Solution Manual for Microwave Circuit Design A Practical Approach Using ADS, 1st Edition

Name: Solution Manual for Microwave Circuit Design A Practical Approach Using ADS, 1st Edition
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Note:
When Figures in the text are cited, it is denoted by Figure and figures in this End-of
chapter solutions are denoted by Figure.
CHAPTER 1 PROBLEMS
1.1 Waveguide generally has lower line loss than microstrip. SIW(Substrate Integrated Waveguide)
can be regarded as the planar version of waveguide. How SIW is configured using substrate?
Solution
A post-wall waveguide (also known as substrate integrated waveguide (SIW) or a laminated
waveguide) is a synthetic rectangular electromagnetic waveguide formed in a dielectric substrate by
densely arraying metallized posts or via-holes which connect the upper and lower metal plates of
the substrate. The waveguide can be easily fabricated with low-cost and mass-production using
through-hole techniques where the post walls consists of via fences. The post-wall waveguide is
known to have similar guided wave and mode characteristics to the conventional rectangular
waveguide with equivalent guided wavelength. The structure is shown below. Since SIW provides
low loss, filters using SIW have studied by many researchers.
Figure 1.1 The structure of SIW
1.2 Find the microwave integrated circuit example built using LTCC in the web-site.
Solution
In the web site http://www. barryltcc.com/, there is a TR (Transmit and Receive) module for radar.
1.3 How the ALC in Figure 1.7 is constructed?
Solution
Figure 1.2 shows a block diagram for ALC. Usually PA module has the terminal for output power
control. If not, a driver amplifier should be designed to control the output power level. A small
portion of output power is sampled through a directional coupler as shown in Fig. 1.2 and the RF
power level is detected through a diode detector. The reference signal is a variable DC voltage

source. Sometimes when the variable voltage source is difficult, it is generated using a lowpass
filter PWM (Pulse width modulation) signal generated from the controller. The average DC voltage
corresponds to the desired power level. Since it is a negative feedback loop, the detector voltage is
equal to the reference voltage at the steady state. Thus the output power level can be controlled by
changing the reference voltage.
Figure 1.2 ALC block diagram
1.4 Refer to SA605 which is used to demodulate the FM signal. Explain its block diagram.
Solution
FM demodulation in SA605 is based on quadrature FM demodulator shown in Fig. 1.3. In coming
FM signal is split in two paths; one is directly applied to the mixer and the other is coupled to the
resonant circuit centered at the FM carrier frequency by a high reactance capacitor. As a result, the
signal coupled to the high reactance capacitor suffers a 90 phase shift. The FM modulated signal
suffers additional phase shift proportional to the resonator slope and AM modulated waveform
proportional to FM modulation appears at V2. As a result of mixing demodulated FM waveform
appears.
Figure 1.3 Quadrature FM demodulator
1.5 Refer the web-site of PLL IC. Explain the synthesizer data bus shown in Fig. 1.7.
Solution
Refer to Analog Devices, RF PLL Frequency Synthesizers ADF4110/4111/4112/4113, August,
2012.

1.6 How the PLL can be modulated? Explain how to set the PLL loop bandwidth.
Solution
For FM signal generation at the VCO output, the modulation signal may be directly injected to
VCO tuning input. The fast varying signal than PLL loop bandwidth can modulate VCO
successfully, however the slow varying one will be compensated by the PLL and does not present
the modulated output.
Figure 1.4 FM modulation using PLL
CHAPTER 2 PROBLEMS
2.1 A company A fabricates a thin film resistor whose sheet resistivity is 50 ohm/square and in the
case of a company B, an identical thin film resistor has a sheet resistivity of 100 ohm/square. What
is the difference between the processes of the two companies? Furthermore, given that the
material’s volume resistivity is ρ and its thickness is t, find its sheet resistivity.
Solution
The thickness of the resistor material of the company A is two times thicker than that of the
company B.
The resistance R with a volume resistivity of ρ and a thickness of t is
S
L L
R R
tW W
ρ= =
Thus
SR t
ρ
=

