In this related post:


I have demonstrated the similarity of FIR filters and their related frequency response to antenna arrays and their beam patterns (replace time with space).

I also included a link describing other “super-resolution” techniques that can be used with phased array beam-forming antennas to achieve even tighter nulls in a spatial direction.

With that in mind, my question is how an actual super-resolution technique could be used with an FIR filter structure to achieve a tighter transition and what the assumed conditions on the source waveform would need to be. I am looking for a specific intuitive example and demonstration to be included showing the utility of the approach for practical filtering applications and what its limitations/ constraints would be.

  • 1
    $\begingroup$ Glad you posted another question as I was planning on addressing this after looking at the related post and this seems more appropriate. I've been short on time recently but I'll have something soon. Spoiler alert, what some of you might not know is that there is indeed a null-placement-filter-design-like method in antenna design called Schelkunoff synthesis. This is where the FIR design mindset meets antenna theory. Also we're conflating beamforming via the electromagnetics/geometry of the array with signal subspace techniques that achieve "super-resolution". $\endgroup$
    – Envidia
    Commented Feb 24, 2023 at 17:40

2 Answers 2


This answer is a continuation of the discussion in https://dsp.stackexchange.com/a/86776.
As written in the comment, it is only an idea, not fully digested.

Yet, since you ignited it and asked for sharing, I think it should be great to have your input as well.

To sketch the idea, let's assert the assumption: We can generate nulls in FIR form which are sharper than the given inherent resolution (Angle / Frequency in 1D Spatial Signal / 1D Time Signals).

For clarity, I will illustrate the idea on the time domain, but it is equivalent to the spatial domain.

Now, assume we have a sampling grid with a given $ {F}_{S} $ sampling frequency / $ {T}_{S} $ sampling interval with $ N $ samples.

Now, assume we have 2 harmonic signals with $ {f}_{1} $ and $ {f}_{2} $ frequencies. Assume that $ \left| {f}_{1} - {f}_{2} \right| $ is smaller than the resolution of the data. Namely, in the DFT we see only a single lobe.

Let's assume that $ \left| {f}_{1} - {f}_{2} \right| = \frac{{F}_{S}}{4 N} $ and both of them are multiplication of $ \frac{{F}_{S}}{4 N} $. What we can do is create a grid with the resolution $ \frac{{F}_{S}}{4 N} $. Then we can throw nulls on each bin and see the effect on the energy of the data.

Once we hit the the frequencies of the signals we'll have a big impact on the energy.

This is like peeling each frequency on its own, yet using the null allows us doing so without knowing the amplitude or phase, just the frequency.

So it is a grid search operation, yet in 1D instead of 3 parameters (Amplitude, Frequency, Phase).

It works in my mind, I will try to sketch a Proof of Concept using a MATLAB Code (Feel free to edit my question if you beat me to it, I have little free time these days).

  • $\begingroup$ Yes very cool, I see what you are getting at. Scanning a signal zero should have the same response (resolution) since passband ripple and stopband ripple are reciprocal, but perhaps scanning with multiple clustered zeros can lead to a finer resolution. It seems similar though to increasing the number of taps in a filter by using lower order filters in a parallel filter construction. (Or cascade structure in series). With an actual example we can compare to that to see if any advantage is realized or if we are just fooling ourselves into thinking that. Interesting and clever nonetheless. $\endgroup$ Commented Feb 24, 2023 at 20:57
  • $\begingroup$ The ideas are just a sketch. Maybe we can talk about making something real out of this. $\endgroup$
    – Royi
    Commented Feb 25, 2023 at 8:39
  • $\begingroup$ @DanBoschen I would think something like an MVDR criterion might work. Set it to have unit gain at a certain frequency and then minimize the output energy - but this would seem to just be an adaptive filter no? $\endgroup$
    – David
    Commented Feb 28, 2023 at 17:46
  • $\begingroup$ @David Adaptive is ok for the case of a stationary waveform if the resulting solution can be shown to have better resolution than a moving average under condition of white noise. I’m starting to conclude that isn’t possible (see my answer) but If you think you have something, please detail it further as another answer— it would be interesting to see the thoughts $\endgroup$ Commented Feb 28, 2023 at 18:39
  • $\begingroup$ @DanBoschen For white noise, I'm thinking you're probably right. Adaptive filters usually require some additional information via the noise/interference correlation matrix, so probably not your white noise case. $\endgroup$
    – David
    Commented Feb 28, 2023 at 18:58

This isn't the answer I am looking for regarding known super-resolution techniques applied to filter construction, but to demonstrate the fallacy that zeros themselves can be used to provide tighter resolution for passband solutions. Ultimately without utilizing other information, the best noise bandwidth that can be achieved is the reciprocal of the length of the filter, and for white noise, the best SNR achieved is with a unity gain moving average filter. This seem to suggest that any technique that doesn't utilize external information or synthetically increase the time duration (by doing multiple scans) cannot improve the SNR further under this white noise condition and therefore cannot increase the resolution achieved with a moving average filter (which for the resolution of a single tone is proportional to both the length and SNR: increasing either allows us to more precisely estimate the frequency of that tone).

