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ApplicationsParseval’s Identity
MA 201: Differentiation and Integration ofFourier Series
Applications of Fourier SeriesLecture - 10
MA 201 (2016): PDE
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
Gibbs’ Phenomenon
Figure : Gibbs Phenomenon with 2 terms
MA 201 (2016): PDE
ApplicationsParseval’s Identity
Gibbs’ Phenomenon
Figure : Gibbs Phenomenon with 4 terms
MA 201 (2016): PDE
ApplicationsParseval’s Identity
Gibbs’ Phenomenon
Figure : Gibbs Phenomenon with 8 terms
MA 201 (2016): PDE
ApplicationsParseval’s Identity
Gibbs’ Phenomenon
Figure : Gibbs Phenomenon with 16 terms
MA 201 (2016): PDE
ApplicationsParseval’s Identity
Gibbs’ Phenomenon
Figure : Gibbs Phenomenon with 32 terms
MA 201 (2016): PDE
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
Figure : Square wave
MA 201 (2016): PDE
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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
ApplicationsParseval’s Identity
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