Let there be a signal f and its low pass filtered signal be fL. Then what can we say about the spectrum of f/fL ?

To be specific I am obtaining fL simply using gaussian blurring in the spatial domain.

If it helps in giving some intuitive explanation, f can be assumed to have large low frequency components and small high frequency components.

Although for me f is a 2D natural image but for simplicity we can take f to be a 1D signal.

From some MATLAB demos it appears that f/fL boosts the high frequency components and dampens the lower frequencies. But cannot find any mathematical or intuitive explanation for this.

  • $\begingroup$ One would assume that the magnitude in passband will be nearly the same, so the quotient should be roughly one. The magnitude in the stopband will be for the filtered signals way smaller than the unfiltered data, so quotient of the magnitudes will go up for those frequencies. As your simulation also shows $\endgroup$ Nov 18, 2022 at 7:52

1 Answer 1


Assuming you mean time (or space) domain division: this is an ill defined problem unless you put some specific constraints on the signal itself. Most signals have zeros and/or very small values and inverting them will create very large values or just "divide by zero" error.

If you the original and the low pass filtered signals are positive than at least the quotient reasonably well defined and the math doesn't blow up. Beyond that it's hard to make a generic statement.

You can make a hand-wavy argument that for low frequencies we have $f \approx f_L$ and the quotient is close to 1. For high frequencies $f \gg f_L$ and hence the quotient would be large.

However division is a highly non-linear operation so this reasoning will only hold if the means of the signals are sufficiently large as compared to the rest, i.e. and you can roughly approximate the inversion around the mean as $\frac{1}{1+x} \approx 1-x$.

If you have signal where $f_L$ gets close to zero, the inverse of that will dominate the entire spectrum making it more or less white.


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