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Scaling the FFT and the IFFT
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What is the correct way to scale results when taking the Fast Fourier Transform (FFT) and/or the Inverse Fast Fourier Transform (IFFT)?
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Rick Rosson
2011-9-13
5 个评论
Jaimin Jha
2017-4-9
I am using ifft for my problem. The thing with it is, if I multiply my ifft results by number of sampling points, I get nearby i.e. expected results. is it right to do so? Also, is it necessary that my sampling points should be of the form of 2^N.
Raunaq Nayar
2020-2-20
What about the case where I perform scaling by factor(x) in time domain perform FFT followed by IFFT and then scale back by factor (1/x) to recover signal. Do I need to take care of the value of x then? I am asking this because in this case we will be scaling up and scaling down both in time domain, so I think for this specific case the Parseval's Theorum should hold true irrespective of the value of x.
In other words, I am following the following sequence of steps-
Scaling by x --> FFT --> IFFT -->Scaling by 1/x
I would also like to confirm the same for scaling by x in frequency domain followed by IFFT then followed by FFT and then scaling by 1/x in frequency domain itself.
In other words, I am following the following sequence of steps-
Scaling by x --> IFFT --> FFT -->Scaling by 1/x
Can x assume any value to recover the signal for the mentioned cases?
更多回答（8 个）
Dr. Seis
2012-1-28
You are right about scaling being unimportant if only the shape of the spectrum is desired. However, if it is necessary that the amplitudes in the frequency spectrum be correct, then there is only one option for scaling - your option #1. In order for Parseval's theorem to hold, the energy in the time domain must equal the energy in the frequency domain. The example below demonstrates this:
> N = 8;
> dt = 0.1;
> df = 1/(dt*N)
df =
1.25
> a = randn(N,1)
a =
0.70154
-2.0518
-0.35385
-0.82359
-1.5771
0.50797
0.28198
0.03348
> b = fft(a)*dt
b =
-0.32813
0.10746 + 0.30519i
-0.080365 + 0.075374i
0.34826 + 0.17802i
0.13866
0.34826 - 0.17802i
-0.080365 - 0.075374i
0.10746 - 0.30519i
> energy_a = sum(a.*conj(a) * dt) % Not necessary to use conj here
energy_a =
0.83314
> energy_b = sum(b.*conj(b) * df) % Necessary to use conj here
energy_b =
0.83314
11 个评论
Jack
2019-9-28
编辑：Jack
2019-9-28
Thank you for this clear answer Dr.Seis, it should be moved to the top since it has the most upvotes, above the unhelpful, flippant accepted answer.
Rick's response does not address a main reason for this question: the common situation where different component signals are processed separately and added together at the end in the frequency domain, hence the need to ensure that they are of the same scale.
Raunaq Nayar
2020-2-20
What about the case where I perform scaling by factor(x) in time domain perform FFT followed by IFFT and then scale back by factor (1/x) to recover signal. Do I need to take care of the value of x then? I am asking this because in this case we will be scaling up and scaling down both in time domain, so I think for this specific case the Parseval's Theorum should hold true irrespective of the value of x.
In other words, I am following the following sequence of steps-
Scaling by x --> FFT --> IFFT -->Scaling by 1/x
I would also like to confirm the same for scaling by x in frequency domain followed by IFFT then followed by FFT and then scaling by 1/x in frequency domain itself.
In other words, I am following the following sequence of steps-
Scaling by x --> IFFT --> FFT -->Scaling by 1/x
Can x assume any value to recover the signal for the mentioned cases?
Jérome
2017-8-1
Hello
I have seen so many subjects related to this issue and so don't know on which to put my answer...I hope this can help some people. I can see there are mainly 3 groups of people, those saying that is not important, those dividing by the number of points of the signal, and those dividing by the sampling frequency. first, for some applications,yes the correct amplitude is important. for example, the magnitude of an earthquake is computed from the amplitude of the spectrum.
then I agree with Dr Seis, the correct way of scaling spectrum is multiplying by dt.
People saying fft has to be divided by the number of points often take the example of the sin wave with amplitude A and want to see 2 peaks with amplitude A/2 on the spectrum. however, this is not the Fourier transform of a continuous sin wave. The Fourier transform has two Diracs. The value of each peak is infinite and the integration over the frequency domain is A/2. So the value of a correctly-scaled discrete spectrum we should have on both peaks is
A/df/2 = Npoints*A/Fs/2
Fs being the sampling frequency, df the step of the frequency vector.
the matlab fft outputs 2 pics of amplitude A*Npoints/2 and so the correct way of scaling the spectrum is multiplying the fft by dt = 1/Fs. Dividing by Npoints highlights A but is not the correct factor to approximate the spectrum of the continuous signal.
