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Unformatted text preview: EE 562a Homework Solutions 7 November 17, 2009 1 1. A valid correlation function should satisfy the following two requirements: • Hermitian symmetric: R x ( m ) = R * x ( m ) • Non negative definite (nnd) ⇐⇒ S x ( f ) ≥ ∀ f . We also proved the CauchySchwartz inequality for correlation functions, which for WSS processes is  R x ( m )  ≤ R x (0) ∀ m ∈ T . This can be proved by using the nonnegative definite (NND) property for correlation func tions and a twopoint sum. So if a function does not satisfy the CS inequality, then the function is not NND and the function is not a correlation function: but the converse is not true, namely, satisfying the CS inequality does not mean that the function is NND and is a correlation function. (a) R a ( m ) = 1 m = 1 , , 1 0 otherwise. This function is clearly Hermitian symmetric; it also satisfies CauchySchwartz. Next, we check the NND property by finding the PSD (i.e. the eigenvalues of the WSS operator) = ⇒ S a ( f ) = ∞ X m =∞ R a ( m ) e i 2 πfm = 1 + 2cos(2 πf ) . Since S a ( f ) is not always greater than or equal to zero (e.g. S a (1 / 2) = 1), R a ( m ) is not non negative definite, and therefore is not a valid correlation function. (b) R b ( m ) =  m  + 1  m  ≤ 3 3  m   3   m  = 4 , 5 , 6 , 7  m  > 7 Since R b ( m ) = R b ( m ), it is Hermitian symmetric. However, R b (0) = 1 and R b (1) = 2 > R b (0), so this function doesn’t satisfy the CauchySchwartz Inequality. Therefore, it is not a valid correlation function. (c) R c ( m ) = ( 1 2 )  m  cos( π 4 m ) . This function is Hermitian symmetric due to the even property of the cosine. The candidate for a PSD is found by taking the F Z R c ( m ), which may be obtained in several ways. Here’s one: R c ( m ) = 1 2  m  cos( π 4 m ) = 1 2 1 2  m  e i π 4 m + 1 2 1 2  m  e i π 4 m 2 EE 562a Homework Solutions 7 November 17, 2009 S c ( f ) = 1 2 1 1 4 5 4 cos(2 πf π 4 ) + 1 2 1 1 4 5 4 cos(2 πf + π 4 ) = 3 8 " 1 5 4 cos(2 πf π 4 ) + 1 5 4 cos(2 πf + π 4 ) # ≥ . So R c ( m ) is nonnegative definite, and thus is a valid correlation function. Here we used the “frequency shift property” of F Z : e i 2 πf n x ( n ) ↔ X ( f f ) . The PSD can also be easily found using the “modulation property” of F Z . Since this is a valid correlation function, we must find the probability density of x ( u,n ). Since the process is Gaussian, x ( u,n ) is a Gaussian random variable for any value of n . We only need its mean and variance to get its density. m x = 0 since random sequence x ( u,n ) has zero mean σ 2 x ( u,n ) = E { x ( u,n ) x ( u,n ) } = R x (0) = 1 = ⇒ f x ( u,n ) ( z ) = 1 √ 2 π e z 2 / 2 ....
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This note was uploaded on 04/03/2010 for the course EE 562a taught by Professor Toddbrun during the Fall '07 term at USC.
 Fall '07
 ToddBrun

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