Question

Problem 01 A Gaussian random voltage having a standard deviation of 2V is applied to a resistance of 40. a) If the mean Voltage is 5VDC., find the probability that the Voltage will be below a Threshold of 2VDC. b) If the mean Voltage is 0.5VDC., find the probability that the Voltage will be above a Threshold of 2VDC. c) If the mean Voltage is zero, what the probability that the power dissipated in the 40 resistor is between 2 W and 2.5 W (use the power density function) d) If the mean Voltage is 6 Volts find 1) The probability that the voltage is between 8 Volts and 9 Volts. 2) The variance of the current through the 40 resistor 3) The probability that the current through the resistor is greater than 2.5 Amps.

Answer 1

Sol - P(x) = . 12102 é - Cam - (x-m) ? 202 X we wually use the shorthand form W(m, 2) to denote the Pd F of Gaussian random variable and write N(1,02), for this random Variable E [x] = m [mean] var [x] = o I variance] A Gaussian random variable with m=0 and a = 1 is called Standard normal . A function related to the craussian random variable is the a function defined as 8()P[w(0,1)><]3 + lentadt we have Introduce the change of variable u= (tm) In general it XnW(m, ²) then Resistor Plx>&J: 0 (2-) P[XCQJ = 8 (-2) (a) mean = 5V , 2v (given) [ Random variable is voltage across P[*> ev]: [Sv-2v): 6 ()=9(5)0-066807 (b) mean.sv, o=2v P [x<2v): 1- 0(-05)1-0 (2-05) 21-01-75) - 1-0.22663 - 0.77337

2) mean = 0 o = 2v v2 =DW:V: Vuxe T8 - 282V 4.2 = 2.5 = .5x4 = 10 = 3.1622 P[a 2.52CX 23:16227 = P [X23-1622] - P[X (2.82] - 14-0051622) J-[-6(3:22)] :-(15811) +8 (141) = 8 (1-41) - 8 (1-5811) *(1-) - 8(16) = 0.088508 -0.0547gg = 0.03352 (d) ma 6v, o=20 (1) P[ 8 (XL9] = P[xL g]- P[X28] - P[*>8] - PCX> 97 20(5-)-0(976) - 8 (1) 8 (15) = 0.158660 -- 0-066807 = 0.091853

2v variance variance to ( 2 40 His1W (3) mean - W - ISA, mean z - )-SA o 20.3 z 0.5 P [x>25= 3(25-15): 812) 2 0.0 22750 05