ECE 650 Fall, /9 Homework Set 2 Solutions

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1 ECE 65 Fall, 14 1/9 Homework Set Solutions 1. Gamma with c = 1, from eq. 3.1a (tetbook): f() = ( / b) ep( / b) b () u() (1/b) ep(-/b) u(), which is eponential since gamma() = 1, either from MATLAB or from the lecture notes, because gamma(n) = (n-1)! for integer n. gamma pdf, with (b,c) = (,), (4,), (,4), (4, 4) Type: >> help gampdf to see which parameter is which, relative to the notation in our book. MATLAB notation: f( a, b) = (1/(b a (a)) a-1 e -/b So MATLAB s b parameter matches the tetbook b, while MATLAB s a parameter matches the tetbook c. Use MATLAB s gui: disttool to view pdf s with given parameters in order to make a decent choice for an appropriate range of values to plot. I choose to plot all 4 pdf s on the same set of values, to be able to compare the various shapes. MATLAB code: % program ece_65_hwset_ clear all = [:5]; y1 = gampdf(,,); y = gampdf(,,4); y3 = gampdf(,4,); y4 = gampdf(,4,4); subplot(4,1,1), plot(,y1) subplot(4,1,), plot(,y) subplot(4,1,3), plot(,y3) subplot(4,1,4), plot(,y4)

2 ECE 65 Fall, 14 /9. (b, c) = (, 4) (b,c) = (4, ) (b,c) = (,4) (b,c)=(4,4) (M&C, 3.31) R: Rayleigh with f R (r) = c r ep(-r ) u(r) a. Find constant c: Form-matching with Appendi D, D.47: f X () = (/ ) ep(- /( ) u() = 1 = ½ c = 1/ = Alternate approach: integrate the pdf, and set the integral equal to 1; solve for c. b. From App. D, D.48, with = ½: F R (r) = 1 ep(-r ) u(r) Alternate approach: integrate the pdf from = to = r. c. Using the definition of conditional probability:

3 ECE 65 Fall, 14 3/9 Pr(R 1, R ) Pr(1 R ) Pr(R > 1 R < ) = = Pr(R ) Pr(R ) Numerator: F() F(1) = (1-e -4 ) (1-e -1 ) = e -1 e -4 =.3496 Denominator: F() = (1- e -4 ) =.9817 Answer:.3496/.9817 = (M&C, 3.35b, with = 1 for the sake of the plot.) Given X: N(, ) f f X X <3 () = X (), X 3 Pr( 3 X 3) 1 ep( / ) =, X 3 (3) ( 3) Denominator: (3)-(-3) =normcdf(3,,1)-normcdf(-3,,1) =.9973, 1/denominator = = ep( / ), X 3 Therefore use = 1, to plot: 1.7 ep( / ), X 3 MATLAB Code: >> = -4:.1 : 4; >> f = (1.7/sqrt(*pi))*ep(-.^/).*(abs()<3); % using logical switch >> plot(,f) Note tails drop off abruptly at = +/- 3, due to given condition

4 ECE 65 Fall, 14 4/9 5. (M&C, 3.36a) (Hint: See Lecture 1, p. 5, and let A be the event: M =. From Lecture 1, p. 5, with A replaced by M = : Pr(M= X=) = fx M () Pr(M ) fx () = f f X M X M () f () X M1 () Total probability on denominator = ep ( ep ( / ) / ) ep ( ( 1) / ) 1 Pr(M = X = ) M&C, 3.39a) V is U(, ); f V (v), unconditional 1 f V V>1 (v) = f V (v)/pr(v>1) for v > 1, else = 1, 1 < v < ( else) ½ v

5 ECE 65 Fall, 14 5/9 7a. rate: 3.57 * 1 1 particles /sec = 35.7 particles/nanosec. k 35 7b. Pr(X = k = 35 particles) = e = e. 668 k! 35! 8. Pr(saturation) = 16 k e 16 k5 k! 4 16 k e 1 16 =.3 k k! 9. X: N( = 3); W db = 1 log 1 (W) = 1 log 1 (1 X/1 ) =1 X/1 = X; But W = 1 X/1 is a RV whose log (base 1) is X/1; log 1 (W) = X/1 = X/1 = X/1 ln(w) =.33 X ln(w) is N((.36), = (.33) 3), or ln(w) is N(.461,.1591) W is lognormal, with parameters of the Gaussian (mean.461, var..1591) f W 1 ln(w).461 (w) ep u(w).1591 w (.1591) (from Lect.1, p. 43) = ep w ln(w) u(w) 1. Pr(T > 1) = 1 F(1) = 1 - gamcdf(1, 1.5,.5) = f ( hot) = 1 ( 5) ep ; f 8 ( cool) = 8 1 ep ; 8 8 Over-all, by total probability: f () = f ( hot) (.) + f ( cool) (.8) MATLAB Code: >> = -8:.1 : 1; >> hotpdf = normpdf(,5,); >> coolpdf = normpdf(,,); >> totalpdf = hotpdf*. +coolpdf*.8; >> hold on >> plot(,hotpdf,'r')

