ch5 fourier series

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Chapter 5

Fourier Series

5.1 Periodic functions

A function f (t) is said to be periodic in t with period 2T if

f (t + 2T ) = f (t)

where T is a positive constant for all t.

Periodic functions occur in many areas of engineering, for example, current and voltage in an AC

circuit, the motion of a piston in an internal combustion engine.

If f (t) is periodic in t and n is any integer, then

f (t + 2nT ) = f (t) for all t

and so 4T , 6T , 8T , · · · are also periods of f (t). The smallest possible value of 2T is called

the (fundamental) period of f (t).



Example 1

(i) cos t and sin t have a fundamental period of 2π.

(ii) cos n(t + 2πn

) = cos(nt + 2π) = cos nt, n > 0.

∴ cos nt has the fundamental period 2π/n.

(iii) f (t) = C , where C is a constant

is periodic but has no fundamental period.

(iv) The fundamental period of tan t is π.

(v) Other example of periodic functions:

f (t) :=

t, 0 ≤ t < 1

1, 1 ≤ t < 2,

with period 2.

0t

21

2

1

-1-2-3 3-4 4

f(t)



(vi)

g(t) :=

2t, 0 ≤ t < 1

2, 1 ≤ t < 3

8− 2t, 3 ≤ t < 4, with period 4.

1 2 3 4 5 6 7t

80-1-2-3-4

4

g(t)

If f (t) and g(t) are periodic, with period 2T , then any linear combination

αf (t) + βg(t)

is also periodic with period 2T .

Example 2

f (t) = a0 + a1 cos t + a2 cos2t + · · ·+ an cos nt + · · ·

+ b1 sin t + b2 sin2t + · · ·+ bn sin nt + · · ·

has period 2π.



5.2 Fourier Series

Let f (t) be defined in the interval (−T, T ), and assumed that f (t) has the period 2T . Then

we may write the Fourier series corresponding to f (t) as

f (t) =a02

+∞n=1

an cos

nπt

T + bn sin

nπt

T

, (5.1)

and we call the constant term a0

2as the mean of f (t) over a period, where a0, an, bn

n = 1, 2, 3, · · · are unknown constants, to be determined. Note that the terms of a trigonometric

series are orthogonal. That is

c+2T

c

sinmπt

T sin

nπt

T dt =

c+2T

c

cosmπt

T cos

nπt

T dt

=

0, if m = n

T, if m = n,

c+2T

c

sinmπt

T cos

nπt

T dt = 0,

for m, n = 1, 2, 3, · · ·,

where c is any real constant.

As a special case, we consider c = −T in this section.



Example 3. Show that

(i)

T −T sin

mπt

T sin

nπt

T dt = T −T cos

mπt

T cos

nπt

T dt

=

0, if m = n

T, if m = n.

(ii) T

−T

sinmπt

T cos

nπt

T dt = 0,

for m, n = 1, 2, 3, · · · .

Proof:

(i) Note that sin A sin B = 12

[cos(A−B)− cos(A + B)]. For m = n,

T

−T

sinmπt

T sin

nπt

T dt =

1

2

T

−T

cos

(m− n)πt

T − cos

(m + n)πt

T

dt

=1

2sin (m−n)πt

T

(m−n)πT

−sin (m+n)πt

T

(m+n)πT

T

−T

= 0.

Since cos A cos B = 12 [cos(A + B) + cos(A−B)],

T

−T

cosmπt

T cos

nπt

T dt =

1

2

T

−T

cos

(m + n)πt

T + cos

(m− n)πt

T

dt

= 0.

For m = n, T

−T

sinmπt

T sin

nπt

T dt =

1

2

T

−T

1− cos

2nπt

T

dt

=1

2

t−

sin 2nπtT

2nπT

T −T

= T.



T

−T

cosmπt

T cos

nπt

T dt =

1

2

T

−T

(1 + cos2nπt

T )dt

=1

2t +sin 2nπt

T

2nπ

T T

−T

= T.

(ii) Note that sin A cos B = 12

sin(A + B) + sin(A−B)

. For m = n,

T

−T

sinmπt

T cos

nπt

T dt =

1

2

T

−T

sin

(m + n)πt

T + sin

(m− n)πt

T

dt

=

1

2− cos (m+n)πt

T

(m+n)πT

+

− cos (m−n)πtT

(m−n)πT

T −T

= −T

2π

cos(m + n)π + cos(m− n)π − cos(m + n)π − cos(m− n)π

= 0.

