{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution_to_problem_5.6

# Solution_to_problem_5.6 - The poles of 1 2 1 2 1 2 1 1 K K...

This preview shows page 1. Sign up to view the full content.

5.6 – 1 Solution to problem 5.6 a) The overall gain of G is G ( s =0) (recall the Final Value Theorem). 1 2 K K K = b) We just need to check the response at 1 2 5 3 8 t τ τ = + = + = : 8 8 5 3 (8) 5 3 1 0.599 0.632 5 3 y e e KM - - - = - = - Therefore, the equivalent time constant is not equal to 1 2 τ τ + . c) y ( t ) will show oscillations for a step input if the poles of the transfer function are complex.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The poles of ( )( ) 1 2 1 2 1 2 1 1 K K G G G s s = ⋅ = + + are 1 1-and 2 1-, which are real numbers, therefore the output cannot show oscillations for a step input....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern