CHAPTER 2

Chapter 2

23

Sections 2-1 to 2-4: Tcansmission-Line Model

~blem 2.UA transmission line of length I connects a load to a sinusoidal voltagesource with an oscillation frequency f. Assuming the velocity of wave propagationon the line is c, for which of the foHowing situations is it reasonable to ignore thepresence of the transmission line in the solution of the circuit:

(a) 1=20 em,f= 10kHz,(b) I =50km,f=60Hz,(c) 1=20 em, f = 300 MHz,(d) 1=1 mm,f= 100GHz.

Solution: A transmission line is negligible when 1II..~ 0.01.

f) !.. _ If _ (20 X 10-2 m) x (10 x HP Hz) _ 6 '67 10-6 ( li 'bl ),8 I.. - Up - 3 x 108 mls -. x neg gt e.

(b) !... = If = (50 x 103 m) x (60 x 10° Hz) _ 0 01 (00 fin)'I 308 -. rdere.r.. Up X I mlsI If (20 x 10-2 m) x (300 x 106 Hz)

(c) l = up = 3 x 1()8 mls =0.20 (nonnegligible).

(d) !.. = If = (l X 10-3 m) x (100 x 109 Hz) _ 0 33 ( r 'bII.. up 3 x lOS mls -. nonneg JgI e).

Calculate the line parameters RI, V, G'" and C' for a coaxial line withdiameter of 0.5 cm and an outer conductor diameter of 1 em,

with an insulating material where J1. =Po, Er = 2.25, and 0' = 10-3 S/m. Theare made of copper with Pc =JlO and Oc = 5.8 x 107 S/m. The operating

is 1GHz.

Given

a = (0.5/2) em =0.25x 10-2 ill,

b = (1.0/2) cm = 0.50x 10-2 m,

Eqs. (2.5) and (2.6) gives

= ~V1tfJlc (.!+.!)21t O'c a b

x(109 Hz)(4x x 10-7 HIm) (1 1)5.8 X 107 Sim 0.25 x 10-2 m + 0.50 x 10-2 mWm.

24

From Eq. (2.7),

From Eq. (2.8),

I Jl (b) 41t x 10-7 HimL = -In - = 2 In2= 139nWm.21t a 1t

G' = 21tO' _ 21t X 10-3 SlmIn(b/a) - In2 =9.1 mS/m.

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(Wm),

From Eq. (2.9),

C' = 21t£ = 2~ _ 21t x 225 x (8.854 x 10-12 F/m) .In (b/a) In(b/a) - 1_" = 181 pF/m.

Problem 2.3 A I-GHz parallel-plate transmission line consists of 1.5-crn-widecopper strips separated by a O.2-em-thick layer of polystyrene. Appendix B givesPc = PO = 41t X 10-7 (Wm) and O'c= 5.8 X 107 (S/m) for copper, and Er = 2.6 forpolystyrene. Use Table 2-1 to determine the line parameters of the transmission line.Assume J.l=Jlo and (j ~ 0 for polystyrene.

Solution:

,2Rs 2 2 (1tX109X41tXIO-7)1/2R = -;- = w ..j1tfPcO'c = 1.5 x 10-2 5.8 X 107 = 1.0

,_Jld_41tX1O-7x2xlO-3 -167 10-7 (HI)L- - 2 -. x m,w 1.5x lO-G' =0 becausea =0,

, ew w 10-9 1.5 X 10-2 10

C = - = EoEr- =~ x 2.6 x 3 = 1.72x 10- (F/m).d d 1t 2xlO-

Problem 2.4 Show that the transmission line model shown in Fig. 2-37 (P2.4)yields the same telegrapher's equations given by Egs. (2.14) and (2.16).

Solution: The voltage at the central upper node is the same whether it is calculatedfrom the left port or the right port:

v(z+ ~~,t) = v(z,t) - !R'L\z i(z,t) - !L'L\z~i(z,t)

= V(z+L\z,t) + !R'L\z i(z+~,t) +1L'~;i(Z+L\z,t).

26

Section 2-5: The Lossless Line

CHAPTER 2

Problem 2.6 In addition to not dissipating power. a lossless line has two importantfeatures: (1) it is dispertionless (p.p is independent of frequency) and (2) itscharacteristic impedance Zo is purely real. Sometimes, it is not possible to designa transmission line such that RI «: ooL' and GI «: «c1, but it is possible to choose thedimensions of the line and its material properties so as to satisfy the condition

R' Cl = L' G' (distortionless line).

Such a line is called a distortionless line because despite the fact that it is not lossless,it does nonetheless possess the previously mentioned features of the loss line. Showthat for a distortionless line,

~'a=R' - =VR'G,D (3 = 00.../DC I ,

Solution: Using the distortionless condition in Eq. (2.22) gives

'{= a+ j~ = J(R' + jml!)(G' + jo:C/)

=VL'Clj(~+jOO) (g+jro)

= VL'C't I (Z + joo) (~: +jro)

(R' ) (CI= VL'C' L' + joo =R'V Ii + jrovL'C'.

Hence,

f3 = :Jm(y) =OO.../L'C',ro 1

up = J3 = VllC' .

Similarly, using the distortion less condition in Eq. (2.29) gives

zo= R'+jroL' = {I!G' +jroCl V Ci RilL' +jro = {ifG'IC' +joo V Ci .

(mNp/m),

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Solution: The product of the expressions for ex and Zo given in Problem 2.6 gives

It =aZo =40 X 10-3 X 50 =2 (Wm),

and taking the ratio of the expression for Zo to that for up = ro/f3 = 1/"; L'CI gives

I ZO 50 -7L =- = 2 5 00 =2 x 10 (HIm) = 200 (nHIm).up • xl

With If known, we use the expression for Zo to find CI:

£1 2 x 10-7

Cl = Z3 = (50'''' =8 x 10-11 (F/m) = 80 (pF/m).

