{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework10_solutions.pdf - ECE 460 Solutions to Homework 10...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 460: Solutions to Homework 10 Instructor: Stan Baek Due: Monday, April 18, at 2:10pm Note 1. Your solutions should be neat and legible; always show your work. Problem 1. Consider the following systems in state space. (i) ˙ x = - 1 1 2 0 - 1 5 0 0 4 x + 0 0 10 r y = [1 0 0] x (ii) ˙ x = 1 3 0 - 2 x + 2 7 r y = [1 0] x (a) (15 points) Find the poles of the systems. Solution: (i) det( λ I - A ) = λ + 1 - 1 - 2 0 λ + 1 - 5 0 0 λ - 4 = ( λ + 1)( λ + 1)( λ - 4) The poles are located at - 1, - 1, and 4. (ii) det( λ I - A ) = λ - 1 - 3 0 λ + 2 = ( λ - 1)( λ + 2) The poles are located at 1, and - 2. 1
Background image of page 1

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

View Full Document Right Arrow Icon
(b) (15 points) Find the corresponding transfer functions. Hint: Use MATLAB to obtain ( s I - A ) - 1 for 3 × 3 matrices. It is not allowed to use MATLAB other than ( s I - A ) - 1 . Solution: (i) ( s I - A ) = s + 1 - 1 - 2 0 s + 1 - 5 0 0 s - 4 Using MATLAB, we can obtain ( s I - A ) - 1 = s 2 - 3 s - 4 s - 4 2 s + 7 0 s 2 - 3 s - 4 5 s + 5 0 0 s 2 + 2 s + 1 ( s + 1) 2 ( s - 4) G ( s ) = C ( s I - A ) - 1 B + D = 1 0 0 s 2 - 3 s - 4 s - 4 2 s + 7 0 s 2 - 3 s - 4 5 s + 5 0 0 s 2 + 2 s + 1 0 0 10 ( s + 1) 2 ( s - 4) = 1 0 0 10(2 s + 7) × × ( s + 1) 2 ( s - 4) = 10(2 s + 7) ( s + 1) 2 ( s - 4) where × represents don’t-care . (ii) ( s I - A ) = s - 1 - 3 0 s + 2 ( s I - A ) - 1 = s + 2 3 0 s - 1 ( s - 1)( s + 2) G ( s ) = C ( s I - A ) - 1 B + D = 1 0 s + 2 3 0 s - 1 2 7 ( s - 1)( s + 2) = 1 0 2( s + 2) + 21 × ( s - 1)( s + 2) = (2 s + 25) ( s - 1)( s + 2) 2
Background image of page 2
(c) (15 points) Convert the transfer functions in phase variable forms. Solution: (i) G ( s ) = 10(2 s + 7) ( s + 1) 2 ( s - 4) = 20 s + 70 s 3 - 2 s 2 - 7 s - 4 We can rewrite the transfer function as G ( s ) = Y ( s ) R ( s ) = N ( s ) R ( s ) Y ( s ) N ( s ) , where N ( s ) R ( s ) = 1 s 3 - 2 s 2 - 7 s - 4 Y ( s ) N ( s ) = 20 s + 70 We have s 3 N ( s ) - 2 s 2 N ( s ) - 7 sN ( s ) - 4 N ( s ) = R ( s ) or ... n = 2¨ n + 7 ˙ n + 4 n + r Let x 1 = n x 2 = ˙ n x 3 = ¨ n Then, ˙ x 1 = ˙ n = x 2 ˙ x 2 = ¨ n = x 3 ˙ x 3 = ... n = 2¨ n + 7 ˙ n + 4 n + r = 2 x 3 + 7 x 2 + 4 x 1 + r A state space representation can be ˙ x = ˙ x 1 ˙ x 2 ˙ x 3 = 0 1 0 0 0 1 4 7 2 x 1 x 2 x 3 + 0 0 1 r 3
Background image of page 3

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

View Full Document Right Arrow Icon
We also have Y ( s ) = 20 sN ( s ) + 70 N ( s ) or y = 20 ˙ n + 70 n = 20 x 2 + 70
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}