Как выбрать гостиницу для кошек
14 декабря, 2021
Consider the integrating circuit of time-constant т, labelled т in Figure 3-1, and consisting of a series resistor and a capacitor to ground. If a pulse of voltage of amplitude E and short time duration compared to т is applied to its input, its output is
s(t) — В e”t,/T. (3-17)
and the time-integral of this voltage must equal the time-integral of the input pulse; i. e.,
.00
(3-18)
so
and
s(t) = m e -‘/т
Also, if the circuit is suppliёd with a succession of short voltage pulses, not necessarily equal, its output is a linear superposition of terms like Equation (3-20) .
t/At
S(t) = ^ S^t-iAt),
І = -00
Or
2
Now the continuously-varying voltage, AV (t)., that is being applied across this circuit can be. considered as supplying a succession of short voltage pulses by dividing the time axis into
1L.
short equal intervals of length At. The amplitude of the iin pulse is •
E{ = Av (iAt),
Equation (3-4) shows that
00 00 .
V2(iAt) = ^ ^ 77t_a 77І-Ь у(аДt)y(bAt).
a = o b = о
so we will compute <8^> .. From Equation (3-27) t) 00 ‘ 00 00 00 X 00 A* — ($) ЕЕШЕ , . c=oa=ob=oh=of=og=o |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
*«-($)’E c = c CAt ___ T "T <*<>2*E E A |
and this can be written as
where it is understood that no two subscripts of i] in any summation are equal.
Equation (3-30) is obtained from Equation (3-29) by performing the following operations:
a. Equation (3-29) is written as the sum of five separate summations. The first has the subscripts of all four it’s equal, i. e., four of a kind. The second has the subscripts of three uf the )j! o equal, i. e., three of a kind. The third has two pairs, the fourth has one pair, and the fifth has none equal. The second, fourth, and fifth summations will go to zero because of Equation (3-11), leaving only the first and third.-
b. All four subscripts are equal only if a = b, f = g, and c + b = h + g. For those terms
in which C>h, we write b for a and b + c — h for f and g, and for those terms in which c<h, we write g for f and g + h — c for a and b; the result is the first summation in Equation (3-30). . . .
c. There are three ways in which there can be two pairs. The first is for a = b and
f =. g. Then, if we write b for a and g for f, the result is the third summation in
Equation (3-30). .
d. The second way to obtain two pairs is for c + a = h + f and c + b = h + g, and the third-way is for c + a = h + g and c + b = h + f. For those terms in which c > h we write a + c — h for f and b + c — h for g to take care of the second way, and we write a + c — h for g and b + c — h for f to take care of the third way; for those terms in which c<h we write f + h — c for a and g + h — c for b to take care of the second way, and we write f + h — c for b and g + h — c for a to take care of the third way. The result is the second summation in Equation (3-30).
e. It is necessary to separate the terms in which c>h from those in which c<h to insure that the argument of v. is always positive [ cf. Equation (3-29)J.
If the average flux of neutrons is Ф and the counting efficienty of the detector is k, then the expected values of the 77 combinations are
<(t])4> = k$At
4n)2(n)2> = (»At)2
о
and the expected value of S (t) is
а ф b cAt hAt • v(bAt + cAt — hAt)e T T (At)4 00 00 00 + ^ ^ ^ ^ ‘ v(fAt)v(gAt)v(fAt + hAt — cAt) f=og=oh=c f * g |
|
K) z z z z
c=ob=oh=og=o
g ф b + c — h
cAt hAt
v2(bAt)v2(gAt)e T T /A*’4
‘ We should now remove the conditions а ф b, f * g, and g ф b + c — h by adding and subtracting the appropriate terms; however, the additional summations so created will go to zero in the limit as At—о because they will contain one more At than summation signs [ cf: Equations (3-7), (3-8), and (3-9)], so we will not write them. As At—o. Equation (3-32) (with the conditions a * b, f * g, and g * b + c — h removed) becomes • 00 ,00 |
•: ■ (^)2 ■» f f v2<y) e . Л) *’0 rx r00 І e‘zA v2(y+x-z)dz + / e’zA v2(y+z-x) dz _Л> . •’x mrrr • ‘ ‘Л ‘Л Г e-z/T. vl — *0 L |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|