FOURIERSERIES - University of Limerick 1. PeriodicFunctions Recall that f has period T if f(x + T) =...
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f T f (x + T )= f (x) x f g T h(x)= af (x)+ bg(x) a b h(x + T )= af (x + T )+ bg(x + T ) = af (x)+ bg(x)= h(x) h T 2π 2π 2π a 0 2 + a 1 cos x + b 1 sin x + a 2 cos 2x + b 2 sin 2x + ... a 0 a 1 a 2 ... b 1 b 2 b 3 ... 2π 2π f (x) f (x)= a 0 2 + ∞ X 1 (a n cos nx + b n sin nx) • a 0 Z π -π f (x) dx = a 0 2 Z π -π dx + ∞ X n=1 a n Z π -π cos nx dx + ∞ X n=1 b n Z π -π sin nx dx = πa 0 +0+0 a 0 = 1 π Z π -π f (x) dx
Transcript of FOURIERSERIES - University of Limerick 1. PeriodicFunctions Recall that f has period T if f(x + T) =...
FOURIER SERIES
1. Periodic Functions
Recall that f has period T if f(x + T ) = f(x) for all x. If f and g are periodic
with period T , then if h(x) = af(x) + bg(x), where a and b are constants,
h(x + T ) = af(x + T ) + bg(x + T )
= af(x) + bg(x) = h(x)
and thus h also has period T .
1.1. Functions of period 2π. We want to represent general functions ofperiod 2π in terms of cosine and sine functions which also have period 2π, i.e., bythe trigonometric series:
a0
2+ a1 cosx + b1 sin x + a2 cos 2x + b2 sin 2x + . . .
where a0, a1, a2, . . . and b1, b2, b3, . . . are real constants, the coe�cients of theseries.Clearly, since each term of this in�nite series has period 2π, if it converges its sumwill also have period 2π.Assume then that we can write f(x) as
f(x) =a0
2+
∞∑1
(an cosnx + bn sin nx)
We have to determine the coe�cients:
• To determine a0:∫ π
−π
f(x) dx =a0
2
∫ π
−π
dx +∞∑
n=1
an
∫ π
−π
cos nx dx
+∞∑
n=1
bn
∫ π
−π
sin nx dx
= πa0 + 0 + 0
Thus, a0 =1π
∫ π
−π
f(x) dx
1
2 FOURIER SERIES
• To determine an:
∫ π
−π
f(x) cos mxdx = a0
∫ π
−π
cosmxdx +∞∑
n=1
an
∫ π
−π
cosmx cosnx dx
+∞∑
n=1
bn
∫ π
−π
cos mx sin nx dx
Remark 1.1.∫ π
−π
cosmx cosnx dx =12
[∫ π
−π
cos(m + n)x dx +∫ π
−π
cos(m− n) dx
]
=12
{[sin(m + n)x
m + n
]π
−π
+[sin(m− n)x
m− n
]π
−π
}
= 0
unless n = m, in which case∫ π
−π
cos mx cosnx dx = π
Similarly
∫ π
−π
cos mx sinnx dx =12
[∫ π
−π
sin(m + n)x dx +∫ π
−π
sin(n−m) dx
]
= −12
{[cos(m + n)x
m + n
]π
−π
+[sin(n−m)x
n−m
]π
−π
}
= 0
(even when m = n Thus∫ π
−π
f(x) cos mxdx = πam
and so
an =1π
∫ π
−π
f(x) cos nx dx
and similarly
bn =1π
∫ π
−π
f(x) sin nx dx
These formulae for the coe�cients a0, an and bn are called the Euler formulae.Note that the formula for a0 is formally the same as that for an and so we canwrite:
an =1π
∫ π
−π
f(x) cos nx dx
bn =1π
∫ π
−π
f(x) sin nx dx
1. PERIODIC FUNCTIONS 3
Note If f(x) is periodic with period 2π then∫ π
−πf(x) dx can be integrated over
any interval of length 2π, e.g.,∫ 2π
0f(x) dx
The expression
a0
2+
∞∑n=1
(an cos nx + bn sin nx)
is called the Fourier series for f(x) with Fourier coe�cients a0, an and bn.Example. Square wave
f(x) =
−k, −π < x < 0
k, 0 < x < πf(x + 2π) = f(x)
k
0
-k
Figure 1. f(x)
a0 = 0
an =1π
∫ π
−π
f(x) cos nx dx
=1π
[∫ 0
−π
(−k) cos nx dx +∫ π
0
k cosnx dx
]
=1π
{[−k
nsin nx
]0
−π
+[
k
nsinnx
]π
0
}
= 0
4 FOURIER SERIES
bn =1π
∫ π
−π
f(x) sin nx dx
=1π
[∫ 0
−π
(−k) sin nx dx +∫ π
0
k sin nx dx
]
=1π
{[k
ncos nx
]0
−π
−[
k
ncos nx
]π
0
}
=k
nπ{cos 0− cos(−nπ)− cos(nπ) + cos 0}
=2k
nπ{1− cos nπ}
=2k
nπ[1− (−1)n]
=
0, neven4knπ , nodd
Thus, b1 = 4kπ , b2 = 0, b3 = 4k
3π , b4 = 0, etc. and the Fourier series is
4k
π
[sin x +
13
sin 3x +15
sin 5x + . . .
