∞
T
∑ ∫
∫
∫
(4.9)
Let us consider the integral
T
∫ cos nω t dt .
0
0
The function cos nω0t has the period
Tn =
2π
nω0
whereas
T=
2π
ω0
.
Hence, T = nTn holds. Since integral of cos nω0t over its period Tn is zero
Tn
∫ cos nω t dt = 0
0
0
then
Tn
T
∫
∫
(4.10)
∫ sin nω t dt = 0 .
(4.11)
cos nω0t dt = n cos nω0t dt = 0 .
0
0
Similarly, we have
T
0
0
123
Substituting (4.10) and (4.11) into (4.9) we obtain
T
∫
T
f (t ) dt = a0dt = a0T
∫
0
0
or
1
a0 =
T
T
∫ f (t ) dt .
(4.12)
0
Formula (4.12) can be modified by adding an arbitrary value t0 to the lower and
upper limit of integration
1
a0 =
T
t 0 +T
∫ f (t ) dt .
(4.13)
t0
Equation (4.13) shows that a0 is the average value of f (t ) over the period.
To compute am we multiply both sides of the Fourier series (4.9) by
cos mω0t and then integrate from 0 to T obtaining
T
∫
T
f (t )cos mω0t dt = a0cosmω0t dt +
∫
0
+
0
(4.14)
∞ T
∑ ∫ (a cos nω tcos mω t + b sin nω tcos mω t )dt .
n
0
0
n
0
0
n =1 0
As explained above, the first term on the right hand side of (4.14) is zero. To
rearrange the other terms we use the following formulas:
T
∫ sin nω tcos mω t dt = 0
0
(4.15)
0
0
⎧⎪ 0
cos nω0tcos mω0t dt = ⎨ 1
T
⎪⎩ 2
0
T
∫
if
m≠n
if
m=n
.
Hence, we have
T
f (t )cos mω0t dt = am cos 2 mω0t dt =
0
0
∫
T
∫
1
amT
2
(4.16)
124
or
2
am =
T
T
∫
0
2
f (t ) cos mω0t dt =
T
t0 +T
∫ f (t )cos mω t dt
m = 1, 2,
0
.
(4.17)
t0
In a similar manner we find
2
bm =
T
T
∫
0
2
f (t )sin mω0tdt =
T
t 0 +T
∫ f (t )sin mω t dt
m = 1, 2,
0
.
(4.18)
t0
Example 4.4
Let us consider the periodic signal with period T=2 having the waveform shown
in Fig.4.3
f (t)
2
0
1
2
t
-1
Fig. 4.3. Signal f (t ) for Example 4.4
We wish to express this signal by a Fourier series. The waveform of f (t ) is
described as follows
⎧2
f (t ) = ⎨
⎩− 1
0 < t <1
.
1< t < 2
for
for
To find a 0 , we apply (4.12)
a0 =
1
2
⎞ 1
1 ⎛⎜
2dt − dt ⎟ = .
⎟ 2
2 ⎜⎝ 0
1
⎠
∫
∫
Next, we calculate the Fourier coefficients a m given by (4.12):
125
1
2
1
2
⎞
2
1
2⎛
sin mω0t −
sin mω0t .
am = ⎜ 2cos mω0t dt − cos mω0t dt ⎟ =
⎟ mω0
2 ⎜⎝ 0
mω0
0
1
1
⎠
∫
Since ω0 =
∫
2π
= π we obtain
T
am = 0
m = 1, 2 ,
.
To determine bm we use (4.18):
bm =
1
2
1
2
⎞
2 ⎛⎜
2sin mω0tdt − sin mω0t dt ⎟ = 2 sin mπt dt − sin mπt dt =
⎟
2 ⎜⎝ 0
1
0
1
⎠
∫
∫
∫
∫
=
1
2
−2
1
1
(− 2cos mπ + 2 + cos2mπ − cos mπ ) =
cos mπt +
cos mπt =
mπ
mπ
mπ
0
1
=
3
(1 − cos mπ ) .
mπ
Hence, it follows
⎧
⎪ 0
⎪
bm = ⎨
⎪ 6
⎪ mπ
⎩
if
m is even
m = 1, 2 ,
if
.
m is odd
Thus, the coefficients bm are
b1 =
6
π
, b2 = 0, b3 =
2
π
, b4 = 0, b5 =
6
6
, b6 = 0, b7 =
,
5π
7π
and the Fourier series has the form
f (t ) =
1 6
2
6
6
+ sinω0t + sin 3ω0t +
sin 5ω0t +
sin 7ω0t +
2 π
π
5π
7π
.
