# Question about PSD calculation using FFT

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Chris Hoer on 5 Jun 2016
Commented: harihara reddy on 7 Apr 2020
Hello everyone,
a short question about the calculation of the PSD using the FFT. I used the example code provided:
rng default
Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*100*t) + randn(size(t));
N = length(x);
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(x):Fs/2;
I understand each step except this one:
psdx(2:end-1) = 2*psdx(2:end-1);
Why do we double entry except the first and the last? Is it about the symmetry of the FFT?
Thanks ahead and best regards, Chris

Jeremy on 14 Jun 2016
The fft results in a spectra that includes positive and negative frequencies. Don't ask me to explain what a negative frequency really means but the negative frequency esults are just the conjugates of the positive frequency results. They are not needed for the PSD calculation which is why they are rmeoved by the following line:
xdft = xdft(1:N/2+1);
You have to account for the enrgy lost when this portion of the spectra is removed so we multiply the rest by 2. The 0Hz and Nyquist frequency results don't have an imaginary portion and are not included in the negative frequency portion so they are not muiltiplied by 2.

Chris Hoer on 14 Jun 2016
Hey Jeremy, I already thought, that it had something to do with the symmetric part. Thanks for making it clear!
harihara reddy on 7 Apr 2020
how the negative frequency is removed by the above line