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If a matched filter is optimal for detecting a known pulse in an AWGN channel, any other filter must have a degraded SNR. Given a discrete pulse shape and an arbitrary non-matched filter, is it possible to calculate how much worse the SNR will be, relative to the matched filter case?

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    $\begingroup$ A matched filter is really also a correlator. One measure of degradation would be how much lower the magnitude of the correlation result is given the same ideal input to the correct MF vs. the incorrect MF when the input pulse is directly lined up with the MF. $\endgroup$ – Andy Walls Jun 14 '17 at 10:18
  • $\begingroup$ @AndyWalls That would require some kind of normalization of the coefficients though, right? $\endgroup$ – Phil Frost Jun 14 '17 at 10:19
  • $\begingroup$ It assumes the pulse matched filter is at some normalized level, yes, for a fair comparison. They are usually normalized to 1. $\endgroup$ – Andy Walls Jun 14 '17 at 10:25
  • $\begingroup$ What's normalized to 1, specifically? The sum of the coefficients? The sum of their squares? The peak? $\endgroup$ – Phil Frost Jun 14 '17 at 10:36
  • $\begingroup$ If matched filter is considered as a correlator, the metric is the correlation result and it is normalized to 1. If matched filter is thought as a filter, the metric is the norm of the sample of the output at instance $T$ where $T$ is the pulse duration (the output duration should have length $2T$); in this case the norm is normalized to 1. $\endgroup$ – AlexTP Jun 14 '17 at 11:30
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Let's consider a received signal

$$Y(t)=Ap(t)+N(t)\tag{1}$$

where $A$ is the information symbol (modeled as a random variable), $p(t)$ is the transmit pulse, and $N(t)$ is additive white Gaussian noise (AWGN, modeled as a random process). For the sake of simplicity let's assume that all signals are real-valued (baseband case).

Filtering with an LTI system with impulse response $h(t)$ and sampling at $t=0$ gives

$$Y_0=\int_{-\infty}^{\infty}Y(t-\tau)h(\tau)d\tau\Big |_{t=0}=\int_{-\infty}^{\infty}Y(-\tau)h(\tau)d\tau\tag{2}$$

With $(1)$ you get

$$Y_0=A\int_{-\infty}^{\infty}p(-\tau)h(\tau)d\tau+\int_{-\infty}^{\infty}N(-\tau)h(\tau)d\tau\tag{3}$$

The power of the desired signal (the first term in $(3)$) is

$$P_S=E[A^2]\left(\int_{-\infty}^{\infty}p(-\tau)h(\tau)d\tau\right)^2\tag{4}$$

where $E[\cdot]$ is the expected value, and $E[A^2]$ is the power of the information symbol $A$. From the second term in $(3)$, the noise power is

$$P_N=\frac{N_0}{2}\int_{-\infty}^{\infty}h^2(\tau)d\tau\tag{5}$$

where $N_0$ is the one-sided power spectral density of the noise. From $(4)$ and $(5)$ the SNR is

$$SNR=\frac{P_S}{P_N}=\frac{E[A^2]\displaystyle\left(\int_{-\infty}^{\infty}p(-\tau)h(\tau)d\tau\right)^2}{\displaystyle\frac{N_0}{2}\int_{-\infty}^{\infty}h^2(\tau)d\tau}\tag{6}$$

We know that $(6)$ can be maximized by a matched filter:

$$h_{opt}(t)=Cp(-t)\tag{7}$$

with some constant $C$. From $(6)$ and $(7)$, the optimal (maximum) SNR is given by

$$SNR_{opt}=\frac{E[A^2]}{\frac{N_0}{2}}\tag{8}$$

The actual SNR for an arbitrary receive filter impulse response $h(t)$ can be written in terms of $SNR_{opt}$:

$$SNR=SNR_{opt}\underbrace{\frac{\displaystyle\left(\int_{-\infty}^{\infty}p(-\tau)h(\tau)d\tau\right)^2}{\displaystyle\int_{-\infty}^{\infty}h^2(\tau)d\tau}}_{\le 1}\tag{9}$$

where the last term in $(9)$ (or its inverse) could be called the "SNR penalty".

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  • $\begingroup$ I think if you introduce the constant $C$ in the matched filter, the constant should be kept in $\mathrm{SNR}_{opt}$. $\endgroup$ – AlexTP Jun 15 '17 at 12:57
  • $\begingroup$ @AlexTP: If you look at Eq. (6) you can see that you get $C^2$ in both the numerator and the denominator, so it cancels out. $\endgroup$ – Matt L. Jun 15 '17 at 15:03
  • $\begingroup$ it is true. I did not see this. $\endgroup$ – AlexTP Jun 15 '17 at 15:37
  • $\begingroup$ Sorry, I'm new at this: what's the $E$ in $E[A^2]$? $\endgroup$ – Phil Frost Jun 17 '17 at 10:25
  • $\begingroup$ @PhilFrost: Sorry for not explaining. $E$ is the expected value; I've assumed that the information symbol is modeled as a random variable. $E[A^2]$ is the power of the random variable $A$, which equals its variance if the mean of $A$ is zero ($E[A]=0$). $\endgroup$ – Matt L. Jun 17 '17 at 10:43
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I'll offer a less rigorous and less general procedure for the math challenged. Let's say you have a matched filter, and some other filter. They are discrete, FIR filters. We're assuming the goal is to find a known pulse in an AWGN channel.

First normalize both filters to unit energy so it's a fair comparison. This means scaling the coefficients such that the sum of the squares of each coefficient = 1. For some filter $h$ with coefficients $h_1, h_2, \dots, h_n$, scale the coefficients such that:

$$ \sum_{i=1}^n h^2(n) = 1 $$

By definition of a matched filter, the matched filter coefficients are the time-reversed, complex conjugate of the pulse shape. For a real and symmetrical pulse, they are equal. Given the normalization above, the correlation between the normalized pulse and the (equal) matched filter is 1 at the optimal sampling point.

Correlate the other filter's coefficients (time reversed, if necessary) with the pulse at the optimal sampling point, and call that $A$. Alternately, run the pulse through the filter and take the maximum: it amounts to the same thing.

Power is proportional to the square of amplitude, and SNR is a ratio of powers, so we'll square the amplitude to give us the ratio of power in the non-matched case ($A^2$) to the matched case ($1$):

$$ A^2 \over 1 $$

This ratio will always be less than one. Convert it to decibels if you like:

$$ 20 \log_{10}(A) $$

This will always be a negative number, and it's the change to SNR as a consequence of an unmatched filter.

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  • $\begingroup$ That's correct. It can get even a bit simpler by realizing that your $A^2_{matched}$ equals $1$ with the given normalization, so you only need to compute $A^2_{other}$. Furthermore, instead of filtering and using only one sample of the output (the maximum), you could directly compute that one value by correlation, as shown in the numerator of Eq (9) in my answer (with the integral being replaced by a sum). $\endgroup$ – Matt L. Jun 17 '17 at 12:35
  • $\begingroup$ Good points, thanks. I've edited to include those simplifications. $\endgroup$ – Phil Frost Jun 17 '17 at 13:00
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In terms of bounds, a mismatched filter can be selected to be orthogonal to the matched filter so an infinite negative SNR is possible.

An energy detector can outperform a mismatched filter, so matched vs energy is a better comparison.

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