Appendix D

Solutions to Problems in Chapter 4

D.1 Problem 4.1

Figure D.1 shows a network of three resistors R1, R2 and R3. Equations 4.80 and 4.81 provideelement values to terminate both, common and differential mode. In this problem we will derivethese formulas.

1I

3R

(b)

(a)

1R

2 1R R=

2 1I I=

3 1 2I I I= +

Common mode

Differential mode

1I

3R

1R

2 1R R=

2 1I I= -

30I =

0I =

Figure D.1: Current definition for common mode and differential mode

1

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 2

Common Mode

The currents I1 and I2 on the signal lines are equal (Figure D.1a). The common ground acts asa return conductor. The differential voltage between the signal lines is zero. Hence, there is nocurrent through resistor R3.

At the end of the line a common mode signal sees R1 and R2 in parallel. In order to avoidreflections at the end of the line the parallel circuit must equal the common mode characteristicimpedance.

Z0,cm = R1 ‖ R2 (D.1)

If we choose R1 = R2 the parallel circuit becomes

R1 ‖ R2 =R1

2=R2

2= Z0,cm (D.2)

and the resistors areR1 = R2 = 2Z0,cm (D.3)

Differential Mode

In differential mode the currents I1 and I2 on the signal lines are of equal magnitude but oppositesign (Figure D.1b). There is no net current in the ground conductor (I3 = 0). At the end of theline a differential mode signal sees R3 in parallel with the sum of R1 and R2.

Z0,diff = R3 ‖ (R1 +R2) (D.4)

With Equation D.3 we get immediately

R3 =4Z0,cm

4Z0,cm − Z0,diff(D.5)

Example

Let us put some numbers in here. Figure 4.29 (book page 153) shows differential and commonmode characteristic impedances of a coupled microstrip line. The characteristic impedance of thesingle ended microstrip line is Z0 = 50 Ω at f = 5 GHz. The transmission line is characterizedby: substrate height h = 635µm, relative permittivity εr = 9.8, trace width w = 600µm,thickness of metallization t = 10µm. The spacing s determines the differential and commonmode characteristic impedances in Figure 4.29.

Let us consider the case s/h = 1. From Figure 4.29 we read

Z0,cm ≈ 28.5 Ω and Z0,diff ≈ 85.0 Ω (D.6)

Hence, the termination network consists of the following three resistors.

R1 = R2 = 57.0 Ω and R3 = 334.1 Ω (D.7)

In order to evaluate our result we perform a circuit simulation with a coupled microstrip line.Figure D.2 shows the schematics for common mode and differential mode excitations. Addi-tionally, we use the transmission line tool LineCalc (Agilent Inc.) to calculate more accuratecharacteristic impedances than those approximate values read from Figure 4.29. The schematicwith the modified values of R1, R2 and R3 are also shown in Figure D.2.

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 3

Common mode

Differential mode

Figure D.2: ADS circuit simulation of line termination: schematics with approximate values forR1, R2 and R3 (left) and schematics with more accurate values (right)

1

0.1

0.01

0.001

0.0001

Reflection coefficient

0.00001

Figure D.3: Simulated s-parameter results (solid lines: approximate values; dashed lines: moreaccurate values)

Figure D.3 shows the reflection coefficients ri = sii for all four circuits. The magnitudes ofthe reflection coefficients are given in dB and in linear scale. Solid lines represent the simulationswith the approximate values. Dashed lines represent simulations with more accurate values. Thecurves indicate appropriate matching (low reflection) of common and differential mode signals.

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 4

Due to the dispersive nature of microstrip lines the characteristic impedances vary with frequency.Hence, small reflections occur.

D.2 Problem 4.2

The cut-off frequencies of higher order modes are given by Equation 4.64 (book page 139).

fc,mn =c0

2

√(ma

)2+(nb

)2(D.8)

Table D.1 lists wave types and cut-off frequencies.

Table D.1: Cut-off frequencies fc,mn in a R260 waveguide (a = 8.636 mm and b = 4.318 mm)

Wavemode H10 H20 H01 H11, E11 H21, E21 H20

fc,mn 17.37 GHz 34.74 GHz 34.74 GHz 38.84 GHz 49.13 GHz, 52.11 GHz

The characteristic impedance of the fundamental wave type (H10 or TE10) is given by Equa-tion 4.62 (book page 138).

