5 - 3/30/10 ECE 483 Homework #5 Solution Problem 1. 6-1....

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ECE 483 Homework #5 Solution Problem 1. 6-1. Solution: (a) The combined plant and data hold transfer function is G ( s ) = 1 - e - Ts s · 0 . 5 s +0 . 5 . Hence, G ( z ) = (1 - z - 1 ) Z b 0 . 5 s ( s + 0 . 5) B = (1 - z - 1 ) z (1 - e - 0 . 5 T ) ( z - 1)( z - e - 0 . 5 T ) = 1 - e - 1 z - e - 1 . The closed-loop transfer function with D ( z ) = 1 is C ( z ) R ( z ) = D ( z ) G ( z ) 1 + D ( z ) G ( z ) = 1 - e - 1 z - 2 e - 1 + 1 . (1) The corresponding step response can be plotted by the following Matlab codes: T = 2; num = 1-exp(-1); den = [1 -2*exp(-1)+1]; [Y,X] = dstep(num,den,11); plot((0:10)*T,Y,’o’); whose result is shown in the left of Figure 1. (b) The analog system has closed-loop transfer function C ( s ) R ( s ) = 0 . 5 s +0 . 5 1 + 0 . 5 s +0 . 5 = 0 . 5 s + 1 , whose step response can be plotted by the step function of Matlab as the solid line in the middle of Figure 1. (c) If D ( z ) = 1 and T = 0 . 4 s instead of 1 s, then G ( z ) = 1 - e - 0 . 5 T z - e - 0 . 5 T = 1 - e - 0 . 2 z - e - 0 . 2 . The closed-loop transfer function is C ( z ) R ( z ) = D ( z ) G ( z ) 1 + D ( z ) G ( z ) = 1 - e - 0 . 2 z - 2 e - 0 . 2 + 1 , (2) whose step response is plotted in the right of Figure 1 as “x”s. As one can see, increasing the sampling frequency (i.e., decreasing T ) makes the analog system (b) a much better approxi- mation of the sampled-data system. Infrequent samplings lead to a sampled-data system (a) that exhibits overshoot in its steps response that is never encountered in the analog system’s step response. 1
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This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.

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5 - 3/30/10 ECE 483 Homework #5 Solution Problem 1. 6-1....

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