Understanding Coverage Area and Decibels

[Fishton]

We buy fish finders to find fish right?  So the coverage area of a cone is actually very important then, because the wider the beam the more area you will be covering to find fish.  If you have been using a dual frequency sonar such as the 83/200kHz for a while, you will know by now that the 83kHz covers 3 times the area of the 200kHz.

You probably thinking to yourself "Boring! Known that for years."  But what do we actually know about the relationship between coverage area and Decibels?  I have been looking into this for quite some time now and have come up with the following conclusion.

When you read the specifications of your sonar they will give you a power output (peak to peak and RMS), frequency (50, 83, 200 or a combination of these), cone angle and then either a -3dB or -10dB at the end.  What does this -3dB and -10dB actually mean?

The Decibel (dB) is the unit which the power or intensity of a sound wave is measured.  In order to determine cone angle, manufacturers generally use the -3dB range, because at this point, according to Airmar Technologies, the sound wave is already at half its full power*.  

Below is the 200kHz sound waves coverage area measured at -3dB
Pro: Wide coverage area = more fish
Con: More clutter

Below is the 200kHz sound waves coverage area measured at -10dB
Pro: Less clutter
Con: Narrow coverage area

*Just think about this statement by Airmar Technologies for a minute “At -3dB the sound wave is already at half of its full potential.”  This got me thinking how important it is to have your transducer set up correctly.

 

[gerritp]

If i understand the working of a sonar correct it is suppose to work the same as RF(radio frequency).

In RF a radio will transmit a signal in a sphere, working with the principle that energy cant be created or destroyed but only modified, the antenna only focuses the signal. By compressing the sphere into a round tube(omni antenna) you will get a higher gain(db). The smaller the beam angle the higher the gain will be. In RF, in a perfect world, every 3db doubles the output power.

Im guessing a sonar work on the same principal as RF. A sonar will transmit the sound wave in a sphere, the transducer will focus that sound wave into a narrow beam, increasing the gain to -3db or -10db depending on the transducer. If the transducer is rated at -3db it means the sound wave will be twice as strong as to a sonar without and transducer. Same goes for the -10db, it will be more that 3 times the power of the -3db and more than 4 times the power of the sonar without an transducer.
With the sonar creating the same sound waves, the only way to get a wide angle would be to have a less powerful transducer, -3db. In order to have more detail and clarity, the only way will be to make the beam narrow, thus increase the gain to -10db.

Regarding the statement that -3db is already at half the strength, sounds about right, if it was compared to the -10db transducer. In perfect conditions(theory) it is supposed to be more that 3 times less, taking all the elements in water into account is should be there and there about.

 

[Fishton]

-3dB = HALF the full potential Gerrit. 83, 50 & 200kHz are cone shaped. 455 & 800kHz are beams, for example 180 degrees port to starboard, but only 1.8 degrees fore to aft.