Effect of zero padding on FFT amplitude

Question

Rahul Kale 2023-1-25

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编辑： Paul 2023-2-11

Many text books and other literature comment on the improvement in FFT resolution due to zero padding but I have not come across a text that comments on the effect of zero padding on FFT amplitude due to change in signal length (effectively the same energy is spread over longer time). In the real world, say for processing of vibration signal, engineers are interested in amplitude at a frequency.

Would zero padding not change original signal and thereby underestimate amplitudes of FFT?

Thanks

Rahul

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Rahul Kale 2023-1-26

Well, by contradiction, I meant that FFT of zero padded signal is the same as the original one except adding values at points on the X axis where they were not present( in the FFT of original signal, thus essentially improving the resolution). However, when we think of its physical interpretation, it means changing the signal in time domain by zero padding would not affect its frequency domain values but would cause different response.

To illustrate this, if we consider a bell (old fashioned one having very less damping and not electronic one) making sound for a strike( it is hit with a hammer) say every 2 sec and compare its sound with a bell that is struck one time and then again struck one time after say every 10-15 min. There is definitely different sound response that we would feel. However, we can consider the two types of strikes the same because from time 2sec to 10 min is equivalent to zero padding the first type of strike to get the second type of strike

I am not sure if I have conveyed properly what I wanted to ask.

Thanks

Rahul

Paul 2023-1-26

编辑：Paul 2023-1-28

Signals go on forever, both into the past, and into the future. However, the DFT is only defined for signals x[n] with finite duration of length N, where the signal is 0 for n < 0 and the signal is 0 for n >= N. It's important to understand that this condition does not preclude a run of zeros for n <= N-1.

One way to interpret the DFT is that the results are always the Discrete Fourier Series coefficients of the periodic extension of x[n] with period Nfft >= N. If Nfft = N, then we are computing the DFT w/o zero-padding and if Nfft > N then we are using zero-padding.

Assuming for this discussion the we are limited to deterministic signals, we can use the DFT in one of two ways.

First, x[n] is one period of a periodic signal with period N, and it is this periodic signal we wish to characterize. In this case, we set Nfft = N and compute the DFT. We now have the DFS coefficients of the periodic signal that we actually care about. Suppose we strike the bell three for three beats, starting from n = 0, and then rest for three beats, and we do this for all time into the past and future. So one period of our striking sequence would look like x1 = [1 1 1 0 0 0], with N=6. Those last three zeros are not zero-padding. They are very important in defining one period of the signal. We can take the DFT of that sequence, without zero-padding, to obtain its DFS coefficients, from which we could compute x[n] for any value of n. These six DFS coefficients are all we need.

Second, x[n] is actually a finite duration signal. In this example, we don't strike the bell until n = 0, then srike it twice more, then stop. So x[n] looks something like this x[n] = [... 0 0 0 0 0 0 1 1 1 0 0 0 ...] where the ellipses indicate never ending and the first 1-entry correponds to n = 0. Now, this signal has frequency content at all frequencies, as could be seen by computing its DTFT. However, we want to use the DFT to compute samples of one period of the DTFT. The way we do that is to start with the sequence x2 =[1 1 1] but take Nfft large, i.e., lots of zero padding, so the DFT of x3 = [x2 zeros(1,Nfft-3)]. Now, when we take the DFT of x3, we are getting the DFS coefficients of the periodic extension of x3, with period Nfft, and these are still frequency domain samples of one period of the DTFT of x[n]. As Nfft >> N, then the periodic extension of x3 becomes closer and closer to actually being x[n] (imagine what happens as Nfft -> inf), and the DFT samples become more and more dense giving us a better picture of one period of the DTFT of x[n].

Rahul Kale 2023-1-27

Paul,

Thanks much for such a detailed explanation. However, I have some followup questions. Hope you would not mind. I apologize for dragging this Post longer.

1) In your first example, the statement,"Those last three zeros are not zero-padding" is very important. So the question is, isn't this(x1) the subset of the seriesx[n] in your second example(x3)? Or, are you trying to say that, since the Nfft>>N in the second case, the zeros(that are not due to zero padding in the first case) do not matter?

2) My confusion is exactly this that how to determinethe period of the signal for FFT scaling when we zero-pad a signal. Whether we need to take the period of the whole signal (original + padded zeros) or just that of the original signal. If its former then wouldn't the FFT become subjective to who does it with how many zeros because the scaling factor and thus the amplitude of FFT depend on the period of the original signal?

3)"Now, when we take the DFT of x3, we are getting the DFS coefficients of the periodic extension of x3, with period Nfft, and these are still frequency domain samples of one period of the DTFT of x[n]"

- Here, x3 is comprized of x1 as the original signal. So how would you scale FFT of x3 if we are interested in absolute amplitude of the physical quantity(say force) of x3 as a funciton of frequency? How much would be the signal length that you would use for dividing(6 which is of x1 or 3 which is of x2)?

Thanks

Rahul

Paul 2023-1-28

1. Yes, x1 in the first case are the first six elements in x3 in the second case. I found a typo in that coment and fixed it, it now says: "x3 = [x2 zeros(1,Nfft-3)]." Maybe that typo confused things. I don't understand the second question here that starts "Or, are you saying ...."

2. I've reread this question many times, and not sure I understand it. We've been saying and showing in this thread that zero-padding doesn't affect the amplitude spectrum of the signal. As I've said and MattJ showed, zero-padding changes the location of the frequency domain samples of the DTFT, but the amplitudes of those samples still follow the amplitude of the DTFT the finite duration signal.

3. Yes, the first six elements of x3 are the same as the elements of x1. As for scaling the results of the DFT, what is the definition of "amplitude of the physical quantity" in either the first or second case?

Rahul Kale 2023-1-30

Paul,

As for scaling the results of the DFT, what is the definition of "amplitude of the physical quantity" in either the first or second case?

- It is the 0-pk magnitude that a signal producing mechanism would produce. e.g. a shaker vibrating a structure.

Thanks

Rahul

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Answer 1

Matt J 2023-1-28

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编辑：Matt J 2023-1-28

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@Rahul Kale It occurs to me now that you are not really trying to compute the Fourier transform of a finite signal. You are instead trying to compute the Fourier series coefficients of a periodic signal, where x(n), n=0,...,N-1 is a sampling of one period of the signal. Note that the Fourier transform is not the usual way to decompose an infinite periodic signal, because it has infinite energy, although there are certainly mathematical relationships between the continuous Fourier transform a continuous Fourier series.

