Variance of the Signal

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

Подпись:(3-18)

image105 Подпись: (3-19)

so

and

Подпись: (3-20)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

Подпись: (3-21) .S(t) = ^ S^t-iAt),

І = -00

Подпись: S(t) Подпись: і = -00 Подпись: E^t T Подпись: g -(t-iAt)/r Подпись: (3-22)

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 •

Подпись: (3-23)Подпись: (3-24)

Подпись: so Equation (3- 22) becomes S(.t)
Подпись: t/At •' A £ V2(iAt)At e-(t-iAt)/T ^ l - -у0

E{ = Av (iAt),

Equation (3-4) shows that

00 00 .

Подпись:V2(iAt) = ^ ^ 77t_a 77І-Ь у(аДt)y(bAt).

a = o b = о

image119

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

image120

(3-26)

 

і = — oo ■ a = о b=o

 

If we let

 

c “ ST "1’

 

image121

(3-27)

 

Now the variance is obtained from

 

„ 2 .„2, 2 a = <S > — <S> ,

 

(3-28)

 

V t 71 t h t

AT“c’a Ж “c_b £F “h“f

 

image122

(3-29)

 

*«-($)’E

c = c

CAt ___

T "T <*<>2*E E A

image123

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

Подпись: (3-31)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

image125
image126

CAt hAt

v(gAt + hAt — cAt) e ^ T (At)4

 

image127

Подпись: 00 00 ao 00Подпись: (At)4. (3-32)

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

image130

dx dy d£ (3-33)

 

By combining Equations (3-14), (3-28), and (3-33), we obtain the variance of the signal

.00 ,00

.Л, л o-x/T

 

dx dy

 

v(?)v(y) e

 

‘o •’o

 

/(y+x-z)v(£+x-z)dz

 

+ I e~ ‘ v(y+z-x)v(£+z-x) dz

x

 

dx dy d£. (3-34)

 

image131

s(t) = — Vi V t v(aAt)v(bAt)Ate’CAt//‘

■ т с = о а — о b — о ЭТ SC — c‘b

 

(3-27)