solutions to problems in chapter 5 · appendix e. solutions to problems in chapter 5 6...

Appendix E

Solutions to Problems in Chapter 5

E.1 Problem 5.1

Properties of the two-port network

• The network is mismatched at both ports (s11 6= 0 and s22 6= 0).

• The network is reciprocal (s12 = s21).

• The network is symmetrical (s12 = s21 and s11 = s22)

• The network is lossless, since its matrix is a unitary matrix. (STS∗ = I).

The unitary matrix condition yields

STS∗ =

(513 j 12

13

j 1213

513

)(513 −j 12

13

−j 1213

513

)(E.1)

=

((513

)2+(

1213

)20

0(

513

)2+(

1213

)2)

=

(1 00 1

)= I (E.2)

Reflection loss and insertion loss

The reflection losses at port 1 and port 2 are

RL1 = 20 lg

∣∣∣∣ 1

s11

∣∣∣∣ = 8.3 dB and RL2 = 20 lg

∣∣∣∣ 1

s22

∣∣∣∣ = 8.3 dB (E.3)

The insertion loss is given by

IL12 = 20 lg

∣∣∣∣ 1

s12

∣∣∣∣ = 0.695 dB and IL21 = 20 lg

∣∣∣∣ 1

s21

∣∣∣∣ = 0.695 dB (E.4)

Renormalization of scattering matrix from 50 Ω to 100 Ω

The set of equations required for the renormalization is given in Section 5.5.3 (S = S50 Ω andSnew = S100 Ω).

1

APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 2

Snew =1

det S

((s11 − r) (1− rs22) + rs12s21 s12

(1− r2

)s21

(1− r2

)(s22 − r) (1− rs11) + rs12s21

)(E.5)

where

det S = (1− rs11) (1− rs22)− r2s12s21 (E.6)

and

r =Z0,new − Z0

Z0,new + Z0(E.7)

Using the given values we get

r =1

3and det S =

100

117(E.8)

Hence, the renormalized matrix is

Snew =1

25

(−7 j24j24 −7

)=

(−0.28 j0.96j0.96 −0.28

)(E.9)

Validation by circuit simulation (ADS)

We will validate our result by using a circuit simulator (ADS from Agilent, Inc.). First, we writeour initial scattering parameter in a snp-file (see Section 5.7).

snp-files (n = number of ports) contain frequency dependent scattering parameters fromsimulation or measurement in ASCII format. Here, we use the file to define scattering parameters.

Figure E.1: s2p-file for s-parameter definition

Figure E.1 shows the s2p-file. Lines with an exclamation mark (!) are comment line. Theline with the sharp symbol (#) defines the format of the subsequent data lines: frequency inGHz (GHz), s-parameters (S) are given with magnitude and phase (MA). The port referenceimpedance (R) is 50 Ω (50).

In our problem the s-parameters do not vary with frequency. In order to calculate and plotthe parameters with ADS an arbitrary frequency range from 1 GHz to 2 GHz is chosen. Thes-parameter data is give in decimal nomination.

We use the previously defined s2p-file in a circuit simulation. Figure E.2 shows the schematic.The two-port network is connected to 50 Ω as well as 100 Ω terminals.

Figure E.3 shows the simulation results. The upper two plots show the s-parameters normal-ized to 50 Ω. The absolute values (magnitude) correspond to the given values of 5/13 ≈ 0.385


Figure E.2: Schematic for renormalization of scattering parameters

and 12/13 ≈ 0.923. The phases are zero degree (positive real value) for the reflection coefficientsand 90 degree (positive imaginary value) for the transmission coefficients.

The lower two plots show the s-parameters for a reference value of 100 Ω. The absolute valuesare 0.28 and 0.96. The faces are 180 degree (negative real values) for the reflection coefficientsand 90 degree (positive imaginary values) for the transmission coefficients. The results agree withour calculated values in Equation E.9.