]
k
0
-k
1 term of series
k
0
-k
2 terms of series
k
0
-k
3 terms of series
k
0
-k
10 terms of series
Figure 2. Fourier series for f(x)
1. PERIODIC FUNCTIONS 5
Formal de�nition. If f(x) is periodic, with period 2π, piecewise continuous in[−π, π], has a left and right-hand derivative at each x0 ∈ [−π, π], then its Fourierseries is convergent to f(x), except at a discontinuity x0, where the sum is 1
2[f(x0+
0) + f(x0 − 0)].
Remark 1.2. The left hand derivative of f at x0 is
limh→0−
f(x0 − 0)− f(x0 − h)h
where f(x0− 0) = limx→x0− f(x) and similarly the right hand derivative of f at x0
islim
h→0+
f(x0 + h)− f(x0 + 0)h
We can in these circumstances write
f(x) =a0
2+
∞∑n=1
(an cosnx + bn sin nx)
Example. Saw-tooth wave.
f(x) =
−x, −π ≤ x ≤ 0
x, 0 ≤ x ≤ π
with period 2π
0
π
0 π 2π 3π 4π−π−2π−3π−4π
Figure 3. f(x)
a0 =1π
∫ π
−π
f(x) dx =1π
[∫ 0
−π
(−x) dx +∫ π
0
x dx
]
=2π
∫ π
0
x dx
=2π
[x2
2
]π
0
= π
6 FOURIER SERIES
an =1π
∫ π
−π
f(x) cos nx dx
=1π
[∫ 0
−π
(−x) cos nx dx +∫ π
0
x cos nx dx
]
=2π
∫ π
0
x cosnx dx
=2π
{[x
nsin nx
]π
0− 1
n
∫ π
0
sin nx dx
}
=2π
{0 +
1n2
[cos nx]π0
}
=2
n2π[cos nπ − 1]
=
0, neven
− 4n2π , nodd
bn =1π
∫ π
−π
f(x) sin nx dx
=1π
[∫ 0
−π
(−x) sin nx dx +∫ π
0
x sinnx dx
]
= 0
Thus
f(x) =π
2− 4
π
∞∑n=1
n odd
cos nx
n2
=π
2− 4
π
[cosx +
cos 3x
9+
cos 5x
25+ . . .
]
π
0 π 2π 3π 4π−π−2π−3π−4π
Figure 4. First 4 terms of Fourier series for f(x)
1. PERIODIC FUNCTIONS 7
Remark 1.3. If f(x) and g(x) each have Fourier series expansions, then the Fourierseries for the sum f(x) + g(x) is just the sum of the two expansions.
For example, the Fourier series of the square wave f(x) =
k, 0 < x < π
−k, −π < x < 0with
period 2π was shown to be4k
π
[sin x +
13
sin 3x +15
sin 5x + . . .
]
Thus the Fourier series for g(x) =
2k, 0 < x < π
0, −π < x < 0with period 2π is
k +4k
π
[sin x +
13
sin 3x +15
sin 5x + . . .
]
1.2. Functions of arbitrary period. Let f(x) have period 2`. If we makethe substitution u =
π
`x, then, whenx = ±`, u = ±π and so f is periodic in the
new variable u with period 2π and thus if f has a Fourier series, then
f(x) = f
(`
πu
)= a0 +
∞∑n=1
(an cos nu + bn sinnu)
where
an =1π
∫ π
−π
f
(`
πu
)cos nudu
bn =1π
∫ π
−π
f
(`
πu
)sin nudu
or, in terms of x, where u =π
`x ⇒ du =
π
`dx
an = 1`
∫ `
−`f(x) cos nπx
` dx
bn = 1`
∫ `
−`f(x) sin nπx
` dx
and
f(x) =a0
2+
∞∑n=1
(an cos
nπx
`+ bn sin
nπx
`
)
Example. Recti�ed sine wave.
f(x) = sin πx 0 < x < 1
f(x + 1) = f(x)
Note: ` = 12 .