To get the discrete Fourier series (DFS) coefficients of a signal, then the way to do that using Matlab's fft() command is

DFS=fft(x)/N

So, 1/N is probably the normalization factor you are looking for. Zero-padding x will indeed change the Fourier series coefficients. However, it is important to understand that padded zeros have a different interpretation when applying the FFT to non-periodic signal analysis than to periodic signal analysis.

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Matt J 2023-1-29

编辑：Matt J 2023-1-30

"However, it is important to understand that padded zeros have a different interpretation when applying the FFT to non-periodic signal analysis than to periodic signal analysis." is very important. Can you please elaborate on this?

Well, imagine for a moment that we have a continuous time signal and that we are going to sample it, but the time sampling rate is fixed and given at

(the exact value won't matter) and we have no control over this.

If the continuous signal is periodic, then we are going to take its Fourier Series. To do that, we sample one period of the signal at intervals

to obtain a sequence

and use the discretized formula I derived above for the Fourier series coefficients,

In this case, because the signal and its time sampling rate are fixed and given, the length N of the sequence

is not under our control. The derivation of c_k required that we integrate over a complete period, so with

fixed, the number of samples N over the period is fixed as well.

On the other hand, suppose instead that the signal is of finite duration. So, we are not going to compute the Fourier series, but instead the Fourier transform which is this integral:

I am now going to discretize this integral, again under the assumption that

second is fixed and given. However, in this case, because the signal is finite duration, I can replace the infinite integral limits with a finite limit D/2 so long as D is large enough to cover the duration of the signal,

I discretize the integral with t=n

, dt=

, f=k/D, and N=D/

leading to

There are two differences to notice between the periodic case and the finite duration case. First, the normalization of the FFT is different when computing

as compared to

. Second, in the finite duration case, the parameter D and therefore N=D/

is under our control even though the sampling rate

is fixed. Increasing or decreasing D results in more or less padded zeros in

. It also has the affect of changing the frequency sampling interval

=1/D, but not the underlying continuous spectrum, as has been demonstrated elsewhere.

Now let's go back and relax the assumption that

is fixed. In the case of the periodic sginal, since we now control

, we now also control N in the calculation of the Fourier series coefficients

. However, the proportion of samples

which are pulse samples and the proportion which are padded zeros remain the same, assuming the period of the given continuous signal has not changed. Conversely, in the case of the finite duration signal, both the truncation interval D and

are freely variable, so the proportion of pulse samples and padded zeros can be controlled arbitrarily.

Matt J 2023-1-31

编辑：Matt J 2023-1-31

@Rahul Kale It sounds like you haven't committed to a decision on whether you are doing periodic signal analysis or non-periodic, finite duration signal analysis. This decision must be made first before anything else, and it must be made in continuous time. Once you know the kind of continuous-time analysis you are trying to approximate, decisions can be made on how best to discretize the analysis using FFTs

Although the signals contain many frequencies, I do not exactly know beforehand what the frequencies are in the two signals

If the signals are periodic, as you have been saying, then you should know. A continuous signal of period T contains a discrete set of harmonic frequencies 1/T, 2/T, 3/T,.... If the two signals have two different periods T1 and T2, then their harmonic frequencies will not coincide and there is no way you can compare them.

This means, based on your comment above, amplitude at each frequency would be scaled differently due to the zero padding length that each frequency component would see depending on how many integer periods fit in the selected total time (keeping aside effect of windowing for the sake of this discussion with zero padding).

I made many "comments above" at this point, so I don't know which one you are referencing. If you are doing periodic signal analysis, as you have been saying, then there is no reason to be windowing the signal down to finite length. The signals should be viewed as infinitely repeating.

If you are windowing your signal because you do in fact want to do finite-duration signal analysis, then in continuous-time you need to be using the Fourier transform to analyse the signals, not the Fourier series. And, if you are to be using Fourier transforms, all of the expectations you've expressed about the amplitude of the spectrum are invalid. In particular, if you take the Fourier transform of 50*sin(2*pi*t), it should not be expected to have a peak of 50.

Rahul Kale 2023-2-1

Yes correct..

Instead of truncating, I select NFFT = 2^nextpow2(L); This results in zero padding.

One question to your earlier comment, "If you are windowing your signal because you do in fact want to do finite-duration signal analysis, then in continuous-time you need to be using the Fourier transform to analyse the signals, not the Fourier series. And, if you are to be using Fourier transforms, all of the expectations you've expressed about the amplitude of the spectrum are invalid"

Any data processing, even of an infinite periodic wave, will have to be done on a finite data only, correct?

In the sine wave with 50 amplitude example, you have used finite data only.

What difference in the processing would it make if the the same two signals containing periodic components were to be processed 1) using infinite data length(hypothetically speaking) 2) using finite data length. Ultimately the end result should be the same i.e. correct amplitude...isn't it? Would fourier transform not give correct amplitude and fourier series give that instead? Or, are you suggesting that if I aim for getting correct amplitude, I should do data processing in such a way that I end up using fourier series and not transform?

Thanks

Rahul

Bruno Luong 2023-2-1

编辑：Bruno Luong 2023-2-1

The Fourier series is defined for periodic signal. It can be approximated by DFT analysic on one period of the signal. The more sample you have the better approximation will be.
The Continuous Fourier transform (CFT) is defined for ANY function that is square integrable, might or might have infinity length.
If the function is periodic of period T, the Fourier transform is NOT defined (the function is not square integrable)
However if we window this infinite periodic function, you can take CFT. Now if you take the window larger and larger such that it covers more and more periods, the CFT will converge to a Dirac-comb weighted by coefficients that are proportional to the Fourier series coefficient described in 1. The frequency of the first dirac is at the fundamental frequency f=1/T.
The shape of the CFT around the fundamental frequency and its harmonic (k*f with k integer) are the CFT of the envelop window function. When the window get wider, the amplitude of those spectrum peaks with go up and the width will shrinking down so that the integral remains (more or less) constant. Therefore the amplitude is hard to interpret. You select the window function by making a compromise between two things: (A) nice decay in frequency domain, but (B) do not cut down too much the original signal in time domain. NOTE: If you truncate abrutely your time signal, meaning using a rectangular window function, you get all kind of strange effect because the CTF of a rectangular is a sinc function that decay slowly to 0, which is not desired.
You can do zero-pad in the FFT during analysis of the windowed signal that will have as effect of increasing the resolution in the Fourier domain. The amplitude of the spectrum does NOT affected by zero padding.

