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Unformatted text preview: EE 562a Homework Solutions 1 21 Jan 2007 1 1. Concave (convex ) functions have the property that every chord of the function lies below the function, as indicated in the defining equation. g ( x + (1 ) y ) g ( x ) + (1 ) g ( y ) for all x,y R It should be clear that a concave function g ( x ) has a first derivative g ( x ) that is a monotone decreasing function of x , and hence the second derivative g 00 ( x ) is nonpositive. This means that tangent lines to the function g ( x ) at any point x o lie on or above the function g ( x ). Hence g ( x ) g ( x o ) + g ( x o )( x x o ) . We can put in x ( u ) for the variable x above and ensemble averaging both sides maintains the sense of the inequality. Then ensembleaveraging gives E { g ( x ( u )) } g ( x o ) + g ( x o )( E { x ( u ) }  x o ) . The result will of course be a function of the choice of x o . This can be simplified somewhat by choosing x o = E { x ( u ) } , yielding the result E { g ( x ( u )) } g ( E { x ( u ) } ) . 2. Since the process y ( u,t ) is defined recursively by y ( u, 1) = (1 p ) 1 / 2 x ( u, 1) , and y ( u,t ) = p 1 / 2 y ( u,t 1) + (1 p ) 1 / 2 x ( u,t ) , t 2 , the first step is to get an explicit expression for the y s in terms of the x s: y ( u,t ) = (1 p ) 1 / 2 t X j =1 p ( t j ) / 2 x ( u,j ) . Recall that E { x ( u,t ) } = 0 and E { x ( u,t 1 ) x ( u,t 2 ) } = t 1 t 2 . Then m y ( t ) = E { y ( u,t ) } = E (1 p ) 1 / 2 t X j =1 p ( t j ) / 2 x ( u,j ) = (1 p ) 1 / 2 t X j =1 p ( t j ) / 2 E { x ( u,j ) } = 0 , so all the random variables y ( u,t ) have zero mean. Without loss of generality, lets assume that t 1 t 2 . The correlation functions are then R y ( t 1 ,t 2 ) = E { y ( u,t 1 ) y ( u,t 2 ) } = E (1 p ) 1 / 2 t 1 X j =1 p ( t 1 j ) / 2 x ( u,j ) (1 p ) 1 / 2 t 2 X k =1 p ( t 2 k ) / 2 x ( u,k ) ! = (1 p ) t 1 X j =1 t 2 X k =1 p ( t 1 + t 2 j k ) / 2 E { x ( u,j ) x ( u,k ) } 2 EE 562a Homework Solutions 1 21 Jan 2007 = (1 p ) t 1 X j =1 t 2 X k =1 p ( t 1 + t 2 j k ) / 2 jk = (1 p ) t 2 X j =1 p ( t 1 + t 2 ) / 2 j = (1 p ) p ( t 1 t 2 ) / 2 t 2 X j =1 p t 2 j = (1 p ) p ( t 1 t 2 ) / 2 t 2 1 X k =0 p k . We can now use the identity t 1 X k =0 p k = 1 p t 1 p , to get the final expression R y ( t 1 ,t 2 ) = (1 p ) p ( t 1 t 2 ) / 2 1 p t 2 1 p = p ( t 1 t 2 ) / 2 p ( t 1 + t 2 ) / 2 . 3. (a) The AM signal is given by z ( u,t ) = x ( u,t ) cos(2 f t + ( u )) , where x ( u,t ) k ( u ) , t , and ( u ) is uniformly distributed on (0 , 2 ]. First, we find the mean = m z ( t ) = E { z ( u,t ) } = E { x ( u,t ) cos(2 f t + ( u )) } = E { x ( u,t ) } E { cos(2 f t + ( u )) } , where we used the fact that, since ( u ) k x ( u,t ) for all t , we have that x ( u,t ) and w ( u,t ) = cos(2 f t + ( u )) are uncorrelated real random processes (i.e. R xw ( t 1 ,t 2 ) = m x ( t...
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 Spring '07
 ToddBrun

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