24 Background B.5 Partial Fraction Expanslon 25 Therefore,F(z)can be expressed as 1213 x-1 =37=-2■ Fe-++2。生+2主 2+4如+3 (B34b) 1111 prope functioe ⊙Example CB.5 A proper function can be further pandedintopartialftatioas .The remaining Using a Computer,solve Example B.7. A=21113-1111:b=371]9 for k=1:3 B.5-1 Partial Fraction Expansion:Method of Clearing Fractions A1(:.k)=h: aemaow5hawc e par D=A15 and cquating the coeclents of simil powers on the twesids This procedure x(k)=det(D)/det(A); demonstrated by the following example. end 2 ■Example B.8 Expand the following rational function F(z)into partial fractions -2⊙ z3+32+4经+6 Fa=+1a+2+评 This function can be exr ed asasum of partial fractions with denominators(1) B.5 Partial Fraction Expansion (任+2，(z+3).and(在+32，as shown below. +3x2+4+6 3 s tha P四产+条+品2+品+中 say Such fu ons are known 在+Eaa斗Po.nd时ine both To det F(e)bmb+be (B.32) 若3+3r3+4红+6-k(3+8x2+21x+18)+k(z3+7x2+15r+9) x0十指-1z-1++412+0 +ks(z3+6r2+11r+6)+ka{x2+3z+2) (B.33) =￥(1+2+南)+x2(81+7肠+6k1+k) +x(21k+15k+11k+3k+(18k:+9+6k+2k) e be mprope Equnting ocefmiar powero both sides yields function.Consider.for example.the function mof&polyno 为+a+的=1 Fa=2a3+9r2+z+2 8+72+6+k=3 r2+4r+3 (B.34a 21k1+15+11ks+3k:=4 18k1+9k2+8k+2k=6 Solution of these fourimutaqutions yields 2x+1 =1,g-2,k的=2，s=-3 x2+4虹+3)2z+92+11r+2 Therefore, 2 3 2z3+8x2+6庭 P四子+异 ◆ x2+5x+2 x2+4红+3 not necessarily the most efficient.We now 龙-1 numerical work considerably
26 Background B.5 Partial Fraction Expansion 27 B.5-2 Partial Fractions:The Heaviside "Cover-Up"Method 2x2+9红=11 1.Unrepeated Factors of Q() 题红-22+3 We ahall first consider the partial fraction expans of F(=)-P()/Q().in Step2:Subatitute-1in the remaining expression to obtain which all the factors of Q(z)are unrepeated.Consider the proper function 2-0-11 Fe)=m+8n-1m-1+.+b1z+0 Similarly,to compute ka,we cover up the factor(z-2)mn F()and let=2 in the m<n remaining function,as shown below. +aa-1-1++a+a0 P() 2z2+9r-11 (B.35a) and 可l-荒 We can show that F()in Eq.(B.35a)can be expressed as the sum of partial 2x2+9肛-11 fractions k 大。 Therefore, Fa=-+++ (B.35b) 2z2+9-11 To deterrine the coeficient we multiply both sides of Eq.(B.35b)by a可“+2品■ and then let=A1.This yields Complex Factors in F(z) -e6++++ (-A2】 The procedure above works regardless of whether the factors of Q()are real or complex.Consider,for example, On the right-hand side.all the terms.Therefore 4z2+2红+18 (B.38) =(-)F(= (B.36) Fa)=G+1r2+4+1可 Similarly,we can show that 4r2+2z+18 =+1a+2-30r+2+河 k,=(红-ArF(lcox. r=l2.n (B37) 1 2 K3 =:+++2-3++2+8 ■Bxa ple B.9 where Expand the olowing ration function F()ntpartial fraction 4z2+2红+181 2x2+9x-11 Similarly, 4x2+2x+18 1+j2=V5e69.4 corresponding to the factor (z+1),we cover up the term (+1)In the denominato 与=a+脸石+2+对=-2+9 ()The 4红2+2z+18 s一[口+1e+2=冠 -2-5 =1-2=V5e4 erm红+)F hen et=- 2x2+9z-1 Therefore, F)=在+1z-2+ V5e83.4g V5e-f3.4r F回)=十+：+2-8++2+打 (B.39) Step 1:Cower up (conceal)the factor (+from P()
28 Background B.5 Partial Fraction Expansion 29 The coefficients o the comple computations Involved.We could Just as well use other convenient values for lly true when the c function are real.In such a case,we need to compute only one of the coecents. such as z1.Consider the case 2x2+4红+5 2.Quadratic Factors Fa)=+2+可 uired to combine the two ter We find k=1 by the Heaviside method in the usual manner.As a result, 4r2+2z+18 1 Ci工十C2 Fe)"a+1e+4红+i可+1+2年4z+s 2x2+4红+5 将+ ci+c2 (B.43) The coefficientk is found by the Heaviside method to be 2.Therefore. 4x2+2红+18 To determine c and if we try letting0in E.(B.43),we obtain on both B40) sides.So let us choose =1.This yields ng the p0)-告1+99 8 fractions on both side .(B.)yields c1+c2=3 42+2x+18=2(2+4红+13)+(c1+c2)+1) We can now choose some other value for z,such as老■2，to obtain one more relationship to use in determining c and ca.In this case,however,a simple method ■(2+c1)x2+(8+c1+c2)z+(26+c2) (B.41) is to multiply both sides of Eq.(B.43)by r and then let zoo.This yields Equating terms of similar powers yields c=2.=-8,and 2=1+c1 4r2+2x+18 2 2x-8 so that +102+4+1"+1+2+4+8 (B42) c1 =1 and c2=2 Therefore, Short-Cuts 里+2 The values of c and c2 in Eg.(B.40)can also be determined by using short- AoE阁7 eebe-0 B.5-3 Repeated Factors in Q(=) 指-2+ If a function F(z)has a repeated factor in its denominator,it has the form Therefore, P() c2=-8 P=e-re-ane-2-o (B.44) To determine e we multiply both sides of Eq.(B.40)by z and then leto mber that when Its partial fraction expansion is given by ony the terms of the high st power r are significant. Gr-l 4=1+C1=2+C1 and a=-万+g-+.+为 c1=2 K1 k2 k1-(B.45 +:a+-a+.+ boe时n othin and let ro and thenmu We use the because they reduce t e number of teee ebe)he
