786 ‘Network Analysis and Synthesis First: When the quantity under the radical sign of equation (1) is zero (double, real root) or negative (complex roots). In other words, (ay + by ~ a,b,)'— 4 aghy < 0 o (ay + by — aby) $A aghy a 4612 (fg ‘Second: When the «?, , in equation (3) is negative so that the roots are imaginary. ‘This situation ours when (a, + by ~ a,b,’ — 4 ab, > 0 2) and (ay + by) a,b, <0 From equation (2) we have 4 by — (ay +b) > 2gfayhy > (ao +b ~ a, A) Thus 44> (ay ~ ss) Therefore, necessary and suficient condition for a biquadratic function to be p.rf. is ib 2( Ju -V) 8.4. SYNTHESIS OF ONE PORT NETWORKS WITH TWO KINDS OF ELEMENTS: In this section, we will study the synthesizing process of one port networks. There are three Oe A Te ee rane rs a function can by synthesized using any two types of passive elements, So, the networks to be synthesized are either R-L, L-C or R-C networks. ‘There are number of methods for synthesizing (or realizing) of a one-port network, Here consider only the four basie forms as follows: (Foster or Fester series form, (i) Foster-I or Foster parallel form. iii) Cauer form i) Cauerl Form, ‘The following table is useful while realizing the network by any of the four forms ‘Table, 8.1. impedance, 20)~ neat Elements in series impedance, Z(3) = R + s+ + sC WW TOIT " 1 Admittance, Ys) = Eloments in para 84.1. Foster-I Form ‘The steps to be followed for Foster form as follows: ‘Step-1. To obtain the partial fraction expansion of the driving point impedance function Z(s). The partial fraction expansion gives the equation of Z(s) as summation of the various impedance functions As) = 269) + Bs) +... 2,3)Network Synthesis, 787 ‘Thus a network can be realized by connecting the impedances Z,(s), Z,(3)...2,(s) in series, Note: For finding the partial fractions, make sure that the degree of numerator should be less than the degree of denominator. Residues must be positive Step-IL. To make numerator portion of partial fraction expansion of Z() unity. Step-III_ To identity the elements. ‘After making numerator unity we get two forms of the impedances. If impedance has denomina- tor as summation of the two parts, then compare the denominator portion with ¥( and then identify the elements and connected in parallel. Otherwise, impedance compare with 3) = B+ sb-+ + and then identify the elemenss and connected in scres sc Suppose, after making numerator of partial fraction expansion of Z(s) unity, we get Z(s) as As) = Z(3) * Zs) * Z0) +..Z,(9) Let Z(s) has denominator postion as summation of the two parts, then compare 1 400 [26 mol with 13) = : tose and then identify the elements and connected in 5 %65) parallel. And zssuming rest impedances are not as summation of the two parts, then compare Z(6) with Zs) = Rsk+ sand then identify the elements and connected in series. Therefore, network connection for Foster first form as shown in Fig. 8.2. ‘one leery Zi” 249) 256) t B (wo paraliel hia passive elements) pee we Oe wee REO Fig. 82 84.2. Foster-II form The steps to be followed for Foster-II form as follows: Step-I. To obtain the partial fraction expansion of the driving point admittance function Ys)= ap The partial fraction expansion gives the equation of Y(s) as summation of the various admittance function Ys) = Yy(s) + ¥{s) +... Ys) Step-II. To make numerator portion of partial fraction expansion of Y(s) unity. Step-I. To identify the elements After making numerator unity we get two forms of the admittances. If admittance has denominator as summation of the two parts, then compare the denominator portion with Z(s) = Rash+ = and then identify the elements and connected in series. Otherwise, admittance compare x AL 4 sc and then identify the elements and connected in parallel 7 with Ys788 ‘Network Analysis and Synthesis Suppose after making numerator of partial fraction expansion of ¥(s) unity, we have obtain Yis) as 13) = Ys) 4¥43) + Y(5) + V0) Let ¥4(s)has denominator portion as summation of the two parts, then compare 29/509) x 5] withZ(s)= R+s1.