Signal Processing and Linear Systems-B.P.Lathi copy

# Consider for example o n t he r ight hand s ide all t

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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 +9x-11 (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 5e-j63.43° x =-2-j3 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.

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