Matt J 2023-2-1

编辑：Matt J 2023-2-1

Any data processing, even of an infinite periodic wave, will have to be done on a finite data only, correct?

Yes, but as I showed you in earlier posts, the Fourier Series integral (for periodic signals) and the Fourier Transform integral (for non-periodic signals) are different, and lead to different discretizations.

What difference in the processing would it make if the the same two signals containing periodic components were to be processed 1) using infinite data length(hypothetically speaking) 2) using finite data length.

It depends how you are truncating the periodic signal to finite length. For the example signal 50*sin(2*pi*t), I was able to get the peak Fourier Series coefficient of 50 because I truncated to exactly one period. I would also have gotten the same peak value if I'd truncated to an integer multiple of the period. In any other case, though, I would not have gotten the same peak coefficient.

This is relevant to you because you have two signals of different periods and might have attempted truncating one of them to a fraction of a period in order to get both of them in a common time window.

Instead of truncating, I select NFFT = 2^nextpow2(L); This results in zero padding.

But do you start padding the shorter-period signal after one period, or several? One might suppose that you are letting the shorter signal run for as many periods as it can complete in t<L before zero-padding starts. An argument can be made for both padding schemes.

Matt J 2023-2-1

编辑：Matt J 2023-2-1

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What argument would you make for both padding schemes w.r.t getting amplitudes right?

We can see the differences in the example below. Signals x1 and x2 both have period T=1. However, T=1 is not the fundamental period of the faster-oscillating signal x2, but rather T2=1/2. If I take the Discrete Fourier Series truncating both signals to a horizon of T=1, I get the result below. The harmonic frquencies of both signals have the expected Fourier Series coefficient amplitudes.

N=100;

t=linspace(0,1,N+1); t(end)=[];

fAxis=(0:N-1);

close all

x1=100*(sin(2*pi*t) + sin(8*pi*t)); %Slower signal

x2= 60*(sin(4*pi*t) + sin(12*pi*t)); %Faster signal

X1=fft(x1,N)/N; %Discrete Fourier Series

X2=fft(x2,N)/N; %Discrete Fourier Series

tiledlayout(1,2)

nexttile;

plot(t,x1,t,x2);

xlabel 'Time'; legend('Slower','Faster');axis square

nexttile

stem(fAxis,abs(X1)); hold on

stem(fAxis,abs(X2)); hold on

xlabel ('Frequency (Hz)'); legend('Slower','Faster');axis square

xlim([0,15])

sgtitle('Multi-Period Truncation')

However, if I do the anaysis of x2 by letting it run only for 1 period (0.5 sec.), and zero-padding the rest, each signal's coefficients have the same amplitudes as before at their respective harmonics. However, at 1 Hz and 4 Hz (which are harmonics of x1 but not x2), the coefficients of x2 get sinc-extrapolated, resulting in a much more continuous spectrum. If continuity is important to the comparison you are doing, this might be the better sampling scheme.

figure;

x2(end/2:end)=0;

X2=fft(x2,N)/(N/2); %Discrete Fourier Series -interpolated

Paul 2023-2-1

编辑：Paul 2023-2-1

Referring to @Bruno Luong's comment

Considering items 3 and 4 together, it's true that the CTFT integral does not converge for a periodic signal, but we still do define a CTFT for a periodic signal as an impulse train spaced by the harmonic frequencies and scaled by exponential CFS coefficients of the signal. This definition can be arrived at by the limiting process as Bruno described, which I'd imagine is the rigorous approach. But it can be intuitively understood by taking that scaled impulse train back through the inverse CTFT integral, which yields the CFS representation of the periodic signal. Off topic, but worth noting, is that a CTFT can be defined for other, aperiodic signals for which the CTFT integral doesn't converge, such as

x(t) = heaviside(t) -> X(f) = dirac(f)/2 - 1i/(2*f*pi)

Considering item 5, it seems to me that as long as the bandwidth of the CTFT of the window is enough less than 1/T, and if the peak of the CTFT of the window is at f = 0, then the spectrum peaks of the CTFT of the product of the periodic signal and the window will be at the harmonic frequencies, and the amplitude of those peaks will the product of the amplitude of the CFS coefficients and the peak of the CTFT of the window, which is known. Basically, I'm trying to understand why "the amplitude [of those spectrum peaks] is hard to interpret," assuming that the window bandwidth is narrow relative to 1/T. (Note: I would have shown an example with code, but the RUN feature isn't working for me.)

Rahul Kale 2023-2-2

Matt,

Yes, you have understood my "terminology" correctly.

Now, here is one important and subtle observation that I would like to point out.

To your comment, "However, if I do the anaysis of x2 by letting it run only for 1 period (0.5 sec.), and zero-padding the rest, each signal's coefficients have the same amplitudes as before at their respective harmonics"- This is true only when you used scale factor as N/2 and not N, for fft.

In your example where you truncated x2 and zero padded it, you divided its fft by N/2 and not N as in earlier case when there was no truncation. This was exactly my initial question that to get right amplitudes, whether I should consider signal length including the length of zero paddings for scaling the spectrum. If you had scaled using N, in second case, the signal amplitude would be half of the actual.

In my case,

N1=(length of sensor output data)

N2= N1 + length of padded zeros due to making NFFT = 2^nextpow2(L);

So, whether I should be scaling by N1 or N2?

According to your example it appears to me that I should be using the actual length of signal even for FFT of zero padded signal and that is N1 in my case.

Do you agree?

Thanks

Rahul

Matt J 2023-2-2

编辑：Matt J 2023-2-2

在 MATLAB Online 中打开

This was exactly my initial question that to get right amplitudes, whether I should consider signal length including the length of zero paddings for scaling the spectrum.

Well, then it appears I've answered your question. I hope you will Accept-click the answer. :-)

According to your example it appears to me that I should be using the actual length of signal even for FFT of zero padded signal and that is N1 in my case.

It really depends on your attitude toward the padded zeros and whether or not they're supposed to be considered part of the period of the wave. We don't always just ignore zeros. Suppose my signal was the double sawtooth below. The period of this wave is 6, and you would have to normalize by 1/6 in a proper Fourier series calculation. If you were tempted to say the period is 3 because the length of the non-zero part of the oscillation is 3, you would be wrong. If you were tempted to say the period is 4 because only the length between the start of the front-facing tooth and the back of the rear-facing tooth matters, you would also be wrong.