126
Since
cn = an2 + bn2 = bn
and for n odd:
an
cosθ n =
then θ n = −
an2
+ bn2
bn
sinθ n = −
=0
an2
+ bn2
= −1
π
for n odd.
2
The amplitude and phase spectra of the signal are shown in Fig.4.4
(a )
cn
6
2 .0
π
1 .5
1 .0
2
π
0 .5
0
(b)
2
1
6
5π
4
3
6
7π
6
5
n
7
θn
0
1
2
3
4
5
6
7
n
π
−
2
Fig. 4.4. Spectra of the signal for Example 4.4
4.4. Properties of Fourier series
In this section we will discuss some waveform properties of periodic
functions.
127
A function f (t ) is said to be even if, for every t, it satisfies the condition
f (− t ) = f (t ) .
Examples of the even functions are:
cos ω0t ,
sinω0t ,
sin 2ω0t .
Note that if a function is even, its graph is symmetric in the vertical axis (see
Fig.4.5).
f (t)
−
0
T
2
t
T
2
Fig. 4.5. An even function
A function f (t ) is said to be odd if, for every t, it satisfies the condition
f (− t ) = − f (t ) .
Examples of odd functions:
sin ω0t
sin 3ω0t .
Note that if a function is odd, its graph is symmetric in the origin (see Fig.4.6)
f (t)
−
T
2
0
Fig. 4.6. An odd function
T
2
t
128
A periodic signal f (t ) with period T is said to possess odd half-wave
symmetry if, for every t, it satisfies the condition
⎛ T⎞
f (t ) = − f ⎜ t + ⎟ .
⎝ 2⎠
Function sinω0t has this property; another example is shown in Fig.4.7.
f (t)
0
T
2
T
t
Fig. 4.7. A function possessing odd half-wave symmetry
Even functions
Let f (t ) be a periodic even function. Then for any t0 the equation
t0
t0
−t0
0
∫ f (t )dt = 2 ∫ f (t )dt
holds.
This equation implies
a0 =
1
T
T
2
T
2
∫ f (t ) dt = T ∫ f (t ) dt .
T
−
2
2
0
If a function is even, then its Fourier series does not contain the terms
(4.19)
129
bnsin nω0t
n = 1, 2,
because they violate the relation f (− t ) = f (t ) . As a matter of fact
bnsin nω0 (− t ) = −bnsin nω0t .
The expression for am can be reduced as follows:
am =
t0 +T
T
2
∫ f (t ) cos mω t dt = T ∫ f (t ) cos mω t dt .
2
T
2
0
t0
0
−
T
2
Since both f (t ) and cos mω0t are even, then
f (t ) cos mω0t = f (− t ) cos mω0 (− t ) .
Consequently, we have
T
2
am =
2
4
2 f (t ) cos mω0t dt =
T 0
T
∫
T
2
∫ f (t )cos mω t dt
0
m = 1, 2 ,
. (4.20)
0
Summarizing, we state that the Fourier series of an even periodic signal contains
only a0 and cosine terms.
Odd functions
For an odd function f (t ) and any t0 the equation
t0
∫ f (t )dt = 0
−t0
holds; hence, we conclude that a0 = 0 .
If a function is odd then its Fourier series does not contain the terms
a0
and
an cos nω0t
because they violate the relation f (− t ) = − f (t ) .
n = 1, 2,
130
Similarly as previously, it can be proved that the formula for bm can be
simplified and has the form
bm =
4
T
T
2
∫ f (t )sin mω t dt .
(4.21)
0
0
Thus, the Fourier series of an odd periodic signal contains only sine terms.