+ 1 and then identify th elements and comected in series. And assuming rest admittances are not as summation of the two parts, then compare Y(s) with Y(s) = £+£+5C and then identify the clemens and connected in parallel, Therefore, nework connection for Foster-II form as shown in Fig. 8.3. (Two series passive [_ elemenss) vs) YAl8) yas) ats) 8.4.3, Cauer-I form Steps to be followed for Cauer-I form as follows: Step-l To arranged both numerator and denominator in descending powers of s, starting fiom highest to lowest power of s of immittance function ‘Step-ll, To obtain continued fraction expansion of the driving point immittance function, ‘The continued fraction expansion can be obtained by the division and inversion procedure. ‘Step-HIL. ‘To identify the elements. If the continued fraction form of Z(), then the quotient of first division gives an impedance ‘which isa series arm and then the quotients give alternately Ms) and 2(s) terms representing shunt and series arms respectively If the continued fiaction form of ¥(:), then the quotient of first division gives an admittance which is a shunt arm and then the quotient give alternately Z(s) and ¥(s) terms representing series and shunt arms respectively. ‘The cauer-I form is basically low pass structure of the network 8.4.4, Cauer-Il Form Steps to be followed for Cauer-II form as follows: ‘Step-L Toarranged both numerator and denominator in ascending powers of s, starting from lowest to highest power of s. ‘Step-Il and II] are identical with Cauer-I form. ‘The Cauer-Il form is basically high pass structure. Note: It is possibility of getting negative quotients in the continued fraction expansion. In such case restart with the inversion of the given function and first quotient in such case represents inverse of the given function ie. if given function is Z(3) then first quotient taken as ‘admittance which is a shunt arm and if given function is ¥(s) then first quotient taken as impedance which is a series arm.Network Synthesis 799 85. L-C IMMITTANCE FUNCTION The L-C driving point immittance function has following properties: 1, The immittance function is the ratio of even to odd or odd to even polynomials. 2, The poles and zeros of immittance function are on the imaginary axis including origin). 3. The poles and zeros interlace (or alternate) on the jomaxis, 4. There must be either a zero oF a pole at the origin and infinity. 5. The highest powers of numerator and denominator must differ by unity 6. The lowest powers of numerator and denominator must also der by unity, 86. R-C IMPEDANCE OR R-L ADMITTANCE FUNCTION The R-C impedance or R-L admittance function has following properties: 1, ‘The poles and zeros lic on the negative real axis (including origin) of the complex s-plane, 2, The poles and zero interlace (or alternate) along the negative real avis 3. The residues of the poles are real and positive. 4. The poles and zeros are simple. There are no multiple poles and zeros. 5. Weknow that the poles and zeros are called critical frequencies of the network. The critical frequency nearest tothe origin is always pole. This may be located at the origi. 6. Theeritical frequency ata greatest distance away from the origin is always a zero, which may be located at infinity also. 7. There is no pole located at infinity and no zero at the origin. 8. Theslope ofthe graph Z (a) against o is always negative, 9. The value of Zpc{s) at s =0 is always greater than the value of Zp(s) als = 2. ie. ZycA0) > Zyl). 8.7. P-L IMPEDANCE OR A-C ADMITTANCE FUNCTION The R-L impedance or R-C admittance function has following properties 1. The poles and zeo ie onthe negative real axis (nluding origin) c the complex s-plane. 