ZH100 =

π2b

8a· ZF0√

1− (fc/f)2= 312 Ω (D.9)

D.3 Problem 4.3

Figure D.4 shows two coaxial lines (Z01 = 75 Ω and Z02 = 125 Ω) matched by a quarterwavetransformer at f = 10 GHz.

3l

a1 a2 a3R R R= =

re

r3e

re

i1R

75-W-line

125- -lineW

i2R

l/4-transformer

Figure D.4: Quarterwave transformer for matching two lines at f = 10 GHz

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 5

a) Relative permittivity

The relative permittivity can be derived from the reduced speed of propagation c.

c =c0√εr

= 81% c0 ⇒ εr =

(1

0.81

)2

= 1.524 (D.10)

b) Inner radii of coaxial lines

The characteristic impedance of a coaxial line is given by Equation 4.12 (book page 115).

Z0 =60 Ω√εr

ln

(Ro

Ri

)(D.11)

From the known outer radii we calculate the inner radii.

Ri1 = Ro1 exp

(− Z0

60 Ω√εr

)= 0.427mm and Ri2 = 0.1528mm (D.12)

c) Characteristic impedance of quarterwave transformer

For a quarterwave transformer (` = λ/4) the load impedance ZA and input impedance Zin arelinked by the following equation

ZinZA = Z20 (D.13)

In our problem we have ZA = Z01 and Zin = Z02. Therefore, the characteristic impedance Z03

of the quarterwave transformer is

Z03 =√Z01Z02 = 96.82 Ω (D.14)

d) Relative permittivity εr3

Inner and outer radii of the quarterwave transformer are given (Ro3 = Ro2, Ri3 = Ri2). In orderto obtain the desired characteristic impedance Z03 we select an appropriate relative permittivityεr3.

Z03 =60 Ω√εr3

ln

(Ro3

Ri3

)=

60 Ω√εr3

ln

(Ro2

Ri2

)⇒ εr3 =

(60 Ω

Z03ln

(Ro2

Ri2

))2

= 2.541 (D.15)

e) Length of quarterwave transformer

The wavelength λ in the quarterwave transformer is

λ =c

f=

c0√εr3f

(D.16)

Hence, the length of the line is

`3 =λ

4= 4.705mm (D.17)

For comparison we use a circuit simulation (ADS) (see Figure D.5). Figure D.6 shows thesimulation result. The reflection coefficient is minimum at f = 10 GHz.

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 6

Figure D.5: Schematic for s-parameter simulation with ADS

0.1

0.01

0.001

Reflection coefficient

Figure D.6: Simulation result shows matching at f = 10 GHz

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 7

D.4 Problem 4.4

The resonance frequencies fR,mnp of a rectangular cavity are given by Equation 4.69 (bookpage 142).

fR,mnp =c0

2

√(ma

)2+(nb

)2+(pc

)2(D.18)

Table D.2 lists the resonance frequencies with increasing frequencies for (a = 5 cm, b = 7 cm,c = 9 cm). Side a is the longest side. The frequency with the lowest frequency is given by m = 0and n = p = 1. The electric field has one maximum (n = p = 1) in each horizontal direction.The electric field in vertical direction is constant (no maximum; m = 0).

Table D.2: Resonance frequenciesm n p fR/GHz

0 1 1 2.7151 0 1 3.4321 1 0 3.6870 1 2 3.9631 1 1 4.046

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 8

m n p=0; = =1

2.715 GHz

m=1 n p; =0; =1

3.432 GHz

a

bc

Figure D.7: Electric field distribution of the first two modes

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 9

m=n p=1; =0

3.687 GHz

m=0 n p; =1; =2

3.963 GHz

Figure D.8: Electric field distribution of the third and fourth mode

APPENDIX D. SOLUTIONS TO PROBLEMS IN CHAPTER 4 10

m=n p= =1

4.046 GHz

Figure D.9: Electric field distribution of the fifth mode

(Last modified: 27.12.2012)