FFT value with Matlab

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claudiocamera

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Hi there!

Sorry for the simple question, I just can't figure out what is going on ... We know that the FT of a cosine is given by:

x(t) = cos(ωot) ----FT---- X(jω) =πδ( ω-ωo) + πδ( ω+ωo).

In this way, I would expect that in the plot given by the code below the amplitude value were 10π instead of 5 as it is showing. Someone can explain-me why this happen?

Fs =1000; % Sample Frequency
T=5;
t= 0:1/Fs:T ; % vector of time
s= 10*cos(2*pi*60*t); % signal itself
% FFT performing block
F=fft(s);
F1=fftshift(F);
F2= abs(F1)/(length(F1)-1);
w=(-2.5:1/Fs:2.5)*Fs/T;
plot(w,F2)
 

Hi friend,


The ans is present in the file.

Thank You
 

Hi

the fourier transform of cos(x) is wrong in your equation, it is missing 1/2 in the FT of cos(x)


Sal
 

Sal said:
Hi

the fourier transform of cos(x) is wrong in your equation, it is missing 1/2 in the FT of cos(x)


Sal

That is another doubt, I took this FT from Signals and Systems Haykin , Van Veen book and it is written exactly as I wrote. In spite of the fact that I have also seen in other books the cos(ωot) FT with the 1/2 factor, what is happening ????

The solution gave by malaya.nath also raise me doubt , why he didnt consider the pi value in the fourier transforms of exp(-jωot)? So my doubt is still hovering...
 

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