# Does a matched filter give a gain in SNR?

I am trying to know if filtering a sequence with a matched filters gives a gain equal to the sequence length, he are the calculus, is it correct ?

In my model $s$ is a deterministic complex sequence (lets say RRC pulse shape samples for a QAM modulation), $v$ is a complex white circular gaussian noise.

If it is correct does it mean that there is an "inband" filter which can enhance $SNR$ ?!

\begin{align} s &= [ s_0,\dots,s_{ N-1} ]^T\\ v &= [ v_0,\dots,v_{ N-1} ]^T \\ x &= s + v \\ R_v &= \mathbb{ E } \left[ v v ^ H \right] = \sigma ^ 2 I \\ \textrm{SNR}_1 &= \frac{ \mathbb{ E } \left[ \lvert s \rvert ^ 2 \right] } { \mathbb{ E } \left[ \lvert v \rvert ^ 2 \right] } = \frac{ \mathbb{ E } \left[ s ^ H s \right] } { \mathbb{ E } \left[ v ^ H v \right]} = \frac{ s ^ H s } { \mathbb{ E } \left[ v ^ H v \right] } = \frac{ s ^ H s } { \mathbb{ E } \left[ \Sigma \lvert v_i \rvert ^ 2 \right] } = \frac{ s ^ H s }{ N \sigma ^ 2 }\\ h &= s \\ y &= h ^ H x = h ^ H s + h ^ H v = s ^ H s + s ^ H v \\ \textrm{SNR}_2 &= \frac{ \mathbb{ E } \left[ \lvert h ^ H s \rvert ^ 2 \right] } { \mathbb{ E } \left[ \lvert h ^ H v \rvert ^ 2 \right] } = \frac{ h ^ H s s ^ H h }{ \mathbb{ E } \left[ h ^ H v v ^ H h \right] } = \frac{ h ^ H s s ^ H h }{ h ^ H R_v h } = \frac{ s ^ H s s ^ H s }{ s ^ H \sigma ^ 2 I s } = \frac{ s ^ H s }{ \sigma ^ 2 } \\ \frac{\textrm{SNR}_2 }{ \textrm{SNR}_1 } &= N \end{align}

• $s^H s$ seems to be a matrix? – MBaz Aug 4 '16 at 19:00
• as @MBaz already declared, you have a simple dimension problem that you shall clarify. In addition, to make things further clear it's better if you also provide details such as noise type being as an additive white (gaussian) noise uncorrelated with the deterministic input signal both being real? so that your probability calculus is justified. Also the matched filter should not be $h=s$ if s is not real. Other than these it seems correct to me... – Fat32 Aug 4 '16 at 21:10
• Sorry I added some details to clarify. In the array model $h^H$ are filter coefficients ( so for me the matched filter ). Let's say that $h$ is it's trans-conjugate. – user19720 Aug 5 '16 at 6:32

Output SNR for matched filter can be analysed in time domain, which is as explained below,

Consider a filter with Impulse response $\enspace h(t)$ and input $\,s(t)+w(t)$, where $\,s(t)$ represents the required signal and $\,w(t)$ represents AWGN noise.

Output signal is given by $$y(t) = s(t) * h(t) + w(t) * h(t)$$

Writing signal component of output in inverse fourier form $$y'(t) = \int_{-\infty}^{\infty} S(f) H(f) \ e^{j2pit}\ df$$ where $S(f)$ and $H(f)$ are spectrums of input signal $s(t)$ and Frequency response of filter.

Output SNR of filter which is sampled at time instant $\,t=T$ is output power with signal component by noise power which can be return as

$$SNR = \frac{\displaystyle\left |\int_{-\infty}^{\infty} S(f) H(f) \ e^{j2piT}\ df \right|^2}{ \displaystyle\left|\int_{-\infty}^{\infty} \frac{N_o}{2} |H(f)|^2 df \right|^2}$$

Note that numerator term is evaluated at time t = T, which is the sampling instant for Matched filter.

To maximise this SNR , we need to cancel the phase term . if we substitute $$H(f) = S^*(f) e^{-j2pif}$$ which is nothing but $$h(t) = s^*(T-t)$$

and simplify the above SNR expression, we will get $$SNR = \frac{\displaystyle\int_{-\infty}^{\infty} \left|S(f)\right|^2 df}{N_o/2}$$ which is nothing but ration of energy of signal $s(t)$ to value of power spectral density of AWGN noise.

In this time domain analysis you can see that there is no filter effect , and it also tells us what should be the filter impulse response to maximise the output SNR.

• I am beginner in latex,please let me know if there is any correction and any suggestions for the answer are welcome. – spectre Aug 5 '16 at 16:38
• Instead of formulas in bold (in the text), pure latex could be fine – Laurent Duval Sep 4 '16 at 17:06