2. The poles and zeros interlace (or alternate) along the negative teal axis 3. The residuces of the poles must real and positive 4. The poles and ze10s are simple. There are no multiple poles and ze10s. 5. We know that the poles and zeros are called critical frequencies of the network. The critical frequency nearest to the origin is always zero. This may be located atthe origin, 6. Theeritical frequency ata greatest distance away from the origin is always a pole, which may be located at infinity also. 7. There isno pole located at origin and no zero at the infinity. 8. The slope of the graph of Z(o) against « is always positive. 9. The value of Zy,(8) at s= 0 is always less than the values of Zy(s) at s = 9. Le. 2 qy(0) < Zyl). 8.8. SOLVED EXAMPLES ON NETWORK SYNTHESIS sQs +8) GHN6+3) Example, 8.22, Show that function F(s) = represents an Rel. impedance. Synthesize the impedance in Foster-I form.790 Network Analysis and Synthesis Solution, ‘Step-. To show the given function F(s) represents on R-L impedance. 563548) FO= ean (5+3) , 8 The zeros are: #=0,-5 ‘The poles are: sa-13 The pole-zero plot shown in Fig. E 8.22 Fig, £822 From pole-zero plot we see that zeros and poles interlace along the negative rea axis and first singularity at the origin, Therefore, given function F(s) is an R-L. impedance oF R-C admittance funetion ‘Step-Il. To obtain partial fraction of F(s) as an impedance function (for Foster-I from). — oy) 835+ 8) R= 29)" Nis43) For obtaining positive residues, we can write ZG), 3548 A,B” s (SFI G43) stl” 543 where, and 5s, 1 os as or 4s)=2-42_ Sal S43 Making numerator portion unity, then we getNetwork Synthesis 791 ‘Step-IIL. To identify the network elements and draw the Foster-I form of network. a 2S A (s)+ Za(s) 1 KOs) Kl) @ For R-L-C in parallel: 1 etetiuc nO Rsk (in series) ‘The synthesized network as per equation (2) is shown in Fig, E8.22(a). FAC) zie) 2s)» WIT WOOT . | beg wet | WW ww 0 ni=$a te=ta ° Fig, €8.22(0), Example. 8.23. For the network shown in Fig. £8.23. L s(s? +3) . then synthesize Y as the L-C admitiance. 2+Y "DS 4s 46541792 Network Analysis and Synthesis 4a. ve y 10d Fig £0 Satution. Step-1. To obtain Ys). 1 s(s2+3) Given 24+ “Ws? +6541 or s(s2+3)Q+ N=29 +324 6541 or S(P+3V=2s9 + 524 65+ 1-299 65 eel 543) ‘Step-I. To obtain partial fraction of ¥(s). 4 MOTD A Bee ms S43 sa 43 2 Bs+C= = (By partial fraction) where, 41 o B+ Cs or -3B+C Comparing left and right sides, we get oA) 22sNetwork Synthesis 793 =K(s) + his) =K(+ (2) |. Z,()) (in parallel) a For parallel R-L-C: Ws) = Bt T_T SC end \—vera—o gfe For series R-L-C: 209) = R+ s+ y Y,)~3 aMO=351 3 L— The synthesized network as per equation (2) is shown in Fig. £8.23 (a). > (In paratiel) Yo Ye oF Lino 20000, a wie ve) Fig. £8.23 (a) Example. 8.24 The input impedance of the network shown in Fig. E8.24 , 2s? +2 29) = Te DP 2s 4d I,Z(9) is an L-C network: (a) find the expression for Z,(s). (b) synthesiz Z,(9) ina Foster Horm. Yal—> 10 240 Fig, £8.24.794 ‘Network Analysis and Synthesis Solution. (a) To obtain Z,(s) From the network, we have 249)= VIE) 749-44) WZ) or Als)= Zs) + Z()] o 2) (1 ~ ZN = Zl) Zql3) or 29) F A) 2s? +2 Putting 2,(0)= TZ, s yess im equation (1), we get 2s? +2 429+ 2542 2s? +2 ss 842s? +2542. 2s? +2 “sh 42s 28+ (+2) AQ) or ox (0) Foster-1 form: ‘Step-l. To obtain partial fraction expansion of Z,(s where, and Bs+C= or B+ Cs=—2 or ~B+Cs=-2 ‘Comparing left and right sides, we get B=1,C=0 r los MS)=Network Synthesis 795 o Z(s) @) set ‘Step-Il, To identify the network elements and draw the Foster-I form of network. 