The distortionless condition given in Problem 2.6 is then used to find G'.

G' = KC' = 2 X 80 X 1~-12 8 X 10-4 (S/m) = 800 (.uS/m),L' 2 x W-

and the wavelength is obtained by applying the relation

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Problem 2.8 Find ex and Zo of a distortionless line whose RI = 4 Wm andGI =4 X 10-4 S/m.

Solution: From the equations given in Problem 2.6.

ex =VR'(JI = 14x 4 x 10-4)1/2 =4 x 10-2 (Np/m),

{if (iii ( 4 ) 1/2Zo=V(J=V(j= 4xlO-4 =100Q .

.Problem 2.9 A transmission line operating at 125 MHz has Zo = 40 11. a = 0.02and p =0.75 rad/m. Find the line parameters RI• If, G'. and CI .

..Solution: Given an arbitrary transmission line, f = 125 MHz, Zo = 40 a,= 0.02 Np/m, and f3 = 0.75 radlm. Since Zo is real and ex:j: O. the line is

From Problem 2.6, 13 =rovL'C' and 20 = .jL'/C', therefore,

LI = 1320 = _ 0.7~ ~ 40 __ =38.2 nH/m.ID

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Then, from Zo = vL'Ie'.

, L' 38.2 nH/me = Z2 = 402 = 23.9 pF/m.o

From a= v'R'G' and R'e' =L'G',

CHAPTER 2

/R' (TfR' = v'RIG'y 0' = v'R'G' V Ci = aZo = 0.02 Np/m x 40 Q = 0.8 Wm

and

a.2 (0.02 Np/mf = 0.5 mS/m.G' = - = 0 8 WmR' .

Problem 2.10 Using a slotted line, the voltage on a lossless transmission line wasfound to have a maximum magnitude of 1.5 V and a minimum magnitude of 0.8 v.Find the magnitude of the load's reflection coefficient.

Solution: From the definition of the Standing Wave Ratio given by Eq. (2.59),

s= IVlmax 1.5IVloon = 0.8 = 1.9.

Solving for the magnitude of the reflection coefficient in terms of S, as inExample 2-4,

In= S-l = 1.9- I =0.3.. , S+ 1 1.9+ 1

@oblem 2J.P Polyethylene with Er = 2.25 is used as the insulating material in aIossless coaxial line with characteristic impedance of 50 Q. The radius of the innerconductor is 1 rom.

(a) What is the radius of the outer conductor?(b) What is the phase velocity of the line?

Solution: Given a Iossless coaxial line, Zo = 50.0, Er = 2.3, a = 1 rom:(a) From Table 2-2, Zo = (60/ fiJln (b/a) which can be rearranged to give

b = a/-Ov'€r/60 = (1 mm)e50v1.3/60 = 3.5 mm.

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(b) Also from Table 2-2,

Up = ~ = 3 X 108 mls.JE; .J23 = 1.98 X 108 mfS.

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(problem 2.122 A 50-Q lossless transmission line is tenninated in a load withimpedance ZL = (30- j60) 0. The wavelength is 5 em. Find:

(a) the reflection coefficient at the load,(b) the standing-wave ratio on the line,(c) the position of the voltage maximum nearest the load,(d) the position of the current maximum nearest the load

Solution:(a) From Eq. (2.49a),

r= Zr. -Zo . (30- j60) -50 = 0.632e-j71.6°.ZL+Zo (30-j60) +50

(b) From Eq. (2.59),

1+Irl _ 1+0.632 =4.44.S = - .. - 1_ 0.632

(c) From Eq. (2.56)

1 _ erA. nA. _ -71.6° x 5 em 1t rad n x 5 emmax - 41t + 2 - 41t 1800 + 2

= -0.50 em + 2.50 em =2.00 em.

(d) A current maximum occurs at a voltage minimum, and from Eq. (2.58),

lmin = lmax-'}../4 =2.oocm-5 cm/4 = 0.75 em.

CProblem 2W On a 150-Q lossless transmission Jine, the following observationswere noted: distance of first voltage minimum from the load = 3 em; distance of firstvoltage maximum from the load = 9 em; S =3. Find ZL.

Solution: Distance between a minimum and an adjacent maximum = A./4. Hence,

9 em - 3 em = 6 em = '}../4,

30 CHAPTER 2

or A. = 24 em. Accordingly, the first voltage minimum is at fmin = 3 cm = ~.Application of Eq. (2.57) with n = 0 gives

21t AaT - 2 x - x - = -1t

A 8 '

which gives aT = -1t/2.

S-1 3-1 2W/=-=-=-=0.5.S+l 3+1 4

Hence, r= 0.5 e- jrr.j2 = - jO.5.

Finally,

Problem 2.14 Using a slotted line, the following results were obtained: distance offirst minimum from the load = 4 em; distance of second minimum from the load =14 em, voltage standing-wave ratio = 2.5. If the line is lossless and Zo = 50 .0, findthe load impedance.

Solution: Following Example 2.5: Given a loss less line with Zo = 50 Q, S = 2.5,

lmin(O) = 4 em, lmin(l) = 14 em. Then

A.

lmin(l) -lmin(O) = 2"

or

11.=2 X (lmin(l) -lmin(o») = 20 em

and

211: 211:rad/cycle = 101t rad/m.J3= T = 20 em/cycle

From this we obtain

aT = 2f3lmin(n) - (2n + 1)1t rad = 2 x 101t rad/m x 0.04 m - 1t rad= -0.21t rad = -36.0°.

Also,

WI = S - I _ 2.5 - IS+] - - - - = 0.429.