THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

Example Problem 115.12 The Sonar Equations 427Find the voltage across the hydrophone terminals exposed to an acoustic rmspressure of 1 μPa if the receiving response is –80 decibels re 1 V.SolutionWe use the relationshipand since SPL =−80 dB,SPL = 20 log(p/1 μPa)p/1 μPa = antilog(−80/20),yielding p = 0.0001 μPa. For the 1 μPa sound field, the voltage is 0.0001 V acrossthe terminals.For a projector, the transmitting current response is the acoustic pressure producedat a point 1 m from the projector in the direction of the axis of its beampattern by a unit current into the projector. This means that the transmitting responseis stated in terms of the number of decibels relative to 1 μPa as measuredat the reference distance, produced by1Aofcurrent into the electric terminals ofthe projector. While the transmitting response are usually referred to a referencedistance of 1 m from the source, a correction of 0.78 dB must be added in order toconvert the transmission response expressed in terms of a reference level of 1 yardinstead of 1 m.Example Problem 2Predict the rms pressure at 1 m when the projector is driven with a current of1 rms ampere, if the response is rated at 100 dB re 1 μPa/A (referred to 1 m).Express the response in terms of a reference distance of 1 yard instead of 1 m.SolutionThe corresponding rms acoustic pressure for 100 dB is 10 5 μPa, which is producedby 1 rms ampere current. The corresponding transmitting response for 1 yard is100 + 0.78 = 100.78 dB.15.12 The Sonar EquationsThe purpose of the sonar equations, originally formulated during World War IIand derivative of similar considerations in radar, is twofold: (a) prediction of thecapabilities of existing sonar equipment with respect to detection probability orsearch rate and (b) design of new equipment to meet preestablished range ofdetection or actuation.