I pursue a simple subtraction technique noting the magnitude of a 2 sample high-pass as $H(z) = 0.5-0.5z^{-1}$ with a low-pass as $1 - |H(z)|$ as plotted below. This was compared to a moving average (mavg) low-pass as $0.5+0.5z^{-1}$ showing in the simplest case the opportunity for tighter resolution (higher Q) if we can do the subtraction properly with preservation of phase. Importantly note how the green curve as $1-|H(z)|$ has a tighter lowpass response over the orange curve representing that as a moving average, contradicting my own intuition that the moving average would be best! This is a fallacy as further detailed below. Before proceeding one may also think that an approach using an absolute value would then be superior, in cases when phase information is not needed. With regards to operating in noise this is not the case as explained in more detail at this post. I only show the possibility of what magnitude response can be achieved if we were to do a phase accurate subtraction.

high Q FIR

To achieve this, the solution would be consistent with the following block diagram where the frequency response of the normalized two-sample high-pass (which by the way is the upper bin of a 2 point DFT, so this analysis applies to the DFT as well). An ideal subtraction would be an all-pass with the matching delay to the two-sample high-pass (yes, this exploration led me to this recently posted "DSP Puzzle"). The two-sample high-pass as $H(z)= 0.5-0.5z^{-1}$ has a zero on the unit circle at $z=1$. The idea is could we utilize the tighter bandwidth in the notch (as demonstrated in the first plot above), to achieve a tighter bandwidth (higher Q) lowpass? And then with that in a general case a higher Q FIR resonator for the construction of higher order filters.

subtraction technique

So I pursued this, noting the 2 sample high-pass has a linear phase with a 1/2 sample delay. We could achieve the half sample delay all-pass easily by decimating a unit sample delay by two (but get noise folding that we would need to deal with). Similarly we could zero-insert upsample the high-pass so that it instead has a 1 sample delay as shown below (then resulting in a comb instead of actual high-pass but still suitable for a simple comparison to a standard comb):

first attempt

This would almost work, but includes a quadrature offset as we see when reviewing the magnitude and phase response of the 2-sample upsampled high-pass compared to the unit sample delay:

frequency response

So it might appear, that if we implemented a complex filter for the case of a complex input, we might be able to achieve a higher-Q resonator, as given by the structure below, which is an FIR filter with coefficients $[-0.5, j, .5]$:

complex solution

Surely the frequency response now appears to confirm we have successfully increased the resolution over what could be achieved with an equivalently up-sampled moving average:

comparison to up-sampled moving average

This appears to defy the initial intuition regarding moving averages in that with it's true phase retaining response that it should result in an SNR improvement exceeding that of a moving average. However, this is instead a good demonstration why it is important, when dealing with complex signals and filters, to consider the complete frequency response from DC to the sampling rate, and not just DC to Nyquist (or equivalently why we should consider the complete response from -Nyquist to +Nyquist). When we view the complete frequency response, we see the "reveal" of the apparent DSP magic trick presented above. Notably this solution and approach that utilizes the narrower bandwidth of a notch from a zero on the unit circle cannot be used to improve a passband in an FIR resonator structure with regards to noise rejection under condition of white noise. Here we do get improved noise rejection of the positive frequencies, but this comes at the expense of noise enhancement in the negative frequencies.

complete frequency response

Perhaps it could be considered to utilized this with a Hilbert created analytic signal (which only contains the positive frequency components). The challenge with realizability is that a Hilbert does not pass DC, and takes a lot of complexity in implementation to approach DC, where here is showing what was an attempt to improve the estimate of the mean (DC).

  • $\begingroup$ It has been a long time since I worked on filters :-). Yet indeed my idea was going through the Hilbert transform since otherwise you can't do with time based signals what you can do with spatial signals. $\endgroup$
    – Royi
    Commented Feb 28, 2023 at 14:56
  • $\begingroup$ @Royi you can- it just has the same symmetry relationship between front lobe and back lobe (assuming omni-patterns in each spatial sensor). Still the implementation of the Hilbert as a filter has complexity that would (I suspect) offset any gain. However if you already have the Hilbert for other purposes anyway perhaps there is something that could be done to take advantage of that $\endgroup$ Commented Feb 28, 2023 at 15:27
  • 1
    $\begingroup$ I am talking about super resolution in the signal we measure. In spatial signal processing you get both components of the signal which you don't for 1D signals. This is why you can't work with the phase easily. This is the magic the methods employ to get super resolution in the context of spatial signals, the angle of view can be narrow by optimized summation. You can't do that in real domain only. I think at least. $\endgroup$
    – Royi
    Commented Feb 28, 2023 at 15:35

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