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Unformatted text preview: e B .ll Expand F (x) into partial fractions i f which agrees w ith t he p revious result.
F (x) = 3 x 2 + 9x x2 A Mixture o f t he Heaviside "CoverUp" and Short Cuts
Here m = n = 2 with bn = b2 = 3. +X  20
6 = 3 x 2 + 9x  20 (x  2)(x + 3) Therefore, I n t he above example, after determining t he coefficients ao = 2 a nd k = 1 by
t he Heaviside m ethod a s before, we have
4 x 3 + 1 6x 2 + 23x + 13
2
al
a2
1
 ;:.,;;,;, =    +    +   +  (x+l)3(x+2)
( x+l)3
( x+l)2
x +l
x +2 T here a re only t wo u nknown coefficients, a l a nd a2. I f we m ultiply b oth sides of
t he above e quation b y x a nd t hen let x  > 0 0, we c an eliminate a 1 . T his yields in which
kl = 3x 2 + 9x  I 20 = 12 + 18 (2 z =2 and
2 k2 = 3x + 9 x  20 ~ I 20 = + 3) 5 = 27  z =3 ~ =2 27  20
( 32) =  20 = 4
5 Therefore,
F (x) Therefore, ( x  2)(x 4 x 3 + 1 6x 2 + 2 3x + 13
2
al
3
1
 ,......,...,,; =    +    +   +  (x + 1 )3(x + 2)
( x+l)3
( x+l)2
x +l
x +2 T here is now o nly one unknown a I, which c an b e r eadily found by s etting x e qual
t o a ny convenient value, say x = O. T his yields ¥ = 2 + al + 3 + ~ =} Improper F (x) with m 8 .56 = 3 + _ 2_ + _ 4_
x 2 x +3 • Modified Partial Fractions F () =
x 5x2 + 2 0x + 18
( x + 2 )(x + 3)2 ...,.,.,= Dividing b oth sides by x yields =n A g eneral m ethod of handling an i mproper function is indicated in t he beginning of t his s ection. However, for a special case where the n umerator a nd d enominator p olynomials of F (x) a re of t he s ame degree (m = n ), t he p rocedure is t he
s ame as t hat for a p roper function. We c an show t hat for 20 + 3) O ften we r equire p artial f ractions of t he form (x~f.)r r ather t han ( x_\.jr. T his
c an b e achieved by e xpanding F (x)/x i nto p artial fractions. Consider, for example, al = 1 which agrees w ith o ur e arlier answer. B .55 = 3 x 2 + 9x  F (x) 5x 2 =
x
x (x + 2 0x + 18
+ 2 )(x + 3)2 E xpansion of t he r ighthand side into p artial f ractions as u sual yields
F (x) = 5 x + 2 0x + 18 = a l + ~ + _ a_3_ + _ a_4_
x
x (x + 2 )(x + 3)2
X
X+2
( x + 3)
( x + 3)2
Using t he p rocedure discussed earlier, we find a l = 1, a2 = 1, a3 =  2, a nd a4 = 1.
T herefore,
2 F (x) = .!.+_1_ _ _ _+_1_
2
x
x
x+2
x+3
( x + 3)2 Now multiplying b oth s ides by x yields
t he coefficients k l' k 2,"" k n a re c omputed a s if F (x) were proper. Thus, x
2x
x
F (x) = 1 +  2  3 +  (
3)2
x+
x+
x+ T his expresses F (x) as t he s um o f p artial f ractions having t he form (x~f.)r.
For q uadratic o r r epeated factors, t he a ppropriate p rocedures discussed in Secs.
B.52 or B.53 s hould b e u sed as if F (x) were proper. I n o ther words, when m = n , 34 Background B .6 Vectors and Matrices
An entity specified by n numbers in a certain order (ordered ntuple) is a n
ndimensional v ector. Thus, an ordered ntuple ( Xl, X 2, . .. , X n) represents an
ndimensional vector x. Vectors may be represented as a row ( row v ector):
X =[XI X2 X nl o r as...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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