Figure E.3: Simulation results: s-parameters for Z0 = 50 Ω (upper plots) and s-parameters forZ0,new = 100 Ω (lower plots)


E.2 Problem 5.2

Antenna input impedance

The reflection coefficient rA of an antenna is determined by its input impedance ZA and the portreference impedance Z0.

rA =ZA − Z0

ZA + Z0(E.10)

Solving Equation E.10 for ZA yields

ZA = Z01 + rA1− rA

(E.11)

We rewrite the given reflection coefficient rA

rA = 0.4e−j20 = 0.4 (cos(20)− j sin(20)) = Re rA+ j Im rA = 0.367− j0.137 (E.12)

Substituting our result in Equation E.11 yields

ZA = Z01 + rA1− rA

= Z01 + Re rA+ j Im rA1− Re rA − j Im rA

(E.13)

= Z0(1 + Re rA+ j Im rA) (1− Re rA+ j Im rA)

(1− Re rA)2 + (Im rA)2 (E.14)

= Z01− (Re rA)2 − (Im rA)2 + j2 Im rA

(1− Re rA)2 + (Im rA)2 = (102.9− j33.51) Ω (E.15)

Reflection coefficient for port reference impedance of Z0,new = 75 Ω

The reflection coefficient with respect to a port reference impedance of Z0,new = 75 Ω is

rA,new =ZA − Z0,new

ZA + Z0,new=

Re ZA+ j Im ZA − Z0,new

Re ZA+ j Im ZA+ Z0,new(E.16)

=(Re ZA − Z0,new + j Im ZA) (Re ZA+ Z0,new − j Im ZA)

(Re ZA+ Z0,new)2 + (Im ZA)2 (E.17)

= 0.1857− j0.1534 (E.18)

Hence, magnitude and phase of the reflection coefficient are

|rA,new| =√

(Re rA,new)2 + (Im rA,new)2 = 0.241 (E.19)

∠rA,new = arctan

(Im rA,newRe rA,new

)= −39.56 (E.20)

Reflected and accepted power

The reflected power is

Pb = |s11|2 Pa = 0.0581W (E.21)

The accepted power is

Pacc =(

1− |s11|2)Pa = 0.9419W (E.22)


Validation by circuit simulation

We apply a circuit simulator (ADS from Agilent, Inc.) in order to validate our results. First,we define a s1p-file for the reflection coefficient with a reference impedance of Z0 = 50 Ω (seeFigure E.4). Once again, we chose an arbitrary frequency range from 1 to 2 GHz for the s-parameter representation.

Figure E.4: s1p-file with reflection coefficient (port reference impedance Z0 = 50 Ω)

Figure E.5: ADS schematic

The schematic of the circuit is shown in Figure E.5. The s-parameter and impedance resultsare displayed in Figure E.6. There is good agreement between circuit simulation and calculation.


Figure E.6: (a) Magnitude and (b) phase of reflection coefficients for port reference impedancesof 50 Ω (red) and 75 Ω (blue). (c) Real and imaginary part of reflection coefficients. (d) Real andimaginary part of input impedance.


E.3 Problem 5.3

Scattering matrix of a two port network with series impedance

Figure E.7 shows a circuit with a two-port network consisting of a series impedance Z. Exam-ple 5.3 (see book page 171) gives us the procedure to calculate the scattering matrix. First, weneed the input impedance.

Zin1 = Z + Z0 (E.23)

The reflection coefficient then becomes

s11 =Zin1 − Z0

Zin1 + Z0=

Z0

2Z0 + Z= s22 (E.24)

Since the network is symmetrical we get s22 = s11.

0Z

in1Z

0Z

0Z

01U 0

Z2

U

Z

Two-port network

Figure E.7: Two-port network with a series impedance Z

The transmission coefficient s21 is given as

s21 =2U2

U01

√Z0

Z0= 2

Z0

Z0 + Z + Z0=

2Z0

2Z0 + Z= s12 (E.25)

The voltage ratio U2/U01 is determined by the voltage divider rule. Due to reciprocity we gets12 = s21. Hence, the scattering matrix S is

S =1

2Z0 + Z

(Z 2Z0

2Z0 Z

)(E.26)

Special cases

We evaluate our result by looking at two special cases. First, we consider a short circuit (Z = 0),i.e. the two-port network becomes a direct through-connection.

The scattering matrix then is

S =

(0 11 0

)(E.27)


As expected there is no reflection (sii = 0) and full transmission (|sij | = 1). Next, we consideran open connection (Z →∞). The scattering matrix now reads

S =

(1 00 1

)(E.28)

As expected there is full reflection (|sii| = 1) and no transmission (sij = 0).

Scattering matrix of T-network

0Z

in1Z

0Z

0Z

01U 0

Z2

U

3Z

T-network

1Z 2

Z

xU

Figure E.8: Two-port circuit with a T-network

Figure E.8 shows the circuit. First, we need the input impedance.

