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DSP Fourier

The document defines a periodic function f(t) that is equal to 2 when t is between 0 and T/2, and equal to -2 when t is between T/2 and T. It then calculates the Fourier series representation of f(t) by determining the cosine and sine coefficients from the discrete Fourier transform and summing the cosine and sine terms. Finally, it plots the original function f(t) and the reconstructed Fourier series fs(t) to compare them.

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radymohamed9876
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37 views5 pages

DSP Fourier

The document defines a periodic function f(t) that is equal to 2 when t is between 0 and T/2, and equal to -2 when t is between T/2 and T. It then calculates the Fourier series representation of f(t) by determining the cosine and sine coefficients from the discrete Fourier transform and summing the cosine and sine terms. Finally, it plots the original function f(t) and the reconstructed Fourier series fs(t) to compare them.

Uploaded by

radymohamed9876
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Define f(t)

Ts = 0.2; %seconds for sample


T = 2; %seconds
t = 0 : Ts : T-Ts

t = 1×10
0 0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 1.4000

((t >= (0) ) & (t< T/2 ))

ans = 1×10 logical array


1 1 1 1 1 0 0 0 0 0

%to find logical values (true(1) or false (0))


%(t < T/2) %to find f(t) coressponding for t
%t(0)=2;
%t(.2)=2;
%t(.4)=2;
%t(.6)=2;
%t(.8)=2;
%t(1), 1 is not smaller than 1 , we want to find t<T/2 , T/2 = 1;

f((t >= (0) ) & (t< (T /2))) = 2

f = 1×10
2 2 2 2 2 -2 -2 -2 -2 -2

f((t >= (T/2) ) & (t< T )) = -2

f = 1×10
2 2 2 2 2 -2 -2 -2 -2 -2

plot (t,f)

1
calcualte real_valued coefficients

N=10;
a = zeros(1,N+1);
b = zeros(1,N+1);
for n = 0 : N
a(n+1)=(2*Ts/T)*sum(f.*cos(2*pi*n*t/T)); %shift by 1 as Array indices must be positive inte
b(n+1)=(2*Ts/T)*sum(f.*sin(2*pi*n*t/T));
end
plot (0:N,a)
hold on
plot (0:N,b)
hold off
legend ('a','b')

2
%plot disrete
N=10;
a = zeros(1,N+1);
b = zeros(1,N+1);
for n = 0 : N
a(n+1)=(2*Ts/T)*sum(f.*cos(2*pi*n*t/T)); %shift by 1 as Array indices must be positive inte
b(n+1)=(2*Ts/T)*sum(f.*sin(2*pi*n*t/T));
end
stem (0:N,a)
hold on
stem (0:N,b)
hold off
legend ('a','b')

3
fourier series sine/cosine form

fs = (a(1)/2)*ones (size(t)) ;

for n =1: N
fs = fs + (a(n+1)*cos (2*pi*n*t/T) +b (n+1)*sin (2*pi*n*t/T))
end

fs = 1×10
0.8000 2.0944 2.5889 2.0944 0.8000 -0.8000 -2.0944 -2.5889
fs = 1×10
0.8000 2.0944 2.5889 2.0944 0.8000 -0.8000 -2.0944 -2.5889
fs = 1×10
1.6000 2.4000 1.6000 2.4000 1.6000 -1.6000 -2.4000 -1.6000
fs = 1×10
1.6000 2.4000 1.6000 2.4000 1.6000 -1.6000 -2.4000 -1.6000
fs = 1×10
2.4000 1.6000 2.4000 1.6000 2.4000 -2.4000 -1.6000 -2.4000
fs = 1×10
2.4000 1.6000 2.4000 1.6000 2.4000 -2.4000 -1.6000 -2.4000
fs = 1×10
3.2000 1.9056 1.4111 1.9056 3.2000 -3.2000 -1.9056 -1.4111
fs = 1×10
3.2000 1.9056 1.4111 1.9056 3.2000 -3.2000 -1.9056 -1.4111
fs = 1×10
4.0000 4.0000 4.0000 4.0000 4.0000 -4.0000 -4.0000 -4.0000
fs = 1×10
4.0000 4.0000 4.0000 4.0000 4.0000 -4.0000 -4.0000 -4.0000

4
plot(t,fs)

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