Question

I want to learn Fourier series and the propertys and solving the problem. it very important. plz post a proper answer guys........

Answer 1

Fourier sine series S(x) = b1 sin x + b2 sin 2x + b3 sin 3x + ··· = ∞
n=1
bn sin nx (1)
If the numbers b1, b2,... drop off quickly enough (we are foreshadowing the im-
portance of the decay rate) then the sum S(x) will inherit all three properties:
Periodic S(x + 2π) = S(x) Odd S(−x) = −S(x) S(0) = S(π)=0
200 years ago, Fourier startled the mathematicians in France by suggesting that any
function S(x) with those properties could be expressed as an infinite series of sines.
This idea started an enormous development of Fourier series. Our first step is to
compute from S(x) the number bk that multiplies sin kx.

Orthogonality π
0
sin nx sin kx dx = 0 if n = k .

Product of sines sin nx sin kx = 1
2 cos(n − k)x − 1
2 cos(n + k)x . (4)

π
0
sin kx sin kx dx =
π
0
1
2
dx −
π
0
1
2
cos 2kx dx = π
2 . (5)


S(−x) = −S(x) bk = 2
π
π
0
S(x) sin kx dx = 1
π
π
−π
S(x) sin kx dx. (6)



Example 1 Find the Fourier sine coefficients bk of the square wave SW(x).
Solution For k = 1, 2,... use the first formula (6) with S(x)=1 between 0 and π:
bk = 2
π
π
0
sin kx dx = 2
π

− cos kx
k
π
0
= 2
π
2
1
, 0
2
,
2
3
, 0
4
,
2
5
, 0
6
,..."
(7)
The even-numbered coefficients b2k are all zero because cos 2kπ = cos 0 = 1. The
odd-numbered coefficients bk = 4/πk decrease at the rate 1/k. We will see that same
1/k decay rate for all functions formed from smooth pieces and jumps.
Put those coefficients 4/πk and zero into the Fourier sine series for SW(x):
Square wave SW(x) = 4
π

sin x
1 +
sin 3x
3 +
sin 5x
5 +
sin 7x
7 + ···
(8)
Figure 4.2 graphs this sum after one term, then two terms, and then five terms. You
can see the all-important Gibbs phenomenon appearing as these “partial sums”

include more terms. Away from the jumps, we safely approach SW(x) = 1 or −1.
At x = π/2, the series gives a beautiful alternating formula for the number π:
1 =
4
π

1
1 − 1
3 +
1
5 − 1
7 + ···
so that π = 4
1
1 − 1
3 +
1
5 − 1
7 + ···
. (9)
The Gibbs phenomenon is the overshoot that moves closer and closer to the jumps.
Its height approaches 1.18 ... and it does not decrease with more terms of the series!
Overshoot is the one greatest obstacle to calculation of all discontinuous functions
(like shock waves in fluid flow). We try hard to avoid Gibbs but sometimes we can’t.
− x x π π
Dashed 4
π
sin x
1
Solid curve 4
π
sin x
1 +
sin 3x
3

5 terms:
4
π
sin x
1 + ··· +
sin 9x
9

overshoot−→ SW = 1
π
2
Figure 4.2: Gibbs phenomenon: Partial sums N
1 bn sin nx overshoot near jumps.

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Answer 2

Let x(t) be a continuous time signal with Fourier transform

X(ω)

.

We have seen the time-shifting property of the Fourier transform:

F(x(t−

t

0

))=

e

−jω

t

0

X(ω)

which gives us an expression for the Fourier transform of

y(t)=x(t−

t

0

)

in terms of

X(ω)

. Note that, to prove this property, one can proceed as follows:

F(x(t−

t

0

))

=

∫

∞

−∞

x(t−

t

0

)

e

−jωt

dt

=

∫

∞

−∞

x(u)

e

−jω(u+

t

0

)

du, (letting u=t−

t

0

),

=

e

−jω

t

0

∫

∞

−∞

x(u)

e

−jωu

du

=

e

−jω

t

0

X(ω).

Using a similar approach as above, derive an expression for the Fourier transform of y(t)=x(3t+7) in terms of

X(ω)