32 CHAPTER 2

Section 2-6: Input Impedance

Problem 2.17 At an operating frequency of 300 MHz. a Iossless 50-0 air-spacedtransmission line 2.5 m in length is terminated with an impedance ZL = (60+ j20) 0.Find the input impedance.

Solution: Given a lossless transmission line, Zo = 50 0, f = 300 MHz, I = 2.5 In,

and ZL = (60+ j20) Q. Since the line is air filled, up = c and therefore, from Eq.(2.38),

ro 2n x 300 x 106P = - = ~ _~n = 21t rad/m.

up

Since the line is lossless, Eq. (2.69) is valid:

~ = Zo (ZL + j20 tan Pl) = 50 (60+ j20) + j50tan(2n rad/m x 2.5 m»)n \20 + jZLtanpl / ,50+ j(60+ j20)tan(2n fad/m x 2.5 m)"

=50(60+j20}+j50XO) =(60 '20)050+ j{60+ j20) x 0 +] .

~oblem 2.ij) A lossless transmission line of electrical length I = 0.35A. isterminated in a load impedance as shown in Fig. 2-38 (p2.1S). Find r,S, and Zin.

I.• 1= O.35A. •. I

Zin-" 20=100.Q I IZL=(60 +j30) Q

Figure P2.18: Loaded transmission line.

Solution: From Eq. (2.49a),

r= ZL-2o = (60+j30)-loo =0.307ej/32.5°.ZL+20 (60+ j30) + 100

From Eq. (2.59),

S= 1+ WI = 1+0.307 = 1.89.1- rn 1-0.307

CHAPTER 2

From Eq. (2.63)

. (ZL+ jZotan~l)2m =Zo Zo + jZL tan f31

((60+ 1'30) +1'100","(¥0.35J..») = (64.8- 1'38.3) Q.= 100 100+j(60+ 1'30) tan (¥0.351)

33

~em 2.1V Show that the input impedance of a quarter-wavelength long losslessline terminated in a short circuit appears as an open circuit.

Solution:

For I =~. /31= z;:. ~ = ¥. WithZL =O. we have

/ ;Z1tan1t/2\ . _ ."

2m = Zo \''' ~ Zo ' ) = jOO (open CIrCUit).

Problem 2.20 Show that at the position where the magnitude of the voltage on theline is a maximum the input impedance is purely real.

Solution: From Eq. (2.56). lmax= (9r+2nx}f2/3. so from Eq. (2.6I). using polarrepresentation for r.

which is real. provided Zo is real.

Problem 2.21 A voltage generator with vg(t) = 5cos(2x X 109t) V and internalimpedance Zg = 50 Q is connected to a 50-a lossless air-spaced transmissionline. The line length is 5 em and it is tenninated in a load with impedanceZL = (100- jIOO)!2. Find

(a) rat the load.(b) Zin at the input to the trarismission line.(c) the input voltage ~ and input eUITeDt~.

38 CHAPTER 2

Section 2-7: Special Cases

Problem 2.24 At an operating frequency of 200 MHz, it is desired to use a sectionof a Jossless 50-Q transmission line tenninated in a short circuit to construct an

equivalent load with reactance X = 25Q. If the phase velocity of the line is O.75c,what is the shortest possible line length that would exhibit the desired reactance at itsinput?

Solution:

J3 = roJu = (21t rad/cycle) x (200 x 106 cycle/s) = 5.59 rad/m.p 0.75 x (3 x 108 m1s)

On a lossless short-circuited transmission line, the input impedance is always purely

imaginary; i.e., Zf; = j;q:. Solving Eq. (2.68) for the line length,

1= .!..tan-I (X~C)= 1 tan-I (25 Q) = (0.464+n1t) rad ,P Zo 5.59 rad/m SO Q 5.59 rad/m

for which the smallest positive solution is 8.3 cm (with n = 0).

bJem 2.25 A lossless transmission line is terminated in a short circuit. How

long m wavelengths) should the line be in order for it to appear as an open circuit atits iuput tenninals?

Solution: From Eq. (2.68), Zf; = jZo tanf3l. If f3l = (1C/2+ rm), then zt; = joo (Q).Hence,

I = ~ (~+n1t) = ~+ nt...21t 2 4 2

This is evident from Figure 2.15( d) ..

Problem 2.26 The input impedance of a 31-cm-long lossless transmission line ofunknown characteristic impedance was measured at IMHz. With the line tenninatedin a short circuit, the measurement yielded an input impedance equivalent to aninductor with inductance of 0.128 .uH,· and when the line was open circuited, themeasurement yielded an input impedance equivalent to a capacitor with capacitanceof 20 pF. Find Zo of the line, the phase velocity, and the relative permittivity of theinsulating material.

