SolSet7 - EE 562a Homework Solutions 7 1 1 A valid...

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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 ) 0 f . We also proved the Cauchy-Schwartz inequality for correlation functions, which for WSS processes is | R x ( m ) | ≤ R x (0) m ∈ T . This can be proved by using the non-negative definite (NND) property for correlation func- tions and a two-point sum. So if a function does not satisfy the C-S inequality, then the function is not NND and the function is not a correlation function: but the converse is not true, namely, satisfying the C-S inequality does not mean that the function is NND and is a correlation function. (a) R a ( m ) = 1 m = - 1 , 0 , 1 0 otherwise. This function is clearly Hermitian symmetric; it also satisfies Cauchy-Schwartz. 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 + 2 cos(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 0 | 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 Cauchy-Schwartz 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
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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 ) # 0 . So R c ( m ) is non-negative definite, and thus is a valid correlation function. Here we used the “frequency shift property” of F Z : e i 2 πf 0 n x ( n ) X ( f - f 0 ) . 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 . Notice that the pdf is not a function of n . To find the pdf of x ( u, n + 1) given x ( u, n ), we can use the general conditional Gaussian density (because x ( u, n ) and x ( u, n + 1) are jointly-Gaussian for all choices of n ). It follows that p x ( u,n +1) | x ( u,n ) ( w | v ) = N 1 ( w ; m ( n +1) | n ( v ); K ( n +1) | n ) , where m ( n +1) | n ( v ) and K ( n +1) | n are the conditional mean and variance respectively. This is a 1-dimensional example of the more general theory.
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