DFT normalization for amplitude estimation
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.