Appendix C
Solutions to Problems in Chapter 3
C.1
Problem 3.1
In Example 3.6 (book page 90) a complex load impedance ZA2 = (37.5+j150) Ω was connected to
a transmission line (characteristic impedance Z02 = 75 Ω; length of line `t /λ = 0.194). By using
the Smith diagram our result for the input impedance was Zin ≈ (15 − j75) Ω. We introduced
the Smith-chart as a graphical tool to avoid complex-valued calculations.
For comparison we use Equation (3.78) to calculate the input impedance analytically. We
repeat Equation (3.78) here for convenience.
Z0
tan (β`t )
ZA
Zin = A
ZA
tan (β`t )
1+j
Z0
1+j
(C.1)
The complex calculations may be performed manually. However, in engineering practice mathematical tools (e.g. MATLAB from MathWorks, Inc.) are available. Using MATLAB we get
Zin = (14.6 − j75.2) Ω. The numerical and graphical results are in good agreement.
C.2
Problem 3.2
In our problem the resistances are given as RI = 15 Ω and RA = 10 Ω. Hence, the reflection
coefficients become
rin =
RI − Z0
= −0.538
RI + Z0
and
rA =
R A − Z0
= −0.667
R A + Z0
(C.2)
Figure 3.24 (book page 102) lists the relevant equations for our problem. The magnitudes of the
forward-propagating and reflected pulses are
Z0
= 769 mV
RI + Z0
= rA U1f = −513 mV
U1f = U0
(C.3)
U1r
(C.4)
U2f = rI rA U1f = 276 mV
(C.5)
2
U2r = rI rA
U1f = −184 mV
(C.6)
U3f =
2
rI2 rA
U1f
1
= 99.1 mV
(C.7)
APPENDIX C. SOLUTIONS TO PROBLEMS IN CHAPTER 3
2
At the transmission line terminals forward-propagating and reflected pulses add up, so we observe
the following time signal at the input terminals
uin (t) = U1f = 769 mV for 0 ≤ t ≤ tp
(C.8)
uin (t) = (1 + rI ) U1r = −237 mV for 2tD ≤ t ≤ 2tD + tp
(C.9)
uin (t) = (1 + rI ) U2r = −85 mV for 4tD ≤ t ≤ 4tD + tp
(C.10)
At the output (load) terminal we see
uA (t) = (1 + rA ) U1f = 256 mV for tD ≤ t ≤ tD + tp
(C.11)
uA (t) = (1 + rA ) U2f = 92 mV for 3tD ≤ t ≤ 3tD + tp
(C.12)
Figure C.1 shows a graphical representation of the time signals. (The numerical simulation was
done with ADS circuit simulation software from Agilent, Inc.)
Figure C.1: Time plots of the voltages at the transmission line terminals (ADS simulation result)
C.3
Problem 3.3
A load impedance ZA = (120 − j80) Ω is matched to a 50 Ω-source impedance by a matching
network. The network starts (seen from the load) with a serial line and includes an open-ended
stub line, both having line impedances of Z0 = 50 Ω. The frequency of operation is f = 1 GHz.
APPENDIX C. SOLUTIONS TO PROBLEMS IN CHAPTER 3
3
The problem may be solved with freeware computerized Smith chart tools available on the
Internet. Figure C.2 shows the result using a commercial tool (ADS from Agilent, Inc.). The
serial line rotates the (normalized) load impedance zA = ZA /Z0 around the center of the diagram
(clockwise rotation) until is reaches the circle of unity normalized conductance. Following this
circle we approach the matching point (center of the diagram).
Figure C.2: Design of a matching network with ADS Smith-chart tool
The software displays the appropriate electrical line lengths in degree. We can calculate
physical line lengths by using frequency and speed of propagation. For an air-filled line (c = c0 )
our result would be
`serial
50.461◦
=
λ
360◦
`stub
35.857◦
=
λ
360◦
50.461◦
50.461◦ c0
λ
=
·
= 4.2 cm
360◦
360◦
f
35.857◦
35.857◦ c0
=
λ
=
·
= 3.0 cm
360◦
360◦
f
→
`serial =
(C.13)
→
`stub
(C.14)
In order to evaluate our results we perform a circuit simulation with ADS. Figure C.3 shows
the circuit for a s-parameter simulation in the frequency range from 0.1 to 2 GHz. The transmission lines are represented by physical line models (TLINP=Transmission LINe Physical). The
transmission line parameters are
• characteristic impedance Z = Z0 = 50 Ω,
• geometric length L and
• relative permittivity K = εr = 1.
APPENDIX C. SOLUTIONS TO PROBLEMS IN CHAPTER 3
4
Figure C.3: Matching circuit of Problem 3.3
The negative reactance of XA = −j80 Ω at f = 1 GHz is represented by a capacitance.
jXA = −j80 Ω =
1
jωC
→
C=
1
ω80 Ω
(C.15)
The resulting s-parameter s11 is shown in Figure C.4. At f = 1 GHz the circuit is matched to
the source impedance of Z0 = 50 Ω.
