MA 201: Differentiation and Integration of Fourier Series Applications of Fourier ... · · 2016-10-21Fourier Series Applications of Fourier Series Lecture - 10 MA201(2016): PDE

ApplicationsParseval’s Identity

MA 201: Differentiation and Integration ofFourier Series

Applications of Fourier SeriesLecture - 10

MA 201 (2016): PDE


Fourier Series: Orthogonal SetsWe begin our treatment with some observations: For m, n = 1, 2, 3, . . .

∫

L

−L

cosnπx

Ldx = 0 &

∫

L

−L

sinnπx

Ldx = 0, (1)

∫

L

−L

sinmπx

Lcos

nπx

Ldx = 0 (2)

∫

L

−L

sinmπx

Lsin

nπx

Ldx =

{

0, m 6= nL, m = n

(3)

∫

L

−L

cosmπx

Lcos

nπx

Ldx =

{

0, m 6= nL, m = n

(4)

Setting x = t − L, observe that

{1, cosnπx

L, sin

nπx

L: n = 1, 2 3, . . .},

is orthogonal on [0, 2L].MA 201 (2016): PDE


Thus, if we have

f (x) =a02

+

∞∑

n=1

{

an cosnπx

L+ bn sin

nπx

L

}

, x ∈ [0, 2L]

then all coefficients an and bn can be determined as

an =1

L

∫ 2L

0

f (x) cosnπx

Ldx , n = 0, 1, 2, . . . and (5)

bn =1

L

∫ 2L

0

f (x) sinnπx

Ldx , n = 1, 2, . . . (6)

MA 201 (2016): PDE


Fourier Series in [0, 2L]

• The infinite series

a02

+

∞∑

n=1

[

an cosnπx

L+ bn sin

nπx

L

]

, (7)

with

an =1

L

∫ 2L

0

f (x) cosnπx

Ldx , n = 0, 1, 2, . . . and (8)

bn =1

L

∫ 2L

0

f (x) sinnπx

Ldx , n = 1, 2, . . . (9)

is called Fourier series (FS) of f (x) in [0, 2L].

RemarkThe Fourier series of a function is defined whenever the integrals in (8)and (9) have meaning.

MA 201 (2016): PDE


Fourier Series in [c , c + 2L]

• The infinite series

a02

+

∞∑

n=1

[

an cosnπx

L+ bn sin

nπx

L

]

, (10)

with

an =1

L

∫ c+2L

c

f (x) cosnπx

Ldx , n = 0, 1, 2, . . . and (11)

bn =1

L

∫ c+2L

c

f (x) sinnπx

Ldx , n = 1, 2, . . . (12)

is called Fourier series (FS) of f (x) in [c , c + 2L].

RemarkThe Fourier series of a function is defined whenever the integrals in (11)and (12) have meaning.

MA 201 (2016): PDE


Gibbs’ Phenomenon• This is about how the Fourier series of a piecewise continuouslydifferentiable periodic function behaves at a jump discontinuity. It isnamed after the American mathematical physicist J. W Gibbs.

• Consider the following periodic function whose definition in oneperiod is:

f (x) =

{

1, 0 < x < 1,0, 1 < x < 2.

(13)

• This function can be represented as

f (x) =1

2+

2

π

∞∑

n=1

sin(2n − 1)πx

2n − 1, x ∈?. (14)

• To see how well this infinite series represents the function, let ustruncate the series after N terms.

• Let the sum of these first N terms of the infinite series be denotedby SN :

SN =1

2+

2

π

N∑

n=1

sin(2n− 1)πx

2n− 1. (15)

MA 201 (2016): PDE


Gibbs’ Phenomenon

• Graphs of equation (15) are shown for N = 2, 4, 8, 16 and 32.

• In the figure the overshoot (overshoot is the occurrence of a signalor function exceeding its target) at x = 1− and the undershoot atx = 1+ are characteristics of Fourier series at the points ofdiscontinuity.

• This phenomenon is known as Gibbs’ phenomenon

• This phenomenon persists even though a large number of terms areconsidered in the partial sum.

• In the approximation of functions, overshoot/undershoot is one termdescribing quality of approximation. Here, the convergence is inpoint-wise sense.

MA 201 (2016): PDE


Gibbs’ Phenomenon

Figure : Gibbs Phenomenon with 2 terms

MA 201 (2016): PDE


Gibbs’ Phenomenon


MA 201 (2016): PDE


Gibbs’ Phenomenon


MA 201 (2016): PDE


Gibbs’ Phenomenon


MA 201 (2016): PDE


Gibbs’ Phenomenon


MA 201 (2016): PDE


Convergence of Fourier series for piecewise continuous

functions

DefinitionLet f (t) : [−L, L] → R be a piecewise C 1-function. Define the adjusted function g(t)as follows:

g(t) =

12 [f (t

+) + f (t−)], −L < t < L,

12 [f (−L+) + f (L−)], t = ±L.

