EEC130A: Practice Problems for Midterm 1

Version 1.1 Feb. 5th 2012

Instructor: Xiaoguang Leo Liu (lxgliu@ucdavis.edu)

P-1. A parallel-wire transmission line is constructed of #6 AWG copper wire (diameterd = 0.162 in., conductivity c = 5.8 107 S/m) with a 12-inch separation in air. As-suming no leakage between the two wires, find R, L, G, and C. Assume a workingfrequency of 1 MHz.

Using Table. 1,Rs =

pi fc/c = 2.61 104

R = 2Rspid

= 4.04 102 /m

L = 0pi

ln2Dd

= 2.0 H/m

C = pie

ln2Dd

= 5.56 pF/m

G = 0 because the problem states that there is no leakage between the two wires.

Figure 1: (Fig.2-4 from FAE) A few examples of transmission lines.

Some problems are adapted from The Schaums Outlines on Electromagnetics and 2008+ Solved Prob-lems in Electromagnetics.

1

Table 1: (Table 2-1 from FAE)Transmission-line parameters R, L, G, and C for threetypes of lines.

P-2. Derive the phasor-form wave equation in terms of current from the telegraphersequations.

The telegraphers equations are:

dV(z)dz

=(R + jL

)I(z), (1)

dI(z)dz

=(G + jC

)V(z). (2)

Taking the derivative of Eqn. 2 with respect to z, we get

d2 I(z)dz2

=(G + jC

) dV(z)dz

. (3)

Substituting Eqn. 1 into Eqn. 3, we get

d2 I(z)dz2

= (G + jC) (R + jL) I(z). (4)Rearranging Eqn. 4 and defining

=

(R + jL) (G + jC),

we get the wave equation

d2 I(z)dz2

2 I(z) = 0. (5)

From now on, we will omit the tilde in the phasors. Any capatalized symbol with a single positionvariable represents a phasor.

2

P-3. In lossline transmission lines, we learned that the input reflection coefficient in isrelated to the load reflection coefficient L by

in = Lej2l,

where l is the length of the tranmission line. We see that in and L have the samemagnitude but differ in phase by 2l. For lossy lines, however, both magnitude andphase are different for in and L. From the general solution to the wave equation,derive an expression for in in terms of L, , l, and attenuation constant .

In the presence of attenuation, the general solution to the wave equation is

V(z) = V+0 ez +V0 e

z

= V+0 ezjz +V0 e

z+jz.

The input impedance at a distance l from the load (z = l) is

Zin =V(l)I(l)

=V+0

[el+jl + Leljl

]I+0[el+jl Leljl

]= Z0

1 + Le2lej2l

1 Le2lej2l= Z0

1 + in1 in , (6)

where the input reflection coefficient is given by

in = Le2lej2l (7)

P-4. A parallel-wire line operating at 100 kHz has Z0 = 557 , = 2.3 105 Np/m,and = 2.12 103 rad/m. For a matched termination at z = 0 and VL = 10/0 V, (a)give a general expression of V(z), (b) evaluate V(z) at a distance of 10 km from the load.

Since the line is matched, there is no reflected wave. Therefore a general expression forthe voltage wave takes the form

V(z) = V+0 ezejz.

At z = 0,V(z) = V+0 = 10/0

At z = 10 km (notice z becomes negative as we move away from the load )V(z) = 10/0 e2.310510103ej2.1210310103 = 12.71/135 V.

3

P-5. A 600- transmission line is 150 m long, operates at 400 kHz with = 2.4 103Np/m and = 0.0212 rad/m, and supplies a load impedance ZL = 300 + j300 . Findthe length of line in wavelength, L, in and Zin. For a received voltage V(z = 0) = 50V, find the total voltage at the input V(z = 150 m).

= 2pi/ = 296.4 m

l = 150/296.4 = 0.51

L =ZL Z0ZL + Z0

=300 + j300 600300 + j300 + 600

= 0.2 + j0.4 = 0.45/116.6

in = Le2lej2l = 0.09 + j0.20Zin = Z0

1 + in1 in = 502/22.8

V(z) = V+0[ejzjz + Lejz+jz

].

At z = 0,V(z = 0) = V+0 (1 + L) = 50, (V)

therefore,V+0 = 56.2/ 26.6 (V).

