How to compute the main lobe width of generalized Hamming windows (i.e. Hann, Hamming, etc.)?

Question

Is there a way to compute the main lobe width of windows of the generalized Hamming window family (i.e. Hann, Hamming, etc.)? By main lobe I mean the first zeros left and right of the center of the Fourier transform of the windows.

I could not find any derivation in any of the standard textbooks. In Oppenheim/Schafer 'Discrete-time signal processing' (2nd Ed., p. 471), only the 'approximate width' of the main lobe is provided without proof.

Just to give you some background: Windows of the generalized Hamming window family are given by the following equation: $w_H(n) = w_R(n) \cdot (\alpha + 2\beta \cdot cos(\Omega_M n))$ (see e.g. here), where $w_R(n)$ is the rectangular window of length $M$, and $\Omega_M = \frac{2\pi}{M}$.

The Fourier transform (FT) of these windows is given by: $W_H(\omega) = \alpha W_R(\omega) + \beta W_R(\omega - \Omega_M) + \beta W_R(\omega + \Omega_M)$. In this equation $W_R(\omega)$ denotes the Fourier transform of the rectangular window.

$W_R(\omega)$ is simply a sinc-function, so $W_H(\omega)$ consists of the sum of three sinc functions with centers at different positions of the frequency axis.

Answer 1

(This used to be a partial answer, but I have edited it to include the requested result at the bottom.)

First of all, you can view a Hamming family window as the sum of a Hann window and a rectangular window: If $W_N$ is the fourier transform of a Hann window, then $W_H(\omega)=(\alpha-2\beta)W_R(\omega)+4\beta W_N(\omega)$. If, for simplicity, we assume unit amplitude ($\beta=\frac{1-\alpha}{2}$), then $x:=4\beta=2-2\alpha$ determines the weighting of the Hann window against the rectangular window ($\alpha-2\beta=1-x$ in that case). In the general case, you just need to factor out the coefficient $\alpha+2\beta$ first and let $x:=\frac{4\beta}{\alpha+2\beta}$.

For example, here is a plot of $W_H$ (green) and its two components for $x=0.5$ (assuming $M=2$, I guess, so that $W_R(\omega)=\frac{\sin \omega}{\omega}$):

Obviously, giving the Hann window more weight by increasing $x$ also increases the main lobe width. However, plotting this relationship reveals an interesting effect:

Since $W_R(2\Omega_M)=W_N(2\Omega_M)=0$, as soon as the Hann window component completely cancels out the first side lobe of the rectangular window component, the main lobe width stays constant at $4\Omega_M$. By calculating the derivatives at this point, it is easy to find out that this is the case for $x\geq\frac{6}{7}$ (equivalently, $\alpha\leq\frac{4}{7}$ in the unit amplitude case, or $3\alpha\leq 8\beta$ in the general case).

Unfortunately, from the plot, I haven't successfully guessed anything about the remaining $\frac{6}{7}$. ;-) Hopefully someone else will figure this out, if it is possible at all.

Edit: Don't know why I didn't go for the simplest route first. Maxima (open-source CAS) gives $\Omega_M \sqrt{2\frac{x-2}{x-1}}$ for the remaining part. So, in general, the main lobe width is:

$\begin{cases} 2\Omega_M \sqrt{\frac{\alpha}{\alpha-2\beta}} & \text{if } 8\beta<3\alpha\\ 4\Omega_M & \text{otherwise} \end{cases}$