The reason the interval of zeros doesn't matter to you when you consider the earlier sinusoidal example is because you apparently don't consider them part of the period of the wave. The zeros there serve a different purpose. If you did consider them part of the wave, then you would be wrong to consider the "correct amplitudes" of the harmonic coefficients of x2 to be 30.

f=@(t) max((abs(t)-0.5).*(abs(t)<=2),0);

x=@(t) f(mod(t,6)-2);

fplot(x,[-0.1,18]); ylim([0,3])

Rahul Kale 2023-2-3

How to do that? I would certainly want to give him credit. Rahul

Bruno Luong 2023-2-3

@Rahul Kale please don't bother. Let's move on.

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Answer 2

Bruno Luong 2023-1-29

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https://ww2.mathworks.cn/matlabcentral/answers/1900240-effect-of-zero-padding-on-fft-amplitude#answer_1158760

编辑：Bruno Luong 2023-1-29

在 MATLAB Online 中打开

If your (windowed) signal dies out to 0 in a smooth way then the padding with 0s on the tail changes little to the FFT spectrum (amplitude), it is equivalent to densify the spectrum (interpolation) without scaling it.

N=70;

x=(0:N-1)-(N-1)/2;

E=exp(-(x./(N/6)).^2);

y = E.*randn(size(x));

f = (0:N-1)/N;

yhat = fft(y);

figure

plot(f,y,'b',f,E,'c');

NPad = 2^nextpow2(N);

ypad=y; ypad(NPad)=0;

fpad = (0:NPad-1)/NPad;

ypadhat = fft(ypad);

figure

plot(f, abs(yhat), 'b-+', fpad, abs(ypadhat), 'r-.')

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Bruno Luong 2023-1-30

编辑：Bruno Luong 2023-1-30

在 MATLAB Online 中打开

"I did try to do resampling by interpolating to get NFFT = 2^nextpow2(L)"

Why you do that? I show zero-padding IS equivalent to interpolating in frequency doman IF you signal decays to 0.

Interpolation frequency does NOT introduce new frequency (how on earth linear interpolation can?), whereas zeros padding with non decay signal will do all sort of effect of the spectrum.

If you are windowing you signal correctly (as you claim) then your signal will decay to 0, then just 0-pad as in my code show, it is more or less interpolation in spectral domain: the amplitude is unchanged, the frequency has to be adjusted with the new array indices.

Afterall that is the main purpose of windowing when study signal in Fourier domain: avoid artefact.

The question of how to scale the FFT to get "physically correct" is a separate question, regardless if you zero pad or not the signal. It seems Matt has address the normalization for your goal (though I haven't read his answer carefully).

The one sided spectrum of y

% yhat = fft(y)*2*sqrt(dt/(N-1))

has unit force/sqrt(Hz) if you original signal is Force, and associated with the corresponding frequency vector f. If you want to get back to force unit you must integrate the yhat^2 on a certain non-zero interval (the width is ~ bandwidth) and then take a square-root of the integration. There are probaby a tone of discussions on this topic. The bandwidth is non-zeros since your signal is finite in length, even for perfect input sine wave, since it is truncated.

% sample frequency and period

dt=0.1;

fs=1/dt;

% signal length and time vector

N=4000;

t=dt*(0:N-1);

% generate tested signal mono frequency

T=10;

A=2023;

y=A*sin(2*pi/T*t);

% Fourier transform and one-sided spectrum

df=fs/(N-1);

yhat=fft(y)*2*sqrt(dt/(N-1));

yhat=abs(yhat(1:ceil((end+1)/2)));

nf = length(yhat);

iseven = mod(N+1,2);

if iseven

duplicatedside = union(1,nf);

else

duplicatedside = 1;

end

yhat(duplicatedside) = yhat(duplicatedside)/sqrt(2);

% associate frequency vector

f=df*(0:nf-1);

plot(f, yhat);

xlabel('f [Hz]')

ylabel('spectrum [V/sqrt(Hz)]')

xline(f(end),'-','Nyquist','LabelHorizontalAlignment','left')

% Retrieve amplitude

fres=1/T;

bw=0.2;

fint = fres+bw/2*[-1 1];

keep = f>=fint(1) & f<=fint(2);

Amplitude = sqrt(trapz(df,yhat(keep).^2))

Amplitude = 2.0235e+03

xlim(fint)

The direct interpretation of the amplitude of the FFT is always problematic, only the energy and RMS value via L^2 integration in fourier domain via and parseval is mathematically correct.

Bruno Luong 2023-1-30

编辑：Bruno Luong 2023-1-31

在 MATLAB Online 中打开

Here is the full code that does windowing and then regroup both facts

zeros padding => amplitude FFT does not change on windowing signal
Correct scaling to get back amplittude after integration

%%

%clear

%close all

% sample frequency and sample period

dt=0.3;

fs=1/dt;

% signal length and time vector

N = 600;

t = dt*(-(N-1)/2:(N-1)/2);

% generate tested signal mono frequency of period T

T=10;

A=2023;

y=A*sin(2*pi/T*t);

% Windows function

Ewidth = t(end)/2;

E = exp(-(t/Ewidth).^2);

% Windowed signal to make them decay to 0 smoothly

ywin = y.*E;

figure

h = plot(t, [y(:) ywin(:) A*E(:)]);

xlabel('t [s]')

ylabel('signal [V]')

legend(h, 'original','windowed', 'window');

% Fourier transform of 0 padded signal and one-sided spectrum

Npad = 8*2^nextpow2(N); % 8 factor for fine resolution in Fourier domain

df = fs/Npad;

yhat = fft(ywin,Npad);

scalefft = 2*sqrt(dt/(N-1)); % N, not NPad

yhat = scalefft*abs(yhat(1:ceil((end+1)/2)));

nf = length(yhat);

iseven = mod(Npad+1,2); % EDIT fix a small glitch here

% The DC and Nyquist are common for both sides

if iseven

duplicatedside = union(1,nf);

else

duplicatedside = 1;

end

yhat(duplicatedside) = yhat(duplicatedside)/sqrt(2);

% associate frequency vector

f = df*(0:nf-1);

figure

plot(f, yhat);

xlabel('f [Hz]')

ylabel('spectrum [V/sqrt(Hz)]')

xline(f(end),'-','Nyquist','LabelHorizontalAlignment','left')

% Retrieve amplitude from Parseval

fres=1/T; % we can also detect the frequency of the peak of yhat, but we know it's 1/T

bw=sqrt(2)/Ewidth*sqrt(2);

fint = fres+bw/2*[-1 1];

keep = f>=fint(1) & f<=fint(2);

RMSE = sqrt(trapz(E.^2)/(N-1));

Amplitude = sqrt(trapz(df,yhat(keep).^2)) / RMSE % Estimate of A

Amplitude = 2.0230e+03

% Zoom in the peak of the spectrum

xlim(fint)

Rahul Kale 2023-1-31

Bruno,

Thanks much for this code..I am still trying to understand each step...will let you know if there are more questions.