At z = 150 m,

Vin = V+0[ej2.410

3(150)j0.0212(150) + 0.45/116.6ej2.4103(150)+j0.0212(150)

]= 75.0/167.3 (V)

P-6. A 15-m length of 300- line must be connected to a 3-m length of 150- line that isterminated in a 150- resister. Assuming all lines are lossless, find the VSWR on the 300- line. In order to match the two sections, a quarter-wavelength line of characteristicimpedance Z0 is added (Fig. 2). Find the appropriate Z0. Assume a working frequencyof 50 MHz.

Since the load and line 2 are matched, the input impedance looking into line 2 is 150 .This is also the load impedance to line 1. The reflection coefficient at A A is

AA =150 300150 + 300

= 1/3.

Therefore,

VSWR =1 + 1/31 1/3 = 2.

4

ZL=150 Z02=150 Z01=300

ZL=150 Z02=150 Z01=300 Z0

/4l1=15 m l2=3 m

A

A

Figure 2: Circuit diagram for Problem. 6.

The quarter-wave length line needs to match between Z01 = 300 and Z0,2 = 150 ,therefore its characteristic impedance needs to be

Z0 =Z01Z02 = 212.1 .

P-7. A 50- slotted line that is 40 cm long is inserted in a 50- lossless line feeding anantenna at 600 MHz. Standing-wave measurements with the antenna in place yield thedata of Fig. 3. The scale on the slotted line has the lowest number on the load side. Findthe impedance of the antenna, the reflection coefficient due to the load, and the velocityof propagation on the line.

Slotted Line

z=0z=8 cmz=20.5 cm

VminVmax

Antenna

ZL

VSWR =2.2

Figure 3: Circuit diagram for Problem. 7.

We know that the separation between adjacent maxima and minima is a quarter of awavelength. Therefore = 4 (20.5 8) = 50 cm. At 600 MHz, the propagation ve-locity (phase velocity) is up = f = 3 108 m/s (this means that the slotted line has airdielectric).

We can calculate the magnitude of the reflection coefficient from the VSWR measurement.

|L| = VSWR 1VSWR+ 1 = 0.375

5

We know that the voltage maxima condition is

2dmax = r + 2npi

Therefore, the phase of the reflection coefficient r is

r = 2dmax 2npi = 22pidmax 2npi = 295.2 (n = 0)

Therefore the refection coefficient is L = 0.375/295.2.The load impedance (antenna impedance) is calculated from the reflection coefficient,

ZL = Z01 + 1 = (52.8 j41.4)

P-8. Show that a short section of a shorted transmission line appears as if its an inductor.

A short circuit is equivalent to a load impedance of ZL = 0.

The input impedance of a short circuited transmission line is

Zin,SC = Z0ZL + jZ0 tan lZ0 + jZL tan l

.

With ZL = 0,Zin,SC = jZ0 tan l.

The input impedance of a capacitor takes a similar form

ZL = jL.

P-9. Show that a short section of an open transmission line appears as if its an capacitor.

An open circuit is equivalent to a load impedance of ZL = .

The input impedance of an open circuited transmission line is

Zin,OC = Z0ZL + jZ0 tan lZ0 + jZL tan l

.

With ZL = ,

Zin,OC =Z0

j tan l.

The input impedance of a capacitor takes a similar form

ZC =1

jC.

6

P-10. A standing wave given by V(z) = V+0 ejz + V0 e

jz exist on a short circuitedtransmission line with characteristic impedance Z0. Sketch the magnitude of the voltageand current on the transmission line.

For short circuit load, the reflection coefficient is 1. Therefore V0 = V+0 . The totalvoltage is then given by

V(z) = V+0(ejz ejz

)= 2V+0 sin z.

And the total current is given by

I(z) =V+0Z0

(ejz + ejz

)= 2

V+0Z0

cos z.

Plots of the voltage and current are given in Fig. 4.

2|V+

0| 2|V

+0|

Z0

|V| |I|

3/4 /2 /4 0 3/4 /2 /4 0

Figure 4: Circuit diagram for Problem. 10.

P-11. A section of lossless coaxial cable having Z0 = 50 and phase velocity up =2 108 m/s is terminated in a short circuit and operated at a frequency of 10 MHz. De-termine the shortest length of the lines such that, at the input terminals, the line appearsto be a 100-pF capacitor.

The input impedance of a short circuited transmission line is

Zin,SC = jZ0 tan l.

The input impedance of a capacitor is

ZC =1

jC.

