assignment7_solutions

# assignment7_solutions - EE 505 B Fall 2011 Assignment 7...

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Unformatted text preview: EE 505 B Fall 2011 Assignment 7 Solutions 1. (a) E { X } = integraldisplay ∞ − ∞ x f X ( x ) dx = integraldisplay b a x b − a dx = 1 b − a bracketleftbigg x 2 2 bracketrightbigg b a = 1 2 ( b − a ) ( b 2 − a 2 ) = ( b − a )( b + a ) 2 ( b − a ) = b + 2 2 (b) We use the direct method to find the probability density function for Y : f Y ( y ) = ∑ i f X ( x ) vextendsingle vextendsingle vextendsingle d dx g ( x ) vextendsingle vextendsingle vextendsingle x = x i . There are two values of x that satisfy the equation y = x 2 : x 1 = − √ y x 2 = √ y f X ( x 1 ) = because x 1 is a negative number, and f X ( x 2 ) = braceleftBigg 1 b − a a < √ y < b otherwise The denominator is vextendsingle vextendsingle vextendsingle vextendsingle d dx g ( x ) vextendsingle vextendsingle vextendsingle vextendsingle x = x 2 = 2 √ y Putting this all together, f Y ( y ) = braceleftBigg 1 2 ( b − a ) √ y a 2 < y < b 2 otherwise 1 We can now evaluate the expected value of Y : E { Y } = integraldisplay ∞ − ∞ yf Y ( y ) dy = integraldisplay b 2 a 2 y 2 ( b − a ) √ y dy = 1 2 ( b − a ) integraldisplay b 2 a 2 y 1 2 dy = 1 2 ( b − a ) bracketleftbigg 2 3 y 3 2 bracketrightbigg b 2 a 2 = 1 3 ( b − a ) ( b 3 − a 3 ) = 1 3 ( a 2 + ab + b 2 ) (c) This method is much easier. E { Y } = E { X 2 } = integraldisplay ∞ − ∞ x 2 f X ( x ) dx = integraldisplay b a x 2 b − a dx = 1 b − a bracketleftbigg x 3 3 bracketrightbigg b a = 1 3 ( b − a ) ( b 3 − a 3 ) = 1 3 ( a 2 + ab + b 2 ) 2. (a) E { Y } = E { a cos ( ω t + Θ ) } = integraldisplay ∞ − ∞ bracketleftbig a cos ( ω t + θ ) bracketrightbig f Θ ( θ ) d θ = a integraldisplay 2 π 1 2 π cos ( ω t + θ ) d θ = a 2 π bracketleftbig sin ( ω t + θ ) bracketrightbig 2 π = 0 . 2 E braceleftBig Y 2 bracerightBig = E braceleftBig a 2 cos 2 ( ω t + Θ ) bracerightBig = E braceleftbigg a 2 2 + a 2 2 cos ( 2 ω t + 2 Θ ) bracerightbigg . Now, because the expectation operator is a linear operator, and because the expected value of a constant is a constant, we have E braceleftBig Y 2 bracerightBig = a 2 2 + a 2 2 integraldisplay ∞ − ∞ bracketleftbig cos ( 2 ω t + 2 Θ ) bracketrightbig f Θ ( θ ) d θ = a 2 2 + a 2 2 integraldisplay 2 π 1 2 π cos ( 2 ω t + 2 θ ) d θ = a 2 2 + a 2 4 π integraldisplay 2 π cos ( 2 ω t + 2 θ ) d θ = a 2 2 + a 2 4 π bracketleftbigg 1 2 sin ( 2 ω t + 2 θ ) bracketrightbigg 2 π = a 2 2 ....
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assignment7_solutions - EE 505 B Fall 2011 Assignment 7...

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