I am not sure why in the scale factor there is N and not Npad.

Thanks

Rahul

Bruno Luong 2023-1-31

编辑：Bruno Luong 2023-1-31

@Rahul Kale "I am not sure why in the scale factor there is N and not Npad."

This has been somewhat explained in my various posts if you follow closely.

What happens here is the Parseval equality, or equivalent in discrete is the so called Rayleigh Energy Theorem and DFT that defined with the factor 1/sqrt(N) so make the transformation symetric in time and Fourier domain on the energy point of view.

Combine with the second fact that 0-padding does not change the amplitude of the FFT as I show in my answer, but just make the frequency vector with finer step (denser) on the same frequency domain [0,Niquist], there is no reason to adjust the amplitude with the length of the zero-pad signal.

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Answer 3

Paul 2023-2-5

2
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https://ww2.mathworks.cn/matlabcentral/answers/1900240-effect-of-zero-padding-on-fft-amplitude#answer_1164320

编辑：Paul 2023-2-11

在 MATLAB Online 中打开

This thread started with one question about zero-padding the Discrete Fourier Transform (DFT) and the answers and comments brought up many points on other related concepts including the Continuous Fourier Series (CFS), Continuous Time Fourier Transform (CTFT), windowing, and possibly many others. As might be expected for a thread with so many comments and additional questions in the comments, it might be hard to follow how all of the concepts relate and flow to each other. Consequently, I thought it might be good to show exactly that, at least for some of these of these concepts.

I'm posting this as its own answer because I wasn't sure it would flow nicely as a comment in any of the other answers.

Also, this answer does not add to any of the material already posted. Again, it's only trying to use Matlab to illustrate some of the points already made by (in no particluar order) @Mathieu NOE, @Star Strider, @Bjorn Gustavsson, @Matt J and @Bruno Luong and me.

Define a simple, continuous-time signal

syms t f real
Pi = sym(pi);
T1 = sym(2);
T2 = sym(3);
x(t) = 12*sin(2*Pi/T1*t) + 4*cos(2*Pi/T2*t);

x(t) is the sum of two periodic signals. Because the ratio T1/T2 is a rational number, we know that x(t) is periodic. By inspection, its fundamental period is

T = sym(6)

which we can verify by

simplify(x(t) - x(t+T))
ans = 0

Because x(t) is periodic, its CTFT is a train of scaled Dirac delta functions at the fundamental frequency (1/6 Hz) and its harmonics. In this case, we only get non-zero harmonics at f = +-1/3 and f = +-1/2

X(f) = expand(fourier(x(t),t,2*Pi*f))

X(f) =

Because x(t) is periodic, it has an exponential CFS represenation. The CFS coefficients as a function of n are computed by

syms n integer

C(n) = int(x(t)*exp(-1j*2*Pi*n/T*t),t,0,T)/T

C(n) =

Or, we could have used fourier to compute them just as well.

% C(f) = fourier(x(t)*rectangularPulse(0,T,t),t,2*Pi*f)/T;
% C(n) = C(n/T);

Because the integrand in int has n as a parameter, it doesn't capture the special cases where abs(n) = 2 or abs(n) =3. Clean that up and C(n) is pretty simple

[~,denC] = numden(C(n));

npoles = solve(denC);

for npole = npoles(:).'

C(n) = piecewise(n == npole,limit(C(n),n,npole),C(n));

end

C(n) = simplify(C(n))

C(n) =

The CFS coefficients are exactly the cofficients in the impulse train that defines X(f), a fact that will be taken advantage of below.

Plot the amplitudes of the four CFS coefficients.

nval = -5:5;

figure

stem(nval/T,abs(C(nval)))

ylim([0 7])

xlabel('Hz')

Because we are using the exponential CFS, the amplitudes of C(n) and C(-n) are each 1/2 of the amplitude of the corresponding harmonic. Therefore, the plot tells us that the amplitude of the harmonic at 1/2 Hz is 12 and at 1/3 Hz is 4, which is exactly what we expect based on the definition of x(t).

Suppose we only have available to us a portion of x(t). The next step shows how Fourier analysis can still extract the amplitude information from the signal.

Define a window function that covers the portion of the signal that we have for processing. The triangular window function (for illustration) is parameterized by the number of periods of x(t) that it covers.

syms numperiod
w(t,numperiod) = triangularPulse(0,numperiod*T,t);

Suppose we have 10 periods of x(t)

nperiod = 10;

The CTFT of the window signal is

W(f) = simplify(fourier(w(t,nperiod),t,2*Pi*f));

Because w(t) covers a significant number of periods, its bandwidth in the frequency domain looks small relative to the frequency spacing between harmonics of x(t)

figure

fplot(abs(W(f)),[-0.3 0.3]);

xline( double( 1/2/T),'r', '1/(2T)');

xline( double(-1/2/T),'r','-1/(2T)');

xlabel('Hz')

ylim([0 31])

The peak of W(f) at f = 0 is

W0 = limit(W(f),f,0)
W0 = 30

which is also area under the window.

int(w(t,nperiod),t,0,nperiod*T)
ans = 30

Apply the window to x(t) and compute the CTFT of the result

xw(t) = x(t)*w(t,nperiod);
XW(f) = fourier(xw(t),t,2*Pi*f);

Multiplication in the time domain corresponds to convolution in the frequency domain. Convolution of W(f) with the Dirac delta functions in X(f) correponds to shifting W(f) to the corresponding harmonics and multiplying by the corresponding Dirac coefficients. But those Dirac coefficients are just the CFS coefficients. Therefore, we can just as easily express the CTFT of xw(t) directly as

nval = [-2 2 -3 3];
XWalt(f) = sum(C(nval).*W(f-nval/T));

Show that XW(f) and XWalt(f) are equivalent.

figure

hold on

% fplot was taking too long.

