Engineering 6, Fall 2006
Midterm Solutions
Lecture Section B (TTh121:30)
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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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 Fall '06
 Lagerstrom
 Elementary algebra, Partial fractions in complex analysis, partial fraction expansion

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