1 249) =5 s4e = 24s) +2") «ay (In series) Fig. € 6.240). Example 828. Realize the admittance function P+ TE 46 Na) = SES iin Foster and Cauer form, Solution: (/) Foster-I1 form ‘Step-l: To obtain partial fraction of Ms). P4746 Loker) Since, s+2Fa TF 6G 42s 5546796 ‘Network Analysis and Synthesis S546 We ya = sp READIES © 5434 2 42 v2 or Ms)= 843444 wo 2s ‘Step-Il : To identify the network elements and draw the Foster-II form of network. eT 2°93 = V6) + VAs) + Y4(0) fy = Ys) + Yois) + J) (in parallel) -Q) For R-L-C in Parallel in parle and ¥3(0) =3 Sopot t > he For R-L-C in series: ' U)= Re sL tse 2x0) ‘The synthesized network as per equation (2) is shown in fig. E 825 (a). Hy) > voy [] Ke, YiNetwork Synthesis 797 1) > Fig, E 825 (a). (i) Cauer-1 form: Step-l: To obtain continued fraction expansion of Y(s) beginning from highest powers of s. Bante O° ‘The continued fraction expansion of Ys) is given by s+ 2) * +6 (se ¥, 9 C,=1F (in shunt) 242s 4 25 28 = )ss+o Ss oy a ey= PF lin shun 5s + 4.2 6) F (sp PA Rega jgM lin series) fo draw the Cauer-/ form of the network, he elements of Z(s) and ¥(s) will be in series and shunt manner respectively. The Caver-l form of the network is shown in Fig. F 8.25 (b) H 1 Rieko Reeko FOF Oe Ye - Fig. E825 (6) unple 8.26, Find Cauer-t and I forms of an impedance function no . WtAdls+6) 49= eay(e+3)798 ‘Network Analysis and Synthesis Solution: (2) Cauer-I form. Step-I : To obtain continued fraction expansion of Zs) beginning from highest powers of s. a9)= (s+4)(s+6) _ s2+10s +24 (s+3)(s+5) +8S 415 1 In series RL-C : Z(s) = R + sl + — s) R= 12 Gin series) s+ B+ 15 aro)eeiseis Lonrg “9 (in shunt) 1 4 4 Teatslos+0) 4 49z, +R, =40 (in series 5 2549 5 2,9 Ra TO ) 60, s+ 3 zs +15 2 ronsgetr (in shun 7 2 15) 2 [4 462,48) =Lainserien 7 \35 ’ 3S 2 7 Step-II: To draw the Cauer-I form of the network, The clemenis of Z(s) and Y%s) will be in series and shunt manner respectively, The Cauer-l form of the network is shown in fig. E 8.26. (i) Cawer-It form: ‘Step-I: To obtain continued fraction expansion of Z(s) beginning with lowest powers of s.Network Synthesis 799 15485457 “ M9 Dae Os +5 Note: Here the continued fraction expansion of Z(s) give negative quotients, therefore obtain the ‘continued fisction expansion of Y(3), The continued fraction expansion of Y(s) is given by 99) 78, = $e Ginsu Tt 4 2 2542 s+ 24-5 oF (in seri Git ge jae tower 24-3 39g F (in series) 36 438 7 77,49 ooo hor a 34” 136 = 136 "19 Q (in shunt) 2{ 34.68 _ 2312 +7 * shunt) Step-Il : To draw the Cauer-II form of the network. ‘The Cauer-II form of the network is shown in Fig, E 8,26 (a).800 Network Analysis and Synthesis Fig. F 8.26 (a). Example 8.27. Synthesize an impedance function SP 4) +2 2) — READE +2) 5 Fostered and Il forms. 3416) Solution: Fosters form: ‘Step-t: To obtain partial fraction expansion of Z(s) (9° +16) a) __ 854 +2325" + 800 “ st4+l6s Since, 3+ 16s R232 + 800(8 Bot + 128s? 104s? + 800 104 s? +800, 2 2) = 54 ASSO * +P albs 104s? +800 s(s? +16) A, Bote = 84445 SF +16 ses = ts? +800 we © F416 leo end m+n 1H teal . HOA (16 +900 or Bs? + Cy=— 864 or 16 B+ Cs= — 854 = 16) ‘Comparing left and right sides, we get Ba 84 54. ¢ 16 ‘ 2) = 854 04 SH S416Network Synthesis 801 Making numerator postion as unity, then we get = Z(s) +2) +29, .Q) ya TO] (in series) = Zs) + Zs) + For R-L-C in series: 1 Ms) = R+ sb + Als) = 8s Ly L+1,-88 >IT « 1 and Z(9) 598 G \ *C\=hFa SE te om (in series) For #-L-C in parallel: 19" R 16 1 1e)> ys +18 The synthesized network as per equation (2) is shown in Fig, E 827. 49 9) A) 1-8 C= TOTS aa Foster-II for Step-L: To obtain partial fraction expansion of Y(3). We know that802 Network Analysis and Synthesis _ 8 +16) © 8 +H +25) By partial fraction, we can write o Hs) sA+B C+D 425 hs Rs where, = yap = ao L828) ~ Bar and a Gs+D= = Comparing left and right sides, we get Making numerator portion unity, then we get — 5 00 3°" 5s Ys) = AL) ‘Step-Hl: To identify the network elements and draw the Foster-II form of the network. 