%fplot(abs([XW(f) XWalt(f)]))

fplot(matlabFunction(abs(XW(f))),[-1 1]);

fplot(matlabFunction(abs(XWalt(f))),[-1 1]);

xlim([-1 1]*3/4)

xlabel('Hz')

Because the bandwidth of W(f) is small relative to the harmonic spacing, XW(f) is essentially composed of replicas of W(f) shifted to the harmonic frequencies and scaled by the CFS coefficients. This result suggests that we can recover the amplitudes of C(n) at the harmonic frequencies by simply dividing XW(f) by W0.

figure

fplot(abs(XW(f)/W0))

hold on

stem(nval/T,abs(C(nval)))

xlim([-1 1]*3/4)

ylim([0 7])

xlabel('Hz')

If all we have is the blue curve, we would surmise that the underlying periodic signal is composed of two sinusoids at 1/2 Hz and 1/3 Hz with amplitudes 12 and 4 respectively.

With that framework in continuous-time, now move to discrete-time.

Define functions to evaluate numerically (not stricly necessary, just runs a bit faster than evaluating in symbolic and converting)

xfunc = matlabFunction(x(t));
wfunc = matlabFunction(w(t,numperiod),'Vars',[t numperiod]);

Assume a sampling period of

Ts = 0.01;

The number of samples to cover that many periods is

N = nperiod*double(T)/Ts;

and the associated time samples are

tval = (0:(N-1))*Ts;

Evaluate the x(t) and w(t) at the sample times.

xval = xfunc(tval);
wval = wfunc(tval,nperiod);

Compute the DFT

XWdft= fftshift(fft(xval.*wval));
nfft = numel(XWdft);
fval = ceil(linspace(-nfft/2,nfft/2-1,nfft))/nfft/Ts;

Now, to plot the DFT with the appropriate scaling, we need to divide out the effect of the window function. In discrete-time, without any other scaling on the output of fft, that corresoponds to dividing by the sum of the window samples

figure
stem(fval,abs(XWdft)/sum(wval));
xlim([-0.75 0.75])
xlabel('Hz')

If we had used a rectangular window, then sum(wval) = N and we'd just divide by N.

Overlay with the scaled version of XW(f)

hold on
fplot(abs(XW(f)/W0),'g')

The scaled DFT samples are essentially frequency domain samples of the XW(f)/W0 (in general, this relationship is only approximate where the approximation gets better as Ts gets smaller and N gets larger).

Zoom in to see things a little better

copyobj(gca,figure);

xlim([0.2 0.6])

Because the DFT samples effectively recover XW(f)/W0, we again know the harmonic frequencies and the corresponding amplitudes.

Four DFT samples lay exactly on the harmonic frequencies because of how N was computed to cover exactly 10 periods for the given the value of Ts. In general, we cannot guarantee the data covers an integer number of periods; after all, the period is what we are trying to determine.

Also, the selection of Ts in this example is such that x[n] = x(n*Ts) is periodic in discrete-time.

Suppose we have bit more than 10 periods of samples.

nperiod = 10 + 0.25;
N = round(nperiod*double(T)/Ts);

Going through the same process

tval = (0:(N-1))*Ts;

xval = xfunc(tval);

wval = wfunc(tval,nperiod);

XWdft= fftshift(fft(xval.*wval));

nfft = numel(XWdft);

fval = ceil(linspace(-nfft/2,nfft/2-1,nfft))/nfft/Ts;

figure

stem(fval,abs(XWdft)/sum(wval));

xline(.5,'r')

xline(1/3,'r')

xlabel('Hz')

xlim([0.2 0.6])

ylim([0 7])

shows that the DFT samples are not so "nice," particularly around 1/3 Hz, which is basically halfway between two DFT samples.

Now, we finally get to the original question about zero padding, which we can use to fill in DFT samples in the frequency domain

N = round(nperiod*double(T)/Ts);

tval = (0:(N-1))*Ts;

xval = xfunc(tval);

wval = wfunc(tval,nperiod);

XWdft= fftshift(fft(xval.*wval,2^14)); % use a lot of padding

nfft = numel(XWdft);

fval = ceil(linspace(-nfft/2,nfft/2-1,nfft))/nfft/Ts;

figure

stem(fval,abs(XWdft)/sum(wval));

xline(.5,'r')

xline(1/3,'r')

xlabel('Hz')

xlim([0.2 0.6])

ylim([0 7])

As has been stated repeatedly, zero padding has no impact on the unscaled amplitude spectrum, and we recover the correct scaled amplitude spectrum as long as we divide by sum(wval), not the number of samples in DFT (2^14 in this instance). The zero padding recovers the information we seek, which is that the harmonics are somewhere very near 1/2 Hz and 1/3 Hz, with amplitudes nearly 12 and 4 respectively.

Although it looks like the DFT samples are samples of XW(f), that's not true because XW(f) is based on a 60 second window and the DFT sampes are based on a 60 + 6/4 = 61.5 second window, which has a slightly different CTFT. However, as long as the peak of CTFT of the window (whatever it may be) is at f = 0, the same process applies, i.e., using the peaks of XWdft(n)/sum(wval) to estimate the harmonic frequencies and amplitudes, as long as the DFT is normalized based on the actual window used, as done in this example.

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Answer 4

Matt J 2023-1-26

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https://ww2.mathworks.cn/matlabcentral/answers/1900240-effect-of-zero-padding-on-fft-amplitude#answer_1156495

编辑：Matt J 2023-1-27

在 MATLAB Online 中打开

Here is an example showing that the amplitude of an FFT does not change due to zero-padding. In all cases, the peak amplitude is 5.

x=[1,1,1,1 1];

N=(1:5)*5;

for n=N

plotFcn(x,n); %zero-pad x by 2*n and plot FFT

end

hold off; legend("n="+N)

function plotFcn(x,N)

z=zeros(1,N);

x=[z,x,z]; %zero pad

M=floor(numel(x)/2);

X=abs(fftshift(fft(x)));

plot((-M:M)/numel(X), X,'o'); hold on

end

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Bjorn Gustavsson 2023-1-27

在 MATLAB Online 中打开

@Matt J: Unfortunately you also removed the fft of the orignal signal (x all ones) which only have a non-zero amplitude at DC - which it corresponds to a Dirac-spike at 0 as it should. When zero-padding such a signal one fundamentally change the signal from a constant-valued signal (that in the fft-sense also is periodic over the length of the signal) to a square-wave with the period of the zero-padded signal-length (which obviously have a sinc as its fft.)

