4-sol

# 4-sol - ECE 483 Homework#4 Solution These problems are from...

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Unformatted text preview: ECE 483 Homework #4 Solution These problems are from the textbook. Problems 4-13 Solution: (a) The transfer function from e ( kT ) to m ( kT ) is D ( z ) = . 2 1 − . 9 z- 1 = . 2 z z − . 9 . Thus, the transfer function from e ( kT ) to c ( kT ) is C ( z ) E ( z ) = D ( z ) G ( z ) , where G ( z ) is the z-transform of the combined plant and data hold transfer function: G ( z ) = Z bracketleftbigg 1 − e- sT s G p ( s ) bracketrightbigg = Z bracketleftbigg (1 − e- sT ) 1 s ( s + 0 . 2) bracketrightbigg = (1 − z- 1 ) Z bracketleftbigg 5 s − 5 s + 0 . 2 bracketrightbigg = z − 1 z · bracketleftbigg 5 z z − 1 − 5 z z − e- . 2 T bracketrightbigg = 5(1 − e- . 2 T ) z − e- . 2 T = . 9063 z − . 8187 . Thus, the system transfer function is C ( z ) E ( z ) = D ( z ) G ( z ) = . 2 z − . 9 · . 9063 z − . 8187 = . 1813 z ( z − . 9)( z − . 8187) . (b) The DC gain of the system is D ( z ) G ( z ) | z =1 = . 1813 (1 − . 9)(1 − . 8187) = 10 . (c) Alternatively, the DC gain can be computed as D ( z ) | z =1 · G p ( s ) | s =0 = . 2 1 − . 9 · 1 . 2 = 10 . (d) The steady-state value of the output under unit step input is 10. 1 (e) Under unit step input e ( t ) = u ( t ), we have E ( z ) = z z- 1 , and C ( z ) = . 1813 z ( z − . 9)( z − . 8187) · z z − 1 = . 1813 z 2 ( z − . 9)( z − . 8187)( z − 1) = − 20 . 07 z z − . 9 + 10 . 07 z z − . 8187 + 10 z z − 1 . Thus, c ( kT ) = − 20 . 07 · (0 . 9) k + 10 . 07 · (0 . 8187) k + 10 , k = 0 , 1 , 2 , . . ., which approaches 10 and k → ∞ ....
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4-sol - ECE 483 Homework#4 Solution These problems are from...

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