timeresponse4

# timeresponse4 - + + (3) For T p = 5 s and the damping ratio...

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

Time Response Q UESTION 4 For each pair of second–order system specifications below, find the pole locations. a. % OS = 12% and T s = 0.6 s b. % OS = 25% and T p = 5 s c. T s = 7 s and T p = 3 s For % OS = 12%, the damping ratio is 2 2 2 2 % 12 ln ln 100 100 0.5594 % 12 ln ln 100 100 OS OS ζ π - - = = = + + (1) For T s = 0.6 s and the damping ratio given in equation (1), the natural frequency is ( 29 4 rad 11.92 0.6 0.5594 s n ϖ = = (2) Therefore, the poles are located at 2 1 6.668 9.880 n n j j ζϖ - ± - = - ± . For % OS = 25%, the damping ratio is 2 2 2 2 % 25 ln ln 100 100 0.4037 % 25 ln ln 100 100 OS OS - - = =

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + + (3) For T p = 5 s and the damping ratio given in equation (3), the natural frequency is 2 rad 0.6868 s 5 1 0.4037 n = =-(4) Therefore, the poles are located at 2 1 0.2773 0.6283 n n j j-±-= -± . Time Response For T s = 7 s, the product ζω n = 4/7 = 0.5714. For T p = 3 s 2 1 3 n p T π ϖ ζ-= = (5) Therefore, the poles are located at 2 1 0.5714 3 n n j j ζϖ-±-= -± . 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

timeresponse4 - + + (3) For T p = 5 s and the damping ratio...

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

View Full Document
Ask a homework question - tutors are online