@Paul, 1: yeah, my rant now definitely comes raging out of nowhere. 2: Yes given that problem-description a rectangular window is natural, but "on an exam" I'd hope the next question is something along the lines of "explain how the implicit periodic assumption of the fft leads to aliasing of high-frequency components due to end-effects, and how to control that problem with windowing".

...and if I think we might kind of argue not so much about how many angels should be on the head of this pin, but what colour their wings should be?

This might be a script that illustrates the windowing issues a bit:

t = linspace(0,5*pi,512);
f1 = fftfreq(mean(diff(t(1:end-1))),numel(t(1:end-1)));
f0 = fftfreq(mean(diff(t)),numel(t));
figure
subplot(4,2,1)
% Signal that's periodic and its spectral amplitudes
plot(t(1:end-1),sin((2)*t(1:end-1)))  
subplot(4,2,2)
plot(f1,log10(abs(fftshift(fft(sin((2)*t(1:end-1)))))),'.-')  
axis([-5 5 -16 1.2*log10(numel(t))])
subplot(4,2,3)
% Signal that's one sample off being periodic
plot(t(1:end),sin((2)*t(1:end)))                            
subplot(4,2,4)
plot(f0,log10(abs(fftshift(fft(sin(2*t(1:end)))))),'.-')      
axis([-5 5 -5 1.2*log10(numel(t))])
subplot(4,2,5)
% signal a quarter period from being periodic
plot(t(1:end-1),sin((2.5)*t(1:end-1)))                      
subplot(4,2,6)
plot(f1,log10(abs(fftshift(fft(sin((2.5)*t(1:end-1)))))),'.-')
axis([-5 5 -5 1.2*log10(numel(t))])
w1 = tukeywin(numel(t(1:end-1)),0.75);
w2 = chebwin(numel(t(1:end-1)),50);
subplot(4,2,7)
% Windowed versions and their spectral amplitudes
plot(t(1:end-1),sin((2.5)*t(1:end-1)).*w1')
hold on
plot(t(1:end-1),sin((2.5)*t(1:end-1)).*w2')
subplot(4,2,8)
plot(f1,log10(abs(fftshift(fft(w1.'.*sin((2.5)*t(1:end-1)))))),'-') 
hold on
plot(f1,log10(abs(fftshift(fft(w2.'.*sin((2.5)*t(1:end-1)))))),'-')
axis([-5 5 -5 1.2*log10(numel(t))])
function f = fftfreq(dt,n)
% FFTFREQ - fft-frequency-array
%   
% Calling:
%   f = fftfreq(dt,n)
% Input:
%   dt - sampling duration (s), double scalar
%   n  - number of frequencies,  scalar int
% Output:
%   f - frequency array (Hz), dobule array
if rem(n,2) == 0
  f = [0:n/2-1,-n/2:-1] / (dt*n); %   if n is even
else
  f = [0:(n-1)/2,-(n-1)/2:-1] / (dt*n); %   if n is odd
end
end

Should produce a figure something like this:

Paul 2023-1-28

在 MATLAB Online 中打开

@Matt J

To the contrary, I said quite the opposite: ".. the results [of the DFT] are always the Discrete Fourier Series coefficients of the periodic extension of x[n] ..." To be clearer, I should have said "DFS coefficients (to within a scale factor)"

We can see this in this example.

Let the signal x1[n] = [1 1 1 0 0 0] for n = 0:(N1-1), N1 = 6, and zero otherwise

Let the signal x2[n] = [1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0] for n = 0:(N2-1), N2 = 18, and zero otherwise.

Let x1p[n] be the periodic extension of x1[n] with period N1.

Let x2p[n] be the periodic extension of x2[n] with period N2.

It should be clear that x1p[n] == x2p[n]

Consequently, the DFTs of one period x1p[n] and x2p[n] should contain the same information

x1 = [1 1 1 0 0 0]; N1 = 6; % one period of x1

x2 = [x1 x1 x1]; N2 = 18;% one period of x2

X1 = fft(x1); f1 = (0:(N1-1))/N1;

X2 = fft(x2); f2 = (0:(N2-1))/N2;

figure

subplot(211);plot(f1,abs(X1),'-o',f2,abs(X2),'-x');

subplot(212);plot(f1,angle(X1),'-o',f2,angle(X2),'-x');

The ampltidues are identical to within a factor of N2/N1 = 3, and the phases are identical. The DFT of one period of x2[n] is the identical (to within that scale factor of 3) as DFT of one period of x1[n], because the periodic extensions of x1[n] and x2[n] are the same. By "identical" I mean wrt the four non-zero points that are encoding all of the information about the underlying periodic signals.

Paul 2023-1-28

编辑：Paul 2023-1-28

在 MATLAB Online 中打开

@Rahul Kale

Sure, we can do that.

x1 = [1 1 1 0 0 0]; N1 = 6; % one period of x1
x2 = [x1 x1 x1];    N2 = 18;% one period of x2
x3 = [x2 zeros(1,6)]; N3 = N2 + 6; % requested by Rahul
X1 = fft(x1); f1 = (0:(N1-1))/N1;
X2 = fft(x2); f2 = (0:(N2-1))/N2;
X3 = fft(x3); f3 = (0:(N3-1))/N3;

Let's also compute the DTFT of x1[n] and x2[n]. The DTFT of x3[n] is exactly the same as that of x2[n], so we don't need to compute it.