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2.2 Given that the sheet resistivity is 50 ohm/(square) and we want to design a 100 ohm resistor. If
the current flowing in this resistor is 2 mA, find its minimum width. Its rated current per width is
0.5 mA/μm.
Solution
The minimum width Wmin theta can sustain 2mA is
min
2(mA) 4(μm)
0.5(mA/μm)
W = =
Taking Wmin into consideration, the width is determined as 10 mm. Thus the length should be 20 mm
to yield 100 ohm resistor.
2.3 The dielectric material of a MIM capacitor is Silicon Nitride and its dielectric constant is 7.2. If
the area is 50 μm2 find the thickness of a 0.53 pF capacitor. In addition, find the capacitance per
unit area in (F/ μm2
).
Solution
0.53( pFC = 8.854( pF
) o r A
t
ε ε
= =
2
/m) 7.2 50(μm× × )
( μmt
6
3
)
8.854 10 7.2 50 6.0 10 (μm) 6.0(nm)
0.53
t
−
−× × ×
= = × =
The sheet capacitance CS is
2
2
0.53(pF) 10.6 (fF / μm )
50(μm )
SC = =
2.4 In the equivalent circuit of an inductor such as that in Fig. 2.10, show that the approximate
magnitude of the impedance at resonance is
( )2
max
o L
Z R
ω
=
1o LC
ω =
Solution
The impedance of Z(ω) of the inductor equivalent circuit
( )
( ) 1
1
R j L j C
Z
R j L C
ω ω
ω ω ω
+
=  
+ − 
 
Near the resonance Z(ω) can be approximated as
( )
( ) ( ) ( ) ( )2
1 1 1
1
o o oo o
R j L R j L j L Lj C j C j C
Z R R R
R j L C
ω
ω
ω
ω
ω
ω
ω
ω ω ω
+ +
= ≅ ≅ =
 
+ − 
 

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2.5 In this text, we have covered the method of extracting the equivalent circuit of a capacitor from
|Z(
ω)| characteristics obtained from impedance analyzer such as that shown in Fig. 2.8. The
impedance of the equivalent circuit of a capacitor is
( ) 1
Z R jX R j L C
ω
ω ω
 
= + = + − 
 
Using this, the real and imaginary parts for frequency can be obtained by graphical representation.
Then show that the approximate impedance near the resonant frequency
ωo is
( ) ( )2 oZ R j L
ω
ω ω≅ + −
Solution
( )
( )( )
( )( ) ( )
2
2 2
2 2
2
1 1
1 1
2
o
o oo
o o
o
Z R jX R j L R j L R j L
C LC
R j L R jL
R jL R j L
ω
ω
ω
ω
ω
ω
ω
ω
ω ω ω ω
ω ω
ω ω
ω
ω ω ω ω
ω ω
ω
    
= + = + − = + − = + −    
     
− + −
= + = + 
 
− +
= + ≅ + −
2.6 Similar to Problem 2.5 above, for a parallel resonant circuit, show that the admittance near the
resonant frequency, Y(
ω) is given by
( ) ( )2 oY G j C
ω
ω ω≅ + −
Solution
The admittance of a parallel resonant circuit can be expressed as
( ) 1
Y G jB R j C L
ω
ω ω
 
= + = + − 
 
This is similar to Z(ω) in problem 2.5. Thus,
( ) ( )2 oY G j C
ω
ω ω≅ + −
2.7 (ADS Problem) In Example 2.2, the equivalent circuit of the 10 pF capacitor from Murata are
obtained using the slope of the reactance versus frequency and resonance frequency such as
4
of
X L
f
π
∂ =
∂ and 1o LC
ω = .
In another way, the equivalent circuit can be obtained using the impedance plot in log-log scale.
Compute the equivalent circuit values using ADS.
Solution
Fig. 2.1 shows the simulation schematic for impedance plot.

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Figure 2.1 Simulation schematic setup for impedance plot. Here log sweep is set to 30 points/decade
The simulated impedance is shown in Fig. 2.2. Using the impedance values at markers, the series
equivalent circuit values can be obtained. From the minimum value, the value of R can be obtained
and R=0.363 Ω. The following equations for C and L are written in the Display window.
Eqn C=1/(2*pi*m2*indep(m2))
Eqn L=m3/(2*pi*indep(m3))
The computed C = 9.9 pF and L=0.96 nH. The values of R, C, and L are close to those computed in
Example 2.2 but shows little differences. The error of R is mainly due to the sparse sweep points.
The differences of the other values are believed to come from different approximations in the
computation of element values.
Figure 2.2 Simulated impedance for the Murata 10 pF capacitor.

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CHAPTER 3 PROBLEMS
3.1 In Example 3.1, a traveling sine wave propagating along z direction is given by cos(
ωt–
βz). For
a constant C, if
ωt–
βz =C is chosen, what is the meaning? Also, when this constant is taken as the
peak of the sine wave, prove that its velocity is v p.
Solution
When
ωt–
βz =C is chosen, the wave propagating along z direction becomes cos(C). This means that
time and space which yields constant phase is chosen. For example, let C=0. Then it represents a
wave at its peak value. This will propagate along z direction with time, which one usually observes
the wave velocity. Thus the phase velocity is
{ } 0 p
d dz
t z v
dt dt
ω β
ω β β− = → = =
pv
ω
β
=
3.2 In the special case of the transmission line when θ is small, find the values of L and C using the
results in Fig. 3.25. In addition, show that the following equivalent circuits yield the same wave
equation as θ→0.
(a) (b)
Figure 3.1 Equivalent circuit of a small length transmission line: (a) Tee and (b) Pi shape
Solution
For the circuit in Fig. 3.1(a), from Tee-equivalent circuit in section 3.6.3.2,
1 tan 2 2 2
a c o o o
p
z
j L Z Z jZ jZ jZ v
θ
θ ω
ω ∆ 
= = = ≅ = 
 
