ECE332Discnotes - Discussion Notes ECE 332 Algebraic Pole...

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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
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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 . We have X Δ + D + M = ( s + 1) 2 + ( s - 4) M ( s ) . If we pick M ( s ) = - ( s + 6), then X Δ + D + M = 25, which has degree 0 (even better than we needed). So this gives G c = 25( s + 4) ( s + 3)( s - 19) and H = 25( s - 3) ( s + 1) 2 .
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