Alright, to start out I am a mechanical engineer with very little circuits and digital signal processing background. So please excuse my ignorance as this is far out of my usual territory.

When looking at the magnitude of an FFT of a signal, I understand that the magnitude is in the units of the time signal (in my case it is Volts). I also understand that you need to divide the FFT results by the transform length in order to normalize.

What I don't completely understand is how the magnitude of the FFT result relates back to the time signal. From what (I think) I understand it should be related to the amplitude of the time signal. From my MATLAB code that I made (posed below) it seems that it returns 1/4 of the amplitude of the signal. Is this correct? What is the physical meaning of it?

And then on a related note, if it is related to the amplitude of a time signal, what is the sampling rate needed to get a decent measure of the amplitude. I know the highest resolvable frequency for an FFT is the nyquist frequency, but I haven't been able to find anything about accurate measurements of a signal's amplitude.

Thanks in advance for any and all help on this.


f_s = 20;              % sampling (hz)

f_w = 2;                % frequency of wave (hz)
A = 5;                  % amplitude

d_t = 1/f_s;
t = 0:d_t:2-d_t;
n = length(t);

% Creating the time signal
time_signal = A*sin(f_w*2*pi*t);

% Generating and applying a hanning window on the time signal
window = hann(n)';
windowed_time_signal = time_signal.*window;

% Taking FFT of windowed time signal
raw_fft = fft(windowed_time_signal);

% Frequency resolution of FFT
d_f = f_s/n;

% Frequency spectrum for full FFT
f_spectrum_full = (0:n-1)*d_f;

% Only first half of FFT contains useful information
half_fft = raw_fft(1:(n/2));

% Frequency scale for half FFT
f_spectrum_half = (0:n/2-1)*d_f;

% Magnitude of the half FFT
magnitude = abs(half_fft);

% Normalized version of the magnitude
normalized = magnitude/(n); 

figure (1)

title('time wave')

title('windowed time wave')

figure (2)


  • 1
    $\begingroup$ The signal energy is distributed between positive frequencies (the first half of the fft) and negative frequencies (the second half). $\endgroup$ Oct 4, 2015 at 5:46
  • $\begingroup$ In addition to what CMDoolittle said, you need to consider spectral leakage. What is your use caes? Are you sure the FFT is the right tool? $\endgroup$ Oct 4, 2015 at 11:55

2 Answers 2


You can calculate the coherent or DC gain of a window with:

coherent_gain = sum(window)/n;

with a Hann window, and n = 40, this comes to 0.4875, as hotpaw2 mentioned.

You can therefore eliminate the effect of the coherent gain of the window on your spectral estimate by replacing windowed_time_signal = time_signal.*window; with windowed_time_signal = time_signal.*window/coherent_gain;.

If your main priority is very accurate amplitude - and you're unconcerned with spectral leakage as Sebastian mentioned, then the flattop window is the way to go. Here's the matlab docs.

Just replace window = hann(n)'; with window = flattopwin(n)';.

Also, if you have the ability to collect more samples, that will likely improve the spectral estimate from your FFT (assuming the underlying process in your application is somewhat stationary/ergodic). You can also take FFTs of successive (possibly overlapping) slices and average them.

Furthermore, as hotpaw2 mentioned, selecting a sample rate depends very strongly on your application, and how many samples you intend to collect. That said, I've often used fs = 2.56 * bw, with good results when it comes to spectral analysis or digital filtering.

  • $\begingroup$ I would appreciate feedback regarding the downvote. Is something incorrect or ill-advised? $\endgroup$ Jan 4, 2016 at 1:45

If you ignore the energy in the negative frequencies, that cuts the FT energy of a real signal in half. If you apply a typical Von Hann window, you cut the energy of a stationary signal roughly in (another) half. Thus 1/4th.

If the width of an FFT is not an exact integer multiple of the period of an exactly periodic signal then windowing artifacts will affect the magnitude estimate. Thus, some interpolation may be required.

  • $\begingroup$ Thank you hotpaw2! This makes a lot more sense now! Do you happen to know the answer to the second part of my question? Basically, what guidelines do I need to follow in order to get an accurate magnitude measurement. I know I can technically measure up to the Nyquist frequency, but as I approach it my results become worse. Is there some general rule I can follow as to how fast I should sample to avoid obtaining wrong results? $\endgroup$ Oct 5, 2015 at 13:40
  • $\begingroup$ The closer a signal gets to Fs/2, the longer you have to sample it to separate it from its complex conjugate image. At Nyquist, the results can be bogus unless you sample for an infinitely long time and do an infinitely long FFT. $\endgroup$
    – hotpaw2
    Oct 8, 2015 at 22:50

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