Odd half-wave symmetry functions
In the case of odd half-wave symmetry the constant term a0 and all even
⎛ T⎞
harmonics cannot exist because they violate the condition f (t ) = − f ⎜ t + ⎟ , i.e.
2⎠
⎝
a0 = 0
am = bm = 0
for m even .
The coefficients am and bm for m odd are given by:
am =
bm =
4
T
4
T
T
2
∫ f (t )cos mω t dt
0
(4.22)
0
T
2
∫ f (t )sin mω t dt .
0
(4.23)
0
The above formulas can be derived using the following relationships for m odd:
⎛ T⎞
⎛ T⎞
f ⎜ t + ⎟ cos mω0 ⎜ t + ⎟ = − f (t ) cos(mω0t + mπ ) = f (t ) cos mω0t
2
⎝
⎠
⎝ 2⎠
⎛ T⎞
⎛ T⎞
f ⎜ t + ⎟ sin mω0 ⎜ t + ⎟ = − f (t ) sin (mω0t + mπ ) = f (t ) sin mω0t .
⎝ 2⎠
⎝ 2⎠
Using these expressions we can reduce the interval of integration twice and
obtain formulas (4.22) and (4.23).
131
The advantages of the symmetry properties of periodic functions are as follows:
(i) We know in advance what coefficients will be zero; consequently, we do not
waste time to compute them.
(ii) The other coefficients can be obtained using simplified expressions.
4.5. Exponential Fourier series
We consider the trigonometric Fourier series
f (t ) = a0 +
∞
∑ (a cos nω t + b sin nω t )
n
0
n
(4.24)
0
n =1
and substitute:
cos nω0t =
sin nω0t =
(
)
(
)
1 jnω 0 t
e
+ e − jnω 0 t
2
1 jnω 0 t
e
− e − jnω 0 t .
2j
As a result, we obtain
f (t ) = a0 +
∑ ⎜⎜⎝ 2 (e
∞
⎛ an
jnω 0 t
)
+ e − jnω 0 t +
n =1
= a0 +
∞
∑
n =1
(
bn jnω 0 t
− e − jnω 0 t
e
2j
⎛ an − jbn jnω 0 t an + jbn − jnω 0 t ⎞
+
e
e
⎜
⎟.
2
2
⎝
⎠
)⎞⎟⎟ =
⎠
(4.25)
Let
c~0 = a0
(4.26)
1
c~n = (a n − jbn )
2
(4.27)
then we have
1
c~− n = (a − n − jb− n ) .
2
Since a− n = an and b− n = −bn (see (4.17) and (4.18))
132
1
c~− n = (a n + jbn ) = c~n∗
2
(4.28)
Hence, we conclude that c~n and c~− n form a pair of complex conjugate numbers.
Inserting (4.26), (4.27) and (4.28) into (4.25) yields
f (t ) = ~
c0 +
∞
∑ c~ e
jnω 0 t
n
+
n =1
∞
∑ c~
− ne
− jnω 0 t
.
(4.29)
n =1
The third term on the right hand side of (4.29) can be rearranged as follows
∞
∑
−∞
c~− n e − jnω 0 t =
n =1
∑ ~c e
jnω 0 t
n
.
(4.30)
n = −1
Using (4.30) we rewrite (4.29) in the compact form
∞
f (t ) =
∑ ~c e
jnω 0 t
(4.31)
n
n = −∞
called the exponential Fourier series.
The coefficients c~n are given by (4.27) where an and bn are specified by (4.17)
and (4.18) repeated below:
2
an =
T
bn =
2
T
t0 +T
∫ f (t )cos nω t dt
0
t0
t0 +T
∫ f (t )sin nω t dt .
0
t0
Hence, it follows
1 2 ⎛⎜
c~n =
2T⎜
⎝
1
=
T
t0 +T
∫
⎞
f (t )sin nω0t dt ⎟ =
⎟
⎠
t0 +T
f (t ) cos nω0t dt − j
t0
∫
t0
t 0 +T
∫ f (t )(cos nω t − jsin nω t ) dt .
0
t0
0
133
Using Euler’s formula e jα = cosα + jsin α , we obtain
1
c~n =
T
Usually we set t0 = −
T
t 0 +T
∫ f (t ) e
− jnω 0 t
n = 1, 2,
dt
.
