stability8 - to obtain 2 s , place this coefficient in the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Stability Q UESTION 8 For the system shown in Figure 8 with ( 29 1 G s K = , ( 29 2 2 3 1 G s s = + , ( 29 3 G s s = and ( 29 4 4 s G s s = + , create the Routh table and determine for what range of K the closed–loop system is stable. Determine the frequency of oscillation at marginal stability. G 2 (s) + - G 3 (s) G 4 (s) G 1 (s) + + R(s) Y(s) Figure 8 Denote the output of the top summing junction by E ( s ). Write one equation for E ( s ) and one for Y ( s ): ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 4 3 E s G s R s G s Y s G s E s = - + and ( 29 ( 29 ( 29 2 Y s G s E s = . Combining equations and rearranging, the closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 2 4 3 2 2 3 4 3 4 5 4 4 1 G s G s K s T s s s s s G s G s G s + = = + + + + + + . The closed–loop characteristic polynomial is 4 3 2 5 4 4 s s s s + + + + .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Stability The Routh table is 4 3 2 1 0 1 5 4 1 4 0 1 4 0 02 0 0 4 0 0 s s s s s . Since there is a row of zeros in the s 1 row, the characteristic equation contains a 2 nd order even polynomial. The even polynomial is in the s 2 row and is s 2 +4 = 0. Differentiate this polynomial
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: to obtain 2 s , place this coefficient in the s 1 row, and continue. Investigate sign changes from the s 1 row to the s row to determine the stability of the 2 nd order even polynomial. Since there are no sign changes and the roots of an even polynomial are symmetric about the origin, both poles are on the imaginary axis. To determine the location of the other poles, investigate sign changes from the s 2 row to the s 4 row. Since there are no sign changes, the remaining poles are in the left splane. Therefore, there are two poles on the imaginary axis and two poles in the left splane. The closedloop characteristic equation does not depend upon K ; therefore, the stability results are independent of K . 2...
View Full Document

This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

Page1 / 2

stability8 - to obtain 2 s , place this coefficient in the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online