In order for them to match,

tan l =1

Z0C= 3.183.

Therefore,(l)min = 1.876 rad,

lmin = 5.97 m.

7

P-12. Fig. shows a microstrip circuit with a shorted stub 3 mm in length.The effectivepermitivity of the substrate is ee f f = 4 F/m and the microstrip lines are all designedto be 50 . (a) Find the equivalent inductance of the shorted stub at 3 GHz; (b) Asfrequency increases, the short stub becomes more capacitive than inductive. Find outthe frequency at which the short stub appears as an open circuit.

Via

Ground

Trace

Substrate eff=4

Short Stub

Figure 5: Circuit diagram for Problem. 12.

The propagation wavelength on the microstrip line is

=upf

=c

fee f f = 50 mm

The input impedance of the short stub is

Zin,SC = jZ0 tan l = jZ0 tan(

2pil)

The equivalent inductance is then given by

L =Z0 tan

(2pil)

=

50 tan(

2pi50 3)

2pi 3 109 = 1.06 nH/m

P-13. In microstrip circuits working at high frequencies, it is difficult to realize an idealshort circuit because the vias have a finite length, which creates a certain inductance.Sometimes microstrip circuit designers use an open stub to realize a short. For a mi-crostrip circuit working at 30 GHz, find out the shortest length of an open stub thatappears as a short. The effective permitivity of the microstrip is ee f f = 4 F/m.

The propagation wavelength on the microstrip line is

=upf

=c

fee f f = 5 mm

8

Open Stub

Ground

Trace

Substrate eff=4

Figure 6: Circuit diagram for Problem. 13.

We know that a quarter-wavelength line transforms an open to a short, so a 5/4 =1.25 mm open stub appears a short.

Alternatively, we could work out the math. We know that the input impedance of anopen stub is

Zin,OC =Z0

j tan l.

In order for it to appear as a short, Zin,OC must be 0, which means that l must be pi/4 +npi, n = 0, 1, 2, . . .. Therefore, the shortest length of an open stub that appears as a shortis l = /4.

P-14. Show that for a lossless transmission line having a purely resistive load RL, showthat

VSWR =max (ZL, Z0)min (ZL, Z0)

.

=ZL Z0ZL + Z0

If ZL > Z0,

|| = ZL Z0ZL + Z0

VSWR =1 + ||1 || =

1 +ZL Z0ZL + Z0

1 ZL Z0ZL + Z0

=ZLZ0

=max (ZL, Z0)min (ZL, Z0)

If ZL < Z0,

|| = Z0 ZLZ0 + ZL

VSWR =1 + ||1 || =

1 +Z0 ZLZ0 + ZL

1 Z0 ZLZ0 + ZL

=Z0ZL

=max (ZL, Z0)min (ZL, Z0)

Note: this provide a quick way to estimate VSWR for real load impedances.

9

P-15. One method of determining the characteristics of a transmission line (with lengthl) is to measure the input impedance Zin,SC when the line is terminated with a shortcircuit and the input impedance Zin,OC when the line is terminated with an open circuit.(a) Show that you can determine Z0 and from Zin,SC and Zin,OC. (b) Given Zin,SC =62.0/37.7, Zin,OC = 141.9/ 84.1 and l = 2 miles, find Z0, , and .

(a) The input impedance of a short circuited transmission line is

Zin,SC = jZ0 tan l.

The input impedance of an open circuited transmission line is

Zin,OC =Z0

j tan l.

We notice thatZin,SCZin,OC = Z20 .

ThereforeZ0 =

Zin,SCZin,OC.

We also notice thatZin,SC/Zin,OC = tan2 l.

Therefore

=arctan

(Zin,SC/Zin,OC

)l

.

(b) For a lossy transmission line, the input impedance transformation expression is slightlydifferent (note the use of tanh instead of tan and the disappearance of j),

Zin = Z0ZL + Z0 tanhlZ0 + ZL tanhl

.

It follows thatZin,SC = Z0 tanhl,

andZin,OC =

Z0tanhl

.

Z0 =Zin,SCZin,OC =

62.0/37.7 141.9/ 84.1 = 93.8/ 23.2.

=

tanh1(

62.0/37.7141.9/ 84.1

)2 1.609 103 = 1.74 10

4 + j7.46 105.

= 7.46 105 Np/m. = 1.74 104 rad/m.