[X1DTFT,f1dtft] = freqz(x1,1,2048,'whole',1);
[X2DTFT,f2dtft] = freqz(x2,1,2048,'whole',1);

First. plot the DFT's, and I'll muliplty the DFT of X1 by 3 to get everything on equal footing.

figure

hold on

plot(f1,abs(3*X1),'-o');

plot(f2,abs(X2),'-x');

plot(f3,abs(X3),'-*');

legend('X1*3','X2','X3')

We see now that the zero-padding on x3 has added more frequency domain samples, which is expected because N3 > N2. Why are those X3 frequency domain samples where they are? Well, they are the samples of the DTFT of x3, which as we stated above is identical to the DTFT of x2. Add the DTFTs of x1 and x2 to the plot.

figure
hold on
plot(f1,abs(3*X1),'-o');
plot(f2,abs(X2),'-x');
plot(f3,abs(X3),'-*');
plot(f1dtft,abs(X1DTFT)*3,f2dtft,abs(X2DTFT));
legend('X1*3','X2','X3','X1DTFT*3','X2DTFT')

Zoom in a bit for a better view

copyobj(gca,figure)

xlim([0 0.4])

legend('X1*3','X2','X3','X1DTFT*3','X2DTFT')

As must be the case, the samples of X1 lie on X1DTFT. Also, as must be the case, the samples of X2 and X3 lie on X2DTFT. As we can see, X2DTFT is sinc-y. The samples of X3 when viewed alone, as in the previous plot, might look kind of funky. But that's only because of where those X3 samples lie on X2DTFT (which is the same as X3DTFT if we were to compute it). If we add add more zero-padding to x3, the resulting X3 samples would look more and more sinc-y, just like the green curve.

Also, as is clear, 3*X1DTFT ~= X2DTFT. However, they are equal to each other at four frequencies, which are, not coincidentally, the frequencies of the non-zero samples of their respective DFTs.

Paul 2023-1-28

编辑：Paul 2023-1-29

在 MATLAB Online 中打开

@Matt J

Responding to this comment ....

Focusing on all but the last sentence, I agree. We can illustrate with a simple example.

syms t w f real

xn = 1:5; n = 0:4;

x(t) = sum(xn.*sinc(t-n));

figure

fplot(x(t),[-3 10])

hold on

stem(n,xn)

x[n] is a five-sample window of the infinite duration signal x(t),

Going to frequency domain

First the CTFT of x(t)

X(w) = simplify(fourier(x(t),t,w)); % CTFT
X(f) = X(2*sym(pi)*f);

Now the DTFT of x[n]

[Xdtft,fdtft] = freqz(xn,1,linspace(-1.5,1.5,500),1);

Now the DFT of x[n]

Xdft = fftshift(fft(xn));
fdft = (-2:2)/5;

Now plot. I'll lift up Xdtft a bit just so we can see what's going on underneath it

figure

hold on

fplot(abs(X(f)),[-1.5 1.5])

plot(fdtft,1.05*abs(Xdtft))

stem(fdft,abs(Xdft))

legend('CTFT','DTFT','DFT')

As expected, because of the specific definition of x(t), the DTFT of x[n] is an exact periodic extension of X(f).

So far, so good. But I don't see how the above relates to anything about inherit periodicity built into the DFT of a finite duration signal.

"So, if the spectrum you are trying to compute is the continuous Fourier transform of a multi-pulse signal, the DFT will not be aware of the presence of multiple pulses unless they are explicitly sampled in x[n]"

Please correct me if I'm wrong, but I think this statement is in regards to using sampling and the DFT to estimate the CTFT of a finite duration, multi-pulse signal (as opposed to an infinite-duration, periodic signal)

So something like this

x(t) = rectangularPulse(-0.5,2.5,t);

x(t) = x(t) + x(t-6) + x(t-12); % finite duration signal of three pulses

figure

hold on

fplot(x(t),[-1, 20]),grid

axis padded

% samples

xn = repmat([1 1 1 0 0 0],1,3); n = 0:17;

stem((0:17),xn);

CTFT of x(t)

X(f) = subs(simplify(fourier(x(t),t,w)),w,2*sym(pi)*f);

figure

fplot(abs(X(f)),[-0.75 0.75])

ylim([0 9])

If we want to get the best estimate of X(f), we'd use xn, which contains samples of all three pulses, and compute the DTFT

figure
hold on
fplot(abs(X(f)),[-0.75 0.75])
[Xdtft,fdtft] = freqz(xn,1,linspace(-0.5,0.5,1001),1);
plot(fdtft,abs(Xdtft))

The DTFT is a pretty good approximation to the central part of the CTFT, i.e., -0.5 <= f <= 0.5. The DTFT is actually the same as what we'd get with fft(xn,1001), i.e., a massively zero-padded DFT of xn, so we wont plot that here.

Now, we know from above that the DFT of xn are frequency domain samples of the DTFT. Add those to the plot

Xdft = fft(xn);
stem((-9:8)/18,abs(fftshift(Xdft)))

As shown, the DFT of xn are frequency domain samples of the DTFT.

But we can get the exact same (to within the factor of 3) DFT information using only the first pulse

Xdft1 = fft(xn(1:6));

plot((-3:2)/6,3*abs(fftshift(Xdft1)),'-x')

legend('CTFT','DTFT','DFT of three pulses','3*DFT of one pulse')

If the objective is to estimate the CTFT of x(t) via a DFT w/o zero-padding, the only additional information we get using all three pulses is the amplitude scaling of the samples at the four frequencies with non-zero amplitude. Does this result align with what you expected from "the DFT [not being aware] of the presence of multiple pulses"? Or should not including those two additional pulses have a different or an additional effect on how the DFT of a single pulse compares to the CTFT of x(t)?

Paul 2023-1-28

Another way to say this is that if c[k] are the exponential CFS coefficients of a periodic signal x(t) with period T, then the CTFT of that periodic signal is

X(w) = 2*pi*sum( c[k]*delta(w - k*w0) ) (w here in rad/sec, non-unitary transform)

where w0 = 2*pi/T and the summation is taken over all integers k.

It's been a good discussion, IMO.

Bjorn Gustavsson 2023-1-30

By taking the weekend off I missed a lot of discussion. Since this is already a rather sprawling stream of fourier-transform-information I will not be ably to catch up and will stand by and read, but first a last comment on @Paul's reply:

"From my perspective, we are not necessarily taking the Fourier transform of a constant-valued function. Matt J has a finite sequence of data: [1, 1, 1, 1, 1]. Before doing any analysis I think about what I'm assuming about the rest of the signal that is unknown. If these five points are a subset of a constant signal, that's one thing. But if these five points are a subset of a rectangular pulse, then that's a different thing. There may be other things that these five points represent. Whatever that is, it guides me to how I would do the analysis, e.g., zero-padding or not, window or not, etc."

That is a somewhat valid point-of-view, but from an Occhamian (Akaike/Bayesian model-selection) view it will be very hard to justifying additional fourier-components outside of the DC one, since that is enough to perfectly explain the available data, extending outside of that is very dodgy.

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Effect of zero padding on FFT amplitude

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采纳的回答

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Effect of zero padding on FFT amplitude

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