(16)

Note:The above definition tells that g(t) coincides with f (t) at all points in (−L, L), wheref (t) is continuous, but g(t) is the average of the left-hand and right-hand limits off (t) at points of discontinuity in (−L, L).

MA 201 (2016): PDE


Convergence of Fourier series for piecewise continuous

functions

TheoremLet f : [−L, L] → R be a piecewise C 1 function and let g(t) be theadjusted function as defined in (16). Then Fourier series of f (t) = g(t),for all t ∈ [−L, L].

Note:

• The existence of Fourier series depends on the evaluation of Fouriercoefficients.

• On the other hand, convergence of the Fourier series is done viaadjusted function g .

• Thus, a highly oscillating function may be decomposed as sum of a(possibly infinite) set of simple oscillating trigonometric functions.

• A discontinuous function is approximated by a continuous function.

MA 201 (2016): PDE


Applications of Fourier SeriesSquare wave-high frequenciesOne application of Fourier series, the analysis of a square wave in terms of its Fouriercomponents, occurs in electronic circuits designed to handle sharply rising pulses.

Physically, square wave contains many high-frequency components.

Figure : Square wave

MA 201 (2016): PDE


Suppose that our modified square wave is defined by

f (x) =

{

0, −π < x < 0,h, 0 < x < π.

(17)

We can easily calculate the Fourier coefficients to be

A0 = h, An = 0, n = 1, 2, 3, . . . , (18)

Bn =

{

2hnπ, n odd,

0, n even.(19)

So, the resulting series is

g(x) =h

2+

2h

π

(

sin x

1+

sin 3x

3+

sin 5x

5+ · · ·

)

, (20)

g is the adjusted function.

MA 201 (2016): PDE


Figure : Square wave

MA 201 (2016): PDE


Parseval’s Identity for Fourier Series• There is a very useful identity obtained from Fourier series which has foundapplications in electrical engineering.

• But before knowing Parseval’s identity, we need the following theorem which ismore general:

TheoremIf f (t) and g(t) are continuous in (−L, L), and provided

∫ L

−L|f (t)|2dt <∞ and

∫ L

−L|g(t)|2dt <∞, and if An,Bn are Fourier coefficients of f (t) and Cn,Dn are

Fourier coefficients of g(t), then

∫ L

−L

f (t)g(t)dt =L

2A0C0 + L

∞∑

n=1

(AnCn + BnDn).

• Proof: We can express f (t) and g(t) in terms of Fourier series as

f (t) =A0

2+

∞∑

n=1

[

An cosnπt

L+ Bn sin

nπt

L

]

,

g(t) =C0

2+

∞∑

n=1

[

Cn cosnπt

L+ Dn sin

nπt

L

]

.

MA 201 (2016): PDE


Parseval’s Identity for Fourier Series• Taking product of f (t) with g(t) we obtain

f (t)g(t) =A0

2g(t) +

∞∑

n=1

[

An g(t) cosnπt

L+ Bn g(t) sin

nπt

L

]

.

• Integrating this series from −L to L gives

∫

L

−L

f (t)g(t)dt =A0

2

∫

L

−L

g(t)dt+

∞∑

n=1

[

An

∫

L

−L

g(t) cosnπt

Ldt + Bn

∫

L

−L

g(t) sinnπt

Ldt

]

• Putting back the values of the Fourier coefficients Cn and Dn:

1

L

∫ L

−L

f (t)g(t)dt =A0C0

2+

∞∑

n=1

[AnCn + BnDn].

MA 201 (2016): PDE


Parseval’s Identity for Fourier Series

• Theorem(Parseval’s Identity) If f (t) is continuous in the range (−L, L) and is square

integrable (i.e.∫

L

−L|f (t)|2dt <∞) and has Fourier coefficients An and Bn, then

1

L

∫ L

−L

[f (t)]2dt =A20

2+

∞∑

n=1

[A2n + B2

n ].

• This result can obtained easily from the previous theorem by taking g(t) = f (t).• The left-hand side represents the mean square value of f (t).• It can, therefore, be thought of in terms of energy if f (t) represents a signal.• What Parseval’s theorem states therefore is that the energy of a signal expressedas a waveform is proportional to the sum of the squares of its Fourier coefficients.

MA 201 (2016): PDE


Parseval’s Identity for Fourier Series• Parseval’s identity can be used to determine the power delivered by an electriccurrent, I (t), flowing under a voltage, E (t), through a resistor of resistance R :

P = EI = RI 2.

