Page:Alan Turing - Proposed Electronic Calculator (1945).pdf/38

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The velocity of sound in the crystal (X-cut quartz) is 5.72 km/sec. and its density is 2.7. The velocity in water is 1.44 km/sec., and the density 1, hence

u(water) = 0.1 abt.

The velocity in mercury is much the same but the density is 13.5. Hence

u(mercury) = 1.3 abt.

These figures suggest that we consider the two cases where u is small and where u is 1. The latter case may be described by saying that the liquid matches the crystal.

It may be assumed for the moment that our object is to make the minimum value of |R(w)| in a certain given band of frequencies as large as possible. If the width of the band is 2Ω and it is centred on w_o and if we ignore the variations in ϑ we shall find that the optimum value of u is of the form ${\mbox{N}}{\frac {\Omega }{w_{0}}}$ where N is some numerical constant probably not too far from 1. The value of Q should be as large as possible. With Ω = 1 Mc/s, w_o = 10 Mc/s this seems to suggest that water (u = 0.1) is very suitable. In practice the differences due to the value of ϑ are more serious than those due to u, and there is in any case plenty of power. We would not in practice take Q as large as we could but would rather try to arrange that |R(w)| was fairly constant throughout the band concerned and arg R(w) fairly linear when plotted against w. If water were used one would probably choose the thicknesses of the crystals and the value of Q to give poles of |R(w)| somewhat as shown in Fig. 41. With this arrangement of the poles the gain corresponding to |R(w)| is 9 dB throughout the range 8 Mc/s and the phase characteristic lies within 5° of the straight line within this range.

With mercury where u is nearly 1 we should put

${\frac {dw_{0}}{2c}}={\frac {\pi }{2}},{\frac {d\ w_{0}}{2c}}={\frac {\pi }{2}},$

and then

${\big |}R(w){\big |}={\frac {2}{\pi }}{\Big (}sin{\frac {\pi }{2}}{\frac {w}{w_{0}}}{\Big )}^{2}{\Bigg |}{\frac {w\ w_{0}}{{\big (}w+w_{0}+{\frac {iw_{0}}{2Q}}{\big )}{\big (}w-w_{0}+{\frac {iw_{0}}{2Q}}{\Big )}}}{\Bigg |}$

We should probably find it desirable to omit the tuned circuit, in which case R(w) would represent a fairly constant loss of 4 dB. One could use a Q of 2 if one wished, giving a gain of 2 dB instead.

We have assumed above that the crystal is longitudinally excited. If it were transversely excited the figures would be much less satisfactory. At the transmitting end a far larger voltage would have to be applied in order to obtain the same field strength, and at the receiving end the stray capacities will have a more serious effect with transverse electrodes, although if the stray capacity were zero transverse electrodes at the receiving end would actually be more efficient.

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