10

P-16. A 70- high-frequency lossless line is used at a frequency where = 80 cmwith a load at x = 0 of (140 + j91) . Use the Smith chart to find the following: L,VSWR, distance to the first voltage maximum from the load, distance to the first voltageminimum from the load, the impedance at |V|max, the impedance at |V|min, the inputimpedance for a section of line that is 54 cm long, and the input admittance.

Fig. 7 is the Smith chart plot for Problem 16.

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ANGLEOFTRANSMISS

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>W

AVELENGTHSTOWARDGENERATOR>

The following problems from the textbook are also considered practice problems.

1.15, 2.6, 2.16, 2.32, 2.34, 2.52, 2.58, 2.61, 2.66

16

Problem 1.15 A laser beam traveling through fog was observed to have an intensityof 1 (W/m2) at a distance of 2 m from the laser gun and an intensity of 0.2(W/m2) at a distance of 3 m. Given that the intensity of an electromagneticwave is proportional to the square of its electric-field amplitude, find the attenuationconstant of fog.

Solution: If the electric field is of the form

E(x, t) = E0ex cos(tx),then the intensity must have a form

I(x, t) [E0ex cos(tx)]2 E20 e2x cos2(tx)

or

I(x, t) = I0e2x cos2(tx)where we define I0 E20 . We observe that the magnitude of the intensity varies asI0e2x. Hence,

at x = 2 m, I0e4 = 1106 (W/m2),at x = 3 m, I0e6 = 0.2106 (W/m2).

I0e4

I0e6=

106

0.2106 = 5e4 e6 = e2 = 5

= 0.8 (NP/m).

Problem 2.6 A coaxial line with inner and outer conductor diameters of 0.5 cmand 1 cm, respectively, is filled with an insulating material with r = 4.5 and = 103 S/m. The conductors are made of copper.

(a) Calculate the line parameters at 1 GHz.

Solution: (a) Given

a = (0.5/2) cm = 0.25102 m,b = (1.0/2) cm = 0.50102 m,

combining Eqs. (2.5) and (2.6) gives

R =1

2pi

pi f c

c

(1a

+1b

)

=1

2pi

pi(109 Hz)(4pi107 H/m)

5.8107 S/m(

10.25102 m +

10.50102 m

)= 0.788 /m.

From Eq. (2.7),

L =2pi

ln(

ba

)=

4pi107 H/m2pi

ln2 = 139 nH/m.

From Eq. (2.8),

G = 2piln(b/a) =

2pi103 S/mln2

= 9.1 mS/m.

From Eq. (2.9),

C = 2piln(b/a) =

2pir0ln(b/a) =

2pi4.5 (8.8541012 F/m)ln2

= 362 pF/m.

(b) Solution via Module 2.2:

Problem 2.16 A transmission line operating at 125 MHz has Z0 = 40 , = 0.02(Np/m), and = 0.75 rad/m. Find the line parameters R, L, G, and C .Solution: Given an arbitrary transmission line, f = 125 MHz, Z0 = 40 , = 0.02 Np/m, and = 0.75 rad/m. Since Z0 is real and 6= 0, the line isdistortionless. From Problem 2.13, = LC and Z0 =

L/C , therefore,

L =Z0

=0.7540

2pi125106 = 38.2 nH/m.

Then, from Z0 =

L/C ,

C = L

Z20=

38.2 nH/m402 = 23.9 pF/m.

From =

RG and RC = LG,

R =

RG

R

G=

RG

L

C = Z0 = 0.02 Np/m40 = 0.6 /m

and

G = 2

R=

(0.02 Np/m)2

0.8 /m = 0.5 mS/m.

Problem 2.32 A 6-m section of 150- lossless line is driven by a source with

vg(t) = 5cos(8pi107t30) (V)and Zg = 150 . If the line, which has a relative permittivity r = 2.25, is terminatedin a load ZL = (150 j50) , determine:

(a) on the line.(b) The reflection coefficient at the load.(c) The input impedance.(d) The input voltage Vi.(e) The time-domain input voltage vi(t).

Solution:

vg(t) = 5cos(8pi107t30) V,Vg = 5e j30

V.

Vg

IiZg

Zin Z0 ZL

~

Vi~~

+

+

-

+

-

-

VL~

IL~

+

-

Transmission line

Generator Loadz = -l z = 0

Vg

IiZg

Zin

~

Vi~~

+

-

150

(150-j50)

l = 6 m

= 150

Figure P2.32: Circuit for Problem 2.32.

