In the fft example on MATLAB help, why do we multiply by 2?

Hi, I found the following 'fft' example on MATLAB help (https://nl.mathworks.com/help/matlab/ref/fft.html)

 Fs = 1000;            % Sampling frequency
T = 1/Fs;             % Sampling period       
L = 1500;             % Length of signal
t = (0:L-1)*T;        % Time vector
X = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
Y = fft(X);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);   % don't know why we have to mutiply by 2 and why the range is from 2 to (end-1)
f = Fs*(0:(L/2))/L;
plot(f,P1)

My question lies in the line #9: P1(2:end-1) = 2*P1(2:end-1); I do not understand why we have to mutiply by 2 and why the range is from 2 to (end-1).

I shall be obliged if anyone could help on this question.

Thank you.

[kamrul islam sharek]

a=imread('grey.jpg');

g=fspecial('gaussian',256,10);

max(g(:))

gi=mat2gray(g);

max(g1(:));

af=fftshift(fft2(a));

agi= a.*gi;

fftshow(ag1)

ag1i=ifft2(ag1);

fftshow(ag1i)

my problem is line no 7.plewase help to getting out this problem

[Star Strider]

Post this as a new Question, not here. Include the image.

Answer 1

[Star Strider]

The Fourier transform (link) by definition is two-sided, symmetrical about 0, with frequencies going from -Inf to +Inf. (In practice, the frequency content is limited by the signal length.) The energy at each frequency is represented equally on both the positive and negative halves of the spectrum at one-half the original signal energy (so the total energy remains the same).

Taking the one-sided Fourier transform, as done here, then requires that the amplitude of half the spectrum be multiplied by 2 in order to represent the signal amplitude accurately.

[David Cho]

Thank you for your answer. Now it is clear why we multiply by 2. But why is the range between 2 and (end-1)? Why do we remove the first and the end data? Thank you.

[Star Strider]

My pleasure.

That code removes the d-c (constant offset, the first element of the one-sided transform) and the end from the one-sided Fourier transform vector. That is the reason I much prefer — and always recommend — the R2015a documentation for fft (link). It’s much clearer and straightforward.

If you want to remove the d-c offset, it is much better to calculate the mean of your data (the offset) so you don’t lose the information, and then subtract it from your signal before doing the fft. You can remove the end of the vector if you want. I have never seen any reason to do that.

[David Cho]

Thank you very much for your kind answer! It helped me a lot to understand the MATLAB 'fft' command more.

[Star Strider]

As always, my pleasure!

Answer 2

[george veropoulos]

Hi.. i have an other question the the quantity P2 = abs(Y/L) must multiply by Ts... i make a code to compare the FFt RESULTS with fourier transform

N=128; t=linspace(0,3,N); f=2*exp(-3*t); Ts=t(2)-t(1); Ws=2*pi/Ts; F=fft(f); Fp=F(1:N/2+1)*Ts;

W=Ws*(0:N/2)/N; Fa=2./(3+j*W);

plot(W/(2*pi),abs(Fa),'--',W/(2*pi),abs(Fp),'-'); xlabel('Frequency, Hz'), ylabel('|F(f)|');

legend('analytical','numerical')

when myltiply by Ts the resluts are same...

thank you George