8 FOURIER SERIES
00.10.20.30.40.50.60.70.80.9
1
0 0.5 1 1.5 2 2.5 3 3.5 4
Figure 5. f(x) = |sinπx|
a0 = 2∫ 1/2
−1/2
f(x) dx = 2∫ 1
0
f(x) dx
= 2∫ 1
0
sin πx dx = − 2π
[cos πx]10
=4π
an = 2∫ 1
0
sin πx cos 2nπx dx
=∫ 1
0
sin(2n + 1)πx dx +∫ 1
0
sin(2n− 1)πx dx
= − 1(2n + 1)π
[cos(2n + 1)πx]10 −1
(2n− 1)π[cos(2n− 1)πx]10
= − 1(2n + 1)π
(−2)− 1(2n− 1)π
(−2)
=2
(2n + 1)π+
2(2n− 1)π
=8n
(4n2 − 1)π
bn = 2∫ 1
0
sin πx sin 2nπx dx
=∫ 1
0
cos(2n− 1)πx dx−∫ 1
0
cos(2n + 1)πx dx
=1
(2n− 1)π[sin(2n− 1)πx]10 −
1(2n + 1)π
[sin(2n + 1)πx]10
= 0
So,
f(x) =2π
+∞∑
n=1
8n
(4n2 − 1)πcos 2nπx
=2π
+8π
[cos 2πx
3+
2 cos 4πx
15+ . . .
]
Remark 1.4. In each of the last two examples f(x) = f(−x) for all x and in eachcase the coe�cients bn turned out to be zero. Was this a coincidence or can somework be saved by making use of this property?
2. HALF-RANGE EXPANSIONS. 9
1.3. Even and odd functions. The function f is said to be even if f(−x) =
f(x) for all x. For example, the functions cosx, 1, x2, x4 are even,Similarly, f is said to be odd if f(−x) = −f(x) for all x. For example, the functionssin x, x, x3 are odd.If f(x) is even then
∫ k
−k
f(x) dx =∫ 0
−k
f(x) dx +∫ k
0
f(x) dx
(u = −x, f(x) = f(−u) = f(u)) =∫ k
0
f(u) du +∫ k
0
f(x) dx
(du = −dx) = 2∫ k
0
f(x) dx
Similarly, if f(x) is odd, them∫ k
−kf(x) dx = 0 Note that if f(x) is even and g(x)
is odd then if h(x) = f(x)g(x), h(−x) = f(x)[−g(x)] = −h(x) and so h is odd.Thus, if f(x) is even, the product f(x) sin nπx
` is odd and hence bn = 0. Similarly,if f(x) is odd, the product f(x) cos nπx
` is odd and hence an = 0, and obviouslya0 = 0 also. Thus, we have the following:The Fourier series of an even function f(x) of period 2` is a Fourier cosine series:
f(x) =a0
2+
∞∑n=1
an cosnπ
`x
where
an =2`
∫ `
0
f(x) cosnπx
`dx
The Fourier series of an odd function f(x) of period 2` is a Fourier sine series:
f(x) =∞∑
n=1
bn sinnπ
`x
where
bn =2`
∫ `
0
f(x) sinnπx
`dx
2. Half-range expansions.
Even though Fourier series were introduced for periodic functions on (−∞,∞), wecan also �nd Fourier series for functions de�ned on an interval [0, `], by �ndingeither even or odd periodic extensions of period 2` as follows:
10 FOURIER SERIES
2.1. Periodic extensions.
Definition 2.1. The even periodic extension f1(x) of period 2` of the functionf(x) de�ned on [0, `] is
f1(x) =
f(x), 0 ≤ x ≤ `
f(−x), −` ≤ x ≤ 0
with f1(x + 2`) = f(x)
f1(x) has a Fourier cosine series:
f1(x) =a0
2+
∞∑n=1
an cosnπ
`x
where
an =2`
∫ `
0
f(x) cosnπx
`dx
This can now be regarded as a Fourier cosine series for f(x) on the interval 0 ≤ x ≤ `
if we restrict x to that interval, i.e.,
⇒ f(x) =a0
2+
∞∑n=1
an cosnπx
`
for 0 ≤ x ≤ `
Definition 2.2. The odd periodic extension f2(x) of period 2` of the function f(x)
de�ned on [0, `] is
f2(x) =
f(x), 0 < x < `
−f(−x), −` < x < 0
with f2(x + 2`) = f(x)
f2(x) has a Fourier sine series:
f2(x) =∞∑
n=1
bn sinnπ
`x
where
bn =2`
∫ `
0
f(x) sinnπx
`dx
This can now be regarded as a Fourier sine series for f(x) on the interval 0 ≤ x ≤ `
if we restrict x to that interval, i.e.,
f(x) =∞∑
n=1
bn sinnπx
`
for 0 < x < `
2. HALF-RANGE EXPANSIONS. 11
Example.
