Solution_problem_4.2

# Solution_problem_4.2 - 4.2 – 1 Solution to problem 4.2 a...

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

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: 4.2 – 1 Solution to problem 4.2 a) A first-order transfer function has the form: ( ) 1 K G s s τ = + Where K is the steady-state gain, and τ is the time constant. In our case, the gain is: 5 K = b) The time constant is: 10 τ = c) Remember the Final Value Theorem (FVT), from Chapter 3, which states that: lim ( ) lim ( ) t s y t s Y s →∞ → = ⋅ The Laplace transform of the output - ( ) Y s- can be calculated using the transfer function ( ) G s , which relates the input and the output: ( ) 5 2 ( ) ( ) ( ) ( ) ( ) 10 1 Y s G s Y s G s U s U s s s = → = = + Now we are ready to apply the FVT, however we must first check whether the conditions for the applicability of the theorem are satisfied. Recall from the solution of Exercise 3.9, that the theorem can only be applied when the poles of ( ) sY s have negative real parts. In this case, 5 2 10 ( ) 10 1 10 1 s Y s s s s s ⋅ = = + + has only one pole, which is negative ( 1 10 p = - ). Therefore the FVT is applicable, and yields:...
View Full Document

## This note was uploaded on 11/20/2009 for the course CHME DTU-abroad taught by Professor Rafiqulgani during the Spring '09 term at Rensselaer Polytechnic Institute.

### Page1 / 3

Solution_problem_4.2 - 4.2 – 1 Solution to problem 4.2 a...

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

View Full Document
Ask a homework question - tutors are online