Page:Alan Turing - Proposed Electronic Calculator (1945).pdf/41

 (d) There is a loss due to the viscosity of the medium. For a water tank with a delay of 1 ms. and a carrier of 10 Mc/s the loss may be 12 dB: with mercury and a carrier of 20 Mc/s it may be 4 dB.

(e) Losses in the walls of the tank. Apparently this should not exceed 10 dB.

(f) The noise voltage may be 4 x 10-6 volts RMS (mercury) or 9 x 10-6 volts RMS (water).

(g) The signal voltage (peak to peak) should exceed the noise voltage (RMS) by a factor of 24 for safety.

These figures require input voltages (peak to peak) of 0.2 volts or 4.5 volts with mercury and water respectively. We could quite conveniently put 200 volts on, so that we have 60 dB (or 53 dB) to spare. There is no danger of breaking the crystals when they are operated with so much damping.

(vii) Phase distortion due to reflections from the walls.— We cannot easily treat this problem quantitatively because of lack of information about the boundary conditions and because the ratio of diameter of crystal to diameter of tank is significant. Let us however try to estimate the order of magnitude by assuming the pressure zero on the boundary and considering the gravest mode. In this case the pressure is of form 1 ry ‘ = j eh Pteint where 2a is the diameter of the tank and k1 = 2.4 is the smallest zero of Jo, and 2 ke w 7 Pre sof of tank is <z07 + If we are using carrier working we are chiefly interested in -—t which tums out to be -—Ha=s where wo is the carrier frequency. If we suppose 4a 2.2, then the greatest phase error which is introduced is &y2.92 of 2 oe a® Let us suppose that the greatest admissible error is 0.2 radians, then we must have

Lo Ok fe eed ce

Taking wo = 10 Mc/s

= 1 Mc/s

c = 1.4 x 105 cm/sec.

= 1.4 x 102 cm.

a = 1 cm. Then/