Fourier Series Representation
Fourier series Fourier series is a technique for expressing a
periodic functions in terms of sinusoids. Once the source function is expressed in terms of sinusoids, we can apply the phasor method to analyze circuits.
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Trigonometric Fourier Series According to Fourier theorem, any practical periodic function of
frequency ωo can be expressed as an infinite sum of sine or cosine functions that are integral multiples of ωo . Thus g(t) can be expressed as
g (t ) = a0 + a1 cosω 0t + a2 cos2ω 0t + + an cosnω 0t + + b1 cosω 0t + b2 cos2ω 0t + + bn cosnω 0t + ∞
g (t ) = a 0 + ∑ (a n cos nω 0 t + bn sin nω 0 t ) n =1
(t 0 < t < t 0 + T )
This Equation is the trigonometric Fourier series representation of g(t) over an interval (t0, t0 +To). Where ωo = 2 π fo= 2 π / To is called fundamental frequency in rad / sec. 3
Trigonometric Fourier Series A major task in Fourier series is the determination of
the Fourier coefficients ao,,an & bn..
an
∫ =
( t 0 +T )
t0
∫
g (t ) cos nω 0 t dt
( t 0 +T )
t0
( t 0 +T )
bn
∫ =
t0
cos 2 nω 0 t dt
g (t ) sin nω0 t dt
( t 0 +T )
∫
t0
sin 2 nω0 t dt
If we put n=0 in above Eq, we get
1 a0 = T
∫
( t 0 +T )
t0
g (t ) dt 4
Trigonometric Fourier Series We also have
∫
( t 0 +T )
t0
cos nω 0t dt = ∫ 2
( t 0 +T )
t0
T sin nω 0t dt = 2 2
2 ( t 0 +T ) an = ∫ g (t ) cos nω 0 t dt t T 0
2 ( t 0 +T ) bn = ∫ g (t ) sin nω 0 t dt T t0
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Trigonometric Fourier Series Example ƒ(t)
A
We shall now expand a
function ƒ(t) shown in figure2.1(a,b,c). Using trigonometric Fourier series over the interval (0,1). It is evident that
0
1
ƒ1(t) A
0
ƒ(t) =At, (0
t
1
t
ƒ2(t) A
Time interval T = 1 0
1
Figure 2.1(a,b,c)
t
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Trigonometric Fourier since 2π ω = = 2π Series Example T o
thus ∞
f (t ) = ao + ∑ (an cos nωo t + bn sin nωo t )
→11
n =1
f (t ) = ao + a1 cos 2πt + a2 cos 4πt + ......... + an cos 2nπt + ......... + b1 sin 2πt + b2 sin 4πt + ..... + bn sin 2nπt + ..... 1 ao = T
t 2 1 A ∫0 Atdt = A 2 0 = 2 1
→12
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Trigonometric Fourier 2 a = ∫ A cos n( 2π ) tdt Series T Example 1
n
0
2 A t sin 2πnt 1 1 sin 2πnt an = − ∫ 1. dt T 2πn 0 0 2πn t sin 2πnt cos 2πnt 1 an = 2 A + 2 2 π n ( 2πn ) 0 1 A an = [ cos 2πnt + 2πnt sin 2πnt ] 2 2 2π n 0 A an = [ (1 − 1) + (1.0 − 0.1) ] 2 2 2π n
a n =0
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Trigonometric Fourier Series Example
2 1 2 A ( − cos 2πnt ) 1 1 ( − cos 2πnt ) bn = ∫ At sin 2πntdt = t −∫ dt 0 T 0 1 2πn 2πn 0
( − t cos 2πnt ) ( sin 2πnt ) 1 2A bn = 2 A − =− 2 2πn ( 2πn ) 0 2πn
A bn = − πn
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Trigonometric Fourier Series Example Hence f(t) can be expressed as
A ∞ f (t ) = + ∑ bn sin 2π nt 2 n =1 A − A f (t ) = + ∑ nπ 2 n =1 ∞
sin 2π nt
1 1 ∞ 1 f (t ) = A − ∑ sin 2π nt 2 π n =1 n
(0
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DIRICHLET CONDITIONS A function that can be represented by a Fourier series must meet certain requirements, because the infinite series may or may not converge. These conditions on g(t) to yield a convergent Fourier series are as follows: g(t) is single valued every where. g(t) has a finite number of finite discontinuities in any one period. g(t) has a finite number of maxima and minima in any one period. The integral ∫ |g(t)| dt < ∞ for any t0 These conditions are called Dirichlet conditions. 11