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() _ P (x)
 Q(x) (B.32) 2
3
F( )
x + 3x + 4x + 6
X = ( x + l )(x + 2 )(x + 3)2 2 x3 + 9 x 2 + H x + 2
F) =        (x
x 2 + 4x + 3 x 3 + 3x2 + 4x + 6 = k4 + Sx 2 + 2 1x + IS) + k2(X 3 + 7 x 2 + 1 5x + 9)
+ k3(x3 + 6 x 2 + l lx + 6) + k 4(X 2 + 3 x + 2)
3
= x (k l + k2 + k3) + x 2(Skl + 7k2 + 6k3 + k 4 )
+ x (21kl + 15k2 + l lk3 + 3k4) + ( lSkl + 9k2 + 6k3 + 2k4)
k l(X 3 Equating coefficients of similar powers on both sides yields
k l + k2 + k3 = 1
+ 7k2 + 6k3 + k4 = 3
2 1kl + 15k2 + l lk3 + 3k4 = 4
18k l + 9k2 + 6k3 + 2k4 = 6 (B.34a) Solution of these four simultaneous equations yields
k3 = 2, 2x x I k3 S kl Because this is a n i mproper function, we divide t he n umerator by t he d enominator
until t he r emainder has a lower degree t han t he denominator. +I
2 + 4 x + 3) 2 x 3 + 9 x 2+llx + 2
x
2 x 3 + 8 x 2+6x
x 2+5x + 2
x 2+4x + 3 k2 = X + 1 + x + 2 + x + 3 + ( x + 3)2 To determine the unknowns k l, k2, k3, and k4 we clear fractions by mUltiplying both sides
by ( x + l )(x + 2 )(x + 3)2 to obtain (B.33) T he function F (x) is i mproper if m 2 n a nd p roper if m < n . An improper
function can always be separated into t he s um of a polynomial in x a nd a proper
function. Consider, for example, t he function kl + 1), k4 = 3 Therefore,
F (x) = x 122
3
+ 1  x + 2 + x + 3  (x + 3)2 • A lthough this method is s traightforward a nd applicable t o all situations, it is
n ot necessarily t he most efficient. We now discuss o ther m ethods which can reduce
numerical work considerably. B ackground 26 B .52
1. B .5 27 P artial F raction E xpansion Partial Fractions: The Heaviside "CoverUp" Method Unrepeated Factors o f Q (x)
Step2: Substitute x =  1 in the remaining expression t o o btain k l: We shall first consider t he p artial f raction e xpansion of F (x) = P (x)/Q(x), in
which all t he f actors o f Q (x) a re u nrepeated. C onsider t he p roper f unction m <n k_ P (x) = 2  9  11
=  18
( 12)(1+3)
6 =3 Similarly, to compute k2, we cover up the factor (x  2) in F (x) and let x = 2 in the
remaining function, as shown below. m + b m_lX m  l + '" + b lX + bo
x n + a n_lX n  l + ... + a lx + a o F (x) = b mx kl ( B.35a) 2 2 X2 + 9 x  11
~ I_
x =2  8 + 18  11  ~  1
(2+1)(2+3)  15 and
18  27  11
= _2_0 =  2
7"(::3:+1::;')7"(:3:2=)
10 We c an s how t hat F (x) i n E q. (B.35a) c an b e e xpressed as t he s um o f p artial
f ractions
Therefore, kl
k2
kn
F ( x ) =   +   + .. · +  x  Al
x  A2
x  An ( B.35b)
F (x) T o d etermine t he coefficient k l, we m ultiply b oth s ides of Eq. (B.35b) b y x  Al
a nd t hen l et x = ) '1' T his y ields = 2x2 + 9x  11
(x + l )(x  2)(x + 3) = _ 3_ + _ 1_ _ _ _
2
x+1 x 2 • x+3 Complex Factors in F (x)
T he p rocedure a bove w orks regardless o f w hether t he f actors o f Q (x) a re r eal
o r c omplex. Consider, for example,
O n t he r ighthand s ide, all t he t erms e xcept k l v anish. Therefore,
F (x) = ( x (B.36) 4 x 2 + 2 x + 18
+ 1 )(x 2 + 4...
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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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