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I'm doing some self-study for an important exam I'll have in late March and came across the following question:

enter image description here

So, using the convolution properties, if I want to find an identity system so that the final output is just the input, I would need one of the following, right?

  1. h(t) * h'(t) = $\delta(t)$

  2. H(w) * H'(w) = 1, so H'(z) = $\frac{1}{H(z)}$

My question is, if we go through option 2, then we are going to have a bunch of 1/0 since the Fourier Transform of the sinc will be a rect, so the system would not be stable.

Also, the sinc function itself is not invertible as we have many points that maps to 0 and the even symmetry of the function.

So does that means that for this particular case it is impossible to get such an identity system?

Thanks for the help!

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    $\begingroup$ can't divide by zero. there is no inverse system. $\endgroup$ – robert bristow-johnson Jan 9 at 2:08
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The operation aimed to restore a source signal from the result of the signal's convolution with a transfer function is called deconvolution. Your question concerns the rather innocuous kind of deconvolution problems, because the transfer function is given as an input. The transfer function being one of the unknowns, the problem is called a blind deconvolution problem.

Deconvolution problems are inherently ill-posed problems, that is, they can have many solutions for a given input, have no solutions at all, or the solutions are unstable -- small variations in input data lead to great changes at the output.

Fortunately, your test question is easily tractable. First, notice that the input to the transfer function hdeconv(t) can be only those signals that are the result of convolution with the sinc(). Therefore, a Fourier transform of these signals is zero at frequencies where sinc's transform is zero, and we can define $$ H_{deconv}(ω) = (H(ω)!=0)?{1\over{H(ω)}}:0 $$ By all means, deconvolution cannot recover the information that convolution completely eliminated from the signal, as is the case of frequencies above the threshold of sinc filter, which in the frequency domain is an ideal brick-wall filter. Therefore, the restoration is not perfect with the x(t) that have the frequencies above the threshold of sinc filter and yes, an (universal) identity system cannot be implemented with the first filter in the chain being a sinc filter.

With the sinc filter at the source, the perfect deconvolution problem has no solutions for high-frequency source signals (recall the ill-posed problem definition). But the signals having only frequencies below the threshold in their spectra enjoy perfect restoration, and the identity system for those signals can be implemented.

Furthermore, with a general transfer function a small noise added to the signal after convolution can lead to great errors in the recovered signal. But, strictly speaking, the test question does not require you to analyze general-case deconvolution.

However, if you do, you can start reading about Wiener deconvolution.

The article on the spectral division technique discusses application of deconvolution in geophysics: http://www.ees.nmt.edu/outside/courses/GEOP505/Docs/deconv.pdf.

Good luck with your exams!

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Your $h(t)$ is an ideal low pass filter with cutoff frequency of $75$ Hertz. At the output, everything above $75$ Hertz will be gone and equal to zero. If you're thinking about the inverse system, you need to think about getting back the original input. What can you say about the original input signal at frequencies greater than $75$ Hertz? Isn't it anybody's guess as to what used to be there?

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