Scaling amplitudes by two for the FFT in MATLAB - matlab

Question

I just read the example of Mablab tutorial, trying to studying the FFT function.
Can anyone tell me that for the final step, why P1(2:end-1) = 2*P1(2:end-1). In my opinion, it is not necessary to multiply by 2.
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 1000; % Length of signal
t = (0:L-1)*T; % Time vector
%--------
S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
%---------
X = S + 2*randn(size(t));
%---------
plot(1000*t(1:50),X(1:50))
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('t (milliseconds)')
ylabel('X(t)')
Y = fft(X);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
plot(f,P1)
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
Y = fft(S);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
plot(f,P1)
title('Single-Sided Amplitude Spectrum of S(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
Matlab sample

Answer 1

The reason for the multiplication by 2 is that the spectrum returned by fft is symmetric about the DC component. Since they are showing the single-sided amplitude spectrum, the amplitude of each point is going to be doubled to account for the contributions of data on the other side of the spectrum. For example, the single-sided amplitude of pi/4 is the amplitude at pi/4 plus the amplitude at -pi/4.
The first sample is skipped since it is the DC point and therefore shared between the two sides of the spectrum.
So for example, if we look at the fft of their example signal with a 50Hz sinusoid of amplitude 0.7 and a 120Hz sinusoid of amplitude 1.
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 1000; % Length of signal
t = (0:L-1)*T; % Time vector
S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
% Compute the FFT
Y = fft(S);
% Compute the amplitudes
amplitude = abs(Y / L);
% Figure out the corresponding frequencies
f = Fs/L*[0:(L/2-1),-L/2:-1]
% Plot the result
plot(f, amplitude)
When we plot this, you'll see that it's symmetric and the original input amplitude is only realized by combining the amplitudes from both sides of the spectrum.
A slightly more explicit version of what they have done would be to be the following which sums the two halves of the spectrum
P1(2:end-1) = P1(2:end-1) + P2((L/2+2):end);
But since by definition the spectrum is symmetric, the opt to simply multiply by 2.