Problem 2.20 A 300- lossless air transmission line is connected to a complexload composed of a resistor in series with an inductor, as shown in Fig. P2.20. At5 MHz, determine: (a) , (b) S, (c) location of voltage maximum nearest to the load,and (d) location of current maximum nearest to the load.L = 0.02 mHZ0 = 300 R = 600 Figure P2.20: Circuit for Problem 2.20.Solution:(a)ZL = R+ jL= 600+ j2 51062105= (600+ j628) . = ZLZ0ZL +Z0= 600+ j628300600+ j628+300= 300+ j628900+ j628= 0.63ej29.6.(b)S = 1+||1||= 1+0.6310.63= 1.67.(c)lmax = r4for r > 0.=

29.6180

604,

= 31085106= 60 m

= 2.46 m4.405(d) The locations of current maxima correspond to voltage minima and vice versa.Hence, the location of current maximum nearest the load is the same as location ofvoltage minimum nearest the load. Thuslmin = lmax + 4,

lmax < 4= 15 m

= 2.46+15 = 17.46 m.Problem 2.23 A load with impedance ZL = (25 j50) is to be connected to alossless transmission line with characteristic impedance Z0, with Z0 chosen such thatthe standing-wave ratio is the smallest possible. What should Z0 be?Solution: Since S is monotonic with || (i.e., a plot of S vs. || is always increasing),the value of Z0 which gives the minimum possible S also gives the minimum possible||, and, for that matter, the minimum possible ||2. A necessary condition for aminimum is that its derivative be equal to zero:0 =Z0||2=Z0|RL + jXL Z0|2|RL + jXL +Z0|2=Z0(RL Z0)2+X2L(RL +Z0)2+X2L= 4RL(Z20 (R2L +X2L))((RL +Z0)2+X2L)2.Therefore, Z20=R2L +X2L orZ0 = |ZL| =

(252+(50)2) = 55.9 .Amathematicallyprecise solutionwill alsodemonstrate that this point is aminimum (by calculating the second derivative, for example). Since the endpointsof the range may be local minima or maxima without the derivative being zero there,the endpoints (namely Z0 = 0 and Z0 = ) should be checked also.Problem 2.26 A 50- lossless transmission line is connected to a load composedof a 75-resistor in series with a capacitor of unknown capacitance (Fig. P2.26). If at10 MHz the voltage standing wave ratio on the line was measured to be 3, determinethe capacitance C.RL = 75 Z0 = 50 C = ?Figure P2.26: Circuit for Problem 2.26.Solution:|| = S1S+1= 313+1= 24= 0.5ZL =RL jXC, where XC =1C. = ZLZ0ZL +Z0||2=

ZLZ0ZL +Z0

ZLZ0ZL +Z0

||2= ZLZL +Z20 Z0(ZL +ZL)ZLZL +Z20 +Z0(ZL +ZL)Noting that:ZLZL = (RL jXC)(RL + jXC) =R2L +X2C,Z0(ZL +ZL) =Z0(RL jXC +RL + jXC) = 2Z0RL,||2= R2L +X2C +Z20 2Z0RLR2L +X2C +Z20 +2Z0RL.Upon substituting |L| = 0.5, RL = 75 , and Z0 = 50 , and then solving for XC,we haveXC = 66.1 .HenceC =1XC=12 10766.1= 2.411010= 241 pF.1.4Problem2.19 A50-lossless transmissionlineis terminatedinaloadwithimpedance ZL = (30 j50) . The wavelength is 8 cm. Find:(a) the reection coefcient 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.(e) Verify quantities in parts (a)(d) using CD Module 2.4. Include a printout ofthe screen display.Solution:(a) From Eq. (2.59), = ZLZ0ZL +Z0=(30 j50) 50(30 j50) +50= 0.57ej79.8.(b) From Eq. (2.73),S = 1+||1||= 1+0.5710.57= 3.65.(c) From Eq. (2.70)dmax = r4+ n2= 79.88 cm4 rad180+ n8 cm2= 0.89 cm+4.0 cm = 3.11 cm.(d) A current maximum occurs at a voltage minimum, and from Eq. (2.72),dmin = dmax/4 = 3.11 cm8 cm/4 = 1.11 cm.(e) The problem statement does not specify the frequency, so in Module 2.4 weneed to select the combination offand r such that = 5 cm. With r chosen as 1,f =c=31088102= 3.75 GHz.The generator parameters are irrelevant to the problem.The results listed in the output screens are very close to those given in parts (a)through (d).Figure P2.19(a)Figure P2.19(b)Problem 2.21 On a 150- lossless transmission line, the following observationswere noted: distance of rst voltage minimum from the load = 3 cm; distance of rstvoltage maximum from the load = 9 cm; S = 3. Find ZL.Solution: Distance between a minimum and an adjacent maximum = /4. Hence,9 cm3 cm = 6 cm = /4,or= 24cm. Accordingly, therst voltageminimumisat dmin= 3 cm=8.Application of Eq. (2.71) with n = 0 givesr228=,which gives r =/2.|| = S1S+1= 313+1= 24= 0.5.Hence, = 0.5ej/2=j0.5.Finally,ZL = Z0

1+1

= 150

1 j0.51+ j0.5

= (90 j120) .Problem2.24 A50-losslesslineterminatedinapurelyresistiveloadhasavoltage standing-wave ratio of 3. Find all possible values of ZL.Solution:|| = S1S+1= 313+1=0.5.For a purely resistive load, r=0 or . For r=0,ZL=Z0

1+1

=50

1+0.510.5

=150 .For r=, =0.5 andZL=50

10.51+0.5

=15 .Problem 2.25 ApplyCDModule2.4togenerateplotsofthevoltagestanding-wave pattern for a 50- line terminated in a load impedance ZL= (100 j50) .Set Vg = 1 V, Zg = 50 , r = 2.25, l = 40 cm, andf = 1 GHz. Also determine S,dmax, and dmin.Solution: