DFT normalization for amplitude estimation After an interesting recent answer, I'm doing some research on proper DFT normalization for sinusoidal peak estimation. It's clear that to get the correct amplitude from DFT peaks we need to normalize by the window gain
\begin{equation}
S_w \triangleq \sum_{n=0}^{N-1} w[k]
\end{equation}
Which becomes $S_w = N$ for the rect window.

Now I'd like to find some rough demonstration, Harris has one (see equations 12 and 13 in the article below), but he chooses the easiest path by making exponentials cancel out. Well he actually does a pretty good job, it's me that thought it could be inferred to some more general relation.

For the rectangular window I've come up with this.
Let's say we have a sampled analytic sinusoidal signal $x[n]$, with frequency of $m/N$ cycles per sample (or $m$ cycles in $N$ samples), and compute the DFT $X[k]$

\begin{equation}
x[n] = A \ e^{j 2\pi m n/N}
\end{equation}

\begin{align}
X[k] &amp; = A\sum_{n=0}^{N-1}e^{j \frac{2\pi n}{N} m} e^{-j \frac{2\pi
  n}{N} k} \\
\\
&amp; =  A\sum_{n=0}^{N-1}e^{-j \frac{2\pi n}{N} (k-m)}\notag \\
\\
&amp; = A \ \dfrac{1-e^{-j2\pi(k-m)}}{1-e^{-j\frac{2\pi}{N}(k-m)}} \\
\\
&amp; = AN\delta \left[ k-m\right]
\end{align}

where $\delta[k]$ is the Kronecker delta.
To get the correct amplitude in freq space we need to divide by $N$.

Now, let's say we have a windowed signal, let me call it $x[n]$ again

\begin{equation}
x[n] = w[n] \ A \ e^{j 2\pi m n/N}
\end{equation}

and its transform, again overloading notation

\begin{equation}
X[k] = A\sum_{n=0}^{N-1} w[n] e^{-j \frac{2\pi n}{N} (k-m)} 
\end{equation}

From here I'm not sure how to proceed, guess the result should be
$$
\require{cancel}\cancel{X[k] \overset{?}{=} A \ S_w \ \delta[k-m]}
$$

Edit: after some discussion with Robert I believe the previous equation is just wrong. It works for the rectangular window only, while for the other windows the only thing we can say is about the peak $k=m$, which can be easily demonstrated just from the sum.


  G. Heinzel et al - Spectrum and spectral density estimation by the
  DFT, including a comprehensive list of window functions and some new
  flat-top windows
  
  F. J. Harris - On the Use of Windows for Harmonic Analysis with the
  Discrete Fourier
  Transform
okay, since this is about scaling, i should be anal about definitions.

discrete-time signal: $x[n]$  where $n$ are only integer values.

DTFT:  $$ X\left( e^{j\omega} \right) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] \ e^{-j \omega n} $$

it is necessarily the case that $ X\left( e^{j(\omega + 2 \pi)} \right) = X\left( e^{j\omega} \right) $ for all real $\omega$.

DFT:  $$ \begin{align} X[k] &amp; \triangleq \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi \frac{k}{N} n} \\ 
&amp; = X\left( e^{j\omega} \right) \Bigg|_{\omega=2 \pi \frac{k}{N}} \quad \quad \text{if } x[n] = 0 \text{ for } n&lt;0 \text{ or } n \ge N\\ \end{align} $$

it is also necessarily the case that $X[k+N]=X[k]$ for all integer $k$.

and another fact is that $ X\left( e^{j0} \right) = X(1) = \sum\limits_{n=-\infty}^{+\infty} x[n] $

and $ X[0] = \sum\limits_{n=0}^{N-1} x[n] $

if x[n] is a "low-pass" function, we will expect $X[k]$ to peak at $k=0$.  and the value of that peak is shown as $X[0]$ above.

so, now consider a window function $w[n]$ that has DFT defined the same way

$$ W[k] = \sum\limits_{n=0}^{N-1} w[n] \ e^{-j 2 \pi \frac{k}{N} n} $$

normally we think of the window function as always non-negative, $w[n] \ge 0$, but it wouldn't necessarily have to be that.  nonetheless, just like $x[n]$, the window function has a DC component:

$$ W[0] = \sum\limits_{n=0}^{N-1} w[n] \triangleq S_w $$

and the DFT of the window function must also be periodic, $W[k+N] = W[k]$, for all integer $k$.



now there is a "translation property" of all forms of the Fourier Transform that has a particular expression for the DFT:

$$ \begin{align} \text{DFT} \left\{ x[n] e^{j 2 \pi \frac{m}{N}} \right\} &amp; = \sum\limits_{n=0}^{N-1} \left( x[n] e^{j 2 \pi \frac{m}{N}} \right) \ e^{-j 2 \pi \frac{k}{N} n} \\
&amp; = X[k-m] \\ 
\end{align} $$

for $m$ equal to an integer. 

same applies for $w[n]$ with an amplitude scaling factor of $A$.

$$ \begin{align} \text{DFT} \left\{ w[n] \ A e^{j 2 \pi \frac{m}{N}} \right\} &amp; = \sum\limits_{n=0}^{N-1} \left( w[n] \ A e^{j 2 \pi \frac{m}{N}} \right) \ e^{-j 2 \pi \frac{k}{N} n} \\
&amp; = A \ W[k-m] \\ 
\end{align} $$



now we do expect the window function to be a "low-pass" function.  that's how we design them, so we also expect $W[k]$ to peak at $k=0$ (or any other integer multiple of $N$) and we expect the expression above to peak when $k-m=0$ (or $k=m$) and the value of that peak is 

$$ A \ W[0] = A \ S_w $$

so if $A$ is the amplitude we're trying to find, we would take the value of the peak, wherever $k=m$, and divide that value by $S_w$ to get the amplitude of the complex sinusoid.

it's as simple as that, but it is also shown to be too simple.  we know that a real sinusoid is the sum of two complex sinusoid, so there will be overlapping interference (a.k.a. "leakage") of one onto the other.  we also expect that the frequency will not be exactly an integer number of cycles $m$ during the time of $N$ samples.  then the math gets a little more involved, but the scaling issues should, if $N$ is large enough, approach the result we have above.