Notes about Complex and Discrete Fourier Series - Whitman People

Notes about Complex and Discrete Fourier Series
Background
D.R. Hundley
February 28, 2013
We have already been discussing the Fourier series for a function f(x), and for the part of
the notes from the text, we will work on the interval [−L, L], but then we’ll need to switch
to periodic functions on the interval from [0, L] when we explain the software (the software
assumes that).
Now, on the interval [−L, L], we have the usual formulas that define the Fourier series:
f(x) ∼ a0 +
Where the constants are given by:
And for the remaining:
an = 1
L
f(x) cos
L −L
From Real to Complex
If we recall Euler’s Formula:
∞
nπx
an sin +
L
n=1
∞
nπx
bn cos
L
n=1
a0 = 1
L
f(x) dx
2L −L
nπx
dx and bn =
L
1
L
f(x) sin
L −L
e iθ = cos(θ) + i sin(θ)
nπx
dx
L
it is easy to show that we can write the sine and cosine in the following way 1 :
cos(x) = eix + e −ix
2
and sin(x) = eix − e −ix
1 By changing x to the complex variable z = x + iy, this is how sin(z) and cos(z) are defined in the
complex plane! Also note the similarity to the hyperbolic sine and cosine
1
2i
(1)

Geometrically, the magnitude of e iθ is 1 (you should check that), so that e iθ is a point on the
unit circle (plotted in the complex plane). We might also note that 1 = −i. Substitution of
i
these expressions into the Fourier series, we get:
a0 +
∞
n=1
an
iθ −iθ e + e
+
2
∞
iθ −iθ e − e
(−i)bn
2
where θ = nπx.
We should be able to verify then that the expression can be re-arranged to
L
get:
a0 + 1
∞
(an − ibn)e
2
iθ + 1
∞
(an + ibn)e
2
−iθ
n=1
Going back to our definitions, we can show that, as a function of n, a−n = an and b−n = −bn.
Substituting a dummy index:
a0 + 1
2
n=1
n=1
−∞
(a−n − ib−n)e −iθ + 1
n=−1
We can re-write using symmetry:
Finally, if we define
a0 + 1
2
−∞
n=−1
(an + ibn)e −iθ + 1
2
2
∞
(an + ibn)e −iθ
n=1
∞
(an + ibn)e −iθ
n=1
c0 = a0 cn = an + ibn
2
then the Fourier series for f(x) can be written compactly as:
Now, the formula for cn is given by:
f(x) ∼
∞
n=−∞
cn = 1
2 (an + ibn) = 1
L
1
f(x) cos
2 L −L
Simplifying, we get:
cn = 1
L
f(x) cos
2L −L
nπx
+ i sin
L
cne −inπx/L
nπx
dx +
L
i
L
f(x) sin
L −L
nπx
dx
L
nπx
dx =
L
1
L
f(x)e
2L −L
inπx/L dx
IMPORTANT NOTE: The Fourier series has a negative exponential in the formula, the
coefficients are using a positive exponential.
Note: c−n = cn (this is an exercise)
2

Discretization
To go from the continuous series to the discrete series, we will need to discretize our function.
That is, we will (by necessity) go from an infinite dimensional vector space to a finite dimensional
approximation (if you use n points in the discretization, you will be working in R n ).
Before discussing the implications to the Fourier analysis, let us discuss the discretization
process, first by means of some Matlab exercises:
Exercises:
1. Consider the functions y1 = cos π
4 t , y2 = cos 7π
4 t with 0 ≤ t ≤ 64. Will 64 samples
be enough to differentiate between these functions? Try the following Matlab script,
and write up your observations:
k=0:63; t=0:0.1:63;
y1=cos((pi/4)*k); y2=cos((7*pi/4)*k);
Y1=cos((pi/4)*t); Y2=cos((7*pi/4)*t);
plot(t,Y1,’b-’,t,Y2,’r-’);
hold on
plot(k,y1,’bo’,k,y2,’r*’);
hold off
%Take a closer look at the first 10 values:
axis([0 10 -1 1]);
2. Do the same exercise 2 , if y1 = sin 7π
8
, y2 = − sin
9π . 8
7π
9π
π
7π
i i i i 3. Geometrically, what is the relationship between e 8 and e 8 ? Between e 4 and e 4 ?
4. Using the previous exercise, come up with other pairs of sines and cosines that will
discretize to the same function (under the given discretization).
5. Give a conjecture about what we should allow as the smallest admissable period if the
distance between two points is ∆x. For example, is it adviseable to have a period that
∆x? Why or why not? Draw a picture to justify your conclusions.
is 1
2
The Discrete Form of the FFT
In this section, we’re going to change the interval for f(x) slightly- That is, we’ll assume that
we’re analyzing f on the interval [0, L]. This will keep the notation aligned with how Matlab
computes the coefficients. We assume that the function f has been sampled uniformly, so
the values of x would be (if we have M samples):
x0 = 0, x1 = 1
M , x2 = 2
, · · ·
M
2 Try the Sampling Rate script file at the Appendix for more!
3