For m = n,

T −T

sinmπt

T cos

nπt

T dt =

1

2 T −T

sin2nπt

T dt

= 0.

Integrating both sides of equation (6.1) with respect to t over the interval [−T, T ], T

−T

f (t)dt =a02

(2T ) +∞n=1

an

T

−T

cosnπt

T dt + bn

T

−T

sinnπt

T dt

giving

a0 =

1

T T −T f (t)dt.

Multiplying both sides of equation (6.1) by cos mπt

T and integrating over [−T, T ], we have

T

−T

f (t)cosmπt

T dt =

a02

T

−T

cosmπt

T dt +

∞n=1

an

T

−T

cosmπt

T cos

nπt

T dt

+bn

T

−T

cosmπt

T sin

nπt

T dt

giving

an = 1T

T −T

f (t)cos nπtT

dt, n = 1, 2, 3, · · · .



Multiplying both sides of equation (6.1) by sin mπt

T and integrating gives

bn =1

T

T

−T

f (t)sinnπt

T dt, n = 1, 2, 3, · · · .

Thus we havef (t) =

a02

+∞n=1

an cos

nπt

T + bn sin

nπt

T

where

a0 =1

T

T

−T

f (t)dt,

an =1

T T

−T

f (t)cosnπt

T dt,

bn =1

T

T

−T

f (t)sinnπt

T dt, n = 1, 2, 3, · · · .

This is known as the Fourier series representation of f (t) and the coefficients an, bn are

known as the Fourier coefficients of f (t).

If T = π, then

f (t) =a02

+∞

n=1an cos nt + bn sin nt

,

where

a0 =1

π

π

−π

f (t)dt,

an =1

π

π

−π

f (t)cos ntdt,

bn =1

π π

−π

f (t)sin ntdt, n = 1, 2, 3, · · · .



5.3 Sufficient conditions for a function to have a Fouri

series (the Dirichlet conditions)

1. f (t) be real-valued and bounded in [−T, T ] and extended to other values of t by theperiodicity relation

f (t + 2nT ) = f (t), n = ±1,±2,±3, · · · .

2. f (t) has only a finite number of maximum and minimum points in [−T, T ].

3. f (t) has only a finite number of finite discontinuous points in [−T, T ].

N.B.:

1. Continuity alone is not a sufficient condition for a function to have a Fourier series.

2. If f (t) satisfies the Dirichlet conditions at t then the Fourier series converges to

(i) f (t), if t is a point of continuity.

(ii) the value 12

f (t−0 ) + f (t+0 )

= 1

2

limt→t

−

0

f (t) + limt→t

+

0

f (t)

, if t = t0 is a point of

discontinuity.

Since f (t) is not equal to the Fourier series everywhere. We write

f (t) ∼a02

+∞n=1

an cos

nπt

T + bn sin

nπt

T

.



Example 4.

Let

f (t) :=

t + 1, −3 ≤ t < −2

t2, − 2 ≤ t < 0

cos t, 0 ≤ t < π2

−1, π

2 ≤ t < 3

.

0

1

2

4

3

321-3 -2 -1t

f (t)

-1

-2

f (t) has a Fourier series in [−3, 3].

Let

g(t) := sin1

t , t ∈ [−π, π].

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1The Graph of sin(1/ t)

g(t)

t

g(t) has an infinite number of maxima and minima in the neighborhood of t = 0. Hence g(t)

has no Fourier series in [−π, π].



Let

h(t) :=1

t2 − 1, t ∈ [−π, π].

−π π

−1

h(t)

t 0

Since there is an infinite discontinuity at x = ±1.

Hence h(t) has no Fourier series in [−π, π].



Example 5.

Find the Fourier series for the square wave

f (t) := 0 − π < t ≤ 0

2, 0 < t < π

with period 2π.

Solution:

t 0

f(t)

−π π−3π −2π 3π 4π2π

2

Graph of f (t).

The Fourier series corresponding to f (t) as

a02

+∞n=1

an cos nt + bn sin nt

,

where

an =1

π π

−π

f (t)cos nt dt

=1

π

π

0

2cos nt dt =2

π

sin nt

n

π0

= 0,

a0 =1

π

π

−π

f (t) dt =1

π

π

0

2 dt =2

ππ = 2,



bn =1

π

π

−π

f (t)sin nt dt

= 1π

π0

2sin nt dt = 2π

− cos nt

n

π0

=2

πn

1− cos nπ

=

2

πn

1− (−1)n

=

0, if n is even

4πn

, if n is odd.

