ECE332Discnotes - Discussion Notes ECE 332 Algebraic Pole...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Discussion Notes - ECE 332 - 12/11/06 Algebraic Pole Placement Design Let’s say we have a plant G ( s ) in a unity feedback system, where we design the compensator G c ( s ) to achieve some desired CLTF H ( s ). The most obvious way to accomplish this is to set G c = H G (1- H ) . The problem with this design is that G c ( s ) cancels all the poles and zeros (including the unstable ones) of G ( s ), which can be a problem if we don’t know the exact locations of the poles and zeros of G ( s ). So instead of doing pole-zero cancellation between G ( s ) and G c ( s ), here’s an algebraic design method that allows us to design a compensator G c ( s ) to acheive a CLTF H ( s ) with specified (stable) denominator. First let G = N + N- D + D- , where N- ,D- are Hurwitz (corresponding to stable, open LHP zeros/poles) polyno- mials and N + ,D + are anti-Hurwitz (corresponding to unstable, closed RHP poles/zeros). With the following controller G c ( s ): G c = D- ( X Δ + D + M ) N- ( Y Δ- N + M ) , we can achieve CLTF H(s): H = N + ( X Δ + D + M ) Δ , where X,Y satisfy the Bezout identity XN + + Y D + = 1 and M is chosen to satisfy deg( X Δ + D + M ) ≤ degΔ- deg N + to ensure that H is proper. In practice, we choose X,Y,M to be as simple as possible. 1 Example 1 : Let G ( s ) = ( s + 3)( s- 3) ( s + 4)( s- 4) . Place the poles of the CLTF H ( s ) at ( s + 1) 2 . Pick an appropriate numerator for H and find the corresponding G c . Make sure that the denominator of G and the numerator of G c do not have any unstable pole-zero cancellations. Answer : Here’s what we have according to the problem statement: N- = s +3 ,N + = s- 3 ,D- = s + 4 ,D + = s- 4 , Δ = ( s + 1) 2 . First let’s find X,Y , which satisfy XN + + Y D + = 1 ⇒ X ( s )( s- 3) + Y ( s )( s- 4) = 1 . The simplest X,Y to do so is X ( s ) = 1 ,Y ( s ) =- 1 (usually we can solve the Bezout identity by inspection). Next we pick M . So we need deg( X Δ + D + M ) ≤ degΔ- deg N + = 2- 1 = 1 ....
View Full Document

This note was uploaded on 03/27/2008 for the course EE ECE 332 taught by Professor Setharus during the Spring '08 term at University of Wisconsin.

Page1 / 7

ECE332Discnotes - Discussion Notes ECE 332 Algebraic Pole...

This preview shows document pages 1 - 3. Sign up to view the full document.

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