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I am learning DPS and I came across the problem of deconvolution and removing the impulse response from a signal? This still does little sense to me.

My understanding of the impulse response is to find out the response, usually in frequency, of a device, a filter, a room etc. If I don't mess things here, to me, the impulse response in signal processing is equivalent to numeric 1. Finding the impulse response is convolving a signal with something that is equivalent to 1 that is the impulse response. So if I need to find what the value of x is, I simply multiply x by 1:

$$y = x\times1$$

Therefore if $y = 0.8$, I know $x = 0.8$ as well.

In DSP, we have the equation

$$Y = XH$$

So to know the impulse response of $X$, I multiply $X$ by $H$ and obtain $Y$ which is equal to $X$. I've gone that far.

I don't understand what deconvolution means. I though by applying the impulse response we have already an answer about the frequency of anything we need. What does deconvolution mean here and why would we look for it?

Thanks.

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A linear time-invariant (LTI) system is completely characterized by its impulse response $h(t)$. The output signal $y(t)$ given an arbitrary input signal $x(t)$ is given by the convolution of this input signal with the system's impulse response:

$$y(t)=(x\star h)(t)\tag{1}$$

where $\star$ denotes convolution. In the frequency domain, Eq. (1) becomes a simple multiplication:

$$Y(f)=X(f)H(f)$$

where $Y(f)$, $X(f)$, and $H(f)$ are the Fourier transforms of the signals $y(t)$, $x(t)$, and $h(t)$, respectively.

Deconvolution now refers to the process of "undoing" the effect of an LTI system. I.e. you're given the signal $y(t)$ and you want to recover the original signal $x(t)$. E.g., you might want to remove the reverberation added to a signal by a given room. In this case you need to undo the effect of the convolution of the room's impulse response with the original signal.

Naive deconvolution can in principle be done by applying an inverse filter to the signal $y(t)$. The inverse filter ideally has a frequency response $G(f)=1/H(f)$. Note that such a filter does not necessarily exist, e.g. if $H(f)$ has zeros at certain frequencies. Another disadvantage of this naive deconvolution is the effect it has on measurement noise. In frequency regions where $H(f)$ has a small magnitude, the magnitude of $G(f)$ becomes large, and the noise is highly amplified. This is why applying an inverse filter is usually not a good solution. If the system's impulse response is known and if the noise can be modeled as (wide-sense) stationary, Wiener deconvolution can be used. In that case, the original signal is estimated using a mean square error criterion, which takes into account the impulse response as well as the noise. If the impulse response is unknown (or can't be estimated), there are methods available referred to as blind deconvolution.

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  • $\begingroup$ You have mentioned reverberation. Is it like a couple of interfered waves of the same origin approached the target (mic) and we want to know what the original wave is? Can I say that if I measure seismic vibrations which may also come directly to apparatus or after reflection, I want to see what's the original wave is and then I have to apply deconvolution? $\endgroup$ – Celdor Mar 16 '15 at 12:02
  • $\begingroup$ Hi. I was thinking about the deconvolution and I have one more question regarding the example of room. Let's say I need to obtain a pure signal from a recording that was done in a room which is very echoic. If I obtain the impulse response of the room by say only recording its explosion or blast which I believe is close to impulse, and if I deconvolve this response from the recording, should I obtain the pure signal? .... or something close depending how close was the impulse response? Am I thinking correctly? Thanks $\endgroup$ – Celdor Mar 18 '15 at 14:42
  • $\begingroup$ @ZikO: Yes, you're right. In practice this is not so easy because of the issues mentioned in my answer (noise, non-existence of exact inverse, etc.), but in principle it can be done, at least approximately. $\endgroup$ – Matt L. Mar 18 '15 at 15:37
  • $\begingroup$ Thanks for all this and pointing out different realizations of deconvolution :) $\endgroup$ – Celdor Mar 19 '15 at 9:05

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