Solution: Now 00 = 21tf = 6.28 x 106 rad/s, so

zt;; = jooL = j21t X 106 x 0.128 x 10-6 = jO.804 Q

CHAPTER 2 39

and z:: = 1/ jroC = 1/(j21t X 106 X 20 X 10-12) = - j8000 Q.From Eq. (2.74), Zo = vZSii~ = V(jO.804 Q)(- j8000 Q) = 80 D. Using

Eq. (2.75).

ro ro[U - - - --:--====

P - J} - tan-I V-Z~/Zfu

_ 6.28 x 106 X 0.31 1.95 x 106

_ --C..,....--- ,..-----------~-----------) = ---- mfstan-I ±.J'- jO.804/{ - j8000) (±0.0l +n1t) ,

where n 2: 0 for the plus sign and n 2: 1 for the minus sign. For n = O.

Up = 1.94 X 108 m1s = O.65c and Er = (c/up)2 = 1/0.652 = 2.4. For other values -of 11. up is very slow and Er is unreasonably high.

@-oblem 2!!:i) A 60-Q resistive load is preceded by a A/4 section of a 50-Q lossless- line. which itself is preceded by another A/4 section of a l00-Q line. What is the input

impedance?

Solution: The input impedance of the 1../4 section of line closest to the load is foundfrom Eq. (2.77):

Zm = Zfi _ 502ZL - 60 =41.7 Q.

The input impedance of the line section closest to the load can be considered as theload impedance of the next section of the line. By reapplying Eq. (2.77), the next

of 'A./4line is taken into account:

~n = Z5 _ 1002ZL - 41.7 =240 Q.

2.28 A 1DO-MHz FM broadcast station uses a 300-Q transmission line

the transmitter and a tower-mounted half-wave dipole antenna. The antennais 73 Q. You are asked to design a quarter-wave transformer to match the

to the line .

..(a) Detennine the electrica11ength and characteristic impedance of the quarter­wave section.

If the quarter-wave section is a two-wire line with d = 2.5 cm. and the spacingbetween the wires is made of polystyrene with Er = 2.6, determine the physicallength of u;e quarter-wave section and the radius of t.l}etwo wire conductors.

46 CHAPTER 2

z~=75 f!(Antenna I)

5004AJ.~VV

r-Oo

A

Zin-B

Generator

Line 1

D

ZL2= 75 Q(Antenna 2)

Figure P2.32: Antenna configuration for Problem 2.32.

1 --. 1 - - (2.86)2 x 37.5Pin = Z9'ie[Ii Vi ] = 29'ie(hli~J = ... = 153.37 (W).

This is divided equally between the two antennas. Hence, each antenna receives

15i37 = 76.68 CW)..

~blem ~ For the circuit shown in Fig. 2-42 (p2.33), calculate the averageincident power. the average reflected power, and the average power transmitted intothe infinite 100-.0 line. The "-/2 line is lossless and the infinitely long line issHghtly lossy. (Hint: The input impedance or an infinitely long line is equal to itscharacteristic impedance so long as c£=1= 0.)

Solution: Considering the semi-infinite transmission line as equivalent to a load(since all power sent down the line is lost to the rest of the circuit), ZL =Zl = 100 Q.Since the feed line is A/2 in length, Eq. (2.76) gives Zm = ZL = 100 .Q andf31 = (21t/l)(A/2) = 1t,so e±j131 = -1. From Eq. (2.49a),

r= ZL-4> = 100-50 =!.ZL+Zo 100+50 3

CHAPTER 2

500 ~

Zo= 500 ---P~V_I_P~VIpr_rav I

Figure P2.33: Line tenninated in an infinite line.

47

Also, converting the generator to a phasor gives 1/g = 2ejOO (V). Plugging all theseresults into Eq. (2.66),

v.+ = Cigl<.)C I ) = ex 100 ) ( I )o Zg +Zm ejf31+re-jfJl 50+ 100 (-1) + H-l)= lej1800 =-I (V).

From Eqs. (2.84), (2.85), and (2.86),

. Jvo+12 IIej1800 /2

~v= 2Zo = 2x50 =1O.0mW,

1,2

2 . 1~v=-frl ~v=- 3" x lOmW=-1.1 roW,

~v = Pay =~v+~~ = 10.0 roW -1.1 mW = 8.9 m\V.

2.34 An antenna with a load impedanceZL = (75+ j25) 12is connected totransmitter through a 50-12 lossless transmission line. If under matched conditions

load), the transmitter can deliver lOW to the load, how much power does itto the antenna? Assume Zg = ~.

48 CHAPTER 2

Solution: From Eqs. (2.66) and (2.61),

y,+= ( Vg~ ) ( 1 )o Zg +Zin ejJ31+re- jf!l

_ VgZo [(1 +re-j2f!1)/(1_re-j2j31)] e-j/31

- (20+20 [(1+re-j2/31)/(1-re-.i2/31)])(' . T""' _MAl)

_ Vge-j/31

- (1- re-j2/3I) +(1+ re-j2(1). - -j/3/

Vge I - - 'Pl

- (1- re-j2/31) +(1+ re-j2/31) 2"Vge J •

Thus. in Eg. (2.86).

