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Scilab Week 7

October 5, 2009

In this assignment, the focus will be on Fourier Transforms to solve various problems

Study the spectra of data

Solve differential equations

Filter data to extract interesting information

1 Fourier Analysis

The Fourier Transform is defined by

F() =1

2

f(t)ejtdt ,

work in frequency domain, solve our problem there, and invert to get the solution

f(t) =

F()ejtd .

Note that you can put the factor of 1/2 in front of either integral. Also note that the signof the exponent is opposite for the two integrals.

You have studied in Maths and in Networks that

F() is unique given f(t)

The Fourier transform is a linear transformation. i.e., the transform of a linear com-bination of f(t) and g(t) is the same linear combination of F() and G().

The transform of a convolution is the product of the transforms, i.e., if

h(t) =

f()g(t)d

thenH() = F()G()

The transforms of a derivitive and indefinite integral are simpler in the Transformdomain:

F

d f

dt,t

= jF()

F

tf d,t

=

F()

j

Many problems are solved using Fourier transforms. In electrical engineering, the follow-

ing are some of the types of ways we use Fourier transforms:

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Harmonic content of nonlinear circuits driven by a phasor source. For instance if adiode switches on and off based on the sign of the voltage across it, what harmonics

are induced?

Fourier spectrum of AM and FM modulated signals.

Frequency response of a circuit. This has been done last time using Laplace trans-froms, but it can be done in the Fourier domain as well.

Studying the transfer function of a system to know if it is stable under feedback.This is a crucial part of control systems theory and analog circuit theory.

The spectrum of noise. Central to communications theory and control systems the-ory.

Filter design. We want to isolate a desired part of the spectrum. How to do it infourier domain, and how to do it in time domain itself.

System identification: An earthquake has occurred and you have the trace. Howto analyse it, knowing the type of trace an earthquake can give to identify where

and when it occurred and what type and how strong it was. This often uses fourier

methods as part of the process of system identification.

Adaptive circuits: You want to listen to a radio station, but its transmission frequencyis not constant (or your receivers tuner is not stable). How to lock the receiver to

the transmitted signal? These are known as phase locked loops. Also analysed and

designed using fourier methods, since what we are tracking is a frequency signal.

We will take some sample problems here to look at.

2 Digital Fourier Transforms

Now, computers are not good at performing continuous integrals over unbounded domains,

and so these equations are not useful as they are. However, we know that the Fourier

Transform is just the limit of the Fourier Series. So we can solve the Fourier Series problem

that is nearest to the exact problem and get an approximate solution. But even that is too

complicated, since the general function has an infinite number of fourier coefficients. So

we keep only a finite number of fourier coefficients to get to the computer approximation

of the Fourier Transform.

Let us take an example:

f(t) =1

1 + t2et

2

The plot of the function looks as follows:

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3 2 1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

t

f

Since the plot goes nearly to zero, we can safely ignore the rest of the function. Even

better, we can make copies of f(t) every 6 seconds to make it a periodic function, g(t). Theplot ofg(t) looks like this

10 8 6 4 2 0 2 4 6 8 10

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

t

g(t)

Then a Fourier series will do the job. We write

g(t) =

n=

cnejnt

where = 2/6. Since the function is even and real, the fourier series is also even andreal, and so it collapses into a cosine series:

g(t) =

n=0

an cos (nt)

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We can compute the coefficients as

a0 =1

6

3

3

g(t)dt

an =1

3

33

g(t) cos (nt) dt

Using the intg function that is built in, we can compute the integrals. The code is as

follows

[branch code0:

4 * 4global count;

function y=g(t)

global count;

count=count+1

y=exp(-t.^2)./(1+t.^2);

endfunctionfunction y=gn(t)

global count;

count=count+1;

y=exp(-t.^2).*cos(n*omega*t)./(1+t.^2);

endfunction

omega=2*%pi/6;

count=0;

a=zeros(21,1);

a(1)=intg(0,3,g)/3;

for n=1:20

a(n+1)=intg(0,3,gn)/1.5;

end

bar(0:20,a);

xtitle("Fourier coefficients","n","a(n)");

mprintf("Number of times function called: %d\n",count);

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]

The coefficients look as follows:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0.05

0.00

0.05

0.10

0.15

0.20

0.25

Fourier coefficients

n

a(n)

Now, while this works, we are actually doing numerical integration to compute each

coefficient. That is quite expensive. If you run the above code, you find that the function

was called 8883 times, or about 450 times per coefficient. In practical situations we often

take fourier transforms of very large data sets (could be millions of elements). We need a

faster method. This faster method is called the Discrete Fourier Transform or the DFT.

