## Scaling the FFT and the IFFT

### Rick Rosson (view profile)

on 13 Sep 2011
Latest activity Commented on by Dr. Seis

on 12 Apr 2016

### Rick Rosson (view profile)

What is the correct way to scale results when taking the Fast Fourier Transform (FFT) and/or the Inverse Fast Fourier Transform (IFFT)?

## Products

### Rick Rosson (view profile)

on 13 Sep 2011

Does scaling matter?

In most situations, scaling is really not all that important. The overall shape of the spectrum matters much more than the absolute scale.

What are the conventions?

But if you really are worried about it, there are several different conventions from which you can choose (see definitions below):

1. Scale by dt for the FFT, and by Fs for the IFFT
2. Scale by 1/M for the FFT, and by M for the IFFT
3. Scale by 1 for the FFT, and by 1 for the IFFT
4. Scale by 1/sqrt(M) for the FFT, and by sqrt(M) for the IFFT.
5. and so on.

I generally use either option #1 or option #2 depending on my mood and whether it's raining outside.

Definitions

Here I am assuming that I have a discrete-time signal x represented as an M x N matrix, where M is the number of samples and N is the number of channels.

`   [M,N] = size(x);`

Furthermore, I am assuming that the sampling rate is Fs and that I have defined the time increment as

`   dt = 1/Fs;`

and the frequency increment as:

`   dF = Fs/M;`

What do all these conventions have in common?

All of these conventions have one thing in common: The product of the two scaling factors is always 1. Please note that the ifft function in MATLAB includes a scaling factor of 1/M as part of the computation, so that the overall round-trip scaling is 1/M (as it should be).

HTH.

Rick

Eugene

### Eugene (view profile)

on 15 Feb 2012

If my scaling is 'dt', for 1-Dimension, is it should by dt^2 for 2-D, please ?

Dr. Seis

### Dr. Seis (view profile)

on 16 Feb 2012

Are you sure you are supposed to get 1. This reminds me of another post: http://www.mathworks.co.uk/matlabcentral/answers/25479-sdof-frf-fft-magnitude-discrepancy

Incidentally, you should probably turn this into another question.

omar thamer

### omar thamer (view profile)

on 19 Dec 2013

answer worked in case the reconstruction has the same length with original signal. What if the reconstruction has more length in order to do some prediction, How would you scale the ifft ?

### Dr. Seis (view profile)

on 28 Jan 2012

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`

omar thamer

### omar thamer (view profile)

on 19 Dec 2013

answer worked in case the reconstruction has the same length with original signal. What if the reconstruction has more length in order to do some prediction, How would you scale the ifft ?

Rick Rosson

### Rick Rosson (view profile)

on 5 Nov 2015

In the continuous case, dt and df are infinitesimal quantities, which approach 0 in the limit that defines the Fourier integral. In the MATLAB code, dt and df are variables that take on values that are strictly non-zero (and positive). So the variable dt in the code is not in any way the same thing as the infinitesimal quantity dt in the continuous Fourier integral (and likewise for the variable df versus the infinitesimal df).

Moreover, the fft function is an implementation of the discrete Fourier transform (DFT), not the continuous time Fourier transform (CTFT).

Dr. Seis

### Dr. Seis (view profile)

on 12 Apr 2016

Johnathan, not sure what I am confusing here. I would suggest you look at the answer from "Chani" below in addition to my comment to their answer. Multiplying the output from Matlab's FFT function with "dt" is the only way to approximate the correct amplitudes (if you care about having physically meaningful amplitudes) AND allow for Parseval's theorem to be satisfied.

Just to follow Rick's comment and clarify my notation - when I say "dt" or "df" it is meant as a "delta-t" or "delta-f" (as in a discrete time or frequency increment), respectively.

### masoud Aghaei (view profile)

on 7 Jul 2015

MVDR in multipath environment with high resolution method hi,how to simulate in matlab doa by mvdr in multipath environment and better performance by high resoultion method(esprit-music-wsf-spatial smoothing....)?

on 16 Sep 2015
Edited by Chani

### Chani (view profile)

on 16 Sep 2015

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?

Best Regards Chani

Chani

on 17 Sep 2015

Image Analyst

### Image Analyst (view profile)

on 10 Nov 2015

Shaui's flag on Rick's comment moved here to be a comment:

Fantastic!

Dr. Seis

### Dr. Seis (view profile)

on 10 Apr 2016

Rick, for Option #2, the graph represents spectral density in units of "volts per sample" (not simply "volts"). Until you supply the correct value used to sample the data as a "scale factor", the amplitudes are physically meaningless.

Also, if you do the actual Fourier Transform of that example SINE wave (with limits from 0 to 10) you will find that the amplitude at f=1Hz SHOULD be 5. So really, option 1 satisfies the amplitude issue and Parseval's theorem.

The point I have been trying to make in this post (and several others around "Matlab Answers") is if you are trying to approximate the Fourier Transform for a periodically sampled time series that is finite in length, then there is only one way to scale the output from the FFT function in Matlab - multiply the result by dt.

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