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Unformatted text preview: Problem 1 The facts we are going to be using here are the following: 1. Every transfer function can be written as H ( s ) = k Q m i =1 s z i Q n i =1 s p i , (1) where k is not the DC gain, just a parameter (try s → 0 to see that), m is the number of zeros, n is the number of poles, z i are the zeros and p i are the poles of H ( s ). Using this representation, the phase and magnitude of H ( s ) (note that this is really just a complex number, you insert s into H ( s )) can be found as ∠ H ( s ) = m X i =1 ∠ ( s z i ) n X i =1 ∠ ( s p i ) (2a)  H ( s )  =  k  Q m i =1  s z i  Q n i =1  s p i  (2b) Note that an empty product defaults to 1, and an empty sum defaults to 0. 2. The vector of s s 1 in the complex plane can be constructed by drawing an arrow from s 1 to s (note the direction!): s 1 s Re Im 1 s s 1 s { 1 s { s 1 s { s 1 µ It is then very easy to read off the angle θ = ∠ ( s s 1 ) and the magnitude  s s 1  . 3. Recall that the root locus of a transfer function H ( s ) is defined by all p ∈ C for which 1 + kH ( p ) = 0 for some positive k , so it really does not matter which k you need in order to have 1 + kH ( p ) = 0, as long as it is positive. Solving for H ( p ), we obtain H ( p ) = 1 k (3) and since k can vary from 0 to ∞ , the right hand side of (3) can be anything between∞ and 0, so any point on the left half of the real axis or in other words any number with a phase angle of 180 ◦ or π ! So the root locus consists of all the points p in the complex plane, for which ∠ H ( p ) = π , and question iii. really only asks you for the root locus of the given transfer function. Refer to the textbook, Chapter 5, in particular Section 2, for more elaboration on this. Now we are ready to start. 1 a) H a ( s ) = 64 ( s 4) 4 i. Drawing the arrows for (0 4) in the complex plane, we get this picture: Im Re 4 µ Using Equation (2a), we find ∠ H a (0) = 0 4 · 180 ◦ = 720 ◦ , and with Equation (2b) we have  H a (0)  = 64 1 4 4 = 1 4 . ii. Now imagine, you connect the point (4 , j ) to a point on the imaginary axis and then move this point all the way up to inifinity. The angle is going to approach 90 ◦ , and the picture looks like this: Im Re 4 µ Again, with Equation (2a), we have ∠ H a ( j ∞ ) = 0 4 · 90 ◦ = 360 ◦ , and with Equation (2b) we have  H a ( j ∞ )  = 64 1 ∞ = 0. iii. Reconsider Equation (2a): To find the phase of H a ( s ) at some point s ∈ C , we connect the pole 4 with s and multiply the angle – let’s call it θ – of that arrow by 4 (because the pole has multiplicity 4). The angle ∠ H ( s ) is thus 4 θ . The assignment tells us that that should be 180 ◦ + k 360 ◦ , so θ = 45 ◦ + k 90 ◦ . If you do the math, you will see that this gives you 4 different angles, for k = 0 , 1 , 2 , 3, after that, the angle wrapped around once, so for k = 4 we have the same angle as for k = 0 and so on. So the s ∈ C we are looking for are all that lie at an angle of...
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This note was uploaded on 08/03/2010 for the course ECE PROF. VOLK taught by Professor Volkanrodoplu during the Spring '10 term at UCSB.
 Spring '10
 VolkanRodoplu

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