1. No category

How to use these equations?
Three methods to evaluate the Fourier
Coefficients of a Fourier Series
Define a function to evaluate:
A
f( t )
10
f0
A.
10
T
t
T
1
f0
ω0
2 . π . f0
This is a sawtooth wave
Time period of wave
T = 0.1
Note: This function is only valid to describe the
sawtooth in the range 0 to T.
First Method
Use mathcad (or equivalent) to calculate the integral
numerically. Use the function f(t) defined above and the
definition of b3 from [10.2c]
b3
2 .
T
T
0
f( t ) . sin 3 . ω 0 . t dt
Mathcad evaluates this as: b3 = 1.061
Second Method
Graph the function multiplied by the appropriate sine
or cosine wave, then count the squares on the graph by
hand.
REMEMBER: Points below the line are NEGATIVE!
Plot the curve in the range: t
Area of each square =
0,
T
.. T
100
0.1 . 20
3
= 1.25 10
40
40
Above the line : 6 + 32 + 58 = 96
Below the line : 18 + 47 + 74 = 139
T
f (t ) sin(3 ⋅ ω0 ⋅ t )dt = [96 − 139] ⋅ [0.00125] = −0.0538
0
b3 =
=>
2
⋅ (−0.0538) = −1.07
T
So we get the same value of b3 as we expect to about
1%
Method 3
Algebraically : solve the integral definite integral
analytically.
2
b3 =
T
T
0
T
A⋅t
2A
sin(3ω0t )dt =
t ⋅ sin(3ω0t )dt
2
T
T
0
Now solve the integral part of the function.
Reminder: Integration by parts uses the identity
udv =uv − vdu
u
du
du
=1
dt
u=t
v
dv
cos(3ω0t )
v=−
3ω0
dv
= sin(3ω0t )
dt
T
T
T
cos(3ω0t )
cos(3ω0t )
t ⋅ sin(3ω0t )dt = −
t − −
dt
3ω0
3ω0
0 0
0
T
cos(3ω0T )
cos(3ω0t )
= −
T −0 +
dt
3ω0
3ω0
0
T
sin(3ω0t )
T
+
3ω0
9ω0 2 0
T
=−
+ [0 − 0]
3ω0
=−
Remember ω0 =
2π
and sin(n2π) = 0
T
So the putting the solution to the integral into the
equation for b3:
b3 =
2A
T
T2 0
2A
t ⋅ sin(3ω0t ) dt =
T2
−T
2A
=−
3ω0T
3ω0
When A = 10, T = 0.1 s so ω0 = 2π / 0.1 = 20π :
b3 = −
20
= −1.06
3 × 20π × 0.1
So b3 is −1.06 from the analytic equation (exact) which
matches the predictions from the two approximations
above.
Mathcad demonstration of the formula
f0
10 Hz
Fundamental frequency
Period in seconds = T
1
f0
Define the amplitude of the periodic function.
A
10
Function
f( t )
A.
t
T
Note when t = 0 , f(t)=0 and when t = T, f(t) = A
Now use the formulae to find the coefficients in the
Fourier Series
Find the constant term:
a0
2.
T
T
f( t ) dt
0
Calculated value of a0 = 10
Set the number of coefficients, n
( should really be infinite! )
nmax = 10
Find the n coefficients for cosine terms
an
2.
T
T
0
f( t ) . cos n . ω 0 . t dt
Find the n coefficients for sine
terms
bn
2.
T
T
0
f( t ) . sin n . ω 0 . t dt
n
1
2
3
4
5
6
7
8
9
10
an
bn
3.309. 10
12
4.388. 10
12
3.307. 10
12
4.663. 10
12
3.31. 10
12
4.39. 10
12
3.304. 10
12
4.82. 10
12
3.312. 10
12
4.39. 10
12
3.183
1.592
1.061
0.796
0.637
0.531
0.455
0.398
0.354
0.318
Define the Fourier Series
a0
g( t )
an. cos n . ω 0 . t
2
bn. sin n . ω 0 . t
n
n
Compare the Fourier series g(t) with n terms
to the original function f(t) over two periods
10
g( t )
f( t )
5
0
0
t
Note this is an ODD function about the MEAN value,
hence only sine terms appear in Fourier expansion
Also note original function f(t) is only valid between 0
at T (blue dotted curve). This is OK as we only
considered that repeat of the function.