(a)

up =cr

=3108

2.25= 2108 (m/s),

= upf =2piup

=

2pi21088pi107 = 5 m,

= up

=8pi1072108 = 0.4pi (rad/m),

l = 0.4pi6 = 2.4pi (rad).Since this exceeds 2pi (rad), we can subtract 2pi , which leaves a remainder l = 0.4pi(rad).

(b) = ZLZ0ZL +Z0

=150 j50150150 j50+150 =

j50300 j50 = 0.16e

j80.54.

(c)

Zin = Z0[

ZL + jZ0 tan lZ0 + jZL tan l

]= 150

[(150 j50)+ j150tan(0.4pi)150+ j(150 j50) tan(0.4pi)

]= (115.70+ j27.42) .

(d)

Vi =VgZin

Zg +Zin=

5e j30(115.7+ j27.42)150+115.7+ j27.42

= 5e j30(

115.7+ j27.42265.7+ j27.42

)= 5e j300.44e j7.44 = 2.2e j22.56 (V).

(e)

vi(t) = Re[Vie jt ] = Re[2.2e j22.56e jt ] = 2.2cos(8pi107t22.56) V.

Problem 2.34 A 50- lossless line is terminated in a load impedanceZL = (30 j20) .

(a)

(b)

Z0 = 50 ZL = (30 j20)

Z0 = 50 ZL = (30 j20)

Zi

dmax

R

Figure P2.34: Circuit for Problem 2.34.

(a) Calculate and S.(b) It has been proposed that by placing an appropriately selected resistor across

the line at a distance dmax from the load (as shown in Fig. P2.34(b)), wheredmax is the distance from the load of a voltage maximum, then it is possible torender Zi = Z0, thereby eliminating reflection back to the end. Show that theproposed approach is valid and find the value of the shunt resistance.

Solution:(a)

=ZLZ0ZL +Z0

=30 j205030 j20+50 =

20 j2080 j20 =

(20+ j20)80 j20 = 0.34e

j121 .

S = 1+ ||1|| =

1+0.3410.34 = 2.

(b) We start by finding dmax, the distance of the voltage maximum nearest to theload. Using (2.70) with n = 1,

dmax =r4pi

+2

=

(121pi180

) 4pi

+2

= 0.33 .

Applying (2.79) at d = dmax = 0.33 , for which l = (2pi/ )0.33 = 2.07 radians,the value of Zin before adding the shunt resistance is:

Zin = Z0(

ZL + jZ0 tan lZ0 + jZL tan l

)= 50

((30 j20)+ j50tan2.0750+ j(30 j20) tan2.07

)= (102+ j0) .

Thus, at the location A (at a distance dmax from the load), the input impedance ispurely real. If we add a shunt resistor R in parallel such that the combination is equalto Z0, then the new Zin at any point to the left of that location will be equal to Z0.

Hence, we need to select R such that

1R

+1

102 =150

or R = 98 .

Problem 2.52 On a lossless transmission line terminated in a load ZL = 100 ,the standing-wave ratio was measured to be 2.5. Use the Smith chart to find the twopossible values of Z0.

Solution: Refer to Fig. P2.52. S = 2.5 is at point L1 and the constant SWRcircle is shown. zL is real at only two places on the SWR circle, at L1, wherezL = S = 2.5, and L2, where zL = 1/S = 0.4. so Z01 = ZL/zL1 = 100 /2.5 = 40 and Z02 = ZL/zL2 = 100 /0.4 = 250 .

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0.11

0.11

0.12

0.12

0.13

0.13

0.14

0.14

0.15

0.15

0.16

0.16

0.17

0.17

0.18

0.18

0.190.19

0.2

0.2

0.21

0.210.22

0.220.23

0.230.24

0.24

0.25

0.25

0.26

0.26

0.27

0.27

0.28

0.28

0.29

0.29

0.3

0.3

0.31

0.31

0.32

0.32

0.33

0.33

0.34

0.34

0.35

0.35

0.36

0.36

0.37

0.37

0.38

0.38

0.39

0.39

0.4

0.4

0.41

0.41

0.42

0.42

0.43

0.43

0.44

0.44

0.45

0.45

0.46

0.46

0.47

0.47

0.48

0.48

0.49

0.49

0.0

0.0

AN

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GR

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