• In most applications I (t) is a periodic function.•

Average Power = Pav =1

2L

∫

L

−L

RI 2(t) dt

=R

2L

∫ L

−L

I 2(t) dt

= R

[

A20

4+

1

2

∞∑

n=1

(A2n + B2

n )

]

,

• Here we have made use of the Fourier expansion of I (t):

I (t) =A0

2+

∞∑

n=1

[

An cosnπt

L+ Bn sin

nπt

L

]

.

MA 201 (2016): PDE


Parseval’s Identity for Fourier Series

• Mean square of the current is:

Iav =1

2L

∫

L

−L

I 2(t)dt =A20

4+

1

2

∞∑

n=1

(A2n+ B2

n).

• Root mean square of the current is:

Irms =

√

√

√

√

A20

4+

1

2

∞∑

n=1

(A2n + B2

n ).

MA 201 (2016): PDE


Application of Parseval’s Identity

• Example: Given the Fourier series t2 =π2

3+ 4

∞∑

n=1

(−1)n

n2cos nt, −π < t < π,

deduce that

∞∑

n=1

1

n4=π4

90.

• Here A0 =2π2

3, An =

4(−1)n

n2, Bn = 0.

• Parseval’s identity is

1

L

∫

L

−L

[f (t)]2dt =A20

2+

∞∑

n=1

[A2n+ B2

n].

so that here L = π.• We get

1

π

∫

π

−π

t4dx =2π4

9+

∞∑

n=1

16

n4

giving us

2π4

5=

2π4

9+

∞∑

n=1

16

n4

• Hence the result follows.

MA 201 (2016): PDE


Finite Vibrating String Problem

• The IBVP under consideration consists of the following:

• The governing equation:

utt = c2uxx , (x , t) ∈ (0, L)× (0,∞). (21)

The boundary conditions for all t > 0:

u(0, t) = 0, u(L, t) = 0. (22)

The initial conditions for 0 ≤ x ≤ L are

u(x , 0) = φ(x), ut(x , 0) = ψ(x). (23)

MA 201 (2016): PDE


Bernoulli’s Solution• In [0, π], Bernoulli gave the solution of (21) as a series of the form

u = b1 sin x cos ct + b2 sin 2x cos 2ct + . . . . (24)

• When t = 0, we should have

φ(x) = b1 sin x + b2 sin 2x + . . . . (25)

This is possible and can be treated as the Fourier sine series of φ(x),and which converges to φ(x).

Few Facts:

As we are expecting u(x , t) as a solution of (21), term-wisedifferentiation of the series given by (24) should exist. In other sense, thelimit function u of the infinite series (24) should be differentiable.

Term-wise differentiation of an infinite series is not always possible.In fact, infinite sum of continuous functions may have discontinuouslimit.

1 +1

1 + x2+

1

(1 + x2)2+ . . . =

1 + x2

x2.

MA 201 (2016): PDE


Differentiation and integration of Fourier seriesThe term-by-term differentiation of a Fourier series is not always permissible.

ExampleRecall that Fourier series for f (x) = x , −π < x < π is

∞∑

n=1

(−1)n+1 sin nx

n,

which converges to f (x) for all x ∈ (−π, π), that is

x =

∞∑

n=1

(−1)n+1 sin nx

n.

Term-by-term differentiation leads to

1 =

∞∑

n=1

(−1)n+1 cos nx , x ∈ (−π, π).

which fails at x = 0. In fact, the RHS series diverges for all x(?)

MA 201 (2016): PDE


Differentiation of Fourier series

Theorem (Differentiation of Fourier series)Let f (x) : R → R be continuous and f (x + 2L) = f (x). Let f ′(x) andf ′′(x) be piecewise continuous on [−L, L]. Then, The Fourier series off ′(x) can be obtained from the Fourier series for f (x) by termwisedifferentiation. In particular, if

f (x) =A0

2+

∞∑

n=1

{

An cosnπx

L+ Bn sin

nπx

L

}

,

then

f ′(x) =

∞∑

n=1

nπ

L

{

−An sinnπx

L+ Bn cos

nπx

L

}

.

MA 201 (2016): PDE


Integration of Fourier series

Termwise integration of a Fourier series is permissible undermuch weaker conditions.

Theorem (Integration of Fourier series)Let f (x) : [−L, L] → R be piecewise continuous function with Fourierseries

f (x) =A0

2+

∞∑

n=1

{

An cosnπx

L+ Bn sin

nπx

L

}

.

Then, for any x ∈ [−L, L], we have

∫ x

−L

f (τ)dτ =

∫ x

−L

A0

2dτ +

∞∑

n=1

∫ x

−L

{

An cosnπτ

L+ Bn sin

nπτ

L

}

dτ.

MA 201 (2016): PDE

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MA 201: Differentiation and Integration of Fourier Series Applications of Fourier ... · · 2016-10-21Fourier Series Applications of Fourier Series Lecture - 10 MA201(2016): PDE