Hence, the Fourier series of f (t) is

f (t) ∼ 1 +4

π

∞n=1

sin(2n− 1)t

2n− 1.

If we write

S n(t) = 1 +4

π

nk=1

sin(2k − 1)t

2k − 1

Then

S 1(t) = 1 + 4π

sin t

S 2(t) = 1 +4

πsin t +

4

3πsin3t

S 3(t) = 1 +4

πsin t +

4

3πsin3t +

4

5πsin5t

By looking at various partial sums of this series we may see how it approaches f (t) as n →∞.

−ππ

t 0

1

2

S 1 = 1 + 4π

sin t



0 0−π π −π π

1

2

1

2

t t

S 2 = 1 +4

πsin t +

4

3πsin t S 5 = 1 +

4

πsin t + · · ·+

4

9πsin9t

N.B.:

1. As more terms are taken the series does approach f (t).

2. At a point of discontinuity of f (t) that is, at t = 0 (and also t = nπ, n = ±1,±2, · · · .

the series converges to 1 which is the average of the left and right hand limits at that point

3. Only sine terms are present in this series and only the odd harmonics (sin t, sin3tm sin5t, etc)

4. the coefficients bn (the amplitudes of the sine terms present) are proportional to 1/n and

hence we may expect fairly slow convergence of this series.



Example 6.

Find the Fourier series for the following function of period 10:

f (t) := 40 + 4t, −5 ≤ t ≤ 0

4t 0 < t ≤ 5,

Solution:

The Fourier series corresponding to f (t) as

a02

+∞n=1

an cos

nπt

5+ bn sin

nπt

5

,

where

an =1

5

5

−5

f (t)cosnπt

5

dt

=1

5

0−5

(40 + 4t)cosnπt

5dt +

50

4t cosnπt

5dt

=1

5

0−5

40cosnπt

5dt +

0−5

4t cosnπt

5dt +

50

4t cosnπt

5dt

=1

540

0

−5

cosnπt

5dt +

1

5 5

−5

4t cosnπt

5dt

= 8

sin(nπt

5)

nπ

5

0−5

+ 0, since

α

−α

F (t) dt = 0 if F (t) is an odd function

= 0,

bn =1

5

5−5

f (t)sinnπt

5dt

= 15

0−5

(40 + 4t)sin nπt5

dt + 50

4t sin nπt5

dt

=1

5

0−5

40sinnπt

5dt +

0−5

4t sinnπt

5dt +

50

4t sinnπt

5dt

=1

540

0−5

sinnπt

5dt +

1

5

5−5

4t sinnπt

5dt

= 8− cos(nπt

5)

nπ

5

0−5

+ 85−t cos(nπt

5)

nπ

5

50

+ 5nπ 50

cos nπt5 dt



= −40

nπ

1− cos nπ

+

8

5

−

5

nπ

5cos nπ

+

5

nπ

sin(nπt5 )

nπ

5

50

= −40

nπ

(1− cos nπ)−40

nπ

cos nπ

= −40

nπ,

a0 =1

5

5−5

f (t) dt

=1

5

0−5

(40 + 4t) dt +

50

4t dt

=1

5

0−5

40 dt +

0−5

4t dt +

50

4t dt

=1

540

0−5

dt +1

5

5−5

4t dt

= 8

t

0−5

+ 0

= 40.

∴ The Fourier series of f (t) is

f (t) ∼ 20−∞n=1

40

nπsin

nπt

5.



Method II

The graph of y = f (t) is shown in figure 1.

0

f (t)

40

-10-20 10t 20 305-5

Figure 1

Hence the function f (t) is equivalent to the function g(t) defined on the interval (0, 10] as

g(t) := 4t, 0 < t ≤ 10, with period 10.

And the Fourier series corresponding to g(t) as

a02

+∞n=1

an cos

nπt

5+ bn sin

nπt

5

,

where

an =1

5

100

g(t)cosnπt

5dt

=1

5 100

4t cosnπt

5dt

=4

5

t sin nπt

5nπ

5

100

−5

nπ

100

sinnπt

5dt

= −4

nπ

−

cos nπt

5nπ

5

100

= 0,



bn =1

5

100

g(t)sinnπt

5dt

= 45

10

0

t sin nπt5

dt

=4

5

−

t cos nπt

5nπ

5

100

+5

nπ

100

cosnπt

5dt

=4

5

−

50

nπ+

5

nπ

sin nπt

5nπ

5

100

= − 40nπ

,

a0 =1

5

100

g(t) dt

=4

5

100

t dt

= 45

t2

2

100

= 40

Hence the required series of f (t) is

f (t) ∼ 20−∞n=1

40

nπsin

nπt

5.



