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• Fourier series
Let f (t) be a periodic function with period T
(frequency of fundamental f0 = T1 ), the Fourier
series of f (t) is
f (t) = A0 +
∞
X
An cos(2πnf0t + φn)
n=1
= A0 +
∞
X
An cos(nω0t + φn )
n=1
where ω0 = 2πf0.
The Fourier series represents the signal f (t)
as a sum of ‘harmonically related’ sinusoids.
Constituent sinusoids are termed harmonics:
– sinusoid with frequency f0 is ‘the first harmonic’ or ‘fundamental’;
– sinusoid with frequency nf0 is ‘the nth harmonic’;
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– A0 is DC level or zeroth harmonic;
– All harmonics have frequencies which are integer multiples of fundamental frequency f0.
Alternative form:
f (t) = a0+
∞
X
an cos(2πnf0t)+
n=1
= a0 +
∞
X
∞
X
bn sin(2πnf0t)
n=1
an cos(nω0t) +
n=1
∞
X
bn sin(nω0t)
n=1
where
1
f (t)dt
a0 =
T T
Z
2
an =
f (t) cos(nω0t)dt
T T
Z
2
bn =
f (t) sin(nω0t)dt
T T
Z
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Relationships between two sets of coefficients:
a0 = A0
an = An cos(φn )
bn = −An sin(φn)
An =
q
2
a2
n + bn
φn = − arctan( abnn )
Note that:
If a function is even, then f (x) = f (−x). e.g.
cos(nω0t) is an even function.
If a function is odd, then f (x) = −f (−x). e.g.
sin(nω0t) is an odd function.
An even function times an odd function is an
odd function, and the integral of an odd function is zero.
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Hence:
If f (t) is an even function, then bn = 0, because f (t) sin(nω0t) is odd.
If f (t) is an odd function, both a0 = 0 and
an = 0 because f (t) cos(nω0t) is odd.
Example: Find the Fourier series coefficients
a0,an,bn for the following periodic rectangular
wave f (t), such that
f (t) = a0 +
∞
X
an cos(nω0t) +
n=1
∞
X
bn sin(nω0t).
n=1
(1)
f(t)
A
T 0
4
T
4
T
2T
t
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where ω0 = 2π
T is frequency (radian/s).
1. Note that for one period of the wave, e.g.
− T2 < t < T2 ,
f (t) =
(
A − T4 < t < T4
0 otherwise
(2)
2. Find a0.
1 T /2
a0 =
f (t)dt
T −T /2
←− Substitute Eq. (2) into left
Z
1 T /4
Adt
=
T −T /4
Z
Z
←−
dt = t + c
A
=
[(T /4) − (−T /4)]
T
A
=
2
(3)
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3. Find an.
an =
=
=
=
=
2
T
Z
T /2
f (t) cos(nω0t)dt
−T /2
←− Substitute (2) into left
Z
2 T /4
A cos(nω0t)dt
T −T /4
Z
1
←−
cos(nω0t)dt =
sin(nω0t) + c
nω0
2A
[sin(nω0 ∗ T /4) − sin(nω0 ∗ (−T /4))]dt
T ∗ nω0
←− ω0 T = 2π
A
[sin(nπ/2) − sin(−nπ/2)]
nπ
←− sin(nπ/2) = − sin(−nπ/2)
2A
sin(nπ/2)
(4)
nπ
4. Find bn.
2 T /2
bn =
f (t) sin(nω0t)dt = 0
T −T /2
Z
(5)
since f (t) is an even function, so the integral
is an odd function.
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Look at an = 2A
nπ sin(nπ/2).
n = 0, 2, 4, .., an = 0.
If n=even, e.g.
If n=odd,
— if n = 1, 5, 9, 4k + 1, an = 2A
nπ .
— if n = 3, 7, 11, 4k + 3, an = − 2A
nπ .
so we have
f (t) = a0 +
∞
X
an cos(nω0t)
n=1
A
2A
1
= +
{cos(ω0t) − cos(3ω0t)+
2
π
3
1
1
cos(5ω0t)) − cos(7ω0t)) + ...}
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Example 2: Find the Fourier series coefficients
a0,an,bn for the following periodic rectangular
wave f (t), such that
f (t) = a0 +
∞
X
an cos(nω0t) +
n=1
∞
X
bn sin(nω0t)
n=1
(6)
where ω0 = 2π
T is frequency (radian/s).
f(t)
A
0
T
2T
t
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1. Note that for one period of the wave, e.g.
− T2 ≤ t < T2 ,
f (t) =
(
−A − T2 ≤ t < 0
A
0 ≤ t < T2
(7)
2. Find a0.
1 T /2
a0 =
f (t)dt
T −T /2
= 0
Z
(8)
Since f (t) is an odd function.
3. Find an.
2 T /2
f (t) cos(nω0t)dt
an =
T −T /2
= 0
Z
since both f (t) and cos(nω0t) are odd functions.
4. Find bn.
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=
=
=
=
=
bn
Z
2 T /2
f (t) sin(nω0t)dt
T −T /2
←− Substitute (7) into left
Z
2 0
(−A) sin(nω0t)dt
T −T /2
Z
2 T /2
+
A sin(nω0t)dt
T 0Z
1
←− sin(nω0t)dt = −
cos(nω0t) + c
nω0
2A
[cos(nω0 ∗ 0) − cos(nω0 ∗ (−T /2))]
T ∗ nω0
2A
−
[cos(nω0 ∗ (T /2)) − cos(nω0 ∗ 0)]
T ∗ nω0
←− ω0T = 2π, cos(0) = 1
A
A
[1 − cos(−nπ))] −
[cos(nπ) − 1]
nπ
nπ
←− cos(−nπ) = cos(nπ)
2A
(1 − cos(nπ))
(9)
nπ
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Look at bn = 2A
nπ (1 − cos(nπ)).
If n=even, e.g. n = 0, 2, 4, .., bn = 0.
If n=odd, e.g. n = 1, 3, 5, .., bn = 4A
nπ .
so we have
f (t) =
∞
X
bn sin(nω0t)
n=1
=
1
4A
{sin(ω0t) + sin(3ω0t)+
π
3
1
1
sin(5ω0t)) + sin(7ω0t)) + ...}
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