Unformatted text preview: x + 13) Similarly, we c an s how t hat
(x
r = 1, 2, '" 1 n (B.37) + 4 x 2 + 2 x + 18
1 )(x + 2  j3)(x + 2 + j 3) kJ + = X • E xample B .9
Expand the following rational function F (x) into partial fractions: F (x)= 2 2x +9x11
(x + l )(x  2)(x + 3) =~+~+~
x+1 x 2 + 9x  11
(x) = ( x + 1 )(x  2)(x + 3)
2 X2 Step 1: Cover up (conceal) the factor (x + 1) from F (x): 4x 2 k2 x +2 j3 + 2 x + 18 k3
+  : x " , ( x 2 + 4x + 13) + 2 + j3
2 x =J  Similarly, x+3 To determine k l, we let x =  1 in (x + I )F(x). Note t hat (x + I )F(x) is obtained
from F (x) by o mitting the term (x + 1) from its denominator. Therefore, to compute
k l corresponding t.o the factor (x + 1), we cover up the term (x + 1) in the denominator
of F (x) and then s ubstitute x =  1 in the remaining expression. (Mentally conceal the
term ( x + 1) in F (x) with a finger and then let x =  1 in the remaining expression.) The
proced ure is explained step by step below.
F +1 where kJ _ [ (B.38) k 2 [ 2 + 2 x + 18
_ 1 +J'2  voe j 63.43°
_ '"5
~X=2+j3
4x = 1  j2 k3 =
___J = J 5ej63.43° x =2j3 T herefore,
2 F (x) =  x +1 + J 5ej63.43° x +2 j3 J 5e j63 .43° + ""::.,.x + 2 + j3 (B.39) I: 28 Background T he coefficients k 2 a nd k3 corresponding t o t he complex conjugate factors are also
conjugates of e ach other. This is generally t rue when the coefficients of a rational
function are real. In such a case, we need t o c ompute only one of the coefficients.
2. 29 B.5 Partial Fraction Expansion c omputations involved. We could j ust as well use other convenient values for x ,
such as x = 1. Consider t he case
F ( x) = ..:2":.x..,,.2....:,+_4..:x_+....:,5.....,.
x (x 2 + 2x + 5) Quadratic Factors k
ClX + C2
= ; ; + x 2 + 2x + 5 Often we are required to combine the two terms arising from complex conjugate
factors into one q uadratic factor. For example, F (x) in Eq. (B.38) can be expressed
as We find k = 1 by the Heaviside method in the usual manner. As a result, 4x 2 + 2x + 18
k1
C IX + C2
F (x) = (x + 1 )(x 2 + 4 x + 13) = x + 1 + x 2 + 4x + 13 T he coefficient k1 2X2 + 4x + 5
1
C IX + C2
.,.. =  + 2
x (x 2 + 2x + 5)
x
x + 2x + 5 is found by the Heaviside method t o be 2. Therefore,
4x 2 + 2x + 18
2
= (x + 1 )(x 2 + 4x + 13)
x+1 ..,.=.,.."..:..;.~.:...., C IX + C2
+ ;:"'"2 (B.40) x + 4x + 13 T he values of C I a nd C2 are determined by clearing fractions and equating t he
coefficients of similar powers of x on b oth sides of the resulting equation. Clearing
fractions on b oth sides of Eq. (B.40) yields
4x 2 + 2x + 18 = 2 (x 2 + 4x + 13) +
= (2 + C I)X Equating terms o f similar powers yields 2 ( ClX + C 2)(X + (8 + CI + = 2, C2 + 1) + (26 + (B.41) To d etermine Cl a nd C2, if we t ry l etting x = 0 in Eq. (B.43), we o btain
sides. So let us choose x = 1. T his yields F (I) 4 x + 2x + 18
2
2x  8
 =,. =   +  =(x + 1 )(x 2 + 4x + 13)
x + 1 x 2 + 4x + 13 C2) 00 on b oth = ~ = 1 + Cl...
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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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