Question

i(z, t) i(z + Az, t) R'Δz 2 L'Az 2 v(2,t) G'Az C'Az v(z+Az, t) R'Az 2 L'AZ 2 Az dvíz, t) R'i(2,4) + 2 (a) Hint: Set up your equations using the appropriate KVL and/or KCL relationships for this circuit model of a transmission line differential section. Derive the following Partial Differential Telegraphy Equations; ai(z,t) (2.14 in Ulaby) az at ai(z,t) av(z,t) G'v(z,t) + C' (2.16 in Ulaby) дz at Sketch the lossless version of the equivalent circuit of the two-conductor transmission line given above. = (6) Now for the lossless version of the equivalent circuit, use Phasor Analysis to derive the two Ordinary Differential Equations from the two Partial Differential Telegraphy Equations. Why are these Ordinary as opposed to Partial differential equations?

Answer 1

Difference between Partial and ordinary difference equation.

• Partial difference equation is a differential equation in which dependent variable is dependent on more than one independent variable.
• Ordinary difference equation is a differential equation in which dependent variable is dependent on only  one independent variable.

(1) To set up KVL and KCL in the given Grcuit, we must know following 2 Concepts. 1) Votloge across inductor L it 4(+) y() L dillt) 1 mm L ət helt helt) C VCLH) at i (2+02, ) ilzit) RAZ L'AZ rimm WWW Z JA + + VIZH) G'az wwn C'az V/z+A2, +) where ¿CH) : Current Parning through induetor VCE) : Voltage across inductor L 2) Current through Capacitor C th a vell) 2) The R, G, and c' are barometer of a trommission line ond defined Per unit Length Ris Reislonce po umil length Krim) Fig 1 Ľ' - Induclonce per unit Length (Hlm) 6' = Conductonce bee unit length (ilm) c = Cobocilence per unit lenglik (Flm) This is the region why we have multiplied each component by oz, small (ve small) length of c Scanned with www t mm L'AZ R'Az 2 very Trommissim Line. CamScanner

(2) Now we will while KVL (Rischhoff's Village Law) which states that tobal voltage drop in a close loob is O. shown by 1 and dt a i(2,6) at = 0 Taking loop as arrow in the fig 1 using eqn 0 V(2,4) – R' Az i (7.1) - 1.92 o ilz 1) – +(2+62, +) R’92 i(2,4) – L'42 Adding some parts &(2,4) R'Az ilz,t) - L'az ə i(2+) - V12+02,+) = 0 Now Keep Vetlage component on on side - [u(2+) + U[Z+82, +)] Now faking 12 on the EHS and Rearranging TV(2+02,4) – V(3,6) R' (2,7) + L at - 12 [R' i(2,1) + L' il24) at i Oil2,1) at 3 AZ Want 2 LHS of the above egh is nothing bull differenhon (pahal) when 12 -> 0. So eq? 3 becomes a 112,6) R'ilz,+) + L' @ i 13,4) Proved 4 . ôz at -> Now we will waile KCL (kecholl's Current Law ) at junchen. Mit shown in 4g 2, which states that all current alm@annjunchon (algebnic Sum) is always zero.

(3) 1(3,4) C'sz U12+62, +) – C'sz QU[Z+02, +) AU(z+021)_ i(2+02, 4) = 0 Taking the current component 1.45 op alove eph - [i{2+62, ) – 1(2,7] sz [c^ v[Z+02,4) + c'2 U2:43,7) Taking az to the LHS and taking Limit AZ-30 [ (z+12,1-2(3,7) G'v{2+82, ) + c' & U/Z+02,47] LHS is partial dulferenhahon wirt z of i(2,4), 10 Lt 12-0 A2>0 AZ Again 5 ai (2,1) az G' [0(2,43] +c' Qulz,+) Proved. ? at 6 le For o line to be loss len Resislonce (R') and conductance (6') should be equal to 0. R' - 6'=0 Pulling this condichon in fog 1 and Redrawing . 2 (2,1) i (Z+03, ) mm losslen version of the equivalent Grcuit 0137) C'az two conductor Trommission Line. mum c Scanned with figz Fig2 shows L'AZ 2 1127021) of L'4212 CamScanner

4 o Now Putting the conduhon of egne of lossles Tv line inb egn @ and 6, we get following 2 eqn a lv (7+)) 8 i (2,1) at L 7 az 8 с 1) ejut i (2,7) } 9 ai (2,7) a v12H aż at Here 1 (26) ond U36) ase imlonlenous current and volleges . el Ilz) and v(z) Corresponding Phaoar of 112,7) and U127) and they are Kelaled as Re[Ilz) e U12, 1) Re [v(z) e jwer] so phasan Jooms are easy to handle and vay with z only, where as mnohonlenows Jorm vang in both Z and t. - jw Home delle renhahon in time domain Ċ multiplied with jw in foreg domain, where w: Angular frequency. 5. Now chonging to phasae form of egno and ® with help of egn © and -) 10 Also we can write colo s CS

5 - dv(z) dz 11 jWL I(2) 2 (12) d I(2) jWC VIZ) dz Dilleren hohon eph Wirt z jWL d I(z) (12 from ean d2v(2) da v(2) dz dz Pulting the value of dilen d?v[22 jWL x jwc V(z) az (jw) PLC V(2) = -6?LC (V(z)) d2 Taking both teim same side d²v(2) 6? Le v(z) = 0 Am dz Similarly for Current Phase I(2) d21(2) w?Le I(z) = 0 An. d2 (13 (14 Equahon (3 ond @ Reforesents Ordinay differental Ewahan. Paritas Dilleren had Equahon has two or more indepeden! i (2, 1) and variable. just in case of Eqn u ond 6 V127) vay with z as well as time t. Ordinary Differential Equahon vary with one variable only. with distance z only. This the notion changes from 92 m2 From equo 0 to Ⓡay, ae dilleven Seal Ewahon and vay CE