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soln8

# soln8 - EE 562a Homework Solutions 8 25 April 2007 1 1(a...

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Unformatted text preview: EE 562a Homework Solutions 8 25 April 2007 1 1. (a) This is a problem from your undergrad systems class. You can drive the system with an eigen-function so that the system e-value can be found = ⇒ x ( t ) = e i 2 πft = ⇒ y ( t ) = M ( f ) e i 2 πft . You can also use the differentiation property of the FT to convert the differential equation into a frequency domain equation = ⇒ [(i2 πf ) 2 + 3(i2 πf ) + 2] Y ( f ) = [3- (i2 πf )] X ( f ) , so that M ( f ) = Y ( f ) X ( f ) = 3- (i2 πf ) (i2 πf ) 2 + 3(i2 πf ) + 2 = 3- (i2 πf ) (1 + i2 πf )(2 + i2 πf ) . (b) From the tables, we have S x ( f ) = F 1 8 e- 4 | τ | = 1 16 + (2 πf ) 2 . Therefore the output PSD is S y ( f ) = | M ( f ) | 2 S x ( f ) = 9 + (2 πf ) 2 [4 + (2 πf ) 2 ][1 + (2 πf ) 2 ][16 + (2 πf ) 2 ] . The cross-PSD between the input and output of the system is S xy ( f ) = M * ( f ) S x ( f ) = 3 + i2 πf (2- i2 πf )(1- i2 πf )[16 + (2 πf ) 2 ] . (c) The first step in any estimation problem is to identify the observable and desirable signals; in this case the observation is y ( u,t ) and x ( u,t ) is desired. Since m x = m y = 0 (why?) the bias in the affine estimator is 0 = ⇒ ˆ x ( u,t ) = g opt ( t ) * y ( u,t ) , where the optimal causal filter is G opt ( f ) = 1 H ( f ) C S xy ( f ) H * ( f ) , where H ( f ) is the min-phase causal factor of S y ( f ). Thus, this problem requires the spectral factorization of S y ( f ). Putting S y ( f ) into pole/zero form and converting to the corresponding Laplace transform = ⇒ S y ( s ) = (3- s )(3 + s ) (2 + s )(2- s )(1 + s )(1- s )(4 + s )(4- s ) . The minimum phase causal factor is obtained by taking the poles and zeros in the left- half-plane (LHP) = ⇒ H ( s ) = (3 + s ) (2 + s )(1 + s )(4 + s ) . 2 EE 562a Homework Solutions 8 25 April 2007 The next step is to find the causal part. Let P ( f ) = S xy ( f ) H * ( f ) , so that P ( s ) = (3 + s ) (4 + s )(3- s ) . Notice that P ( s ) has a pole in the RHP, so p ( τ ) is not a causal signal (i.e. this means that the causal part operator affects the answer and that the causal solution is different than the non-causal solution). Now you can perform the partial fraction expansion (PFE) and use the result given in class: C 1 a + s = , Re { a } < 1 a + s , Re { a } > . Again, I find residue theory works best for me (actually residue theory and PFE are the same if you think about it). In this case we only need to find p ( τ ) for τ ≥ 0 since C { P ( f ) } = F { p ( τ )u( τ ) } . Since the FT of p ( τ ) exists we know that the ROC contains the imaginary axis = ⇒ p ( τ ) = X poles in LHP Res[ e sτ P ( s )] τ ≥ . This is easy to compute since e sτ P ( s ) has only one pole in the LHP (i.e. at s =- 4). It follows that p ( τ ) =- 1 7 e- 4 τ τ ≥ , so that C { P ( f ) } =- 1 7 F { e- 4 τ u( t ) } =- 1 7(4 + i2 πf ) ....
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soln8 - EE 562a Homework Solutions 8 25 April 2007 1 1(a...

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