The second point is the parseval equation. I have seen many people saying the fft can not respect this relation or is not applicable in discrete mode. first, in discrete mode, if should tends to the continuous value. if can not be Npoints lager or smaller. And, if the fft is multiplied by dt, the energy of the input signal equals the energy of the spectrum.
I have seen quite often people using the Parseval equation for discrete signal like this, which is incorrect
sum(abs(xi).^2) = sum(abs(Xi).^2) with X = fft(x)
The correct discrete form of the Parseval relation is:
sum(abs(xi).^2)*dt = sum(abs(Xi).^2)*df
the relation is satisfied if the fft is multiplied dt and df is correctly defined.
Moreover, there are many simple typical Fourier transforms such as exponential decay, triangle function.. you can model the temporal signals and the known continuous transforms and check that fft*dt is the correct way of approaching the continuous transform.
1 个评论
Alex Hanes
2020-5-14
编辑：Alex Hanes
2020-5-14
In MATLAB, your statement is basically correct (e.g. the correct discrete form of Parseval's theorem is:
P_avg_TD = sum(abs(x).^2)*dt;
P_avg_FD = sum(abs(X).^2)*df;
provided that the Fourier transform result is appropriately scaled by a factor of dt (your samling period). You can see this by:
where 
and 
so the above equation becomes:
and so the above equation becomes: 
Canceling terms on the left and right, you get the "incorrect" expression you wrote above:
where 
is just defined as the fft result (
). For reference, Numerical Recipes uses the "incorrect" definition as the discrete form of Parseval's theorem, so maybe there is confusion for varying definitions and conventions.
is just defined as the fft result (). For reference, Numerical Recipes uses the "incorrect" definition as the discrete form of Parseval's theorem, so maybe there is confusion for varying definitions and conventions. Here's an example code that shows the correct Fourier transform scaling with comparison to the analytical result for the Fourier transform of a Gaussian. The analytic solution to the integral:
clear; clc;
%% Analytic Function: Gaussian
sigma = 1; % "width" of Gaussian
% Define the time-domain function, |f(t)|^2:
fun_t = @(t) exp(-t.^2/(2*sigma^2)).*conj(exp(-t.^2/(2*sigma^2)));
% Define the analytic Fourier transform of f(t), e.g. |f(w)|^2
fun_w = @(w) (sigma*sqrt(2*pi)).*exp(-(sigma*w).^2/2).*conj(...
(sigma*sqrt(2*pi)).*exp(-(sigma*w).^2/2));
int_t = integral(fun_t,-inf,inf); % P_avg of f(t)
int_w = integral(fun_w,-inf,inf)/(2*pi); % P_avg of f(w)/2pi
%% Numerical Example: Gaussian
% Define time domain for numerical signal
N = 1e7 ; % no. points in domain
domain = [-5000 5000]; % domain to evaluate f(t)
L = domain(1,2) - domain(1,1); % length of domain (e.g. [-5e3,+5e3])
dt = L/N; % step size, dt
t = min(domain):dt:max(domain)-dt; % t_n = t_min + n*dt (n = 0, ..., N-1)
% Define the function in the time-domain
sigma = 1; % "width" of Gaussian
ft_num = exp(-t.^2/(2*sigma^2)); % f(t), numerical
fw_num = fft(ft_num); % f(w), numerical (unscaled)
% Consider Parseval's Theorem (Numerical Recipes, no dt or df factors)
E1_timedomain = sum(abs(ft_num.^2))
E1_freqdomain = sum(abs(fw_num.^2))/N
% Consider Parseval's Theorem (based on "Riemann" integration)
df = 1/(N*dt); % calculate frequency step
fx = fw_num.*dt; % scale fw_num according to dt (redefine as fx)
E2_timedomain = dt*sum(abs(ft_num.^2))
E2_freqdomain = df*sum(abs(fx.^2)) % factor of 2*pi is dropped by
% differences in FT definitions*
I should also point out that the "incorrect" method satisfies Parseval's theorem (e.g. P_avg_TD = P_avg_FD in the first line of code), however, you get the incorrect average power content, as shown by the example code for E1_timedomain versus E2_timedomain.
