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

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David Cho 2017-10-31

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https://ww2.mathworks.cn/matlabcentral/answers/364256-in-the-fft-example-on-matlab-help-why-do-we-multiply-by-2

评论： Star Strider 2019-9-20

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.

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kamrul islam sharek 2019-9-20

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 2019-9-20

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

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Answer 1

Star Strider 2017-10-31

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https://ww2.mathworks.cn/matlabcentral/answers/364256-in-the-fft-example-on-matlab-help-why-do-we-multiply-by-2#answer_288677

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.