250 Lecture 13 
where ¢« is the dielectric constant and «9 is a constant whose value depends on 
the choice of units. Expressions involving Eqs. (3) through (6) and the electro- 
mechanical transformation ratio a may be substituted in Eq. (2) to give the fol- 
lowing expression for the efficiency, 
porai(1 —o)k 2 
Se een Nar eee 8 
2vbE(1 — Ke tan d + po a(1 = o)k?2 ‘) 
7 
13.3.2. The Free-Field Voltage Response and Equivalent Noise Pressure 
It is now desired to obtain expressions for the free-field voltage response 
and the equivalent noise pressure in terms of the same parameters. As a con- 
sequence of the reciprocal properties of the transducer, the former is given by 
fp =) ee (9) 
™pv 
where A is the wavelength of sound in the medium and Dis the directivity factor, 
which will be taken equal to unity for the assumed conditions. 
It is usually necessary to amplify the low-level output of a hydrophone, and 
no matter how well an amplifier is designed, there still remains a residual 
output produced by thermal noise in the conductors of the input circuit. In the 
ideal case, this residual output will be entirely due to the hydrophone. 
The open-circuit voltage in the hydrophone circuit developed by thermal 
agitation in a 1-cps band is given by Johnson [14]: 
Ey =V4kTRr 
where k is the Boltzmann constant and T is the absolute temperature. 
Now, the equivalent noise pressure is defined as the equivalent acoustic 
pressure produced by sound energy, in a 1-cps band, that will generate an open- 
circuit voltage in the hydrophone just equal to that produced by the thermal 
noise at the same frequency, also for a 1-cps band. Since the sensitivity in Eq. 
(9) is E,/P, where E, is the open-circuit voltage across the terminals of the hy- 
drophone resulting from an acoustic pressure P, it is possible to solve for E, and 
to equate this solution to the expression for the noise voltage above. Solving this 
equation for the pressure, we obtain the equivalent noise pressure 
Bee) eae (10) 
2 
A“nD 
The desired expression for the equivalent noise pressure is obtained by sub- 
stituting Eq. (8) in this expression. This result is 
pie kT[2bE(1 - k?)? tand + pwra‘(1 = a)k2/v] (11) 
: moa(1 — a)k? 
From Eq. (1) it is easily seen that the real part of the series impedance that 
is due to radiation is 
QR, — ee 12 
(Ger (2) 
