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Periodic driving forces• Principle of Superposition
– Particular solution for a sum of cosine waves
• Fourier Series– Periodic functions
• The fundamental frequency
– Orthogonal functions
– Obtaining Fourier coefficients
• Analogy with vector components
• Fourier Series examples– The square wave
– The saw tooth wave
2
What will we do in this chapter?We will consider the problem of driving a damped harmonic oscillator with a periodic but non-sinusoidal driving force. A periodic driving force has the property that f(t +) = f(t) where is the period. This is an interesting problem , but more importantly, it provides a great excuse for discussing the Fourier series.
We begin by discussion the principle of superposition which is true for linear systems which obey linear differential equations. This allows us to obtain particular integrals for sums of sinusoidal driving forces.
We next develop the technique of Fourier Series where we can write any periodic function of as a sum of sine and cosine functions with frequencies which are integral multiples of the fundamental frequency of =. Our demonstration exploits the fact that sine and cosine function are orthogonal functions which means that integral over the trig functions over a complete period vanishes unless they have have are the same multiple. There is analogy between finding Fourier components and vector components. We conclude by finding Fourier series for square and saw tooth waves.
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Superposition20
20 1 1 2 2
In the previous chapter we worked out
solutions to 2 cos
What if we had two driving frequencies?
2 cos cos
Because the differential equ
linear in x and it
ations are
s
x x x f t
x x x f t f t
2
21 1 0 1 1 1
22 2 0 2 2 2
1 2
(no x terms, for example) we have the
which
allows us to simply add solutions. If
2 cos and
2 cos then
2
Principle of Superpositi
time derivat ves
on
i
x x x f t
x x x f t
x x
21 2 0 1 2
1 1 2 2cos cos
x x x x
f t f t
2 22 20
12 00
We could easily generalize our
"particular" solutions from last time. If
we can write the force as ( ) cos
then by the Superposition Princ
cos
2
e
t
l
2an
ip
n n np
nn
nn
n
n nn
n
a t
f a t
x
t
Although this allows us to generalize the result quite a bit, it seems just like a curiosity, it can’t really be true that that many periodic forces can be written as a sum of sinusoidal terms? But interestingly enough it is due to the Fourier Theorem
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Fourier Series
0
0
0
Let 2 /
( ) cos sin2
2( ) cos
2( ) sin
o
o
n nn
t
n t
t
n t
af t a n t b n t
a dt f t n t
b dt f t n t
2
2
( )f t
t
( )f t
t
Let us consider making an expansion of the periodic driving force depicted below. The pattern repeats every seconds.
The Fourier Theorem says that we can write a “reasonable” periodic function as sum of sine and cosine terms with frequencies which are integer multiples of the fundamental frequency of = These functions have the same periodicity as the driving force.
( ) ( )
cos ( ) cos
sin ( ) sin
f t f t
t t
t t
Because the force is periodic, the coefficients can be determined from one representative part. For convenience we use this piece.
We will show that the force can be represented by this series.
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Orthogonal functionsWe will show how to get the coefficients an and bn which we will call the Fourier coefficients. The trick is to exploit an orthogonality property of the sin and cosine functions.
0
0
0
0
Lets prove one of these results
cos cos
1
41
4
1cos cos
2
o
o
o
o
t
t
t in t in t im t im t
t
t i n m t i n m t
t
i n m t i n m t
t
t
dt n t m t
dt e e e e
dt e e
e e
dt n m t n m t
-1.5
-1
-0.5
0
0.5
1
1.5
-10 -5 0 5 10
0
0
0
For =2 / and n,m integers
cos cos 0 if n m
and if n=m=0
sin sin 0 if n m
sin cos 0 always
o
o
o
t
t
t
t
t
t
dt n t m t
dt n t m t
dt n t m t
If we integrate over one complete or multiple complete periods of the cosine curve we will get zero. This will happen for both terms as long as n and m are not equal or both equal to zero.
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Orthogonal functions continued
0
0
0
0
Lets prove another result
cos sin
1
41
4
1sin sin
2
o
o
o
o
t
t
t in t in t im t im t
t
t i n m t i n m t
t
i n m t i n m t
t
t
dt n t m t
dt e e e ei
dt e ei
e e
dt n m t n m t
This vanishes for any integral trig argument since one always integrates over a full multiple of a sine oscillation.