f(x) =
x, 0 ≤ x ≤ 1
2− x, 1 ≤ x ≤ 2
Fourier cosine series
f(x) =a0
2+
∞∑n=1
an cosnπx
20 ≤ x ≤ 2
where
a0 =∫ 2
0
f(x) dx
={∫ 1
0
x dx +∫ 2
1
(2− x) dx
}
=
{[x2
2
]1
0
+[2x− x2
2
]2
1
}
={
12
+ (4− 2)− (2− 12)}
={
12
+12
}= 1
an =12
∫ 2
0
f(x) cosnπx
2dx
=∫ 1
0
x cosnπx
2dx +
∫ 2
1
(2− x) cosnπx
2dx
u = x dv = cosnπx
2dx | u = (2− x) dv = cos
nπx
2dx
du = dx v =2
nπsin
nπx
2| du = −dx v =
2nπ
sinnπx
2
⇒ an =2
nπ
[x sin
nπx
2
]1
0− 2
nπ
∫ 1
0
sinnπx
2dx
+2
nπ
[(2− x) sin
nπx
2
]2
1+
2nπ
∫ 2
1
sinnπx
2dx
=2
nπsin
nπ
2+
4n2π2
[cos
nπx
2
]1
0
− 2nπ
sinnπ
2− 4
n2π2
[cos
nπx
2
]2
1
=4
n2π2
(cos
nπ
2− 1− cosnπ + cos
nπ
2
)
=4
n2π2
(2 cos
nπ
2− 1− cosnπ
)
12 FOURIER SERIES
Note that
cosnπ
2=
0, n odd
1, n/2 even
−1, n/2 odd
Thus, n odd ⇒ an = 0 n even and n/2 even ⇒ an =4
n2π2(2− 2) = 0 n even and
n/2 odd ⇒ an =4
n2π2(−2− 2) = − 16
n2π2
3. Fourier series solutions of di�erential equations
There is an immediate application of Fourier series in the solution of constantcoe�cient second order linear di�erential equations, i.e., equations of the form
y′′ + ay′ + by = R(x)
where a and b are constants. The solution of this equation can be written as
y(x) = yh(x) + yp(x)
the sum of the solution of the homogeneous equation y′′h + ay′h + byh = 0 and aparticular integral or solution yp(x) of the original di�erential equation. Recallthat if R(x) is a constant or a cosine or sine term, the particular solution of thisequation can be found by the method of undetermined coe�cients. Thus, if R(x)
can be expanded in a Fourier series, the solution can also be found in same way.Example
y′′ + y′ = r(x)
where
r(x) =
x, 0 ≤ x ≤ 1
2− x, 1 ≤ x ≤ 2
To �nd yh the solution of the homogeneous equation
y′′h + y′h = 0
let yh = eλx; substituting we obtain (λ2 + λ)eλx = 0 and thus the characteristicequation λ(λ + 1) = 0 which has roots 0 and −1. Thus yh = A + Be−x where A
and B are arbitrary constants.To �nd now the particular solution yp corresponding to the RHS r(x), note thatwe have already found a Fourier cosine series expansion for r(x), i.e.,
r(x) =a0
2+
∞∑n=1
an cosnπx
20 ≤ x ≤ 2
where
3. FOURIER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS 13
a0 = 1
an =4
n2π2
(2 cos
nπ
2− 1− cosnπ
)
The particular solution will take the form
yp = y0 +∞∑
n=1
yn
where y0 is the particular solution corresponding to the constant term a02 and yn
is the particular solution corresponding to an cos nπx2 for each n.