Example 7.

f is a periodic function of period π and f (x) = x2 + π

2x, x ∈ (−π

2, π2

).

(a) Find the Fourier series for f (x).

(b) Use the result in (a) to evaluate

∞n=1

1

n2.

Solution:

t 0

f(x)

2π

2

−π−3π π 3π

2 2 2 2

−π π

graph of f (x)

The Fourier series corresponding to f (x) as

a02

+∞n=1

an cos2nx + bn sin2nx

,

where

an =2

π

π2

−π

2

(x2 +π

2x)cos2nx dx

=4

π

π2

0

x2 cos2nx dx

= −4

nπ

(x)

−

cos2nx

2n

π

2

0

+1

2n

π2

0

cos2nx dx

=2

n2π(

π

2)cos nπ =

(−1)n

n2



bn =2

π

π2

−π

2

(x2 +π

2x)sin2nx dx

= 22

π

π

2

π2

0

x sin2nx dx

= 2

(x)(−

cos2nx

2n)

π

2

0

+1

2n

pi2

0

cos2nx dx

= −1

n

π

2

cos nπ = −

(−1)nπ

2n

a0 =2

π

π2

−π

2

(x2 +π

2x) dx

= 22

π

π2

0

x2 dx =4

π

x3

3

π

2

0

=π2

6.

Hence,

f (x) ∼π2

12+

∞n=1

(−1)n

n2cos2nx +

(−1)n+1π

2nsin2nx

.

At x = ± π

2, our series converges to 1

2

π2

2+ 0

= π2

4, by Dirichlet’s conditions.

Putting x = π

2, the series becomes

π2

12+

∞n=1

1

n2.

Hence,∞n=1

1

n2=

π2

4−

π2

12=

π2

6.



5.4 Even and Odd Functions

A function f (t) is said to be an even function of t if f (−t) = f (t), and to be an odd function

of t if f (−t) = − f (t).

For example,

t4 + 3t2 + 7, cos t, et2

, |t|, · · · .

are even functions of t,

t5 + 2t3 + 7t, sin t, tan t, t cos t, · · · .

are odd functions of t.

Properties:

(i) If f (t) is a even function of t and integrable on [−a.a], then a

−a

f (t) dt = 2

a

0

f (t) dt.

(ii) If f (t) is an odd function of t and integrable on [−a, a], then a

−a

f (t) dt = 0.

(iii) If f (t) and g(t) are both even, the product of f (t) and g(t) is even, since

f (−t)g(−t) = f (t)g(t).

Similarly, if f (t) and g(t) are both odd, the product of f (t) and g(t) is even, since

f (−t)g(−t) = (−f (t))(−g(t)) = f (t)g(t).

If f (t) is odd and g(t) is even, the product of f (t) and g(t) is odd, since

f (−t)g(−t) = (−f (t))g(t) = − f (t)g(t).

Now consider finding the Fourier series of an even function f (t) i n [−π, π]:

an =1

π

π

−π

f (t)cos nt dt =2

π

π

0

f (t)cos nt dt,

since the product of f (t) and cos nt is even.

bn = 1π

π−π

f (t)sin nt dt = 0,



since the product of f (t) and sin nt is odd.

Hence the Fourier series of an even function f (t) i n [−π, π] is:

f (t) = a0

2 +∞

n=1

an cos nt,

where

an = 2π

π

0 f (t)cos nt dt, n = 0, 1, 2, 3, · · · .

This is called a Fourier cosine series for f (t). You may of course continue to use the full form

of the Fourier series if f (t) is even, but the above form required less work and you are less likely

to make an error in the integration.

As expected, an even function f (t) requires only even functions in its representation.Similarly if f (t) is an odd function of t and integrable on [−π, π], all the coefficients an are

zero and we find,

f (t) =∞n=1

bn sin nt,

where

bn = 2π π

0

f (t)sin nt dt, n = 0, 1, 2, 3, · · · .

which is called a Fourier sine series, for f (t) consisting of only odd terms.



Example 8.

