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

Such functions are known as r ational f unctions a r

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Unformatted text preview: :::~""':""":"'::.!:......'=--"'-""':"-""':"-':'-~ () _ 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 .5-2 1. B .5 27 P artial F raction E xpansion Partial Fractions: The Heaviside "Cover-Up" 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 ( -1-2)(-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 ight-hand 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.

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