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Solutions_hw5

# Solutions_hw5 - Homework 5 ECE147A November 2 2009 Problem...

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Homework 5 ECE147A November 2, 2009 Problem 1 Problem 5.6 (d) and (e) from the text book, solution see at the end of the document. 8 points for each root locus sketch, deductions if you didn’t compute asymptotes (centroid included), arrival/departure angles or imaginary axis crossings. Problem 2 Problem 5.7 (b) from the text book, solution see at the end of the document. 9 points for that one, deductions as above. Note that the problem formulation didn’t ask for collisions, so you couldn’t know that the two comp. conj. poles actually meet on the real axis and depart later on, so a sketch that doesn’t include that is still correct. Problem 3 Problem 5.8 (a) from the text book, solution see at the end of the document. 8 points, deductions as above. Problem 4 Problem 5.13 from the text book, solution see at the end of the document. 6 points. 1

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Problem 5 5.1 Which root locus? Consider the plant to be given in the general zero-pole-gain form ( zpk in Matlab) P ( s ) = A · Q n z i =1 ( s - z i ) Q n p i =1 ( s - p i ) . (1) Now, we know that there is exactly one zero with positive real part (so one positive real zero, a complex one would need its conjugate complex to be a zero, too) and the plant is stable, so Re( p i ) < 0 i = 1 . . . n p , and we can refine (1) to P ( s ) = A ( s + z r ) Q n z - 1 i =1 ( s - z i ) Q n p i =1 ( s - p i ) z r , Re( z i ) , Re( p i ) < 0 (2) and get P (0) = A z r Q n z - 1 i =1 ( - z i ) Q n p i =1 ( - p i ) . (3) We are also given that P (0) > 0, which together with (3) implies that A < 0. To see that, convince yourselves that the two products are positive and remember that z r < 0. Please compare with Equation (5.4) and (5.6) from the text to understand, that the characteristic equation whose roots we’re interested in thus has to be 1 + ( - K )( s + z r ) Q n z - 1 i =1 ( s - z i ) Q n p i =1 ( s - p i ) (4) where I absorbed the negative plant gain into the controller gain K . It follows that we need to look at the 0 root locus, because the roots of (4) are given by all s that satisfy ( s + z r ) Q n z - 1 i =1 ( s - z i ) Q n p i =1 ( s - p i ) = 1 - ( - K ) = 1 /K. (5) and for positive K , the phase of the right hand side is 0 . As an easy example, look at P ( s ) = - s - 5 ( s +1) 2 , which has all the required properties. The closed loop denominator is s 2 + (2 - k ) s + 5 k + 1, and the poles are at s = k/ 2 - 1 ± p k ( k - 24) , (6) so they are complex for k = 0 . . . 24. Figure 1 shows the 0 and 180 root locus. Obviously, the 0 root locus is the correct one, since the 180 root locus is entirely on the real axis. 8 points for a good explanation, 3 more for an example 5.2 K < 0 That amounts to inserting a third negative sign in Equation (5), so making the right hand side a negative real number with phase 180 again, thus in that case we would have to look at the 180 root locus again. In the example, just insert negative numbers for k in (6). 3 points 2
-15 -10 -5 0 5 10 15 20 25 -8 -6 -4 -2 0 2 4 6 8 Root Locus Real Axis Imaginary Axis 0° rootlocus 180° rootlocus Figure 1: Problem 5: Root locus of the example plant.

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