You will learn the mathematics in your ADSP course. The main thing about it is that

the integral to compute the coefficients collapses to a sum. We first sample the function

f(t) at N times over a time period t0, t0 +t, etc to get N samples f0, f1, . . . , fN1. TheDFT is then defined by:

fn =N1

k=0

Fkej2kn/N

Fk =1

N

N1

n=0

fnej2kn/N

The coefficients Fn are closely related to the coefficients an in the exact Fourier series. We

can go from Fn to a cosine series

fn = A0 +

N/21

k=1Akcos (2nk/N)

Comparing we can see that

2kn

N nt

or,2

N t

which is reasonable, since Nt should correspond to a period, i.e., 2/.The DFT is a much better algorithm than the integration method above, since only

N evaluations are done. To get 21 coefficients, we would only evaluate the function 21

times. Let us compute the coefficients using the DFT and compare them to the fourier

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series values. Note that we have to compute 41 coefficients, since each cosine is equivalent

to two complex exponentials.

6a * 4+N=41;

t=linspace(-3,3,N+1);t=t(1:N);

This peculiar piece of code is to get 41 samples between 0 and 6, not counting both

t= 0 and t= 6, since they are the same point. So I get 42 points including the end points

and discard the 42nd point.

6b * 4+f=fftshift(g(t));

F=dft(f,-1)/N;

A=zeros(21,1);

A(1)=F(1);

A(2:21)=F(2:21)+F(41:-1:22);

clf

bar(1:21,[a real(A)]);

legends(["exact";"DFT"],2:3);

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The result is plotted below

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

0.05

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

exact

DFT

It is obvious that the coefficients are not the same. In fact, they should not be. After all,

the DFT is computing the integral using 21 uniformly spaced samples of f(t). Only if theintegral can be exactly calculated from those samples, will the coefficients agree. Which

means that the more samples we take, the more accurate will be the approximation. So let

us take (say) 256 samples.

7 * 4+N=256;

t=linspace(-3,3,N+1);t=t(1:N);ff=fftshift(g(t));

FF=dft(ff,-1)/N;

AA=zeros(21,1);

AA(1)=FF(1);

AA(2:21)=FF(2:21)+FF(N:-1:N-19);

clf

bar(1:21,[a real(AA)]);

legends(["exact";"DFT-256"],2:3)

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Now we get the following:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

0.05

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

exact

DFT256

The errors are as follows:

8 * 4+[(1:21) real(A)-a real(AA)-a]

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The answer that Scilab gives is shown below

1. - 4.781E-08 - 1.242E-09

2. - 0.0011045 2.485E-09

3. - 0.0026768 - 2.485E-09

4. - 0.0027972 2.485E-09

5. - 0.0019422 - 2.486E-09

6. - 0.0010921 2.487E-09

7. - 0.0005517 - 2.487E-09

8. - 0.0002619 2.488E-09

9. - 0.0001192 - 2.490E-09

10. - 0.0000525 2.491E-09

11. - 0.0000225 - 2.492E-09

12. - 0.0000095 2.494E-09

13. - 0.0000039 - 2.496E-09

14. - 0.0000016 2.497E-09

15. - 0.0000006 - 2.500E-0916. - 0.0000003 2.502E-09

17. - 3.128E-08 - 2.504E-09

18. - 0.0000001 2.507E-09

19. 6.934E-08 - 2.509E-09

20. - 9.917E-08 2.512E-09

21. 0.0000001 - 2.515E-09

These errors we get for small N are due to the fact that the integral cannot be properly eval-

uated given only a few points. The technical name for this problem is aliasing, something

about which you will learn a lot in your ADSP course.