What happens when the trig integrals do not vanish?
0
0
0
0
cos cos
1cos cos
2If any of the arguments of the cosine function
1vanish we will get / 2
2This happens "twice" when n=m=0 and once
when just n=m
cos
o
o
o
o
t
t
t
t
t
t
t
t
dt n t m t
dt n m t n m t
dt
dt n t
0 if n m
cos /2 if n=m 0
if n=m=0
m t
7
Obtaining Fourier coefficients
0
0
0
0 if n m
cos cos /2 if n=m 0
if n=m=0
0 if n msin sin
/2 if n=m 0
cos sin 0
o
o
o
t
t
t
t
t
t
dt n t m t
dt n t m t
dt n t m t
0
0
0
0 0
0 0
In more compact form:
2cos sin 0
2cos cos
2sin sin
o
o
o
t
t
t
nm n mt
t
nm n mt
dt n t m t
dt n t m t
dt n t m t
0
0
0
0
0
0
0
0
0
0
0
0
( ) cos sin2
2( ) sin
2sin
22
sin cos
2sin sin
2( ) sin
n nn
t
t
t
t
t
nn t
t
nn t
t
n mn mnt
af t a n t b n t
dt f t m t
adt m t
a dt m t n t
b dt m t n t
dt f t m t b b
We thus can get the Fourier sine coefficients by integrating over one period with the sine of the given frequency.
8
Completing the coefficients
0
0
0
0
0
00
Similarly we have:
For m 0
2( ) cos
2For m=0 ( ) cos(0)
2( )
t
n mn mnt
t
t
t
t
dt f t m t a a
dt f t t
dt f t a
0
0
0( ) cos sin2
2( ) cos
2(
If ( ) ( ) and
We have thus demonstrated that:
) i
/
s
2
n
o
o
n nn
t
n t
t
n t
af t a n t b n t
a dt f t n t
b dt f t n t
f t f t
0
n m
The way we find the Fourier components is
analogous to the way we find vector
components. Consider the analogy
2ˆ ˆe e sin sin
The dot product is analogous to the integral.
Here
o
t
nm nmtdt n t m t
mm
is the way we find vector components
ˆ ˆ ˆ ˆ;
ˆn'th component is obtained using :
Consider an odd function: ( ) sin
To get the n'th component we "dot"
ˆ V
m m n m n m m
n n
nm nm m m
n
V V e e V V e e V V
e
f t b
V
t
e
m
0
f(t) with
the sin "unit vector"
2( ) sin
o
t
n t
n t
b dt f t n t
This is a very deep analogy in linear algebra called Hilbert Space
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Evaluate the square wave Fourier coefficients
/ 2
/ 2
/ 20
0
/ 2
0 0
0
0
0
0
2( ) sin
4sin
4 4cos1 cos
2
Substitute in = 2 /
41 cos when n is even
2cos 1 and when n is odd cos 1
44
1 cos2
n
n
n
n
b dt f t n t
fb dt n t
f fn t n
n n
fb n
nn n
ff
b nn
if n is odd
0 if n is even
n
/ 2
/ 2
/ 2
/ 2
2( ) cos
2( ) sin
n
n
a dt f t n t
b dt f t n t
Here is a sketch of the function which is an odd function. We use - < t <
The integrals over cos vanish by symmetry. Thus an=0. This leaves the sin integrals and bn terms. We exploit symmetry and double the sine integral from 0 to /2
2
2
( )f tcos t
sin tt
0f
0f
10
Square wave series
-1.5
-1
-0.5
0
0.5
1
1.5
0 10 20 30 40 50
15
3
t
terms
erms
0
0
odd n
0
odd n
0 0
0
( ) cos sin2
4( ) sin where or
4( ) sin where = 2 /
4 4( ) sin sin 3
34
sin 5 ...5
n nn
n n
af t a n t b n t
ff t b n t b
n
ff t n t
n
f ff t t t
ft
( )f t
t
0f
0f
This plot shows how well this series matches our square wave using 3 terms: f(t) = b1sin t + b3sin 3t + b5sin5t and 15 terms. Clearly we get more fidelity with 15 terms but there are still some vicious overshoots. This is often called the Gibbs phenomenon.
