SolSet6 - EE 562a 1(a Homework Solutions 6 November 5 2009...

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EE 562a Homework Solutions 6 November 5, 2009 1 1. (a) G z ( z ) = X t = -∞ g ( t ) z - t = - 1 X t = -∞ y ( - t ) z - t = X t 0 =1 y ( t 0 ) z t 0 = - y (0) + X t 0 =0 y ( t 0 ) 1 z - t 0 = - y (0) + Y z 1 z The subtraction of y (0) does not affect the convergence of Y z ( 1 z ) , which converges when 1 z > a , or equivalently, when | z | < 1 a . (b) H z ( z ) = X t = -∞ h ( t ) z - t = - 1 X t = -∞ y ( - t - 1) z - t = X t 0 =0 y ( t 0 ) z t 0 +1 = z X t 0 =0 y ( t 0 ) 1 z - t 0 = z Y z 1 z Multiplication by z does not affect the convergence of Y z ( 1 z ) , which (as in (a)) converges when | z | < 1 a . (c) Assuming that the z -transform of g c ( t ) has a non-empty region of convergence, G c z ( z ) = X t = -∞ [ g ( t ) + y ( t )] z - t = G z ( z ) + Y z ( z ) = - y (0) + Y z 1 z + Y z ( z ) . For this to be a two-sided z -transform, z must be in the range a < | z | < 1 a so that both transforms exist in the same region of the z -plane, and hence we must have a < 1 for the two-sided transform G c z ( z ) to exist within a non-empty region of convergence. (d) X z ( z ) = z τ X t =0 y ( t ) z - t = X t =0 y ( t ) z - t + τ = X t 0 = - τ y ( t 0 + τ ) z - t 0 and hence x ( t ) = y ( t + τ ) , t ≥ - τ 0 , otherwise. Certainly | X z ( z ) | = | z | τ | Y z ( z ) | and and therefore the sum in X z ( z ) converges if and only if the sum in Y z ( z ) converges. Therefore z is in the region of convergence of X z ( z ) if and only if | z | > a . 2. Solution: From the definition of the discrete-time Fourier transform, X ( f ) = X t = -∞ x ( t ) e - i 2 πft . Using this we have Z 1 / 2 - 1 / 2 X ( f ) Y * ( f ) df = Z 1 / 2 - 1 / 2 X t = -∞ x ( t ) e - i 2 πft ! Y * ( f ) df
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2 EE 562a Homework Solutions 6 November 5, 2009 = X t = -∞ x ( t ) Z 1 / 2 - 1 / 2 e - i 2 πft Y * ( f ) df ! = X t = -∞ x ( t ) Z 1 / 2 - 1 / 2 e i 2 πft Y ( f ) df ! * = X t = -∞ x ( t ) y * ( t ) 3. Solution: (a) z ( u, t ) = ax ( u, t ) + by ( u, t ) = m z ( t ) = m z = am x + bm y , and using centered notation, K z ( t + τ, t ) = K z ( m ) = E { z o ( u, t + τ ) z * o ( u, t ) } = E { [ ax o ( u, t + τ ) + by o ( u, t + τ )][ a * ( x * o ( u, t ) + b * y * o ( u, t )] } = | a | 2 K x ( τ ) + | b | 2 K y ( τ ) + a * b E { x * o ( u, t ) y o ( u, t + τ ) } + ab * E { x o ( u, t + τ ) y * o ( u, t ) } = | a | 2 K x ( τ ) + | b | 2 K y ( τ ) e K z ( t + τ, t ) = e K z ( τ ) = E { ( z ( u, t + τ ) - m z )( z ( u, t ) - m z ) } = a 2 e K x ( τ ) + b 2 e K y ( τ ) . where the fact that x ( u, t ) and y ( u, t 0 ) are independent for all integers t, t 0 , was used factor the expectations in the cross-covariance and psuedo-cross-covariance derivations. The PSD is found by taking the discrete-time Fourier transform of the correlation function: S z ( f ) = F { R z ( τ ) } = F { K z ( τ ) + | m z | 2 } = F {| a | 2 K x ( τ ) + | b | 2 K y ( τ ) + | am x + bm y | 2 } = F {| a | 2 ( K x ( τ ) + | m x | 2 ) + | b | 2 ( K y ( τ ) + | m y | 2 ) + 2 Re ab * m x m * y } = | a | 2 S x ( f ) + | b | 2 S y ( f ) + 2 Re ab * m x m * y δ D ( f ) , where the above holds for f [ - 1 / 2 , 1 / 2] (periodic with period 1). (b) This is the product of two random sequences m z ( t ) = m z = E { x ( u, t ) y ( u, t ) } = E { x ( u, t ) } E { y ( u, t ) } = m x m y R z ( t + τ, t ) = R z ( τ ) = E { ( x ( u, t + τ ) y ( u, t + τ ))( x ( u, t ) y ( u, t )) * } = E { x ( u, t + τ ) x * ( u, t ) } E { y ( u, t + τ ) y * ( u, t ) } = R x ( τ ) R y ( τ ) K z ( τ ) = R z ( τ ) - | m z | 2 = R x ( τ ) R y ( τ ) - | m x m y | 2 = ( K x ( τ ) + | m x | 2 )( K y ( τ ) + | m y | 2 ) - | m x | 2 | m y | 2 = | m y | 2 K x ( τ ) + | m x | 2 K y ( τ ) + K x ( τ ) K y ( τ ) . Similarly, e K z ( τ ) = m 2 y e K x ( τ ) + m 2 x e K y ( τ ) + e K x ( τ ) e K y ( τ ) .
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EE 562a Homework Solutions 6 November 5, 2009 3 Since the correlation function of z ( u, t ) is the product of the correlation functions for x ( u, t ) and y ( u, t ), the PSD is found using the modulation property of the transform: S z ( f ) = S x ( f ) * S y ( f ) = Z 1 / 2 - 1 / 2 S x ( f λ ) S y ( λ ) dλ.
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