Note it is normal to plot the magnitude of the Fourier
Coefficients in a bar chart, for example for the saw tooth
above:
4 4
b 2
n
0 0
0
0
5
10
15
20
25
n
30
30
This emphasises that the sign of the terms depends on
the phase of the wave being decomposed into its Fourier
Components and that an and bn have discrete values for
each integer value of n (they are not continuous
functions)
More comments on Fourier Analysis
1. We have confirmed that the first term in the
Fourier series (a0/2) is the mean value of the
function f(t) over one period:
a0 1
=
2 T
T
f (t )dt =
0
Area under curve
T
e.g. for a sawtooth wave:
1 TA
a0
A
= 2
= = Mean value
2
T
2
A
Area
0
0
T
2. We have considered waves that are periodic in
time [f(t)=f(t+nT)]. Structures that are periodic
in space obey the same maths. For example a
diffraction grating is a regular series of slits. To
do this:
Replace
ω0t by k0 x
k0 =
Where
2π
λ0
and λ0 is the spatial period of the structure in
meters.
λ0
x
The rest of the maths is the same:
a0 ∞
+ bn sin( nk0 x)
f ( x) =
2 n =1
and
bn =
2
λ0
λ0
f ( x) sin( nk0 x)dx
0
[Choose x = 0 point to have only sine terms]
3. Sometimes it is useful to do the integration over
an interval –T/2 to +T/2 rather than 0 to T
4. If the function is complicated the integration
might be easier if split into many parts:
T
e.g.
dt =
0
TA
T
0
TA
dt +
dt
5. A negative value for a coefficient simply means
the phase of the wave is changed by π .
e.g. − bn sin( nω0t ) = +bn sin( nω0t + π )
Also the energy carried by the wave depends on
the value of bn 2 the sign of the coefficients is of
no real interest.
A graph of the energy density verse frequency is
called the power spectrum.
Fourier analysis of real waveforms
Most real problems, like the example of the white
dwarf, have a series of measurements not a function. The
method of finding coefficients by numerical integration
works but is very slow, especially for high frequency
terms.
There is a clever technique called the Fast Fourier
Transform (FFT) that reduces the time required
dramatically.
Implementations of the FFT exist in many
computer programs: for example in Mathcad,
even in Excel!
It can be thousands of times quicker for large data
sets
It works by accepting regularly spaced discrete
data points, these can be calculated from a
continuous function if required.
Technical note: The FFT algorithm requires that there
must be exactly 2n data points (n is an integer) e.g.: 256,
512, 1024 etc.
Mathcad example of FFT
Remember there must be 2n points: i
0 .. 255
Use these to count across 2π, ie one cycle:
φi
2. π. i
255
Function (two “real terms” + noise):
Fi
sin 7 . φi
cos 3 . φi
FFT
0
1
2
F= 3
4
5
6
7
8
0
1.029
1.108
1.171
1.353
1.74
1.684
1.806
1.871
1.993
0.2 . ( rnd( 2 )
1)
fft ( F )
0
0
1
2
FFT = 3
4
5
6
7
8
0.204
0.093+0.159i
0.052-0.092i
8.053-0.234i
-0.02+0.186i
-0.137+0.247i
-0.031+0.37i
0.713+7.937i
-0.115-0.226i
NOTE: Only the first 8 terms of the data (F) and
Fourier transformed data (FFT) are shown. There are 256
data points and 128 FFT point in total.
The real parts of FFT correspond to sine terms in F,
imaginary to cosine terms.
Plot of data:
2
1
F
0
i
1
2
0
50
100
150
200
250
i
Plot of power spectrum of the above data
j
0 .. 128
100
10
FFT
j
2
1
0.1
0.01
3
1 10
0
20
40
60
80
j
See peaks only at 3 and 7 as expected
Note: log y scale!.
100
120