Is the sum of windowed sinusoids equal a single windowed sinusoid?

Question

I am working on a signal processing problem which requires some property to be true, and now I am not even sure if it is true or not. I appreciate help in finding the proof/disproof for the property.

Problem setup

Suppose we have a sum of $K$ sinusoidal signals, where I represent them in the complex domain: $$r[n] = \sum_{k=0}^{K-1} \widetilde{\alpha}_k g[n - \tau_k] \exp[j2\pi \omega n - j2\pi\tau_k]$$

$\widetilde{\alpha}_k$ is the amplitude and phase of the $k^{th}$ sinusoid,

$g[n]$ is some window function in discrete time with $n$ the discrete time value $n \in \{0...N-1\}$.

$\tau_k$ is the time delay for the $k^{th}$ sinusoid.

Required Proof

I need to find out whether or not $r[n]$ can be represented as:

$$r[n] = \tilde{\beta} h[n] \exp[j2\pi\omega n]$$

That is, in words: is the sum of windowed sinusoids with different amplitudes, phases and time delays equal to one sinusoid with some amplitude $\tilde{\beta}$ and arbitrary window function $h[n]$?

Answer 1

No.

Take $\sin(x)$ and $\cos x$, and window $\cos x$ on $[-2 \pi,0]$ using a rectangular window and $\sin x$ on $[0, 2 \pi]$.

There isn't a $h(x)$ such that the construction above is of the form $A h(x) \sin(x+\phi)$ for some non-negative $h(x)$.

Answer 2

$$x(n) = \sum_{k=1}^K A_k e^{i \omega n} w(n-t_k) = e^{i \omega n} (\sum_{k=1}^K A_k w(n-t_k)) = e^{i \omega n} W(n)$$

So yes if you allow the window $W(n) = \sum_{k=1}^K A_k w(n-t_k)$ to be complex valued.

If $w(n)$ is band-limited then $W(n)$ will be band-limited too. But if it is finitely supported, it is only $\epsilon$-approximatively band limited, and $W(n)$ will be $K\epsilon$-approximatively band limited.