The sinusoidal voltage E (t) = sin t is passed through a full-wave rectifier to produce

E (t) = | sin t|, −π ≤ t ≤ π

Find its Fourier series.

t 02π −π 2ππ

1

E(t)

E (t) is an even function and so

E (t) =1

2a0 +

∞n=1

an cos nt,

where

an =2

π

π

0

E (t)cos nt dt

=2

π

π

0

sin t cos nt dt

=1

π

π

0

(sin(1 + n)t + sin(1− n)t) dt

=1

π

− cos(1 + n)t

1 + n−

cos(1− n)t

1− n

π0

=1

π− cos(1 + n)π

1 + n −cos(1− n)π

1− n +1

1 + n +1

1− n

=

0 if n is odd, n = 1

4π(1−n2)

, if n is even,

a0 =2

π

π

0

sin t dt =2

π

− cos t

π0

=4

π,

a1 = 2π

π0

sin t cos t dt = 2π

π0

sin td(sin t) = 2π

sin

2

t2

π0

= 0.



Hence in [−π, π],

E (t) =2

π+

4

π

∞n=1

cos2nt

1− 4n2=

4

π

1

2−

cos2t

1 · 3−

cos4t

3 · 5−

cos6t

5 · 7− · · ·

.



5.5 Half Range Series

In various applications (such as the solution of partial differential equations by the separation of

variables method), it is necessary to express a function f (t) defined only on [0, π] as a series o

cosines or sines.

We may construct an even function f 1(t) by

f 1(t) :=

f (t) 0 < t < π

f (−t) − π < t < 0

and f 1(t + 2π) = f 1(t) for all t, is known as the even periodic extension of f (t). This has

a Fourier cosine series which converges to f (t) i n ( 0, π).

Similarly we may construct an odd function f 2(t) by

f 2(t) :=

f (t) 0 < t < π

−f (−t) − π < t < 0

and f 2(t + 2π) = f 2(t) for all t, is known as the odd periodic extension of f (t). This has

a Fourier cosine series which converges to f (t) i n ( 0, π).

f (t)2f (t)1

f(t)

0 π 3π2ππ0 3ππ−π −π 0−2π 2π −2πt t t

graph of f (t) graph of f 1(t) graph of f 2(t)



Example 9.

Find Fourier cosine and sine series for

f (t) = t, 0 ≤ t ≤ π.

Solution:

(i) Fourier cosine series

The even periodic extension of f (t) corresponds to the ‘triangular’ wave

0

f(t)

−π−2π π 2πt 3π

π

f (t) =1

2a0 +

∞n=1

an cos nt

where

an =2

π

π

0

t cos nt dt

=2

πt sin nt

nπ

0

−1

n

π

0

sin nt dt=

2

π

cos nt

n2

π0

=2

n2π((−1)n − 1)

=

− 4

n2π, if n is odd

0, if n is even



a0 =2

π

π

0

t dt

= 2π

t22

π0

= π

Hence in [0, π],

f (t) =π

2−

4

π

∞n=1

1

(2n− 1)2cos(2n− 1)t

=π

2−

4

π

cos t +

cos3t

9+

cos5t

25+ · · ·

.

Putting t = 0 in this series gives

0 =π

2−

4

π

∞n=1

1

(2n− 1)2,

∴

∞n=1

1

(2n− 1)2=

π2

8.



(ii) Fourier sine series

The odd periodic extension of f (t) corresponds to the ‘saw-tooth’ wave

0

f(t)

−π−2π π 2πt 3π

π

f (t) ∼∞n=1

bn sin nt

where

bn =2

π

π

0

t sin nt dt

=2

π

−t

cos nt

n

π0

−

π

0

− cos nt

ndt

=−2(−1)n

n

Hence in (0, π),

f (t) = − 2∞

n=1

(−1)n

n

sin nt.



5.6 Fourier Approximation

Suppose we approximate f (t) by a finite trigonometric series

S N (t) =1

2α0 +

N n=1

(αn cos nt + β sin nt)

such that the squared error

E =

π

−π

[f (t)− S N (t)]2 dt

is minimized.