Zin1 = Z1 + Z3 ‖ (Z2 + Z0) = Z1 +Z3 (Z2 + Z0)

Z3 + Z2 + Z0(E.29)

After a short calculation the input reflection coefficient is

s11 =Zin1 − Z0

Zin1 + Z0(E.30)

=Z1Z2 + Z1Z3 + Z2Z3 + Z1Z0 − Z2Z0 − Z2

0

Z1Z2 + Z1Z3 + Z2Z3 + Z1Z0 + Z2Z0 + 2Z3Z0 + Z20

(E.31)

By swapping the indices 1 and 2 we get the output reflection coefficient.

s22 =Z1Z2 + Z1Z3 + Z2Z3 − Z1Z0 + Z2Z0 − Z2

0

Z1Z2 + Z1Z3 + Z2Z3 + Z1Z0 + Z2Z0 + 2Z3Z0 + Z20

(E.32)

The calculation of the transmission factors involves the voltage ratio U2/U01. With an auxiliaryquantity Ux we first determine the following voltage ratios.

U2

Ux=

Z0

Z2 + Z0and

Ux

U01=

Z3 ‖ (Z2 + Z0)

Z0 + Z1 + Z3 ‖ (Z2 + Z0)(E.33)

Finally, we get

s21 =2U2

U01

√Z0

Z0= 2

U2

Ux· Ux

U01= 2

Z0

Z2 + Z0· Z3 ‖ (Z2 + Z0)

Z0 + Z1 + Z3 ‖ (Z2 + Z0)(E.34)

=2Z3Z0

Z1Z2 + Z1Z3 + Z2Z3 + Z1Z0 + Z2Z0 + 2Z3Z0 + Z20

= s12 (E.35)

Due to reciprocity we find s12 = s21.


Special case Z3 = 0

In order to validate our formula we consider the case Z3 = 0. In this case we expect zerotransmission sij = 0 and indeed Equation E.35 is zero.

Equation E.31 becomes

s11 =Zin1 − Z0

Zin1 + Z0=Z1Z2 + Z1Z0 − Z2Z0 − Z2

0

Z1Z2 + Z1Z0 + Z2Z0 + Z20

(E.36)

=Z2 (Z1 − Z0) + Z0 (Z1 − Z0)

Z2 (Z1 + Z0) + Z0 (Z1 + Z0)=

(Z2 + Z0) (Z1 − Z0)

(Z2 + Z0) (Z1 + Z0)=Z1 − Z0

Z1 + Z0(E.37)

Since the input impedance is Zin1 = Z1 our result is correct.

Scattering matrix of a π-network

0Z

in1Z

0Z

0Z

01U 0

Z2

U3

Z

p-network

1Z

2Zx

U

Figure E.9: Two-port circuit with a π-network

Figure E.9 shows the π-network. First, we calculate the input impedance.

Zin1 = Z1 ‖ (Z3 + Z2 ‖ Z0) (E.38)

The input reflection coefficient becomes

s11 =Zin1 − Z0

Zin1 + Z0(E.39)

=Z1Z2Z3 + Z1Z3Z0 − Z2Z3Z0 − Z1Z

20 − Z2Z

20 − Z3Z

20

Z1Z2Z3 + 2Z1Z2Z0 + Z1Z3Z0 + Z2Z3Z0 + Z1Z20 + Z2Z2

0 + Z3Z20

(E.40)

In order to calculate the output reflection coefficient we swap indices 1 and 2.

s22 =Z1Z2Z3 − Z1Z3Z0 + Z2Z3Z0 − Z1Z

20 − Z2Z

20 − Z3Z

20


0 + Z3Z20

(E.41)

The calculation of the transmission factors involves the voltage ratio U2/U01. With an auxiliaryquantity Ux we first determine the following voltage ratios.

U2

Ux=

Z2 ‖ Z0

Z3 + Z2 ‖ Z0and

Ux

U01=

Z1 ‖ (Z3 + Z2 ‖ Z0)

Z0 + Z1 ‖ (Z3 + Z2 ‖ Z0)(E.42)


Finally, we get

s21 =2U2

U01

√Z0

Z0= 2

U2

Ux· Ux

U01(E.43)

= 2Z2 ‖ Z0

Z3 + Z2 ‖ Z0· Z1 ‖ (Z3 + Z2 ‖ Z0)

Z0 + Z1 ‖ (Z3 + Z2 ‖ Z0)(E.44)

=2Z1Z2Z0


0 + Z3Z20

= s12 (E.45)

Due to reciprocity we find s12 = s21.