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Our particular “square” solution
2 22 2odd 0
12
0
odd n
00
0
4( ) sin
4 sin
2
2tan
n np
nn n
nn
n
ff t n t
n
f
n
tx
( )f t
t
0f
0f
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1 3 5 7 9
11
13
15
17
19
21
23
25
27
29
nnA
We can depict the series by the Fourier “spectrum”. I would call this the amplitude spectrum. Only odd
frequencies are excited. When function has sharp edges -- as in the square wave-- high frequencies are required or the resultant distribution will be rounded.
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How to find a Fourier Series(1) Is the function f(t) periodic? If so find the smallest such that ( ) ( ).
We will expand in multiples of n where n=0,1,2,3... and =2 /
(2) Find a suitable interval of width t
f t f t
o represent the function.
If possible use an from
even functi
odd functi (- ) -
on: If possible use an from
(3) Evaluate th
- / 2
e Fourier c
(- ) (
oef
on: / 2
- / 2 / 2
(
)
) t
f
f t
t f t
f
t
t
0
0
0 0
0
0
1
ficients over the interval from
2 If the function has a non-zero integral : a ( )
If the function has even co
The series will be
m
( ) cos sin2 n n
t
t
n
af t a n t b n
t t t
dt f t
t
0
0
0
0
n
n
2ponents :a ( ) cos (where n 1)
2 If the function has odd components :b ( ) sin (where n 1)
t
t
t
t
dt f t n t
dt f t n t
13
Saw tooth wave example
2
( )f t
0
2
( ) / 2 0<t<f t t
This function is neither even or odd about t=0 so we will have both sin and cosine terms. We begin by evaluating the an cosine terms. This time we integrate from 0 torather than to as we did last time.
2
2
cos / sin cos /
sin / cos sin /
dx x ax x a ax ax a
dx x ax x a ax ax a
0 0
2 20
2 2
2 2/ 2 cos cos
2 cossin
2 cos 1sin
Since 2 , sin 0, cos 1 n
0
a d
n
n
n
a dt t n t dt t n t
t n tn t
n n
na n
n n
n n
a
We actually get no cosine terms even though the saw tooth has mixed symmetry. This is actually because all cosine functions are even about /2 and the saw tooth is odd about this point, but you can always work it out the hard way as we did. On to the sine coefficients..
14
The Saw Tooth continued...
n
2
0
2 20
2 2
Lets calculate the b coefficients using
sin / cos sin /
2/ 2 sin
2 sincos
2 sincos
Since 2 , sin 0, cos 1 and
2 and
n
n
dx x ax x a ax ax a
b dt t n t
t n tn t
n n
nn
n n
n n
bn n
n
0
1
recall a 0
( ) cos sin2
sin( )
n nn
n
af t a n t b n t
n tf t
n
Again we get a fairly good representation with 15 terms although
the Gibbs overshoot is quite apparent.
( )f t
2
-4
-3
-2
-1
0
1
2
3
4
0 2 4 6 8 10
15
3
t
terms
erms
1
2
2
15
A shifted saw tooth
/ 2
( )f t t
/ 2
( )f t
2
2
/ 2
0
/ 2/
0
2
2
20
The shifted form of triangle wave is an oddfunction and hence can only have sin n t.
4Thus ( ) sin s
sin / cos sin /
in
4 cos 2 sin
n nn
n
dx x ax
f t b n t b dt t n t
t n t n tb
n n
x a ax ax a
2
1 1
4 cos / 2 2 sin / 2
2
cos 22 1 1n n
n
n n
n n
nb
n n n
-4
-3
-2
-1
0
1
2
3
4
0 2 4 6 8 10
15
3
t
terms
erms
1
2
1
1
sinThus ( ) ( 1 )
sin 2 sin 3( ) sin ...
2 3
n
n
n tf t
nt t
f t t
1
1
1 1
1
1
sinReconcile this with ( )
sin ( / 2)( )
2
sin sin1
sin( ) 1
n
n
n
n n
n
n
n tf t
nn t
f t f tn
n t n n t
n nn t
f tn