Chani
2015-9-16
编辑：Chani
2015-9-16
Hi everyone,
I tried both options mentioned above, namely option #1 and option #2 with a simple sine curve. The results of both options confuse me and I am hoping you can help clear things up.
The code I am referring to is the following:
dt = 0.05; %Time step
t = [0:dt:10]; %Time vector
s = sin(2*pi*t); %Signal vector
N = length(s); %Number of data points
f_s = 1/dt; %Sampling frequency
df = f_s/N; %Frequency increment
NFFT = N; %FFT length used as second argument of fft()
y_option1 = fft(s,NFFT); %Compute FFT using sampling interval as scaling factor
y_option1 = y_option1*dt;
y_shiftOption1 = fftshift(y_option1);
y_option2 = fft(s); %Compute FFT using signal length as scaling factor
y_option2 = y_option2/N;
y_shiftOption2 = fftshift(y_option2);
if mod(N,2) == 0 %N is even
f = -f_s/2 : df : f_s/2-df;
else %N is odd
f = [sort(-1*(df:df:f_s/2)) (0:df:f_s/2)];
end
Plotting the sine signal and both result vectors y_shiftOption1 and y_shiftOption2 leads to the following Figure:
From my point of view the result of option #2 exhibits the correct amplitude of 0.5V, since the time signal has an amplitude of 1V and this value is split into the two impulses on the negative and the positive frequency scale.
However, regarding Parseval's Theorem, the energy is only preserved using option #1:
sum(abs(s).^2)*dt = 5
sum(abs(y_shiftOption1).^2).*df = 5
sum(abs(y_shiftOption2).^2).*df = 0.0495
Does this mean, if I am interested in the real amplitude of the signal, I have to use option #2 and if I want to preserve the energy of the signal option #1 has to applied?
Thanks in advance for your help!
Best Regards Chani
8 个评论
Kenny
2021-5-14
Hi there Matthias. I think I found my answer for the scaling ----- which at that time I didn't understand what it was for. But it looks like that if scaling is done for either or BOTH the FFT and the IFFT, then the scaling factors applied to both FFT and IFFT need to have their product become equal to 1/N. So if no scaling is applied to the IFFT, while the FFT has a 1/N, scaling factor ..... then that's ok. And if the IFFT and the FFT are both scaled with the same factor of 1/sqrt(N), then that's ok too. There is some discussion about it at this link here ..... https://www.dsprelated.com/thread/4705/fft-ifft-scaling-revisited Thanks Matthias.
Rahul Kale
2023-1-25
Hello,
If I need to compare two signals with different sampling rate, total duration and number of samples, what should be the absolute criterion for comparison? whether amplitude(option#2 above) or energy(option#1 above) is to be used? Practicing engineers often think in terms of amplitude but is that correct mathematically?
Thanks
RSK
Bruno Luong
2018-9-21
编辑：Bruno Luong
2018-9-21
IMHO, if the input signal is x(t) is real, only about half of X(f) = fft(x)(f) is relevant (index 1:ceil((N+1)/2))) the other half is just conjugate of the flip of the first part, so the Parseval would be matching the integration of X(f)^2 on [0,1/(2*dt)] (up to Nyquist frequency) and one must multiply by an extra factor of abot sqrt(2) on top of dt on fft(x(t_i)) to get the Parseval equality up to Nyquist frequency (and not beyond that).
Some subtle consideration: if N is even, the frequency ceil((N-1)/2)*df count only once so it should not be multiplied by sqrt(2). The same for static term (f=0) for all the cases.
A factor of sqrt(2): it can make a bridge resists or fallen down.
0 个评论
Michele Marconi
2017-10-24
How is it possible that the fft computed with matlab (no particular scaling!) and the FFT obtained by utilizing Xilinx module for SystemGenerator, have a 2000x factor scaling difference? i.e. Xilinx one is 2000 times lower amplitude?
(I thought this topic was spot-on regarding amplitude issues, sorry for gravedigging!)
1 个评论
Hossam Fadeel
2019-5-23
I have a situation where the values of the spectrum from the MATLAB FFT is double the values of the Xilinx FPGA FFT for the same input data.