Special case Z1 = 0

Let us consider the case Z1 = 0 (short circuit). Now, the input reflection coefficient in Equa-tion E.40 becomes

s11 =−Z2Z3Z0 − Z2Z

20 − Z3Z

20

Z2Z3Z0 + Z2Z20 + Z3Z2

0

= −1 (E.46)

This represents the short circuit (r = −1) at port 1. The output reflection coefficient in Equa-tion E.41 is now

s22 =Z2Z3 − Z2Z0 − Z3Z0

Z2Z3 + Z2Z0 + Z3Z0=Z2Z3 − (Z2 + Z3)Z0

Z2Z3 + (Z2 + Z3)Z0(E.47)

=

Z2Z3

Z2 + Z3− Z0

Z2Z3

Z2 + Z3+ Z0

=Z2 ‖ Z3 − Z0

Z2 ‖ Z3 + Z0(E.48)

This result is correct since the output impedance is Z2 ‖ Z3. Finally, Equation E.45 gives us zerotransmission which is correct.

E.4 Problem 5.4

Figure E.10 shows us the signal flow graph of the circuit in Figure 5.25 (book page 185).

,11xs,22xs

,21xs

,12xs2a

2b1a

1b

1je

b- l

1je

b- l

2je

b- l

2je

b- l

Figure E.10: Signal flow graph


The signal flow graph gives us the following scattering parameters.

s11 =b1a1

∣∣∣∣a2=0

= sx,11e−jβ(`1+`1) = sx,11e−j2β`1 (E.49)

s22 =b2a2

∣∣∣∣a1=0

= sx,22e−jβ(`2+`2) = sx,22e−j2β`2 (E.50)

s21 =b2a1

∣∣∣∣a2=0

= sx,21e−jβ(`1+`2) (E.51)

s12 =b1a2

∣∣∣∣a1=0

= sx,12e−jβ(`2+`1) (E.52)

The transmission lines only modify the phase of the scattering parameters. If the lengths of thelines are known the effect can be compensated.

E.5 Problem 5.5

We will derive the equation for the wave source in Figure 5.14 (book page 177). We repeat therepresentation in Figure E.11 for convenience.

1r

1b

1a

EquationsSignal flow graph

1 01 1 1b b r a= +

01 0

1

01 0

Z Zr

Z Z

-=

+

Voltage source

0Z

01Z

01U

1b

1a

01b 1

01 0

01

0 01

U Zb

Z Z=

+

Figure E.11: Signal flow graph of a voltage source

Reflection coefficient r1

The reflection coefficient is determined by the input impedance of the circuit. The ideal volt-age source represents a short circuit. Therefore, the input impedance is Z01 and the reflectioncoefficient is

r1 =Z01 − Z0

Z01 + Z0(E.53)

Wave source

The circuit in Figure E.11 includes an ideal voltage source and is therefore an active circuit.Consequently, there is an outgoing wave b1 even though there is no incoming wave a1. This isexpressed by the term b01 in the following equation.

b1 = b01 + r1a1 (E.54)


0Z

01Z

01U

1b

1a 0

Z

1I

1U

Figure E.12: Voltage, current and power waves

In order to calculate the term b01, we consider the relation between voltage U and current I andpower waves a and b (see Figure E.12).

The relation between power waves a and b and voltage U and current I is given as

b1 =U1 − Z0I1

2√Z0

and a1 =U1 + Z0I1

2√Z0

(E.55)

The voltage is determined by the voltage divider rule.

U1 =Z0

Z0 + Z01U01 (E.56)

The current is given by Ohm’s law.

I1 = − U01

Z0 + Z01(E.57)

Therefore, we get

a1 =U1 + Z0I1

2√Z0

=Z0U01 + (−Z0U01)

2√Z0 (Z0 + Z01)

= 0 (E.58)

and

b1 =U1 − Z0I1

2√Z0

=Z0U01 − (−Z0U01)

2√Z0 (Z0 + Z01)

=

√Z0U01

Z0 + Z01= b01 (E.59)

The term b01 represents the active part in the signal flow graph.

E.6 Problem 5.6

Figure E.13 shows a network with two two-port networks: the first network has a forward gainGf and the feedback network has a gain of Gr.

From Figure E.13 we derive the following two relations.

