////Sonar Equation Example: Active Sonar
Sonar Equation Example: Active Sonar2018-08-09T15:02:43+00:00

Sonar Equation Example: Active Sonar

Submarine Search Sonar: Signal-to-Noise Ratio

Although the characteristics of submarine search sonars vary substantially for different systems, typical sizes of the terms in the sonar equation can be obtained by working through an example for a hypothetical mid-frequency sonar operating at 8,000 Hz. The active sonar equation is:

SNR (decibels) = SL -2TL +TS – (NL – AG)
 

Where SNR is signal-to-noise ratio, SL is the source level, TL is the transmission loss, TS is the target strength, NL is the noise level, and AG is the array gain. Typical operating parameters for such a sonar might be:

SL = 220 dB re 1 µPa at 1 m
 

AG = 20 dB
 

The transmission loss TL includes both sound spreading loss and attenuation. Sound spreading loss assuming spherical spreading for a range R of 10,000 m is:

Spreading loss (dB) = 20 log R = 20 log (10,000) = 80 dB
 

The attenuation due to sound absorption calculated using the absorption coefficient α (alpha) of about 0.5 dB/km at a frequency of 8,000 Hz is:

Attenuation (dB) = αR = (0.5 dB/km x 10 km) = 5 dB
 

The total transmission loss is then:

TL = 80 + 5 = 85 dB
 

Measured target strengths TS for a submarine, as measured broadside to the submarine, vary widely, but a typical number is:

TS = 25 dB
 

The noise level for a search sonar mounted on a destroyer is typically dominated by noise generated by water flowing past the ship and noise generated by the ship itself, which together are called self-noise. For a destroyer traveling at 15 knots, a typical self-noise level in a 1 Hertz wide frequency band at 8,000 Hz might be:

NL = 63 dB re 1 μPa/√Hz
 

A typical sonar receives signals over frequency bands greater than 1 Hertz wide. The frequency band over which the receiver operates is called the bandwidth BW, given in Hertz. The total noise is:

NLtotal (dB re 1 µPa) = NL + 10 log BW
 

Shorter sonar pulses have broader bandwidths, which means that the receiver bandwidth must be greater. The relationship between the sonar pulse length and receiver bandwidth is roughly:

BW (Hz) = 1 / T, where T is the sonar pulse length in seconds.
 

For this example we assume that the sonar transmits a 0.1 second long pulse. The bandwidth of the signal is:

BW = 1/T = 1/(0.1) = 10 Hz
 

The total receiver noise level is then:

NLtotal = NL + 10 log BW = 63 + 10 log (10) = 73 dB re 1 μPa
 

Combining all of these values the signal-to-noise ratio at the receiver is:

SNR (dB) = SL – 2TL + TS – (NL – AG) = 220 – 2(85) + 25 – (73 – 20) = 22 dB
 

This is a reasonable, but not large, signal-to-noise ratio for detection. One caveat is that spherical spreading is a rough approximation for computing spreading loss in the ocean, and actual transmission losses often exceed those computed for spherical spreading. Modern search sonars therefore often have higher source levels than the value of 220 dB re 1 μPa at 1 m used here.