I used below MATLAB Code,
clc;
clear all;
close all;
NFFT=2048; %NFFT-point DFT
fileID = fopen('Sine2048_test.txt','r');
A_Data = fscanf(fileID, '%d');% 2048 Sample points
%% Generate Sine with different freucies frpom the samples
mem_len = length(A_Data)
Sine0=[];% Base Frequency
Sine1=[];% Required Frequency
index=1;
N=32;% represend the frequency multiplier
for i=1:mem_len
Sine0 = [Sine0 A_Data(i)]; % get sin value, in range 0.0-1.0
if (index >= mem_len) %|| (index > mem_len)
index =1;
else
Sine1 = [Sine1 A_Data(index)];
index = index +N;
end
end
figure;
subplot(2,2,1);
plot(Sine0)
title('Original Sine Wave 1 Hz')
subplot(2,2,2);
plot(Sine1)
title('Sine Wave with Higher Frequency (5 Hz)')
%%
X_Sine0=fftshift(fft(Sine0,NFFT)); %compute DFT using FFT
fVals=(-NFFT/2:NFFT/2-1)/NFFT; %DFT Sample points
subplot(2,2,3);
% plot(fVals,abs(X_Sine0));
plot(fVals,abs(X_Sine0));
title('Double Sided FFT - with FFTShift');
xlabel('Normalized Frequency')
ylabel('DFT Values');
X_Sine1=fftshift(fft(Sine1,NFFT)); %compute DFT using FFT
fVals=(-NFFT/2:NFFT/2-1)/NFFT; %DFT Sample points
subplot(2,2,4);
plot(fVals,abs(X_Sine1));
title('Double Sided FFT - with FFTShift');
xlabel('Normalized Frequency')
ylabel('DFT Values');
%%
figure;
subplot(2,1,1);
plot(Sine1)
title('Sine Wave with Higher Frequency (5 Hz)')
X_Sine1=fftshift(fft(Sine1,NFFT)); %compute DFT using FFT
fVals=(-NFFT/2:NFFT/2-1)/NFFT; %DFT Sample points
subplot(2,1,2);
plot(fVals,abs(X_Sine1));
title('Double Sided FFT - with FFTShift');
xlabel('Normalized Frequency')
ylabel('DFT Values');
%%
X_Sine1=fft(Sine1,NFFT)/NFFT; %compute DFT using FFT --(/NFFT) scale factor for FFT
%fVals=(-NFFT/2:NFFT/2-1)/(NFFT); %DFT Sample points
FFT_imag = imag(X_Sine1)';
FFT_real = real(X_Sine1)';
figure;
plot(abs(FFT_imag));
figure;
plot(abs(FFT_real));
figure;
plot(abs(X_Sine1));
title('Double Sided FFT - with FFTShift');
xlabel('Normalized Frequency')
ylabel('DFT Values');
===========
The output of MATLAB - Last figure
The amplitude is 8000
The output of the Xilinx FPGA is:
The amplitude is 4000
Any explaination for this will be appreciated.
RSK
2018-4-19
Dr. Seis,
I think here and at other places where you have mentioned the word "amplitude", you meant the amplitude of the resulting transform after multiplying by dt, correct? (and that is not going to be the amplitude of the sine wave at that frequency and having that transform). However, if people are interested in the amplitude of sine wave that generated the transform, should option#2 in Chani's example be used(although that is not the real amplitude of the transform)? If I have something, say, machine bed(for the sake of example) vibrating at 1hz with unit magnitude and I am presented with the transform generated by "option#1" as the measured vibration, I should certainly not conclude that the machine bed is vibrating with the amplitude shown by "option#1". So I guess dividing by N (option#2) would tell me the magnitude with which the machine bed is vibrating. Is this understanding correct? Thanks
0 个评论
Frantz Bouchereau
2021-7-29
Here is a popular MATLAB doc page that explains FFT scaling and the relationship between FFT and true power spectra: Power Spectral Density Estimates Using FFT.
1 个评论
NSmith
2022-10-25
编辑：NSmith
2022-10-25
I found this link you mentioned on my first time through and I have read carefully but I don't agree with it. It also uses a functions that are not generally available (pow2db). I have been using FFT for quite some time for the purpose of detecting relative energy of frequency content but I have never bothered to dive into scaling it until now because I need to measure how specific distortion of my waveform impacts neighboring bands above and below the fundamental frequency. I started in with reading many threads like this and the Matlab helps then decided to prove that I was achieving the expected result with this approch. That didn't work so I tried all the approaches discussed in my code below which was also not satisfactory.
