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# 6 - ECE 483 Homework#6 Solution Problem 1 7-10(a(d 01873z 0...

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4/6/10 ECE 483 Homework #6 Solution Problem 1. 7-10 (a)-(d). Solution: Let G ( z ) = 0 . 01873 z +0 . 01752 ( z 1)( z 0 . 8187) . (a) The characteristic equation is 1 + KD ( z ) G ( z ) H = 1 + 0 . 07 K 0 . 01873 z + 0 . 01752 ( z 1)( z 0 . 8187) = 0 . (1) (b) Using the bilinear map z = 1 + ( T/ 2) w 1 ( T/ 2) w = 1 + 0 . 05 w 1 0 . 05 w = 20 + w 20 w , The characteristic equation is transformed to 1 + 0 . 07 K 0 . 01873 20+ w 20 w + 0 . 01752 ( 20+ w 20 w 1)( 20+ w 20 w 0 . 8187) = 0 1 + 0 . 07 K 0 . 01873(400 w 2 ) + 0 . 01752(20 w ) 2 2 w (1 . 8187 w + 3 . 626) = 0 (2) (3 . 6374 0 . 000084 K ) w 2 + (7 . 252 0 . 0491 K ) w + 1 . 015 K = 0 . From the Routh array, the condition for stability is that 3 . 6374 0 . 00084 K , 7 . 252 0 . 0491 K and 1 . 015 K need to be of the same sign. It can be checked that this implies 0 <K < 147 . 6986 . (c) The characteristic equation (1) is equivalent to z 2 + (0 . 0013 K 1 . 8187) z + 0 . 0012 K + 0 . 8187 = 0 . (3) Let Q ( z ) = z 2 + (0 . 0013 K 1 . 8187) z + 0 . 0012 K + 0 . 8187. By the Jury test, for stability we need to have Q (1) = 0 . 0025 K > 0 K > 0 ( 1) 2 Q ( 1) = 0 . 0001 K + 3 . 6374 > 0 K < 36374 | a 0 | = | 0 . 0012 K + 0 . 8187 | <a 1 = 1 103 . 8071 <K < 151 . To sum up, the range of K for stability is 0 <K < 151 147 . 6986. (d) Let K = 147 . 6986. Thus, equation (3) becomes z 2 1 . 6249 z + 1 = 0, which has two roots at 0 . 8125 ± j 0 . 5831. On the w -plane, they correspond to (using the inverse of the bilinear map): w = 2 T z 1 z + 1 = 20 0 . 8125 ± j 0 . 5831 1 0 . 8125 ± j 0 . 5831 + 1 = ±

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6 - ECE 483 Homework#6 Solution Problem 1 7-10(a(d 01873z 0...

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