HW01 - University of Illinois Spring 2011 ECE 361: Problem...

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Unformatted text preview: University of Illinois Spring 2011 ECE 361: Problem Set 1: Solutions Decision-Making and Matched Filtering 1. [Crosscorrelations and Autocorrelations] (a) R x, y ( ) = Z - x ( t + ) y * ( t ) dt = Z - x ( t- + ) y * ( t- ) dt = Z - x ( + ) y * ( ) d = R x,y ( ) on substituting = t- . (b) Writing x ( t + ) = Z - X ( f )exp( j 2 ( t + ) f ) df in the definition of R x,y ( ), we get R x,y ( ) = Z - x ( t + ) y * ( t ) dt = Z - Z - X ( f )exp( j 2 ( t + ) f ) df y * ( t ) dt = Z - X ( f )exp( j 2 f ) Z - y ( t )exp(- j 2 ft ) dt * df = Z - [ X ( f ) Y * ( f )]exp( j 2 f ) df that is, R x,y ( ) X ( f ) Y * ( f ). (c) Since R x,y ( ) = 0 for all , we have that F{ R x,y ( ) } = 0 for all f . Hence it must be that X ( f ) Y * ( f ) = 0 for all f , that is, for each f , at least one of X ( f ) and Y ( f ) must be zero. In other words, the supports of X ( f ) and Y ( f ) are disjoint sets. In more engineering terms, the signals x ( t ) and y ( t ) are uncorrelated if and only if they occupy non-overlapping frequency bands . (d) R x (0) = Z - x ( t ) x * ( t ) dt = Z - | x ( t ) | 2 dt is the energy in the signal x ( t ) and is thus greater than zero except for (uninteresting) zero-energy signals. Next, if x ( t ) is real-valued, then x * ( t ) = x ( t ), and so R x ( ) = Z - x ( t + ) x * ( t ) dt = Z - x ( t + ) x ( t ) dt is a real-valued function. Fur- thermore, R x (- ) = Z - x ( t- ) x ( t ) dt = Z - x ( ) x ( + ) d = R x ( ) (upon making a change of variable = t- ), and thus R x ( ) is a real-valued even function of . Next, note that F{ R x ( ) } = S x ( f ) = | X ( f ) | 2 is a real-valued function of f , and since Fourier trans- forms of real functions have conjugate symmetry (that is, X (- f ) = X * ( f )), we have that S x (- f ) = | X (- f ) | 2 = | X * ( f ) | 2 = | X ( f ) | 2 = S x ( f ), and so S x ( f ) is a real-valued even func- tion of f . (e) Obviously, Z - [ x ( t ) x ( t + )] 2 dt 0. On expanding out the integrand and using the properties developed above, we get 2[ R x (0) R x ( )] 0, that is,- R x (0) R x ( ) R x (0)....
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HW01 - University of Illinois Spring 2011 ECE 361: Problem...

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