6 ECE 65 Fall, 14 6/9 >> plot(,coolpdf,'g') >> plot(,totalpdf) MATLAB Plot(s): PDF s cool hot Overall 1. X: N(, 4). f X X > () = f X < X < (). Use eq. on p. 45 of Lect. 1, with a =, b = : From MATLAB: f () f () f () X X X normcdf(,, ) FX ( ) FX () 1 FX () ans =.8413 f () 6.31 f (), MATLAB Code: >> = -8:.1 : 8; >> y1 = normpdf(,,); >> y = 6.31*normpdf(,,).* ( > ); >> plot(,y1,,y,'r') MATLAB Plot: X X X, else

7 ECE 65 Fall, 14 7/ f X > (). f X () X (u) = E{e ux 1 4 4u 4u u u e e }= e f () d e d 8 4 8u X () = E{e jx j4 j4 1 e e sin(4) } = M (u) uj 4 j X (u) = E{e ux }= ½ e u + ½ e 4u ; X () = E{e jx }= ½ e j + ½ e 4j 15. For problem 1: E[X] = d du e 4u e 8u 4u u u(4e 4u 4e 4u 8u ) (e 4u e 4u ) u = / Use L Hospital s Rule: d/du (numerator of ) = u(16e 4u - 16e -4u ) d/du (denominator) = 16u Take the quotient & evaluate at ; E[X] = /16 = For problem : E[X] = {(d/du)[ ½ e u + ½ e 4u ]} u= = {e u + e 4u } u= = 3 E[X ] = {(d /du )[ ½ e u + ½ e 4u ]} u= = {(d/du)[e u + e 4u ]} u= = [e u + 8e 4u ] u= = 1 Check for problem : E[X] = 3 (symmetry about the mean in the pdf) E[X ] = (1/) 4 + (1/) 16 = + 8 = For problem 1: var[x] = 8 /1 = 64/1 = 16/3

8 ECE 65 Fall, 14 8/9 For problem : var[x] = E[X ] - = 1 (3) = X: N(, = ) is input to halfwave rectifier, so output y =, >= ( else) Note: the standard method: f Y (y) = f X () / dy/d works for y >= ; But for y = : Pr(Y = ) = Pr(X < ) = ½; so f Y (y) will need a delta function at the origin with area = ½. (This etra step is necessary whenever there is a flat region in the device characteristic. f Y (y) = f X () +.5 (y) y >= ( else) = 1 ep( y /8)u(y).5(y) Note: based on MATLAB plot, for < < 1, y ranges from infinity to. For y, y = -ln() has a single solution ( = e -y ). For y <, y has no roots. (y can t be negative) Also: dy/d = -1/ = -e y. And f () = 1, < < 1 f () 1 Thus, for y >, f (y) X Y U(y) dy / d y e = 65, = 3 P{ X 65 5 } /5 = 9/5 P{X 7} 9/5. X: U(, 5) a. Pr{X 1} /1 = 5/1 =.5 Note: This is a useless bound in this eample, since all probabilities are less than 1. b. Pr{X 4} /4 = 5/4 = 5/8 =.65 c. Pr{X 4} = 1/5 =..65 (This checks, since the actual probability is less than the upper bound on the probability.) 1a. Markov: Pr(X > 15) = Pr(X 151) E(X)/151 = 1/151 =.663 1b. Chebyshev: Pr( X -1 51) var(x)/51 = 5/51 =

9 ECE 65 Fall, 14 9/9. (mostly from Mi) Label the statements correct as is, or modify the statements as needed to make them correct. a. If events A and B are independent, then Pr(A B) = False should be: if A, B m.e., then Pr(A B) = b. If RV s are independent, then they are uncorrelated. True as is c. RV s X and Y are uncorrelated if E(XY) =. False should be: if X, Y uncorrelated, then E(XY) = E(X)E(Y) Or: if X, Y orthogonal, then E(XY) = d. The conditional cdf F X Y () = F( y) is the derivative (with respect to ) of the conditional pdf f X Y () = f( y). False should be the other way around: little f is the derivative of capital F e. The pdf for the n-dimensional Gaussian random vector can be written down if we know the means, the variances, and the covariances. True even better, we only need the means and the covariances (see tet p. 357) f. The central limit theorem says that the sum of a large number of RV s is Gaussian, even though none of the RV s may be Gaussian. Almost true most proofs require that the RV s be independent, but that is not always necessary; see discussion, p. 38 g. To apply Chebyshev s inequality to a RV X, we need to know the mean and variance of X. True as is h. The linear transformation of a Gaussian RV is Gaussian. True as is i. The sum of n Gaussian RV s is Gaussian. True as is j. If RV s are uncorrelated, then they are independent. False should be the other way around: independent implies uncorrelated Also: true as is if the RV s are Gaussian