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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 eigenfunction so that the system evalue can be found = x ( t ) = e i 2 ft = y ( t ) = M ( f ) e i 2 ft . You can also use the differentiation property of the FT to convert the differential equation into a frequency domain equation = [(i2 f ) 2 + 3(i2 f ) + 2] Y ( f ) = [3 (i2 f )] X ( f ) , so that M ( f ) = Y ( f ) X ( f ) = 3 (i2 f ) (i2 f ) 2 + 3(i2 f ) + 2 = 3 (i2 f ) (1 + i2 f )(2 + i2 f ) . (b) From the tables, we have S x ( f ) = F 1 8 e 4   = 1 16 + (2 f ) 2 . Therefore the output PSD is S y ( f ) =  M ( f )  2 S x ( f ) = 9 + (2 f ) 2 [4 + (2 f ) 2 ][1 + (2 f ) 2 ][16 + (2 f ) 2 ] . The crossPSD between the input and output of the system is S xy ( f ) = M * ( f ) S x ( f ) = 3 + i2 f (2 i2 f )(1 i2 f )[16 + (2 f ) 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 = 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 minphase 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 halfplane (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 ( ) 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 noncausal 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 ( ) for 0 since C { P ( f ) } = F { p ( )u( ) } . Since the FT of p ( ) exists we know that the ROC contains the imaginary axis = p ( ) = X poles in LHP Res[ e s P ( s )] . This is easy to compute since e s P ( s ) has only one pole in the LHP (i.e. at s = 4). It follows that p ( ) = 1 7 e 4 , so that C { P ( f ) } = 1 7 F { e 4 u( t ) } = 1 7(4 + i2 f ) ....
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 Spring '07
 ToddBrun

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