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8 or 8 We c an now choose some other value for x , such as x = 2, t o o btain one more
relationship to use in determining Cl a nd C2. I n this case, however, a simple method
is t o multiply b oth sides of Eq. (B.43) by x a nd t hen let x  > 0 0. This yields =  8, and CI C 2)X (B.43) 2 =1+Cl 2 (B.42) so t hat
a nd
Therefore, ShortCuts
T he values o f Cl and C2 in Eq. (B.40) can also be determined by using shortcuts. After c omputing k l = 2 by the Heaviside method as before, we let x = 0 on
both sides of Eq. (B.40) to eliminate Cl. T his gives us
18 _ 2 C2
13  + 13 Therefore, C2 = 8 To determine Cl, we multiply b oth sides of Eq. (B.40) by x a nd t hen let x > 0 0.
Remember t hat w hen x > 0 0, only the terms of the highest power are significant.
Therefore,
4 a nd = k l + Cl = 2 + C!
Cl = 2 I n the proced u re discussed here, we let x = 0 t o determine C2 a nd then multiply
b oth sides by x a nd let x ...... 0 0 t o determine Cl. However, nothing is sacred about
these values (x = 0 or x = 0 0). We use them because they reduce the number of 1
F (x) = x B .53 x +2 + x 2 + 2x + 5 R epeated Factors in Q(x) I f a function F (x) h as a repeated factor in its denominator, it has the form
F (x) = P (x)
(x  >Jr(x  C"l)(x  (B.44) (x  aj) (2)'" I ts p artial fraction expansion is given by
ao F (x) = (x _ >)r + (x  a rl al >VI + . .. + (x  >)
kl k2 kj +   +   + . .. +  x  a1
x  0!2
x  aj (B.45) T he coefficients k l' k2, . .. , k j corresponding t o t he unrepeated factors in this equation are determined by t he Heaviside method, as before [Eq. (B.37)]. To find the 30 Background coefficients ao, a i, a2, . .. , a rl, we multiply b oth sides of Eq. (B.45) by (x  >.)".
T his gives us
(x  >.)"F(x) = ao + a l(x  >.) + a2(x  >.)2 + . .. + a rl(x  >.)"1
(  >.)"
(x  >.)"
(  >.)"
+ k1x+ k 2x+"'+ k n   x al x (B.46) x  an a2 I f we let x = >. o n b oth sides of Eq. (B.46), we o btain >.)"F(x)) 1 x=>' = al T hus, a l is o btained by concealing t he factor (x  >.)" in F (x), t aking the derivative
of t he remaining expression, a nd t hen l etting x = A. Continuing in this manner, we
find
aj = ~ ~ [(x  J. dx J >')"F(x))1 (B.47b) x=>' Observe t hat ( x  >')" F (x) is o btained from F (x) by omitting t he factor (x  A)"
from its denominator. Therefore, t he coefficient a j is o btained by concealing t he
factor (x  >.)" i n F (x), t aking t he j th derivative of t he remaining expression, a nd
t hen l etting x = A (while dividing by j !).
• 2 (B.47a) Therefore, ao is o btained by concealing t he factor (x  >.)" in F (x) a nd l etting
x = >. in t he r emaining expression (the Heaviside "cover up" m ethod). I f we t ake
t he derivative ( with respect to x) of b oth sides of Eq. (B.46), t he r ighthand side is
a 1+ t erms containing a factor ( x  >.) in their numerators...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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