Solution9

# Solution9 - Solution9 Ch6_18,19,37,39 6.18 Solution: (a) 1...

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(a) G ( s )= 1 s 2 Note that the portion of the Nyquist diagram on the right side below origin. (b) G ( s )= 1 s 2 + ω 2 0 Note here that the portion of the Nyquist plot coming directly from 180 o arc that arose because of the detour around the pole on the imaginary axis.

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(a) N =0 ,P =0= Z = N + P =0 The closed-loop system is stable for any K> 0 . (b) The Bode plot shows an intial phase of 0 o hence the Nyquist starts on the positive real axis at A&. The Bode ends with a phase of - 270 o hence the Nyquist ends the bottom loop by approaching the origin from the positive imaginary axis (or an angle of -270 o ).
The magnitude of the Nyquist plot as it crosses the negative real axis is 0.00174. It will not encircle the 1 /K point until K = 1/0.00174 = 576. i. 0

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## This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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Solution9 - Solution9 Ch6_18,19,37,39 6.18 Solution: (a) 1...

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