Problem 3.1 a) Derive 3.19 from binomial distribution using stirlingβs approximation
average number of events = N = np β p = N and therefore the binomial distribution is
using stirlings approximation. If x is small i guess whatever applies to n! will also apply to (n-x)!
The first term - if you divide the numerator and denominator by n and realize the n is a large number while x is small it just becomes 1
= 1 β N x x ! e x ( 1 β p ) ( n β x ) = N x x ! e x ( 1 β p ) n ( 1 β p ) x = 1*\frac{N^x}{x!e^x}(1-p)^{(n-x)} \\ = \frac{N^x}{x!e^x}\frac{(1-p)^{n}}{(1-p)^x} = 1 β x ! e x N x β ( 1 β p ) ( n β x ) = x ! e x N x β ( 1 β p ) x ( 1 β p ) n β p = N/n and since x is small we assume xβ0 that means e^x and (1-p)^x both are 1
(1-p)^n is the taylor expanision for e^-N
which is the 3.19 we wanted to derive.
b) Deriving 3.18 - factorial moments
so the second term would be zero for anything that is less than m so
and
and p(x) is the poisson distribution we derived in 3.1 and based on Based on 3.17 - normalization for indeendent occurrences
= β x = m + β ( e β N ) N ( x β m ) ( x β m ) ! N m = N m = \sum_{x=m}^{+\infty}(e^{-N})\frac{N^(x-m)}{(x-m)!}N^m \\ = N^m
= β x = m + β β ( e β N ) ( x β m )! N ( x β m ) β N m = N m ο»Ώ
c) so to find the mean <x> we can write it as a factorial moment from earlier but m = 1 so basically it will be only one term. That way <x> = N or we can write the whole thing over again and start the summer from x=1 and splitting N into NxN^(x-1)
To find the variance = Ο 2 = < x 2 > β < x > 2 \sigma^2 = <x^2>-<x>^2 Ο 2 =< x 2 > β < x > 2 ο»Ώ
Starting with finding the first term of the variance
= β x = 0 + β ( x 2 ) e β N N x x ! = β x = 0 + β ( x ) e β N N ( x β 1 ) N 1 ( x β 1 ) ! = N e β N β x = 1 + β ( x β 1 + 1 ) N ( x β 1 ) ( x β 1 ) ! = N e β N ( β x = 1 + β ( x β 1 ) N ( x β 1 ) ( x β 1 ) ! + β x = 1 + β ( 1 ) N ( x β 1 ) ( x β 1 ) ! ) = N e β N ( N e N + e N ) = N 2 + N = \sum_{x=0}^{+\infty}(x^2)e^{-N}\frac{N^x}{x!} \\ = \sum_{x=0}^{+\infty}(x)e^{-N}\frac{N^{(x-1)}N^1}{(x-1)!} \\ = Ne^{-N} \sum_{x=1}^{+\infty}(x-1+1)\frac{N^{(x-1)}}{(x-1)!} \\ = Ne^{-N} (\sum_{x=1}^{+\infty}(x-1)\frac{N^{(x-1)}}{(x-1)!}+\sum_{x=1}^{+\infty}(1)\frac{N^{(x-1)}}{(x-1)!}) \\ = Ne^{-N}(Ne^{N}+e^N) \\ = N^2+N = x = 0 β + β β ( x 2 ) e β N x ! N x β = x = 0 β + β β ( x ) e β N ( x β 1 )! N ( x β 1 ) N 1 β = N e β N x = 1 β + β β ( x β 1 + 1 ) ( x β 1 )! N ( x β 1 ) β = N e β N ( x = 1 β + β β ( x β 1 ) ( x β 1 )! N ( x β 1 ) β + x = 1 β + β β ( 1 ) ( x β 1 )! N ( x β 1 ) β ) = N e β N ( N e N + e N ) = N 2 + N so the variance is Ο 2 = N 2 + N β N 2 = N \sigma^2 = N^2+N-N^2 = N Ο 2 = N 2 + N β N 2 = N ο»Ώ Ο = ( N ) \sigma = \sqrt(N) Ο = ( β N ) ο»Ώ
so Ο < x > = 1 ( N ) \frac{\sigma}{<x>} = \frac{1}{\sqrt(N)} < x > Ο β = ( β N ) 1 β ο»Ώ
Problem 3.2 3.19 we just derived above - estimate of the expected error in a counting measurement
For 1% , N = (1/0.01)^2 = 10^4
For 1 part in million , N = (1/10^-6)^2 = 10^12
Energy per photon = h*f = 6.626x10^(-34) x 4x10^14 = 26.504x10^(-20)
So for 1% = 26.504x10^-16
For 1 part in million = 26.504x10^(-20)x10^12 = 26.504x10^(-8) W
Problem 3.3 a) SNR = 20dB
20 = 20 l o g 10 v s i g n a l v n o i s e = 10 v n o i s e = v s i g n a l = v s i g n a l R M S = 10 β ( 4 k T R d F ) = 10 β ( 4 β 1.38 β 1 0 ( β 23 ) β 293 β 1 0 4 β 20 β 1 0 3 = 1.7985 β 1 0 β 5 20 = 20 log_{10}\frac{v_{signal}}{v_{noise}} \\ = 10v_{noise} = v_{signal} \\ = v_{signalRMS} = 10*\sqrt(4kTRdF) \\ = 10* \sqrt(4*1.38*10^(-23)*293*10^4*20*10^3 \\ = 1.7985*10^{-5} 20 = 20 l o g 10 β v n o i se β v s i g na l β β = 10 v n o i se β = v s i g na l β = v s i g na lRMS β = 10 β ( β 4 k TR d F ) = 10 β ( β 4 β 1.38 β 1 0 ( β 23 ) β 293 β 1 0 4 β 20 β 1 0 3 = 1.7985 β 1 0 β 5 ο»Ώ
b) PE energy in circuit with capacitance = c v 2 / 2 = k T / 2 = cv^2 /2 = kT/2 = c v 2 /2 = k T /2 ο»Ώ
v 2 = k T / C C = 1.38 β 1 0 β 23 β 293 / 1.79 β 1.79 β 1 0 β 10 = 1.26 β 1 0 β 12 v^2 = kT/C \\ C = 1.38*10^{-23} *293 / 1.79*1.79*10^-10 \\ = 1.26*10^{-12} v 2 = k T / C C = 1.38 β 1 0 β 23 β 293/1.79 β 1.79 β 1 0 β 10 = 1.26 β 1 0 β 12 ο»Ώ
c) \sqrt(2*q*I*df) = 1/100*I β I^2*10^{-4} = 2*q*I*20 β I = 2*q*20*10^3*10^4 = 3.2*10^{-19}*20*10^7 = 6.4*10^-9