3 The Fast Fourier Transform

Well that was good. We could compute a decently accurate Fourier series using a few

samples. No need for integration. But to do a good job we needed a lot of samples, and the

computation of the sum requires a large number (roughly N2) multiplications of complex

numbers. That is still very expensive. Instead of 8000 function evaluations we are now

doing 64000 complex multiplications!

There is a better way, through an algorithm known as the Fast Fourier Transform or

FFT. It computes exactly the same DFT but uses only about 3Nlog2N computations. For

N = 256, this works out to be about 3000 complex multiplications. As N increases, thesavings become huge. As far as Scilab is concerned, the only changes required to use the

FFT are:

N must be a power of 2.

Instead of a command like F=dft(f,-1) we instead have a command like F=fft(f,-1).

Thats it. Nothing else changes, but the speed up is huge, and the memory utilization is

far less. If you did the dft calculation with a few thousand samples, you would run out of

memory. No such problems with fft.

4 Filtering and Data Analysis

Suppose we have data whose frequencies of interest lie in a range. We can extract that data

by suitable filtering. For example, consider

f(t) = A(t) cost+B(t) cos2t

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where A(t) and B(t) are signals which vary very slowly. Then, we can filter out the cos2tterm and analyse only A(t). The steps are as follows:

Obtain the fourier transform of f(t).

Plot the magnitude spectrum and identify the frequencies you want and the frequen-cies you want to reject.

Zero the entries of F() for those frequencies you want to reject. This operation iscalled filtering, and in this case, we are low pass filtering since we are throwing away

the high frequencies.

The main thing to keep in mind is that you must keep both the positive low frequen-

cies and the negative low frequencies, i.e., keep terms corresponding to n = 0 to kand n = Nkto N1 while throwing away the rest.

Get back the filtered f(t) by inverse fourier transforming.

Multiply all the terms of the signal with cos t.

Fourier transform the signal again. Now we are taking the fourier transform of

A(t) cos2 t = A(t)

1 + cos2t

2

Again filter to reject the cos2t term.

Inverse transform to get A(t) itself.

5 The Assignment

1. Compute the exact fourier series and the DFT for the following periodic function

f(t) =1

1 + cos2 t

How exact are the solutions if you use 16 terms. How many terms were needed for

the error to become negligible? Do you understand this result?

2. Compute the sine fourier series and extract the corresponding terms from the DFT

for the function. How many terms were needed for the result to become exact?

f(t) = sin3 (t+ 1)

3. Solve the differential equation

d

2

ydt2

+y = cos(3t)

Note that this is not an initial value problem - we would use Laplace Transforms for

that.

You will have to Fourier transform the equation, solve it in frequency domain, and

then invert back to get the function in real time. To do that, use y=fft(Y,1).

Note: You have to figure out how to implement the derivitive in the DFT. Basically

a term like

ej2kn/N ejnt

and the derivitive will pull out jn. Here your is unity, which gives you the

factor to use.

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Note: Another problem you will face is that of a zero in the denominator. Handle it

by adding a tiny number (say 1014) to eliminate the singularity.

Note: You will have to play around with the factors ofN. fft often has the factor of

N built in. Look at the fourier transform and decide if the answer is reasonble.

How many samples are required for an accurate answer? Verify the answer by direct

differentiation. How does the error scale with N? Do you understand this?

4. Given a signal

f(t) = J0(0.01t) cost+ cos2t

where J0(x) is the the Bessel function of zeroth order. You can compute it in Scilabby besselj(0,x). Create a Scilab function to generate the function f(t), Sampleit from t = 0 to t = 500 with 8 samples per 2. Filter as explained above to extractthe coefficient of cos t. Plot the function extracted along with J0(0.01t) itself andcompare to see if the extraction worked. This is nothing but an example of AM

modulation.

There are many uses of the Fourier transform, and you should get very familiar with it.