Electronic – 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.