1 1 O° gS, y 3 Gs = Y0) +0) 22) 1 Ufa Z(s)_ Za(s) J (in parallel) For R-L-C in series: 2 t As)~ Re skNetwork Synthesis 803 azgor=t4s1 _1 tc, Cake L___.)-14H Bh aw (in parallel) 3s 3 eal =28 L___,1,=%84 The synthesized network as per equation (2) is shown in Fig. E 8.27 (o) > no [uo [> : NW)» Fig. E827 (@) Example 8.28. Which of the following functions are L-C driving point impedance? Why? s(s? +4) (s? +16) AUD = EEO) #2) (8° +1) (5? +8) Z)- > s(x" +4) And, synthesize the realizable impedances in a Foster and Cauer forms. Solution: To obtain LC driving point impedance. For Z,(s) :The zeros are P+ The poles are : sl) $= +3, + /5 ‘The pole-zero plot of Zs) is shown in Fig. E 8.28. Ip (OF io) B +4 +6 Re (ora) Fig. E 8.28, ce, poles and zeros are not interlacing on the jimaxis, therefore Z,(s) is not LC driving point impedance, Sis For Z,(s):The zeros are : s = + fl, /2y2 ‘The poles are : s = 0, + 204 ‘Network Analysis and Syrthesis ‘The pole-zero plot of Z,(s) is shown in fig. F 8.28 (a), i” +R +2 Fig. E 8.28 (a) Since, poles and zeros are interlacing on the jomaxis, therefore Z,(s) is L-C driving point impedance. Foster-1 form: Step-l: To obtain partial fiaction expansion of Zs). _ (P+) (s7 +8) 20)" an +957 48, tds P+ 4s F ITF E ly s+4ae 5248 2 +48 stds 5248 + sts? +4) es Ayes where, and or or (8=-4 Comparing left and right sides, we get ~4B=-120 893 and ceoNetwork Synthesis 805 , 2.3 s = st2e. Als) sD Making numerator portion unity, then we get, 14 20° +7-+T a “ Data Step-ll: To identify the network elements and draw the Foster-I form of network. Ld 29° 87+ gh yttse = Z4(s) + Zils) +2565) = Zi(s)+Z5ls) + For R-L-C in series: (in series) Lb wot ~{7 + G The synthesized network as per equation (2) is shown in fig. E 828 (4), |. bey vo zy et -1H Cyn $F Bley Bley zt A ZO) > Md 248) Fig. E 8.28 (4) Foster-Il form: Step-I: To obtain ¥,(s) and partial fraction expansion of ¥,(s).‘806 ‘Network Analysis and Syrthesis We know that 10" ZG e+4) (8 +) (87 +8) Using paral fraction expansion, we get s Ys) As+B C+D yys)= SAB, 4D 8 Sar Pas cpe HH) 3 where, fee Comparing left and right sides, we get 3 A=>, bso 2 and cee +9 Sat lag 777 Making numerator portion unity, then we get Lt 742 Ty 8 ° ty ay Step-Hl: To identify the network elements and draw the Foster-Il form of Network Ys) = =) 2y(s) 23 (s) “| For R-L-C series: (in parallel) ) i Ms) = Resh od Ho) - E+ BONG as Liege + (n paral) and Z4o)~ 504Network Synthesis 807 ‘The synthesized network as per equation (2) is shown in fig. F 828 (c) Tos 1 tp ul ¥)— Ys) ry T teln a Ts) Fig. E 8.28 (c) ‘Cauer-I form: Step-l: of 5. obtain continued fraction expansion of Z,(s) beginning from highest powers (2 +67 +8) _ 54 +957 +8 32 +4) sds Zs) 1 In series R-L-C : Z(s) = R+sL-+ As) x roa R* sh The continued fraction expansion of Zs) is given by S44 SFOS Bs OZ, HL, = 1H (Series) eae S32 + 8}s% + as| In parallel R-L-C : Ys) = s enrcete (shunt) O44 9b au (Series) DB! AMG AF (Shunt) ‘Step-ll: To draw the Cauer-I form of the network, ‘The elements of Z(3) and ¥(s) will be in ind shunt manner respectively. ‘The Cauer-I form of the network is shown in Fig. E 8.28 (d). 2% Ly til d= qqH o—"00000" T aera AAs)808 ‘Network Analysis and Synthesis (Cauer-II form: Step-I: To obtain continued fraction expansion of Z,(s) beginning with lowest powers os. I)(s? +8) _ 84957 +54 sts? +4) asts® Zs) 1 in series R-L-C : 6) = Rosh In series R-L-C 2) C 1d in parallel RL-C: YG) = + 4sC In parallel RL-C:Y6) = RT The continued frection expansion of Z,{s) is given by a 1 a&ss)se FOLIC =z F Geries 8428 "1 Is + st] 4s + 5 9% Ly = FH (Shunt) 4s+ _—— ishave[ St oF (eri Fe) 18+ 8 PZ 90.2 SF (cries) 12 ON L2H (Shun ‘Step-Il: To draw the Cauer-II form of the network, The elements of 23) and 1(s) will be in series and shunt manner respectively The Cauer-l form of the network is shown in fig, E 8.28 (e), Fig. E 8.28 (e)