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Unformatted text preview: ECE 535 DISCRETE TIME SYSTEMS SPRING 2010 HOMEWORK ASSIGNMENT #5 Solutions 1. (a) We determine values of = 0 . 6901 and n = 3 . 3328. We then get H ( z ) = . 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 ), Ragazzinis 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- . 4663 z 2- . 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 ) = . 1544 z + 0 . 1256 z 2- 1 . 1185 z + 0 . 3985 . Note that having done this, Ragazzinis method simplifies because n H ( z ) = n G ( z ): D ( z ) = d G ( z ) n H ( z ) n G ( z )( d H ( z )-...
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This document was uploaded on 02/08/2012.
- Spring '09