This answer says that there is a tradeoff between SLL (sidelobe level) and BW (main lobe width). I am able to verify this and I understand that we cannot have our cake and eat it too, but why is there a tradeoff between the two?

My guess is energy conservation, but I am unable to see what the formula would be. Something like $BW * SLL = constant$, perhaps.


1 Answer 1


There are two main mathematical principles to keep in mind when dealing with this tradeoff: the uncertainty principle (specifically the Gabor limit) and the Gibbs phenomenon.

The uncertainty principle states that there is a limit to how well pairs of properties can be simultaneously known. With respect to frequency analysis, the idea is that you cannot precisely localize a function in both time and frequency. Specifically the time-bandwidth product has to satisfy \begin{equation} \sigma_{e,t}\sigma_{e,f} \geq \frac{1}{4\pi} \end{equation} with equality achieved for a Gaussian pulse. Specifically with the Fourier transform, the idea is that the larger the window used for the DFT, the less time resolution you have (you don't know where in that window each frequency occurs), but you gain finer frequency resolution.

For a fixed window length, the narrowest mainlobe in the Fourier domain is achieved by a rectangular window. If you add a taper to the window, the mainlobe broadens. The reason for this is that you are weighting the values at the edge of the window less and less, and primarily taking into account the values near the center of the window. This has the effect of improving your time domain resolution (not a ton, as you are still taking all the values within the window into account to a certain extent), which is why the mainlobe broadens in the Fourier domain, ie you achieve worse resolution in the Fourier domain.

As for the sidelobes, the Gibbs phenomenon describes the oscillatory behavior in the Fourier domain near a jump discontinuity in the time domain. The larger the jump discontinuity, the more oscillatory the Fourier response is. Sidelobes are a direct result of the oscillatory behavior resulting from the Gibbs phenomenon. So, in order to reduce the oscillatory behavior, you have to reduce the jump discontinuity at the ends of the time-domain windows. The way to do this is to taper the response.

So, in order to reduce sidelobes due to the Gibbs phenomenon, we have to taper the time domain function, which improves our time resolution, decreasing our Fourier resolution, broadening the mainlobe.

Hopefully this is a sufficient response even though I didn't provide many equations. It's important to note that while there is a connection between mainlobe width and sidelobe level, they are technically related to different concepts. This is why certain windows seemingly have the ability to drastically lower your sidelobe levels without significantly altering your mainlobe width.

EDIT Why the paper reference shows a narrower mainlobe

I think you may be getting confused on windowing vs. filtering. You can have an arbitrarily narrow filter response, but windowing only allows the tradeoff between mainlobe width and sidelobe level.

To better explain this, the periodogram is a filterbank. Specifically, you are convolving your input signal $x(t)$ with $e^{-j\omega t}$. Windowing is different, you are multiplying. The paper you reference is using an alternate filterbank, which is a pseudo-capon filterbank. The periodogram filter is, as I said, $e^{j\omega t}$, which in vector form is \begin{equation} \underline{h}(\omega) = \underline{a}(\omega) = \begin{bmatrix} 1 & e^{j\omega} & \cdots & e^{j\omega(N-1)} \end{bmatrix} \end{equation} The capon filter, similar to what they use, is \begin{equation} \underline{h}(\omega) = \frac{\mathbf{R}^{-1}\underline{a}(\omega)}{\underline{a}^{H}(\omega)\mathbf{R}^{-1}\underline{a}(\omega)} \end{equation} This filterbank will have a narrower mainlobe than the periodogram filterbank. Here's a picture demonstrating that. Periodogram vs. MVM 3-dB Width

As you can see, the one-sided 3-dB beamwidth of the MVM filterbank is roughly 13 samples narrower than that of the Periodogram, which corresponds to $\approx 4.6^{o}$ for 1024 DFT points. So, it is possible to obtain a narrower mainlobe than the DFT using an alternate filterbank, but for a given filterbank you cannot achieve a narrower mainlobe than the rectangularly windowed version of that filterbank.

  • $\begingroup$ Thanks for the comment. Regarding your remark "For a fixed window length, the narrowest mainlobe in the Fourier domain is achieved by a rectangular window", apparently according to this paper, we can have narrower beamwidth than rectangular taper. I am trying to implement this but got stuck. I tried to raise a discussion here and it seems the paper is sloppy. $\endgroup$ Commented May 6 at 10:42
  • $\begingroup$ @RajaKrishnappa I’m honestly not seeing in the paper where they say that. I guess from a theory standpoint you could have a triangular window that increases out from the origin, but that seems silly. If you’re asking how the paper got their result, it looks like a pseudo capon variant. Capon is known to have tight mainlobes and low sidelobes. I’m not really sure where your confusion lies, though. $\endgroup$
    – Baddioes
    Commented May 6 at 13:52
  • $\begingroup$ I am clear on your answer to this question. Section III (Simulation analysis) shows the results of their algo. Looking at figures 2, 3 and 4, we see it has narrower beam than uniform taper. $\endgroup$ Commented May 7 at 2:10
  • $\begingroup$ @RajaKrishnappa see my edit. Hopefully that makes sense. $\endgroup$
    – Baddioes
    Commented May 8 at 3:14
  • 1
    $\begingroup$ @RajaKrishnappa filtering in this case does what filtering does normally. It windows the spectrum around a center frequency, and the shape of that window depends on the filter characteristics. "Steering" is a type of filtering, where you filter as narrowband as possible for a given center frequency, or look angle. I'm happy to paste this over there, but as stated my answer doesn't really answer your question over there. It's up to you. $\endgroup$
    – Baddioes
    Commented May 8 at 5:06

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