(4.32)
t0
or t0 = 0 , then c~n is
2
1
c~n =
T
T
2
∫ f (t ) e
T
−
2
− jn ω 0 t
1
dt =
T
T
∫ f (t ) e
− jnω 0 t
n = 1, 2,
dt
. (4.33)
0
The coefficients c~n are generally complex and can be expressed in the polar form
c~n = c~n e jφ n
n = 1, 2 ,
.
(4.34)
c− n = c~n and φ− n = −φn .
Using (4.28) we obtain the relationships: ~
The plot of c~n versus n is called the amplitude spectrum and the plot of φn
versus n is called the phase spectrum.
Note that for n = 0
1
c~0 =
T
t 0 +T
∫ f (t ) dt = a
0
.
(4.35)
t0
Since
− jtan
a − jbn 1 2
c~n = n
=
an + bn2 e
2
2
−1 bn
an
then, using (4.5) and (4.6), we can express c~n in terms of cn and θ n
1
c~n = cn e jθ n .
2
(4.36)
Assuming an arbitrary integer n = n̂ in the exponential Fourier series, we prove
that the sum of two terms corresponding to n = n̂ and n = − n̂ gives the n-th
harmonic:
134
(
c~n̂ e jn̂ω 0 t + c~− n̂ e − jn̂ω 0 t = ~
cn̂ e jn̂ω 0 t + ~
cn̂∗ e jn̂ω 0 t
) = 2Re(c~ e ) =
∗
jn̂ω 0 t
n̂
⎛1
⎞
= 2Re⎜ cn̂ e j(n̂ω 0 t +θn̂ ) ⎟ = cn̂ cos (n̂ω0t + θ n̂ ) .
⎝2
⎠
To derive the above relation, equations (4.28) and (4.36) have been applied.
Example 4.5
Let us consider a periodic function f (t ) , shown in Fig.4.8, specified by the
equation
1
f (t ) = Asin ω0t
2
where ω0 =
for
0≤t ≤T
2π
. For this function we find the coefficients c~n .
T
f(t)
A
0
T
2T
t
Fig. 4.8. Periodic function for Example 4.5
To determine c~n we use (4.27). Since f (t ) is even then bn = 0 and formula
(4.27) reduces to
1
c~n = an
2
(4.37)
where an is specified by (4.20) repeated below for convenience
an =
4
T
T
2
∫ f (t )cos nω t dt
0
0
n = 1, 2 ,
.
135
Hence, we have
T
2
an =
4A
1
sin ω0tcos nω0t dt
T 0
2
∫
n = 1, 2,
and apply the trigonometric identity
sinαcosβ =
1
(sin(α − β ) + sin(α + β ) )
2
finding
⎛T
⎞
2
⎟
2 A ⎜⎜ ⎛
1
1
⎞ ⎟
an =
⎜ sin( ω0 − nω0 )t + sin( ω0 + nω0 )t ⎟ dt =
T ⎜0⎝
2
2
⎠ ⎟
⎜
⎟
⎝
⎠
∫
T
⎡
⎤2
−1
1
1
1
2A ⎢
⎛
⎞
⎛
⎞ ⎥
=
cos ⎜ ω0 + nω0 ⎟ t ⎥ =
cos ⎜ ω0 − nω0 ⎟ t −
⎢1
1
T ⎢ ω − nω
⎝2
⎠
⎝2
⎠ ⎥
ω 0 + nω 0
0
0
2
⎣2
⎦0
⎤
⎡
⎛1
⎞ T
⎛1
⎞ T
cos ⎜ − n ⎟ω0
cos ⎜ + n ⎟ω0
⎥
⎢
− 2A
1
1
⎝2
⎠ 2 −
⎝2
⎠ 2 −
⎥.