Under the matched condition, WI = 0 and PL = 10 W. so /VgI2/82o = 10 W.When ZL = (75 +j25) a. from Eg. (2.49a).

r= ZL-Zo = (75+j25) n-50n =0 277ei33.6°ZL+Zo (75+ j25) Q+50 Q. ,

SoPav= lOW (1-lr/2) = lOW (1-0.2772) =9.23 W.

Section 2-9: Smith Chart

(f;oblem 2.3sJ Use the Smith chart to find the reflection coefficient correspondingto a load impedance:

(a) ZL = 320.(b) ZL = (2 - 2j)Zo,(c) ZL = -2jZo.(d) ZL =0 (short circuit).

Solution: Refer to Fig. P2.35. °(a) Point A is ZL = 3 + jO. r= O.5eD

(b) Point B is ZL = 2 - j2. r= 0.62e-29.7°°(c) Point C is ZL = 0 - j2. r= I.Oe-53.!

(d) Point D is ZL = 0 +jO. r= 1.0e180.0o

·CHAPTER 2

FigureP2.35: Solution of Problem 2.35.

49

~m 2:3i) Use the Smith chart to find the normalized load impedancecorresponding to a reflection coefficient:

(a). r= 0.5,(b)r=0.5/60°,(c) r= -1,(d) r=0.3/-30" ,

(e) r=o,(f) r= j.

Solution: Refer to Fig. P2.36.

50

Figure P2.36: Solution of Problem 236.

(a) Point A' is r= 0.5 at ZL = 3+jO.

(b) Point BJ is r= 0.5ej6Qo at ZL = 1+ jl.l5.(c}PointCI is r= -1 atzL =0+ jO.

(d) Point D' is r= O.3e-j300 at ZL = 1.60- j0.53.(e) PointE' is r= 0 atZL = 1+jO.

(f) PointF' is r= j atzL =0+ j1.

CHAPTER 2

Problem 2.37 On a IossIess transmission line teiminated in a load ZL = 100 Q,the standing-wave ratio was measured to be 2.5. Use the Smith chart to find the twopossible values of Zoo

CHAPTER 251

80]ution: Refer to Fig. P2.37. S = 2.5 is at point Ll and the constant SWRcircle is shown. ZL is real at only two places on the SWR circle, at LI, where

ZL = S = 2.5, and L2, whereZL = lIS = 0.4. so Zm = ZL/zLl = 100 0./2.5 = 40 0and 202 =ZL/ZU = 100 010.4 = 250 O.

Figure P2.37: Solution of Problem 237.

A 10ssJess 50-Q transmission line is tenninated in a load with

Q. Use the Smith chart to find the following:(a) the reflection coefficient r,

.(b) the standing-wave ratio,(c) the input impedance at 0.35A.from the load,

52 CHAPTER 2

(d) the input admittance at 0.35A.from the ioad.(e) the shortest line length for which the input impedance is purely resistive,(f) the position of the first voltage maximum from the load.

Figure P2.38: Solution of Problem 2.38.

Solution: Refer to Fig. P2.38. The nonnalized impedance

(50+ j25) Q = 1+jO.5ZL = 50 Q

is at point Z-LOAD. o

(a) r= 0.24eJ76.0 The angle of the reflection coefficient is read of that scale atthe point ST'

CHAPTER 2 53

(b) At the point SWR: S = 1.64.(c) Zm is 0.350J... from the load, which is at 0.144A. on the wavelengths to generator

scale. So point Z-IN is at 0.144A.+0.350A = OA94A on the WTG scale. At point

2m = ZinZo= (0.61 - jO.022) X 5012= (30.5 - jI.09) Q.

(d) At the point on the SWR circle opposite Z-IN,

v. _ Yin _ (1.64+ jO.06) - (327 '117) S.Lin--Zo ---so-n--- . +]. m.

(e) Traveling from the point Z-LOAD in the direction of the generator (clockwise),SWR circle crosses the XL =0 line first at the point SWR. To travel from Z-LOAD

one must travel 0.250A -0.14411. =0.10611.. (Readings are on the wavelengthsgenerator scale.) So the shonest line length would be 0.1 06A..

(I) The voltage max occurs at point SWR. From the previous part, this occurs at= -O.I06A.

2.39 A lossless 50-Q transmission line is tenninated in a short circuit.the Smith chart to find

(a) the input impedance at a distance 2.3A. from the load,

the distance from the load at which the input admittance is Yin=- jO.04 S.

Refer to Fig. P2.39.For a sbort. Zin = 0+ jO. This is point Z-SHORT and is at O.OOOl on the WTG

Since a 10ssIess line repeats every A/2. traveling 2.3l toward the generator isto traveling 0.31 toward the generator. This point is at A : Z-/N. and

Zm =ZinZo = (0 - j3.08) x 50 12=- jI54 Q.

The admittance of a short is at point Y -SHORT and is at 0.2S0l on the WTG

Yin=YinZo=- jO.04 S x 50 Q = -:-j2,

point B : Y -IN and is at 0.324l on the WTG scale. Therefore. the line length-0.250A =0.074l. Any integer haIfwavelengths farther is also valid.