Therefore the pole locations are 1933 2202 1 5609

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Therefore the pole locations are: ) 1933 . 0 ( 2202 . 1 ) 5609 . 0 ( 1007 . 1 ) 8736 . 0 ( 8736 . 0 ) 1007 . 1 ( 5609 . 0 ) 2202 . 1 ( 1933 . 0 9 1 0 2354 . 1 10 , 9 8 , 7 6 , 5 4 , 3 2 , 1 20 ) 1 2 ( 2 j s j s j s j s j s ,..., , k e e s k j j k ± - = ± - = ± - = ± - = ± - = = = + π π The transfer function is given by
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) 5262 . 1 4404 . 2 )( 5262 . 1 2015 . 2 ( 1 ) 5262 . 1 7471 . 1 )( 5262 . 1 1217 . 1 )( 5262 . 1 3865 . 0 ( 2809 . 8 ) ( ) ( ) ( ) ( 2 2 2 2 2 9 0 9 0 + + + + × + + + + + + = - = - = = = s s s s s s s s s s s s s s G s H k k N c k k The digital filter is obtained by replacing s by 1 1 2 + - = z z s , which has the transfer function given by ) 1611 . 0 1127 . 0 1 )( 1106 . 0 1195 . 0 1 ( 1 ) 0046 . 0 135 . 0 1 )( 2223 . 0 1642 . 0 1 )( 6404 . 0 2204 . 0 1 ( ) 1 ( 00071343 . 0 ) ( 2 1 2 1 2 1 2 1 2 1 10 1 - - - - - - - - - - - - - - - × + - + - + - + = z z z z z z z z z z z z H (2)Answer Since we are using the impulse invariance method, there is no need for us to prewrap the edge frequencies. Therefore adopting the same specification for the digital lowpass filter, we can write the requirement for the Butterworth analog filter as = = = = 5708 . 1 2 01 . 0 ) ( 0472 . 1 3 0 1 ) ( 89126 . 0 s p π π j H j H Using the two frequency edges, we have 4 2 2 2 2 10 ) 01 . 0 ( ) / 5708 . 1 ( 1 2589 . 1 ) 89126 . 0 ( ) / 0472 . 1 ( 1 = + = = + - - N c N c . By dividing one equation by the other, we obtain the order and cutoff frequency of the filter as 099 . 1 ) 1 2589 . 1 /( 0472 . 1 14 023 . 3 1 N 1 10 1 2589 . 1 ) 5708 . 1 / 0472 . 1 ( 28 / 1 c 4 2 = - = - - N .
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