X ′ = X +GrY and Y = GfX′ (E.60)

This gives us

Y =Gf

1−GclX (E.61)

where Gcl = GfGr is the closed loop gain.If we apply the result to the signal flow graph in Figure 5.12 (book page 175), we get

b =sx

1− sxsya (E.62)


X

fG

rG

Y'X

Figure E.13: Network with feedback

E.7 Problem 5.7

We will calculate reflection coefficients and impedances of the circuit in Figure E.14. We assumethe following parameters.

f = 1GHz (E.63)Z0 = 50 Ω (Port reference impedance) (E.64)ZA = 100 Ω + jωL = 100 Ω + j100 Ω where L = 15.92 nH (E.65)

Z =1

jωCwhere C = 5 pF (E.66)

The input impedance Zin1 is

Zin1 = Z0 ‖1

jωC=Z0(1− jωCZ0)

1 + (ωCZ0)2= (14.42− j22.65) Ω (E.67)

The reflection coefficient is

r1 =Zin1 − Z0

Zin1 + Z0=

Re Zin12 + Im Zin12 − Z20 + j2 Im Zin1Z0

(Re Zin1+ Z0)2 + Im Zin12(E.68)

= 0.6176 e−128.15 (E.69)

2r

Z 0 ,Z AZ0Z

01U

1r

in1Z in2Z

b

0.363l=l

Figure E.14: Network

In order to determine the reflection coefficient r2 we first consider the reflection coefficient r′2at the end of the line.

r′2 =ZA − Z0

ZA + Z0(E.70)

=Re ZA2 + Im ZA2 − Z2

0 + j2 Im ZAZ0

(Re ZA+ Z0)2 + Im ZA2= 0.62 ej29.75 (E.71)


At the input of the transmission line we get

r2 = r′2 e−j2β` = 0.62 ej29.75 ej98.64 = 0.62 ej128.4 = r∗1 (E.72)

If we compare the reflection coefficients r1 and r2 we see that r2 = r∗1 (complex conjugatematching).

Using a Smith-Chart gives us the input impedance Zin2 (see Figure E.15). We apply acommercial circuit simulator (ADS by Agilent, inc.). The electrical line length is 0.363λ =130.68. The load impedance is represented by a black diamond the input impedance is depictedby a red square. We read from the diagram

Zin2 = (14.28 + j22.56) Ω = Z∗in1 (E.73)

Impedances Zin1 and Zin2 – as well as reflection coefficients r1 and r2 – show complex conjugatevalues.

Figure E.15: Smith chart and input impedance Zin2

Circuit simulation

We will validate our result by performing a circuit simulation. In our previous calculation weconsidered a frequeny of f = 1GHz. Now we extend the frequency range from 100MHz to 2GHz.Figure E.16 shows the schematic. The results are given in Figure E.17.

The upper diagrams in Figure E.17 show reflection coefficients (magnitude and phase) aswell as input impedances (real and imaginary part). In order to compare the results with oursimulation we have to look at a frequency of f = 1GHz (indicated by markers). The simulationresults agree with our previously calculated values.


Figure E.16: Schematic

Figure E.17: Simulation results: s-parameters and impedances


E.8 Problem 5.8

A three-port network shall have the following three properties:

• matching at all ports (sii = 0),

• reciprocity (sij = sji) and

• no losses (STS∗ = I).

From these requirements the matrix equation 0 s12 s13

s12 0 s23

s13 s23 0

0 s∗12 s∗13

s∗12 0 s∗23

s∗13 s∗23 0

=

1 0 00 1 00 0 1

(E.74)

gives us the following set of equations

|s12|2 + |s13|2 = 1 (E.75)s13s

∗23 = 0 (E.76)

s12s∗23 = 0 (E.77)

s23s∗13 = 0 (E.78)

|s12|2 + |s23|2 = 1 (E.79)s12s

∗13 = 0 (E.80)

s23s∗12 = 0 (E.81)

s13s∗12 = 0 (E.82)

|s13|2 + |s23|2 = 1 (E.83)

From Equations E.76 and E.77 we get (for s23 6= 0):

s12 = s13 (E.84)

From Equation E.80 we then conclude:

s12 = s13 = 0 (E.85)

This result contradicts Equation E.75 since

|s12|2 + |s13|2 = 0 6= 1 (Conflict for s23 6= 0) (E.86)

The conflict can be solved for s23 = 0. This implies (Equation E.79)

|s12| = 1 (E.87)

and (Equation E.83)

|s13| = 1 (E.88)

This result contradicts Equation E.75 since

|s12|2 + |s13|2 = 2 6= 1 (Conflict for s23 = 0) (E.89)

The set of equations shows that a three-port network cannot satisfy the following conditionssimultaneously


• matching at all ports (sii = 0),

• reciprocity (sij = sji) and

• losslessness.

(Last modified: 02.01.2013)

solutions to problems in chapter 5 · appendix e. solutions to problems in chapter 5 6...

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