# Sonar single transducer element SNR vs array SNR

Background:

I'm taking a flexible learning unit on radar and sonar systems in a vocational college for an assoc. degree level course. The teacher has given us a reading assignment: "Chapter 15 Underwater Acoustics", source unknown, and we are to answer the end-of-chapter questions. I have searched online and off, unsuccessfully, for the source reference, and there are no texts on sonar systems in the college library system in my state.

My question is a general systems engineering design question, I believe:

• Given a single sonar transducer element SNR (signal to noise ratio in dB) of X dB, what would be the SNR of an array of M elements?

My question relates to calculation of the array gain, AG, where, stated in the reading material,

$$AG = 10~ log ~{ \left \lbrace {{(S/N)_{array}} \over {(S/N)_{element}}} \right \rbrace }.$$

I've been given the element SNR, but I have no information on the array SNR.

In the absence of any information on multiple SNR elements, I've gone ahead and guessed a possible answer.

In the problem I'm trying to solve, $~ SNR_{element} = 40 ~dB$ $$\therefore decimal ~ratio ~ (S/N)_{element} = 10^{40 dB ~/~ 20} = 100$$

Question now is, to get the $~SNR_{array}$, do I add or do I multiply? Taking the RMS doesn't make sense, since it is implicit.

Given M = 25 elements in the array. Raising $(S/N)_{element}$ to the 25th power, i.e. $~100^{25}$ doesn't make sense.

But adding together: ${~(S/N)_{array} = (S/N)_{element} \times M = 100 \times 25 ~ elements = 2500 ~}$ puts the answer in the realms of possibility:

$$So, ~ AG = 10 ~log {2500 \over 100} = 10 ~log~ 25 = 14 dB$$

Am I close? A reference would be appreciated.

@drfried, the problem is assuming an ideal solution. The book problem is originally stated as:

• If an element of an array has a signal to noise ratio of 40dB, what would be the array gain of 25 similar elements in such an array?

No further information or diagram is given.

@Andy, From the response you've provided, and a response I received on the linkedin.com "Antenna Solutions" group, https://www.linkedin.com/grp/post/2232865-6001739735423270914 , without source reference, I'll see if I understand you correctly. The answer I've been given there is $SNR_{array} ~=~ 10~log~M ~+~ X ~ [dB]$. Let's see if they coincide. Apologies for the mathematical massacre that follows.

Firstly, let's replace the subscript "array" with $a$ , and the subscript "element" with $e$.

If I understand @Andy correctly, we can write his statements as,

$$(S/N)_a ~=~ { {\sum\limits_{i = 1}^M S_{ei} } \over {\left( \sum\limits_{i=1}^M N_{ei} ^2 \right )^{ 1 \over 2 } } }$$

where, for my problem, $S_{ei} ~=~ S_{ej} ~=~ 100~ units ~(ie~ \mu V,~ mV, ~etc.)$ and $N_{ei} ~=~ N_{ej} ~=~ 1~ unit ~(ie~ \mu V, ~mV,~ etc.)$.

$$\therefore ~~~ SNR_a ~=~ 20~ log ~ \left \lbrace {M S_e} \over { ( M ~ N_e^2 )^{1 \over 2} } \right \rbrace$$

$$=~ 20~ log ~ \left \lbrace {M S_e} \over { M^{1 \over 2}~ N_e } \right \rbrace$$

$$=~ 20~ log ~ M^{1 \over 2} (S / N)_e$$

$$=~ 10~ log ~ M ~+~ 20 ~log ~(S / N)_e$$

$$=~ 10~ log ~ M ~+~ X ~[dB]$$

Looks like both responses coincide. I'm inclined to accept @Andy's reasoning and experience, so I think my question may have been answered.

To finish the whole problem for posterity, as stated above, M = 25 elements and X = 40 dB.

So, $(S/N)_a$ = 14 + 40 = 54 dB.

$$\therefore ~~~~AG ~=~ 10~ log \left \lbrace {(S/N)_a} \over { (S/N)_e } \right \rbrace ~=~ 10~ log \left \lbrace {10^{54~dB ~/~ 20}} \over { 10^{40~dB ~/~ 20} } \right \rbrace ~=~ 10~log~(10^{(2.7 ~-~ 2) }) ~=~ 7dB$$

Don't even need a calculator!!!

Thanks to all who helped.

I'm going to take a stab at this though I don't know if it applies to sonar. If you have a signal received on a transducer there will be both signal and noise. If you have two transducers (in an array) it can be presumed that the same signal is received on both but the noise on each is likely gaussian and not coherent.

The upshot of this is that when you add the two signals from the two transducers the signal will double (+6dB) but the noise will add as per: -

Noise = $\sqrt{A^2+B^2}$ i.e. it will only increase by 3dB

This means the SNR has risen by 3dB

Each time you double the number of transducers, the SNR increases by 3dB

Hope this helps.

• There are a lot of caveats to this, for example relative placement of the transducers, but this thinking is generally correct and good enough for deciding if arrays should be considered as a solution to your problem. – DrFriedParts May 9 '15 at 5:01