# Is inverse matched filter useful?

I am aware of matched filter and its application. Now, wondering if there is any application of inverse matched filter? What I mean my inverse matched filter is that convolution of matched filter and the inverse matched filter would lead to close to delta function.

• Regardless of what the answer by MBaz says, the "inverse" matched filter is a dreadful idea because its frequency response is $[X^*(f)]^{-1}$ which has infinite gain wherever the signal spectrum $X(f)$ has a null in it. Signal spectrum has no null, you say? Even then, since $X(f)\to 0$ as $|f|\to\infty$, you cannot really implement it with any degree of precision, let alone apply it usefully. Worse, the "inverse" filter enhances exactly those frequency bands where the signal has least energy and (in comparison) suppresses those bands where the signal has most energy. – Dilip Sarwate Oct 23 '16 at 21:45
• Oh, God! Give it up, Creator' (pun intended) It does not matter diddlysquat whether you are working with digital signals or continuous-time signals, the issues are the same whether you are working with integral or sums. – Dilip Sarwate Oct 25 '16 at 12:28
• @DilipSarwate It appears the idea is applicable to channel estimation, equalizer as well? Is not it? If such concepts exists, one can design signal to have an inverse? Is it wrong? I understand when you mean sum and integral are same. Here we talking about close to delta function. – Creator Oct 25 '16 at 16:31

In your question, you postulate the existence of two filters, let's say $p(t)$ and $g(t)$, such that $p(t) \star g(t)=\delta(t)$. This assumption is problematic because it implies that $G(f)=1/P(f)$, which requires $P(f) \neq 0$ for all $f$. Furthermore, for small values of $P(f)$, $G(f)$ will take arbitrarily large values (see Dilip Sarwate's comments above and below this answer).
Assuming (for the sake of discussion) the existance of such a pair of filters, there may not be any advantage to using them. I'll describe a possible application from digital communications. Let's say you want to transmit a number $A$ over an analog channel. You choose an appropriate analog pulse $p(t)$ and transmit $$s(t) = A\delta(t)\star p(t).$$ (The pulse $p(t)$ could be chosen to fit the channel bandwidth, and/or to have a certain energy). The receiver could perform the "inverse filter" operation on the signal $s(t)$: $$s(t) \star g(t)=A\delta(t),$$ where $g(t)$ is your "inverse matched filter". However, in practice it turns out that detecting $A$ in this way is not optimal, because it does not have the best signal-to-noise ratio. In other words, when you add noise to the transmitted signal, the received signal becomes $$r(t)=s(t)+n(t).$$ In this case, filtering $r(t)$ with the "inverse matched filter" $g(t)$ results in a worse signal-to-noise ratio than filtering with the filter matched to $p(t)$.
• $g(t)$ in this example is not the inverse of matched filter. Matched filter has a time-reversed impulse response (i.e. $h(t)=p(T-t)$). Here $g(t)$ is just the inverse of the pulse shaping filter $p(t)$. With that, one cannot maximize the SNR. – msm Oct 23 '16 at 1:19
• @DilipSarwate Please help me understand your objection. In the scenario I described, the filter output is $A\delta(t)$, and it is in that sense that the filter output is maximized (since no function has a value larger than $\delta(t)$ at $t=0$. Note that, good or bad idea, this is the kind of filter the OP is asking about. I explained one reason why using it is a bad idea (it does not maximize the SNR). You give another excellent reason in your comment. What is the problem? – MBaz Oct 23 '16 at 22:29
• MBaz, $A\delta(t)\star p(t) = Ap(t)$ "smears" $A\delta(t)$ across time, and no actual system can "unsmear" it back into $A\delta(t)$. The claim that $\delta(t)$ has a "value" at $t=0$ that is larger than the value that any function can take on at $0$. The notation $\delta(t)$ looks very much like an ordinary function of time, but $\delta(t)$ is not a function, and it does not take on a numerical value anywhere. $\delta(0)$ does not "equal" $\infty$ any more than $\delta(x)$ have value $0$ for any $x \neq 0$. – Dilip Sarwate Oct 23 '16 at 23:02
• @DilipSarwate I agree with everything you say. In my answer, I'm starting from the OP's assumption that there are a couple of filters such that their convolution is $\delta(t)$. Even though you need to proceed with caution, such a thing can be studied in theory. I'm more concerned that I'm apparently implying that $\delta(0)=\infty$, which I've never intended to claim. I'll edit my answer to make it clearer in that sense. Thank you for the feedback. – MBaz Oct 23 '16 at 23:14