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e6exam1secBsolnsF06 - Engineering 6 Fall 2006 Midterm...

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Engineering 6, Fall 2006 Midterm Solutions Lecture Section B (TTh12-1:30) Regrade requests: Regrade requests must be submitted in writing to Dr. Lagerstrom no later than the lecture section on Tuesday, November 21. Regrades will only be considered in cases where it looks like the grader missed something. That is, requests along the lines of "I think I deserve more points" will not get very far, because the the grading scale on each problem was applied consistently for all students. Note also : Your exam may have been photocopied before it was returned, so please don't risk your engineering career here at UCD by changing an answer and submitting it for a regrade. EXAM VERSION A (blue and green copies) Problem 1. (12 points) As discussed in lecture, in circuit analysis we sometimes use something called a Laplace transform, which gives us integrals involving polynomial ratios. These polynomial ratio integrals are not easy to calculate in general. Assuming that the ratio is A(x)/B(x), where A and B are both polynomials, describe in detail how we would use Matlab to make our task simpler, including the code involved (but don’t discuss doing the actual integration). Take into account that the order of A could be anything, and the order of B could be anything. Solution: First key part (3 points): This problem is essentially asking about the partial fraction expansion. (If the student tried to do the problem using the deconv function, give maximum partial credit of 4 points, depending on how well they explained the use of the deconv function.) If the order of A is less than the order of B (meaning that we have a “proper rational function”), then we can write a polynomial ratio A/B as n n r x c r x c r x c x B x A - + + - + - = ... ) ( ) ( 2 2 1 1 where the c’s are constants called “residues” and the r’s are the roots of the denominator polynomial B(x). This makes doing the Laplace transform integral much easier. Second key part (3 points): If, however, there is a repeated root, then the expansion has a slightly different form. Assuming the repeated roots are the second and third roots, and there are four roots total, then 4 4 2 2 3 2 2 1 1 ) ( ) ( ) ( r x c r x c r x c r x c x B x A - + - + - + - =
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Note that the residues are still all different. If there were three repeated roots, then you would have a term with a 3 2 ) ( r x - denominator, and so on. Third key part (3 points): In Matlab, the residue function calculates the residues and roots for us, i.e., [c r] = residue(a,b) where a and b are the coefficient polynomials of A(x) and B(x) and c is a vector with the residues and r is a vector with the roots. Fourth key part (3 points): If the order of A is greater than or equal to the order of B (an “improper rational function”), then we can’t use the partial fraction expansion directly. Instead we have to write the ratio as a quotient plus a remainder term, and then rewrite the remainder term (which is guaranteed to be a proper rational function) as a partial fraction expansion. That is: ) ( ) ( ) ( ) ( ) ( x B x R x Q x B x A + = , where the second term is the remainder term
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e6exam1secBsolnsF06 - Engineering 6 Fall 2006 Midterm...

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