If I take two analog, band limited signals $ x(t) $ and $ y(t) $ with non-zero $y(t)$ I can define the division: $$ z(t) = \frac{x(t)}{y(t)} $$

In general $z(t)$ may have a very wide bandwidth, also if the two signals had a small bandwidth.

I ideally may lowpass filter out a limited band part of $z(t)$ and make a discrete version $ z[n] $ without aliasing.

If I instead make division from the discrete versions $ x[n] $ and $ y[n] $ (they are band limited so I can also have a sampled discrete version that completely describe them) the result can be very different from the other discrete version $ z[n] $ after the lowpass antialias filter, in fact I have aliasing.

Is there a way to estimate in digital domain the $ z[n] $ that I would obtain making the division in analog domain and filtering out the lowpass part?

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    $\begingroup$ as you've noticed, $z$ can be quite problematic. The harsh truth is that you very rarely find the ratio of two signals actually be calculated in signal processing, because that's a very unstable operation ($z$ is undefined whenever $y(t)=0$, and it is extremely sensitive to noise when $y$ is small). So, I doubt your low-pass filtered $z$ makes any sense in practice – can you elaborate where you meet that problem? maybe from that context, conditions arise that make answering your question easier. $\endgroup$ Commented Jan 19, 2021 at 14:24
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    $\begingroup$ Your $z$ might not even have defined energy/mean. For example, if $x$ and $y$ are filtered noise, their values will follow a Gaussian distribution, so the ratio would follow a Cauchy distribution, which doesn't have mean/variance: en.wikipedia.org/wiki/Cauchy_distribution#Properties $\endgroup$
    – cloudfeet
    Commented Jan 19, 2021 at 14:46
  • $\begingroup$ Marcus Muller, I know this, but I often found this situation, for example from data derived from sensors that phisically should be divided, I often see aliasing in these operations so I'm looking for a quite general solution for these situations. We can also neglect noise for a theoretical answer. $\endgroup$
    – Andrea
    Commented Jan 19, 2021 at 14:58
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    $\begingroup$ @Andrea also, I've never (literally, never) seen an analog division; that's something that is pretty hard to implement (Probably, taking the logarithms of the absolute input signals, then subtracting then, then an exponential function, then if necessary sign reversal). It's really not too much of a useful operation, overall, and I'd be suprised if it makes all too much sense to model that directly. The cases where we do see operations that turn out to be divisions, it's usually control loops controlling the product of two things, and that's something you'd model differently altogether. $\endgroup$ Commented Jan 19, 2021 at 16:09
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    $\begingroup$ //with analog I means the mathematics and ideal signal $\mathbb{R} \to \mathbb{R}$// ............. Andrea, Analog Devices (same company that makes the SHArC) used to do analog signal process (probably still do, but now it's a small part of the company) and had these multiplier chips that you can put in the feedback loop of an op amp and make a divider out of it. But they hit the rails with $y(t)=0$. $\endgroup$ Commented Jan 21, 2021 at 5:04

1 Answer 1


I've found an estimator that can work for a lot of similar situations, the idea is that you can rewrite the division: $$ x(t) = z(t) \cdot y(t) \tag1$$ The problem now is to estimate $ z(t) $ that solves the equation. One can start to find a constant coefficient that minimize the 2-norm. This coefficient will be the linear regression slope without intercept term ($\beta$), hence you can write an approximation of $ z(t) $ that minimize the 2-norm: $$ x(t) \approx \beta \cdot y(t)\tag2 $$ This estimator is constant, hence it is band limited, and you can estimate the same $ \beta $ from the sampled data, in fact be valid that $$ x[n] \approx \beta \cdot y[n]\tag3 $$

Obviously this estimation is quite useless but you can extend this approach in a kind of FIR way , first. You could make a rectangular window and estimate the linear regression slope on $N$ elements.

$$ x[n] \approx \beta(x[(n-N),..,N], y[(n-N),..,n]) \cdot y[n] \tag4$$

I don't like a lot this estimator but should limit aliasing if $ N $ is large enough.

I instead prefer an IIR like approach, using an exponential window. You can do this using a 0-order RLS algorithm. You can choose a bandwidth choosing the forgetting factor $ \lambda = e^{-\frac{1}{W_s}}$, where $ W_s $ is the width in samples of the exponential window, that can be expressed in function of cut off pulsation ($\omega_c$) using the relation

$$ W_s = \frac{f_s}{\omega_c} \tag 5$$ where $f_s$ is the sampling frequency.

With RLS you can estimate the slope $\beta$ in a time-variant way, with a limited band in discrete domain, where the bandwidth is limited by $ \omega_c $. Note that in this case (0-order) RLS becomes a scalar problem so it is very easy to implement.

  • $\begingroup$ "but should limit aliasing": your problem is that the error between $\beta$ and $z[n]$ can get arbitrarily large in your equation $(4)$, because the ratio of two band-limited signal can (and will) have singularities at isolated points, and you can't predict where they lie, and how close you get to them with your sampling instants! $\endgroup$ Commented Jan 22, 2021 at 15:09
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    $\begingroup$ Yes, but the fact is that when someone divides signals in reality it's probable that he would like to estimate the signal that solves the product equation. And if there is a phisical meaning this is the correct solution. $\endgroup$
    – Andrea
    Commented Jan 22, 2021 at 21:26
  • $\begingroup$ Then it's true that a band limited signal that make division can have an arbitrary large error, but it's inevitable, and in discrete domain you can't do better $\endgroup$
    – Andrea
    Commented Jan 22, 2021 at 21:33
  • $\begingroup$ Good news! You can be better, namely by specifying the metric for which you optimize – it's very literally the difference between a zero-forcing frequency domain equalizer and an MMSE equalizer :) That's why I was pestering you so much about what you use this for, what this signal is: your approach on its own has little utility. Your approach with a good metric is what we do, but the metric depends on the application. $\endgroup$ Commented Jan 22, 2021 at 21:40
  • $\begingroup$ I propose 2-norm because it's the most natural and it's easy to minimize, you can change the norm if you know what you exactly need. For physical data from sensors typically 2-norm works very well $\endgroup$
    – Andrea
    Commented Jan 23, 2021 at 7:00

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