Background B.5 Partial Fraction Expansion 31 e muliply both sides of E (B.45)by 在-0F(x)=a0+a1(e-A+a2e-2+.+4r-i任-1 -器特 Z-a1 1-02 军一am If we let=A on both sides of Eq.(B.46),we obtain -分牌儿， Therefore, (B,47a an+中p+备+中 2 to =of b o( e right-hand s B.5-4 A Hybrid Method:Mixture of the Heaviside "Cover-Up"and c-)in their numerators. Letting=A on both Clearing Fractions For multiple roots,especially of higher order,the Heaviside expansion method, 云e-yFe which requires repeated differentiation,can become cumbersome.For a function =a1 I-A which contains several repeated and unrepeated roots.a hybrid of the two pro- the facto cedures proves the best.The simpler coefficients are determined by the Heaviside r()in F(s),taking the derivative method,and the remaining coefficients are found by clearing fractions or short-cuts. then letting=A.Continuing in thi find 8 manner,we thus incorporating the best of the two methods.We demonstrate this procedure by solving Example B.10 once again by this method. B.47b) Therefore, Observe that (-A)"F()is obtained from F()by omitting the factor (-A)" from its denominator efore.the coefficie is obtain ed by cor aling the 4x3+16r2+23x+13 41 factor (in F().taking the jth derivative of the (+1(+2) 中现++0+中2 then letting=(while dividing byj). We now multiply both sides of the above equation by (+1)(+2)to clear the ■Example B.10 fractions.This yields Expand F(z)into partinl fractions if 4r3+1622+23z+13 =22+2+a1(红+10(e+2)+a2(红+12（红+2）+（红+）3 x+)(江+2) The partial fractions are =(1+z3+(a1+4a2+3)r2+(5+3a1+5a2)x+(4+2a1+2a2+1) 丽+++中 Equating coefficients of the third and second powers of on both sides,we obtain Thecietobtained by concealing the factor(+)nP()and then substituting 1+42=4】 a1=1 =-2 in the remaining expression: 1+4a,+3=16}→ 42=3 k=+162+23+3 We may stop here if we wish because the two desired coefficients,andare =1 红+- now determined.However,equating the coefficients of the two remaining powers of edontheinth thed 6=3+162+2+13 23=5+3a1+5G2 13=4+2a1+2a2+1
33 Background B.5 Partial Fraction Expansion 32 r casc is the appearance of an =1 and az =3.found earlier, extra constant bn in the la providing an additional check for our answers. proof is left as an exereise for the reader. =p+中+是+动 2 ■Example B.11 Expand F(a)into partial fractions if which agrees with the prevlous result. P)=32+9 在-2)z+3 A Mixture of the Heaviside"Cover-Up"and Short Cuts Here m=n2 with on=-3.Therefore, In the above exampl after determining the coefficients ao=2 andk-I by the Heaviside method as before,we have 阳二得-+气+品 in which t=*+号+市 2 41 3x2+9x-20 ”0号=2 x+1(+2) =+列 and 3x2+9r-20 =- ”0碧 4=82+1-中a2■3 Therefore, 2 Therefore -+列3++ =+*有t中 2 3 任+13在+2) B.5-6 Modified Partial Fractions which can be readily found by setting equal Often we require partial of the form There is now ony to ay comvenient value,ay This ylelds 5x2+20z+18 号=2+a1+3+多=a1=1 F()= (红+2（红+3 which agrees with our earlier answer Dividing both sides byr yields -0+e中 53+20x+18 B.5-5 Improper F()with m=n nd den Expansion of the right-hand side into partial fractions as usual yields ning of 四-+龈-g+++ 3 Using the procedure discussed earlier,we find=,21s-2,and=1. Fe)=g”+b-1z-1++b2+如 Therefore, x+an-1x++a12+a0 型-+中年市*+评 .1 1 kL Kn Now multiplying both sides by yields the computed as if F(z)were proper.Thus 2T F)=1++2+3++3那 k=(仁-A)F(elx=x ThisexpreF(a)themf partial fractions having the form For quadratic or repeated factors the appropr riate procedures discussed in Secs B.5-2o B.5-3 should be used as if F()were proper.In other words,whenm