why should I scale the fft using 1/N?

Question

I am writing a report, and my advisor asked me to explain why I scale the fft by a factor 1/N (where N is the length of the array).

I used to use the scaling convention of multipling the fft by the time increment (dt), this convention was good for me, because it ensures the check of the Parseval theorem. Unfortunately I had discussion with one of my advisors, because, since this convention does not give you the right amplitude, he thinks it is wrong. As I have read online, there are not right convention or wrong convention. If I use the factor 1/N the amplitude is checked, if I use the factor dt then the parseval identity is preserved. I have 2 question now:

1. Why, doing the fft, I cannot have both: amplitude checked and energy checked?
2. I have already demonstrated (in my report) that if I scale the fft by a factor 1/N, I obtain the right amplitude, since the first value of my fft is equal to the time average of my function. Now I would like to show with formulas why this convention gives me the right amplitude? I have already searched online and here on the forum, but I did not find any good answer that explain every passage.

Answer 1

OK, let us go for a 2-point DFT. It should be noted that, depending on the software used, scaling can be different, and ought to be checked. The standard unscaled version does multiply the input vector by:

$$\left[\begin{matrix} 1 & 1\\1 & -1\end{matrix}\right]$$

We have two more options: preserve the average, thus we need:

$$\frac{1}{{2}}\left[\begin{matrix} 1 & 1\\1 & -1\end{matrix}\right]$$

as $\frac{1}{{2}}\left[\begin{matrix} 1 & 1\\1 & -1\end{matrix}\right]\left[\begin{matrix} x_1\\x_2\end{matrix}\right]$ has the same average as $\left[\begin{matrix} x_1\\x_2\end{matrix}\right]$, or

$$\frac{1}{\sqrt{2}}\left[\begin{matrix} 1 & 1\\1 & -1\end{matrix}\right]$$

to preserve energy (thus orthogonality) in both domains. More generally, you have three standard options:

• avoid scaling
• scale by $\frac{1}{\sqrt{N}}$
• or by $\frac{1}{{N}}$,

depending on the invariance purpose. But you can scale columns with other quantities, as needed for your computational purpose. The following code aims at showing that the $1/\sqrt{N}$ scaling could work (if I am not too tired) to preserve Parseval-Plancherel or Rayleigh's energy theorem.

nSample = 2^1 ; % One can change the 2^* to other powers
data = randn(nSample,1) ; 
dataFFT = fft(data)/sqrt(nSample); 
ratioParseval = (norm(data)-norm(dataFFT))/norm(data)

Answer 2

Your $dt$ has an implicit $1/N$ in it:

$$ dt \frac{time}{sample} = \frac{ T_{DFT} }{ N } \cdot \frac{ \frac{time}{frame} }{ \frac{samples}{frame} } $$

That's why it works.

My preference is strongly to use a $1/N$ normalization. The principal reason is that it makes the magnitudes of the bin values of pure whole integer tones independent of how many sample points you chose to fill a given duration.

It also turns the DFT into an average instead of a sum. See this: DFT Graphical Interpretation: Centroids of Weighted Roots of Unity

Technically there is no energy in the DFT itself. That's a usage. The sum of squares (energy) is preserved across the transform when a $1/\sqrt{N}$ normalization is used.

Not normalizating can be more efficient to calculate, which is why it makes sense that most FFT library functions don't normalize. It's just a rescaling, after all.

Answer 3

FFT is a fast way to compute DFT. Hence the scale factor $1/N$ belongs to the DFT (specifically the inverse DFT in MATLAB ifft() function).

As Marcus has already pointed out; it's arbitrary to put the scale factor either into the forward or to the inverse DFT.

However, the concept of energy equivalence in time and frequency domains (i.e., norm be preserved by the transform) requires that the scale factor be symmetrically distributed into both forward and inverse transforms. i.e;

$$ X[k] = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x[n] e^{-j\frac{2\pi}{N} n k} \tag{1} $$

$$ x[n] = {\sqrt{N}} \sum_{k=0}^{N-1} X[k] e^{j\frac{2\pi}{N} n k} \tag{2} $$

otherwise you won't have a unitary transform.

MATLAB utilization of the fft() and ifft() funtions to obtain the symmetric DFT pairs (Parseval checks) will be the following:

N = 16;               % sequence length
x = randn(1,N);       % time-domain signal

X = sqrt(1/N)*fft(x,N);  % forward DFT 
xi = sqrt(N)*ifft(X,N);  % inverse DFT

% Check Parseval. 
sum(x.^2)           % Energy in time domain
sum(abs(X).^2)      % Energy in freq domain

Note that a linear transform (linear mapping) is shown like:

$$y = A x \tag{3}$$ where $x$ and $y$ are $N \times 1 $ vectors and $A$ is an $N \times N$ transformation matrix.

The energy of the transformed vector $y$ can be given as :

$$ \mathcal{E_y} = ||y||^2 = y^H y = (Ax)^H (Ax) = x^H(A^H A) x \tag{4}$$

If we desire a transform which would preserve energy in both domains; i.e., $\mathcal{E_y} = \mathcal{E_x}$, then we look for the equality

$$ ||y||^2 = ||x||^2 \tag{5} $$

which implies from Eq.4 that we have :

$$ A^H A = I \tag{6}$$

In other words the linear transformation matrix $A$ has the property that $$A^{-1} = A^H \tag{7}$$ Such matrices are called unitary matrices (aka orthonormal) and such transforms are called unitary transforms.

For DFT being a unitary transform, you need to have the symmetric scaling as in Eqs 1 and 2. Note that if you use the asymmetric scaling, then you will still have an orthogonal transform, but not unitary (orthonormal).