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ECE535_10S_hw5soln

# ECE535_10S_hw5soln - ECE 535 DISCRETE TIME SYSTEMS HOMEWORK...

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ECE 535 DISCRETE TIME SYSTEMS SPRING 2010 HOMEWORK ASSIGNMENT #5 Solutions 1. (a) We determine values of ζ = 0 . 6901 and ω n = 3 . 3328. We then get H ( z ) = 0 . 1615 z + 0 . 1185 z 2 - 1 . 1185 z + 0 . 3985 . With H ( z ) = n H ( z ) /d H ( z ) and G ( z ) = n G ( z ) /d G ( z ), Ragazzini’s method specifies D ( z ) = d G ( z ) n H ( z ) n G ( z ) ( d H ( z ) - n H ( z )) . For our values we get D ( z ) = 9 . 8356 z 3 - 7 . 8198 z 2 - 5 . 7468 z + 3 . 8834 z 3 - 0 . 4663 z 2 - 0 . 7615 z + 0 . 2278 , The step response is plotted in Figure 1, and we can see that we have achieved the specifications (quite tightly, as per the design). (b) We now replace the numerator of H ( z ) with a scaled version of the numerator of G ( z ), with the scale factor chosen so that the DC gain of H ( z ) is unity. This makes H ( z ) = 0 . 1544 z + 0 . 1256 z 2 - 1 . 1185 z + 0 . 3985 . Note that having done this, Ragazzini’s method simplifies because n H ( z ) = α n G ( z ): D ( z ) = d G ( z ) n H ( z ) n G ( z ) ( d H ( z ) - n H ( z )) = αd G ( z ) d H ( z ) - n H ( z ) . The design results in D ( z ) = 9 . 4031 z 2 - 14 . 3775 z + 5 . 0584 z 2 - 1 . 2729 z + 0 . 2729 , which is lower order than D ( z ) in (a). The step response is plotted in Figure 2, and it still meets the specifications.

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ECE535_10S_hw5soln - ECE 535 DISCRETE TIME SYSTEMS HOMEWORK...

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