Page:Alan Turing - Proposed Electronic Calculator (1945).pdf/36

 and therefore the velocity at a is

$$\frac{-iB \sin\beta a}{\zeta} = \frac{iE_2\sin\beta a}{\zeta\cos\beta a + i \zeta_1\sin\beta a}$$

i.e. the velocity at the inside edge of the crystal is

$$\frac{iB \epsilon}{\zeta}\cdot\frac{1}{\cot\frac{dw}{2o} + iu}$$

where $$u = \zeta l / \zeta$$.

Assuming that the exciting voltage is longitudinal we may say that

$$\frac{\mbox{Velocity}}{\mbox{Exciting voltage}} = \frac{i\epsilon}{\zeta d} \cdot \frac{1}{\cot\frac{dw}{2o} + iu}$$

The effect of the medium between the two crystals we will not consider just yet. Let us simply assume that

$$\frac{\mbox{Velocity at inside edge of receiving crystal}}{\mbox{Velocity at inside edge of transmitting crystal}} = \vartheta$$

We have now to consider the effect of the receiving crystal. Fortunately we can deal with this by the principle of reciprocity. When applied to a mixed electrical and mechanical system this states that the velocity produced at the mechanical end by unit voltage at the electrical end is equal to the current produced at the electrical end by unit force at the mechanical end. Hence

$$\frac{\mbox{Current at receiving end}}{\mbox{Force on receiving crystal}} = \frac{i\epsilon}{d'\zeta}\frac{1}{\cot\frac{d'w}{2o} + iu}$$.

To these equations we may add that the ratio of force to pressure is the area A’ of the receiving crystal, and that the ratio of pressure to velocity is the mechanical characteristic impedance ζ1. Combining we obtain

$$Y = \mbox{Transfer admittance} = \vartheta\frac{A^'\eta^2\zeta_1}{dd'\zeta^2}\frac{1}{\big(\cot\frac{dw}{2o} + iu\big)\big(\cot\frac{d'w}{2o} + iu\big)}$$.

Let us now assume that the input to the valve from the receiving crystal consists of a tuned circuit with a fairly low ‘Q’ as in Fig. 41. Then/