⎢
=
+
1
1
1
T ⎢ ⎛1
⎞
⎛
⎞
ω0 − nω0
ω0 + nω0 ⎥
⎜ + n ⎟ω0
⎥
⎢ ⎜⎝ 2 − n ⎟⎠ω0
2
2
2
⎝
⎠
⎦
⎣
Since ω0
T
= π and ω0T = 2π , then
2
⎞
⎛
⎛1
⎞
⎛1
⎞
⎟
⎜ cos ⎜ − n ⎟π
cos ⎜ + n ⎟π
− 2A ⎜
1 ⎟
1
2
2
⎝
⎠
⎝
⎠
an =
=
−
+
−
⎟
1
1
1
1
2π ⎜
n
n
n
n
−
−
+
+
⎟
⎜
2
2
2
2
⎠
⎝
⎛
⎞
π⎞
π⎞
⎛
⎛
⎜ cos ⎜ nπ − ⎟
⎟
cos ⎜ nπ + ⎟
A⎜
1 ⎟
1
2⎠
2⎠
⎝
⎝
.
=
−
−
+
1
1⎟
1
1
π⎜
n
n
n
n
−
−
+
+
⎜
⎟
2
2⎠
2
2
⎝
136
Next we use (4.37) and calculate:
1
2A
~
c1 = a1 = −
2
3π
1
2A
c~2 = a2 = −
2
15π
1
2A
c~3 = a3 = −
2
35π
………………… .
To find c~0 = a 0 we apply equation (4.12)
1
c~0 = a0 =
T
T
∫
0
1
cos ω0t
A
1
A
2
f (t ) dt =
sin ω0t dt = −
T 0
2
T 1ω
0
2
T
T
∫
=
2A
π
.
0
Example 4.6
Let us consider a rectangular-wave signal f (t ) indicated in Fig.4.9.
f (t)
1
T
2
0
T
-1
Fig. 4.9. Rectangular-wave signal f (t ) for Example 4.6
To find the coefficients c~n , we apply (4.33) repeated below
1
c~n =
T
where
T
∫ f (t ) e
0
− jnω 0 t
dt
t
137
⎧1
f (t ) = ⎨
⎩− 1
and ω 0 =
for
for
0<t <
T
2
T
<t <T
2
2π
. Hence, we have
T
T
2
T
1 − jnω 0 t
1 − jnω 0 t
e
dt −
e
dt =
c~n =
T 0
TT
∫
∫
2
T
1⎛
1 ⎞ − jnω 0 t 2 1 ⎛
1 ⎞ − jnω 0 t
⎟e
⎟⎟ e
− ⎜⎜ −
= ⎜⎜ −
T ⎝ jnω0 ⎠
T ⎝ jnω0 ⎟⎠
0
T
.
T
2
Using relationship ω0T = 2π we obtain after simple rearrangements
(
)
j
c~n =
2e − jnπ − 1 − e − jn 2π ,
2πn
n = ±1, ± 2,
.
(4.38)
Note that the signal f (t ) is odd and has odd half-wave symmetry property;
hence, c~0 = a 0 = 0 as well as a n = 0 and for every n bn = 0 for an even n.
Consequently c~n = 0 for an even n. Taking into account the above statements
and relationships (4.38) and (4.28) we obtain:
π
π
2 2 −j
c~1 = − j = e 2
2 2 j
c~−1 = c~1∗ = j = e 2
c~2 = 0
c~− 2 = c~2∗ = 0
π
π
π
π
π
π
2
2 −j2
=
e
c~3 = − j
3π 3π
2
2 j2
=
e
c~− 3 = ~
c3∗ = j
3π 3π
c~4 = 0
c~− 4 = c~4∗ = 0
π
2
2 −j2
=
e
c~5 = − j
5π 5π
……………………
π
2
2 j2
=
e
c~− 5 = ~
c5∗ = j
5π 5π
……………………
.
138
Thus, the exponential Fourier series expansion of f (t ) is
2⎛
f (t ) = j ⎜
π⎝
⎞
⎟.
⎠
1
1
1
1
+ e − j5ω 0 t + e − j3ω 0 t + e − jω 0 t − e jω 0 t − e j3ω 0 t − e j5ω 0 t −
5
3
3
5
The amplitude and phase spectra are indicated in Figs.4.10 and 4.11
~
c
n
2
2
π
π
2
2
5π
π
-5
-4
-3
-2
0
-1
1
2
3
4
n
5
Fig. 4.10. Amplitude spectrum of the signal for Example 4.6
π
φn
2
-5
-4
-3
-2
0
-1
−
1
2
3
4
5
n
π
2
Fig. 4.11. Phase spectrum of the signal for Example 4.6
Since c~− n = c~n , the amplitude spectrum is symmetric in the vertical axis.