E =

π

−π

[f (t)−1

2α0 −

N n=1

(αn cos nt + β n sin nt)]2 dt

=

π

−π

f 2(t) dt +1

4α20

π

−π

12dt

+N n=1

(α2n

π

−π

cos2 nt dt + β 2n

π

−π

sin2 nt dt)

−α0

π

−π

f (t) dt− 2N n=1

αn

π

−π

f (t)cos nt dt + β n

π

−π

f (t)sin nt dt

(all other cross terms are zero due to the orthogonality of

1, cos nt, sin nt)

E =

π

−π

f 2(t) dt +π

2α20 +

π

2α20 + π

N n=1

(α2n

+ β 2n

)

−α0πa0 − 2

N n=1

(αnπan + β nπbn)

For E to be a minimum we must have

∂E

∂α0= πα0 − πa0 = 0 =⇒ α0 = a0

∂E

∂αn

= 2παn − 2πan = 0 =⇒ αn = an, n = 1, 2, · · · , N

∂E ∂β n

= 2πβ n − 2πbn = 0 =⇒ β n = bn, n = 1, 2, · · · , N



i.e. the coefficients α0, αn, β n of the finite series are exactly the coefficients taken from the Fourier

series. Thus a truncated Fourier series is the best approximate to f (t) in the sense that it minimizes

the squared error.

Now, for this choice of the coefficients

E =

π

−π

f 2(t) dt−π

2a20 − π

N n=1

(a2n

+ b2n

)

or, since we must have E ≥ 0,

1

2π π

−π

f 2(t) dt ≥1

4a20 +

1

2

N

n=1

(a2n

+ b2n

)

which is known as Bessel’s inequality.

As N →∞, we have1

4a20 +

1

2

∞n=1

(a2n

+ b2n

) ≤1

2π

π

−π

f 2(t) dt

but in fact the ≤ sign may be replaced by an = sign, since,

π

−π

f 2(t) dt =

π

−π

1

2a0 +

∞

n=1an cos nt + bn sin nt

2dt

=1

4a20 · 2π + π

∞n=1

(a2n

+ b2n

)

(using the orthogonality properties)

∴1

2π

π

−π

f 2(t) dt =1

4a20 +

1

2

∞n=1

(a2n

+ b2n

)

This result is known as Parseval’s Theorem.

If f (t) is an electrical signal, 12π

π

−πf 2(t) dt is proportional to the total mean power in the

signal. This result may be interpreted as ‘the mean power of a signal equals the sum of the mean

powers in each of the frequency components of that signal’.



5.7 Further Results

5.7.1 Differentiation and Integration of Fourier Series

Under certain conditions the Fourier series of f (t) may be differentiated or integrated term byterm to give f (t) and

f (t) dt respectively.

5.7.2 Frequency Spectrum

Each term of a Fourier series has a physical interpretation. The first term

1

2a0 =

1

T

c+T

c

f (t) dt

is the average value of f (t). It is time independent and is known as the dc (direct current) term

cos ωt and sin ωt are simple harmonic (single frequency) oscillations with angular frequency ω

being the fundamental frequency of f (t). a1 and b1 are the amplitudes of these terms.

The terms cos nωt and sin nωt are the harmonic terms, with frequency an integer multiple of the

fundamental frequency, with amplitudes an and bn.

Plots of the coefficients an and bn as functions of ω form the frequency spectrum of f (t)

For a periodic function only the frequencies nω occur and hence the spectrum consists of discrete

lines spaced ω apart. It is a discrete or line spectrum.

nb

0ω

an

0 ω 2ω 3ω 4ω 5ωangular

frequency5ω

angular

frequency3ω2ω 4ω

Many electronic devices are required to pass signals with little distortion. Using Fourier series the

signal may be analyzed into its harmonics and the effect of the device on each harmonic calculated

For example, high frequency waves may get distorted most. By knowing how much of each harmonic

component is present in a signal we may calculate the degree of distortion expected.



The smoother f (t) is, the faster will be the convergence of the Fourier series and so the signal wil

be better represented by a small number of the lower harmonics. If

(i) f (t) is only piecewise-continuous, an, bn ∼ 1/n

(ii) f (t) is continuous but f (t) has discontinuities, an, bn ∼ 1/n2

(iii) f (t) and all derivatives up to f (k−1)(t) are continuous but f (k)(t) is not, an, bn ∼ 1/nk+1

This behavior can be observed in the examples.

5.7.3 General Orthogonal Functions

The results of this section generalize to give expansions of f (t) in terms of other orthogonafunctions (not necessarily trigonometric).

If a set of functions φ0(t), φ1(t), φ2(t), φ3(t), · · · . defined on an interval [a, b] has the property

that b

a

φm(t)φn(t) dt = 0, if m = n,

they are orthogonal.

The generalized Fourier series of f (t) is

f (t) =∞n=1

cnφn(t)

where

cn =

b

af (t)φn(t) dt b

aφ2n

(t) dt

(the coefficients cn are derived as in section 6.2).

ch5 fourier series

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