Revision as of 17:13, 27 February 2013

<Question>

From 555 Signal Counter 3V binary signal is being sent out. The counter has noise levels from -2 V to 2 V with 1V difference. After the counter has sent out a random signal, each noise level has probability of {1/10,2/10,4/10,2/10,1/10}. The signal goes through a filter, Z=2X^2+1. Find E[Z] and Var[Z].

<Solution>

O: Original Signal from the counter(send out either 3V or 0V)
N: Noise level
X: output from the counter
Z: output after the filter

O N X Z

3 -2 1 3
3 -1 2 9
3 0 3 19
3 1 4 33
3 2 5 51
0 -2 -2 9
0 -1 -1 3
0 0 0 1
0 1 1 3
0 2 2 9

First We need to find the Probability for each cases of X

P[X=-2]=P[N=-2]=1/10
P[X=-1]=P[N=-1]=2/10
P[X= 0]=P[N= 0]=4/10
P[X= 1]=P[N=-2]+P[N= 1]=1/10+2/10=3/10
P[X= 2]=P[N=-1]+P[N= 2]=2/10+1/10=3/10
P[X= 3]=P[N= 0]=4/10
P[X= 4]=P[N= 1]=2/10
P[X= 5]=P[N= 2]=1/10

To find the expected value of Z we need to find the expected value of X first.

$ E(X)= \sum_{k} g(xk)px(xk) $

E[X]=(-2)*1/10 +(-1)*2/10 +(0)*4/10 +(1)*3/10 +(2)*3/10 +(3)*4/10 +(4)*2/10 +(5)*1/10=3

To find the expected values of X^2
E[X^2]=4*1/10+1*2/10+0*4/10+1*3/10+4*3/10+9*4/10+16*2/10+25*1/10=11.4

E[Z]=E[2x^2+1]=2*E[X^2]+1=23.8

To find the variance value of Z We have to find values of E[Z^2] and E[Z].
E[X^4]=16*1/10+1*2/10+0*4/10+1*3/10+16*3/10+81*4/10+256*2/10+625*1/10=153

E[Z^2]=E[(2*X^2+1)^2]=E[4X^4+4X^2+1]=4E[X^4]+4E[X^2]+1=4*153+4*11.4+1=658.6

Var[Z]=E[Z^2]-E[Z]^2=658.6-23.8^2=92.16

Questions

Back to third bonus point opportunity, ECE302 Spring 2013