Similarly, since φ− n = −φn , the phase spectrum is symmetric in the origin.
4.6. Convergence of Fourier series
Not every periodic signal can be represented by a Fourier series and the
question arises under what conditions the Fourier series exists. This question is
answered by the Dirichlet conditions as follows:
139
(i) Over any period f (t ) must be absolutely integrable, i.e.
T
∫ f (t )
dt < ∞ .
0
(ii) In any period f (t ) is of bounded variation, i.e. f (t ) has at most a finite
number of maxima and minima.
(iii) In any period f (t ) has at most a finite number of discontinuities and each of
these discontinuities is finite.
Under these conditions the Fourier series of f (t ) converges to f (t ) at all points
where f (t ) is continuous and the series converges to
[ ( ) ( )]
1
f t i+ + f t i−
2
at any point ti of discontinuity.
Note that the condition (i ) guarantees that each coefficient c~n is finite, because
1
c~n =
T
T
∫ f (t ) e
0
− jnω 0 t
1
dt ≤
T
T
∫ f (t ) e
0
− jnω 0 t
1
dt =
T
T
∫ f (t ) dt < ∞ .
0
A periodic function that violates this condition is depicted in Fig.4.12
f (t)
−
3
T
2
-T
−
T
2
T
2
T
3
T
2
Fig. 4.12. Function that violates the first Dirichlet condition
An example of a function that does not satisfy condition (ii) is
t
140
⎛ 2π ⎞
f (t ) = e t −1sin ⎜
⎟
⎝1− t ⎠
1
0 ≤ t <1
with T = 1 as depicted in Fig.4.13.
f (t)
0
2
1
t
Fig. 4.13. Function that violates the second Dirichlet condition
An example of a function that violates condition (iii) is indicated in Fig.4.14.
f (t)
0
t
T
T
2
2
Fig. 4.14. Function that violates the third Dirichlet condition
−
Dirichlet conditions do not require that a function f (t ) be continuous in order to
possess a Fourier expansion. Thus, its waveform may consists of a number of
disjointed arcs as depicted in Fig.4.15.
141
f (t)
c
b
a
t1
0
T
2T
Fig. 4.15. Signal with discontinuity points
t
The signal shown in Fig.4.15 has two points of discontinuity at t1 and T over the
period. Setting t = t1 into the Fourier series of this signal we obtain
1
(a + c ) .
2
Similarly, setting t = T we obtain
1
b.
2
It can be shown that Fourier coefficients tend to zero as n approaches infinity.
Therefore the Fourier series can be truncated after a finite number of terms. The
truncated series including N terms will be called the partial sum and labeled
s N (t ) .
Let us consider a periodic signal f (t ) having the waveform depicted in
Fig.4.16. This signal satisfies the Dirichlet conditions.
f(t)
1
−
T
2
−
T
4
0
T
4
T
2
Fig. 4.16. Rectangular pulse train signal
t
142
T
the points of discontinuity occur. Figure 4.17 shows the partial sums
4
s N (t ) of this signal for N = 7 , N = 11 , N = 15 and N = 25 . Note that the partial
sums exhibit ripples and oscillations. Furthermore, the peak value of these ripples
At t = ±
(a)
s7(t)
T
−
2
T
−
4
(b)
0
T
4
T
2
T
4
T
2
t
s11(t)
0
T
−
2
T
−
4
t
143
(c)
s15(t)
0
T
−
2
(d)
T
4
T
−
4
T
2
t
s25(t)
0
T
T
T
T
−
−
2
4
2
4
Fig. 4.17. An illustration of the Gibbs phenomenon
t
in the neighborhood of the discontinuity points does not decrease as N increases
and remains approximately 9% of the height of the signal for any finite N. Thus,
the deviations in the vicinity of the discontinuities are rather large. This behavior
of s N (t ) is known as the Gibbs phenomenon.