and therefore, we define fk = f(xk).
Then, the input to the FFT algorithm is the length M vector containing f, and the
output is a length M vector whose kth element is given by:
M
Yk = fjω (j−1)(k−1) with ω = e −(2π/M)i
j=1
In matrix notation, if we define the Fourier matrix by F , where the (k, j) element (with
k = 1 to M, and j going from 1 to M) is given by:
Fk,j = ω (k−1)(j−1)
= exp − 2πi
· (k − 1)(j − 1)
N
then F is a square, invertible matrix, and the output to the FFT is:
Matlab Code
Y = F f
To clarify, in Matlab, we could compute the FFT of the vector f directly as:
Y=exp( (-2*pi*i/N)*(0:(N-1))’*(0:(N-1)) )*f;
And the inverse transform (f=ifft(F)) is computed in a similar manner:
fn = 1
N
Yke
N
2πi
N ·(k−1)(n−1) , 1 ≤ n ≤ N
or
k=1
f=(1/N)*exp( (2*pi*i/N)*(0:(N-1))’*(0:(N-1)) )*Y;
Important Note from Cleve Moler, founder of Matlab: We should point out that
this is not the only notation for the finite Fourier transform in common use. The minus
sign in the definition of ω sometimes occurs instead in the definition of ω used in the inverse
transform. The 1/N scaling factor in the inverse transform is sometimes replaced with 1/ √ N
scaling factors in both transforms.
Interpreting the Matlab Results
As we stated earlier, let’s assume that f is defined on the interval [0, L]. If we cut the interval
into N equal subintervals, then let’s use the usual definition for ∆x,
∆x = L
, or L = N · ∆x
N
Recall that our Fourier series is given as:
∞
2πk
f (x) = a0 + ak cos
L x
∞
2πk
+ bk sin
L x
k=1
Now, inserting L = N · ∆x, we get the final discretized form summarized below.
4
k=1

Summary
If we have N points with spacing ∆x, and these points are labeled
with x ∈ [0, L], then the Matlab coefficients:
F=fft(f)
correspond to:
via the relations:
N/2
fn = a0 +
k=1
{f0, f1, . . . , fN−1}
2πk
ak cos
N ∆x xn
2πk
+ bk sin
N ∆x xn
a0 = 1
N F1
ak = 2
· Real (Fk+1)
N
bk = − 2
· Imag (Fk+1)
N
Note the periods of the functions in Equation 2, as k = 1, 2, . . . , N
per Cycle):
2π
2πk
N ∆x
= 2π
1
· N∆x
2πk
= N
k ∆x
Listed out, the periods of the functions we’re using are:
N · ∆x, N
2
· ∆x, N
3
or as frequencies, we would take the reciprocals of those quantities.
EXAMPLE: Counting the periods
2
(2)
are (measured in Time
N
· ∆x, . . . , · ∆x = 2∆x (3)
N/2
Suppose we have N = 16 samples, each at ∆x = 1 time unit (so that the function f is defined
on the interval [0, 16]). What are the possible periods that the discrete Fourier Transform
will pick up?
SOLUTION:
There are 8 possible periods: 16, 8, 5 1
3
NOTES:
, 4, 3.2, 2 2
3
3 , 2 , 2. 7
• The shortest period is 2, twice the sampling distance ∆x. The longest period is 16
time units long- as long as the entire interval.
• While there are 8 periods, there are 16 sample points- The vector of FFT values (for
real functions) that is output by Matlab is symmetric.
5

The Frequency Content of a Function
We define the frequency content by the following, where Y is the vector output from the
FFT command:
freq(k) = a2 k + b2 2
k =
N |Yk+1|, k = 1, 2, . . . , N
2
from which we get the Power Spectrum, which is a plot of the magnitudes of the complex
Fourier coefficients. This plot reveals the frequency content of the signal, and we will explore
this in the next section.
Examples: Applying the FFT
In this section, we will apply the FFT to various signals and look at ways of manipulating
the output.
Analyzing a Known Signal
Let y = sin( π
8 xk) + 2 cos( 7π
8 xk), with xk coming from 64 evenly spaced points in [0, 64].
Before going into Matlab, let’s check the periods involved in this function:
First, the sine function will have period 2π/(π/8) = 16 and the cosine function will have
period 2π/(7π/8) = 16/7.
Using our previous formula, we said that the periods were N∆x/k, or:
N∆x
k
= 16 ⇒ 64
k
= 16 ⇒ k = 4
And
N∆x 16 64 16
= ⇒ = ⇒ k = 28
k 7 k 7
Therefore, in the spectrum, we expect spikes at frequencies 5 and 29 (add one because of
the constant term a0).
1. Set up the domain and range information:
k=0:63;
y=sin((pi/8)*k)+2*cos((7*pi/8)*k);
2. Take the Fourier Transform:
Y=fft(y);
3. Look at the Spectrum:
plot(abs(Y),’k-*’);
4. We should see 4 spikes in the spectrum, and the spectrum is symmetric. The symmetry
is due to the fact that the function is real. The spikes of interest are at positions 5 and
29, as we predicted!
6

Sunspot Analysis
This example has a built-in script. You can type: sunspots in the command window to see
the slideshow. We’ll do the same example here:
The problem: We’re given data corresponding to the number of sunspots counted on
the sun every year between 1700 and 1987. We’d like to analyze it to see if there is some
periodicity in the number of sunspots. We assume the domain to be 0 to 288 (years), and
the data below corresponds to one sample per year (therefore, L = N = 288, and ∆x = 1).
load sunspot.dat
year=sunspot(:,1); sun=sunspot(:,2);
Y=fft(sun);
Y(1)=[]; %Remove the mean
plot(abs(Y(1:143))); %We’ll only plot the relevant half.
[a,b]=max(abs(Y(1:143)));
We see that the 26th position gave the max- how do we translate that into years? From
the formulas for the “Time per Cycle”, (Equation 3) we get:
N · ∆x
k
= 288 · 1
26
≈ 11.0769 years
from which comes the adage that we get a maximum number of sunspots approximately
every 11 years.
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