Here is my process. 1) Define time domain sample size and generate a waveform that repesents a measurable voltage visible into a 50ohm load. 2) Take the FFT of said waveform. 3) Scale the FFT 4) Integrate the scaled FFT to produce total power equal to the power of the input waveform.
I specifically selected a sine wave of magnitude 10V so that in a 50ohm system it represents 1W. P = (Vpk/sqrt(2))^2 / 50 = 1W. Thus I expect the integral of my correctly scaled FFT to equal 1.0W or peaks of roughly half that at +/-10Hz.
One important difference in my example is that I am not letting the length of the signal vector define the FFT size. I have set them equal in a previous test and that did not work either though. The following example produces the following five integration values for the associated scaling.
1) (1/N) * abs(FFT.^2)/50 ==> 100.14
2 (1/N) * abs(FFT) ==> 3.35
3) (1/sqrt(N)) * abs(FFT) ==> 335.1
4) (1/(N*fs)) * abs(FFT) ==> 0.034
5) (1/fs) * abs(xF) ==> 33.51
%% Timebaseand waveform setup
Ts = 1e-3; % Sampling Period
fs = 1/Ts; % Sampling Frequency
t = 0:Ts:1; % Time Vector
xT = 10*sin(2*pi*10*t+1); % Waveform (consider it Voltageon a 50ohm load)
%% FFT of Timebase
xF = fftshift(fft(xT,1e4)); % FFT of Voltage waveform
N = length(xF); % Length of FFT
f = linspace(-fs/2,fs/2,N); % frequency vector for x axis of specrtal plot
fsi = fs/N; % Frequency Spacing increment
%% Rescaling of FFT to represent real power
xF1 = (1/(N)) * abs(xF.^2)*1/50; % FFT scaling attempt 1 Power conversion assuming units of V. Also
xF2 = (1/N) * abs(xF); % FFT scaling attempt 2 Scaled by FFt length
xF3 = (1/(sqrt(N))) * abs(xF); % FFT scaling attempt 3 Scaled by sqrt of fft length (Pareseval's theorum)
xF4 = (1/(N*fs)) * abs(xF); % FFT scaling attempt 4 Scaled by sampling frequency*fft length
xF5 = (1/fs) * abs(xF); % FFT scaling attempt 5 Scaled by time sampling frequency
%% Verification of scaling options
IxT = xT./50; % xT is a voltage waveform measured into a 50ohm load. IxT represents the current waveform.
iPxT = IxT.*xT; % iPxT is the instantaneous power waveform iPower = V * I
aPxT = mean(iPxT) % Numerical calculation for averave of instantaneous power
Power_xT = (max(xT)/sqrt(2))^2/50 % Numerical calculation of RMS Power = Vrms^2 / R
AUC1 = fsi*trapz(xF1) % Area under the curve (integration) of xF1. Should equal aPxT & Power_xT
AUC2 = fsi*trapz(xF2) % Area under the curve (integration) of xF2. Should equal aPxT & Power_xT
AUC3 = fsi*trapz(xF3) % Area under the curve (integration) of xF3. Should equal aPxT & Power_xT
AUC4 = fsi*trapz(xF4) % Area under the curve (integration) of xF4. Should equal aPxT & Power_xT
AUC5 = fsi*trapz(xF5) % Area under the curve (integration) of xF5. Should equal aPxT & Power_xT
%% Figures
figure(1)
subplot(311),plot(t,xT,t,IxT),legend('Voltage Waveform','Current Waveform'),grid
subplot(312),plot(t,iPxT,'m'),legend(['Instantaneous Power Waveform']),grid
subplot(3,5,11),plot(f,xF1,'g'),axis([-20, 20,0,max(xF1)*1.05]),title('1/(N) * fft^2/50'),grid
subplot(3,5,12),plot(f,xF2,'g'),axis([-20, 20,0,max(xF2)*1.05]),title('1/(N) * fft'),grid
subplot(3,5,13),plot(f,xF3,'g'),axis([-20, 20,0,max(xF3)*1.05]),title('1/(sqrt(N)) * fft'),grid
subplot(3,5,14),plot(f,xF4,'g'),axis([-20, 20,0,max(xF4)*1.05]),title('1/(Fs*N) * fft'),grid
subplot(3,5,15),plot(f,xF4,'g'),axis([-20, 20,0,max(xF4)*1.05]),title('1/(fs) * fft'),grid
Abu Taher
2021-8-13
Perform IFFT of the multiplied result of two FFT sequences obtained from x and
0 个评论
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