Note that y0 would usually be a constant (as a0 is) but, as one of the solutions ofthe homogeneous equations is constant, in this case
y0 = c0x
for some constant c0. Substituting into the equation
y′′0 + y′0 =a0
2
we obtainc0 =
a0
2=
12
Similarly yn satis�es the equation
y′′n + y′n = an cosnπx
2
yn takes the form
yn = An cosnπx
2+ Bn sin
nπx
2and has derivatives
y′n = −nπ
2An sin
nπx
2+
nπ
2Bn cos
nπx
2
y′′n = −(nπ
2
)2
An cosnπx
2−
(nπ
2
)2
Bn sinnπx
2
Substituting for y′′n and y′n then gives[−
(nπ
2
)2
An +nπ
2Bn
]cos
nπx
2−
[(nπ
2
)2
Bn +nπ
2An
]sin
nπx
2= an cos
nπx
2
and equating the coe�cients of cos nπx2 and sin nπx
2 on each side of the equation weobtain
−(nπ
2
)2
An +nπ
2Bn = an
(nπ
2
)2
Bn +nπ
2An = 0
14 FOURIER SERIES
The second equation givesAn = −nπ
2Bn
and substituting in the �rst equation we obtain[(nπ
2
)3
+nπ
2
]Bn = an
orBn =
an[(nπ2
)2 + 1]
nπ2
and consequentlyAn = − an[(
nπ2
)2 + 1]
Thus, the general solution of the di�erential equation is
y = A + Be−x +x
2+
∞∑n=1
[An cos
nπx
2+ Bn sin
nπx
2
]
with An and Bn as above.Example. RLC circuit
Ld2i
dt2+ R
di
dt+
1C
i =dv(t)dt
where L = 10 henrys, R = 100 ohms, C = 0.01 farads; i.e.,
10d2i
dt2+ 100
di
dt+ 100i =
dv(t)dt
We �rst �nd the solution ih of the homogeneous equation
10d2i
dt2+ 100
di
dt+ 100i = 0
by setting i = eλt and substituting to obtain the charactertistic equation
10λ2 + 100λ + 100 = 0
which has two (negative) roots λ1,2 = −5 ±√15 Thus ih = Aeλ1t + Beλ2t → 0 as
t →∞.
Now, consider the case where v(t) = 100t(π2 − t2) volts for −π < t < π and let v
have period 2π.We now can �nd a particular integral ip of the di�erential equation by expandingv′(t) = dv(t)
dt in a Fourier series and then �nding a particular solution correspondingto each term of the series.
v′(t) = 100(π2 − 3t2)
and since v′(−t) = v′(t), v′(t) is even and so has a Fourier cosine series:
v′(t) =a0
2+
∞∑1
an cosnt
3. FOURIER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS 15
where
a0 =2π
∫ π
0
v′(t) dt
=2π
∫ π
0
100(π2 − 3t2) dt
=200π
[π2t− t3
]π
0
=200π
(π3 − π3
)
= 0
Or we could simply note that∫ π
0
v′(t) dt = [v(t)]π0 = v(π)− v(0) = 0
an =2π
∫ π
0
v′(t) dt
=2π
{∫ π
0
100π2 cos nt dt−∫ π
0
300t2 cos nt dt
}
=2π
{100π2
n[sinnt]π0 − 300
([t2
nsin nt
]π
0
− 2n
∫ π
0
t sin nt dt
)}
=1200nπ
∫ π
0
t sin nt dt
=1200nπ
{−
[t
ncosnt
]π
0
+1n
∫ π
0
cosnt dt
}
= −1200n2
cos nπ +1200n3π
[sinnt]π0
= (−1)n+1 1200n2
That isv′(t) = 1200
(cos t− 1
4cos 2t +
19
cos 3t− . . .
)
Let in be the particular integral corresponding to cos nt
n2.This takes the form
in = An cosnt + Bn sinnt
where An and Bn are constants (which depend on n)
i′n = −nAn sin nt + nBn cos nt
andi′′n = −n2An cosnt− n2Bn sin nt
Substituting into the equation for in, i.e.,
10d2indt2
+ 100dindt
+ 100in =cos nt
n2
16 FOURIER SERIES
gives
(−10n2An+100nBn+100An) cos nt+(−10n2Bn−100nAn+100Bn) sin nt =cosnt
n2
Equating the coe�cients of cosnt and sin nt on each side of this equation (anddividing by 10) gives
(10− n2)Bn − 10nAn = 0
(10− n2)An + 10nBn =1
10n2
Thus the �rst equation ⇒An =
10− n2
10nBn
and substituting this for An in the second equation gives[(10− n2)2
10n+ 10n
]Bn =
110n2
or [(10− n2)2 + 100n2
10n
]Bn =
110n2
and soBn =
1n [(10− n2)2 + 100n2]
and consequentlyAn =
10− n2
10n2 [(10− n2)2 + 100n2]Therefore the particular integral for v′(t) is
ip = 1200∞∑
n=1
(−1)n+1 [An cos nt + Bn sin nt]
with An and Bn de�ned as above.