Figure C.4: Reflection coefficient s11 showing matching at f = 1 GHz
APPENDIX C. SOLUTIONS TO PROBLEMS IN CHAPTER 3
C.4
5
Problem 3.4
a) Speed of propagation
According to Equation (2.102) the speed of light in dielectric material is (µr = 1)
c= √
c0
m
= 2.5 · 108
= 83.3% c0
εr µ r
s
(C.16)
b) Characteristic line impedance
With
C
`
C0 =
and Equation (3.68) we get
and µr = 1
(C.17)
√
Z0 =
εr µ r
= 32 Ω
c0 C 0
(C.18)
c) Inductance per unit length
In order to calculate the inductance per unit length we start with Equation (3.65)
r
L0
Z0 =
C0
(C.19)
By rearranging Equation (C.19) we get
L0 = Z02 C 0 = 128
nH
m
(C.20)
d) Propagation constant
The propagation constant γ is
√
1
γ = α + jβ = jω L0 C 0 = j6.28
m
Alternatively, we can directly calculate the phase constant β
β=
2π
λ
with λ =
c
c0
=√
= 1m
f
εr f
(C.21)
(C.22)
Consequently, the propagation constant is
γ = jβ = j
2π
1
= j6.28
1m
m
(C.23)
e) Input impedance
In our example the wavelength λ is
λ=
c
c0
=√
= 1m
f
εr f
(C.24)
The geometric length is ` = 0.25 m = λ/4. As shown in Section 3.1.8.3 an an open-ended
quarter-wave line has a zero input impedance (short-circuit).
Zin = 0 Ω
(C.25)
APPENDIX C. SOLUTIONS TO PROBLEMS IN CHAPTER 3
C.5
6
Problem 3.5
a) Line length
The voltage uA (t) shows that the delay time of the transmission line is tD = 200 ns. By using
the basic cinematic equation v = s/t we calculate the line length ` as
` = c0 t0 = 60 m
(C.26)
where c0 is the speed of light in vacuum.
b) Calculation of circuit elements RAS , RAP and C
At t = 200 ns the voltage pulse reaches the end of the line. The voltage across the capacitance is
steady, hence, the capacitance behalves like a short-circuit. At that moment the load impedance
ZA0 is given by a parallel circuit of RAS and RAP .
ZA0 = RAP k RAS
(C.27)
The corresponding reflection coefficient is given by
rA0 =
ZA0 − Z0
ZA0 + Z0
(C.28)
According to Figure 3.22 the voltage at the load terminal is
u(t0 ) = U1f (1 + rA0 )
(C.29)
From Figure 3.27 we read a value of 0.8 V at t0 .
u(t0 ) = U1f (1 + rA0 ) = 0.8 V
(C.30)
The voltage divider rule at the input terminals (RI = Z0 ) yields a value of 1 V for the forward
propagating voltage (Uf1 = 1 V). So, reflection coefficient and load impedance become
rA0 = −0.2
⇒
ZA0 = Z0
1 + rA0
= 16.67 Ω = RAP k RAS
1 − rA0
(C.31)
Next, we look at the voltage uA (t) for t → ∞. The capacitance is fully charged, there is no
(DC) current through the capacitance. The capacitance behaves like an open circuit. Hence, the
load impedance now becomes
ZA∞ = RAP
(C.32)
For t → ∞ we see steady (DC) conditions. The loss-less transmission line represents a simple
through-connection. From the voltage divider rule we get
ZA∞
uA (t → ∞)
=
2V
RI + ZA∞
(C.33)
Now, we can calculate the first circuit element as
ZA∞ = RAP = 80 Ω
(C.34)
APPENDIX C. SOLUTIONS TO PROBLEMS IN CHAPTER 3
7
Next, we can derive the second unknown resistance
1
1
1
+
=
RAP RAS
RAP k RAS
→
RAS = 21 Ω
(C.35)
In order to determine the capacitance we take a look at the time plot of uA (t). For t ≥ t0 the
output voltage is given by the following relation.


t − t0
−
τ  for t ≥ t0
uA (t) = 0.8 V + (1.524 V − 0.8 V) 1 − e
(C.36)
From uA (tx ) = 1.182 V and Equation (C.36) we determine the time constant τ .
τ = 100 ns
(C.37)
The exponential time constant τ depends on the capacitance and the effective resistance R.
τ = RC
(C.38)
From basic circuit theory we derive the resistance R as
R = RAS + RAP k Z0 = 40 Ω
Finally, the capacitance C is
C=
(Last modified: 13.08.2012)
τ
= 2.5 nF
R
(C.39)
(C.40)