Replacing Zo and v p in the above equation by L and C given by (3.6) and (3.14),
1
2
L z
L ∆
=
This is phaysically obvious. Similar procedure leads to
1C C z= ∆
For the circuit in Fig. 3.1(b), by inspection, one can obtain the following results as

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2L L z= ∆ 2
1
2
C C z= ∆
For the circuit in Fig. 3.1(a), define v(z,t) and i(z,t) as shown in Fig. 3.2 then
( ) ( ){ } ( ) ( ), ,1 1
, , 2 2
i z t i z z t
v z z t v z t L z L z
t t
∂ ∂ + ∆
− + ∆ − = ∆ + ∆
∂ ∂
Figure 3.2 Definition of voltages and currents
Dividing both sides by
∆z,
( ) ( ) ( ) ( ), , , ,1 1
2 2
v z z t v z t i z t i z z t
L L
z t t
+ ∆ − ∂ ∂ + ∆
− = +
∆ ∂ ∂
In the limit
∆z →0, the above equation becomes
( ) ( ), ,v z t i z t
L
z t
∂ ∂
= −
∂ ∂
This is equal to the result in (3.2). The other equation can be similarly derived. Thus, the wave
equation in (3.4) is obtained irrespective of the equivalent circuit forms.
3.3 Considering the 2-port circuit in Fig. 3.3 as a transmission line, determine its equivalent
characteristics impedance using open and short-circuit method presented in Section 3.2.4.
Figure 3.3 LC circuit
Solution
The open circuit impedance is

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1 andopen shortZ j L Z j L
C
ω
ω
ω
 
= − = 
 
The equivalent characteristic impedance is
2
2
1o open short
o
L
Z Z Z C
ω
ω
 
= ⋅ = − 
 
Where
2 1
o
LC
ω =
3.4
(1) If the electrical length at a frequency of 1 GHz is 90°, what is the electrical length at a frequency
of 3 GHz?
Solution
The electrical length is
1GHz
1GHz
90f
p f
l
v
ω
θ =
=
= = ° .
Since the electrical length is proportional to frequency, the electrical length at 3 GHz is 270°.
(2) Also, if the phase velocity is equal to the speed of light, what is its length?
Solution
Since v p=c,
1GHz
1GHz1GHz
90
2f
fp f
l l
v c
ω
ω
π
θ =
==
= = = = °
Hence, the length is
2
c
l
π π
ω
= =
8
3 10
2 2
π
× 9
3 0.1(m) 7.5(cm)
410 = × =
×
(3) Finally, what is its propagation constant?
Solution
Since the propagation constant is defined as the phase delay per length (rad/m), it can be computed
as
2 20.9(rad / m)
7.5(cm) 2 0.075c l
π
ω θ
π
β = = = = =
×
3.5 In the definition of the reflection coefficient shown below, when the real part of Z L is positive,
prove that |ΓL| ≤1.
L o
L
L o
V Z Z
V Z Z
−
+
−
Γ = = +

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Solution
Define the normalized impedance zL by Zo as
for > 0Lz r jx r= + ∞ ≥
Then proving the following relation is the problem.
1 1 1 1
1
L
L L L
L
z z z
z
−
Γ = ≤ → − ≤ +
+
Squaring both sides and subtracting, one can find that
{ }2 2 2 2
( 1) ( 1) 4 0r x r x r+ + − − + = ≥
Thus |ΓL| ≤1 and the equality is satisfied for r=0.
3.6 Figure 3.4 shows a slot line measurement. Voltage standing wave minimum points are shown by
the numbers for two measurements. When the load is replaced by a short, the first minimum occurs
at 90cm. The distance dmin from the first minimum to the short is 15 cm (=90–75cm in Fig. 3.4) at a
frequency of 375 MHz. Find the load impedance.
Figure 3.4 Slot line measurements
Solution
This can be solved using Smith chart. Since VSWR=5, one can draw a circle in the Smith chart
corresponding to VSWR=5. Furthermore the half wavelength is found to be 40 cm (=75–35) from
Fig. P3.3. Thus, the wavelength is 80cm. The dmin normalized by wavelength is 0.1875 (=15/80).
Thus the load impedance ZL is found by reading the impedance at point B shown in Fig. 3.5. ZL
reads as
56.5 94.5LZ j= −

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Figure 3.5 Slot line measurements
3.7 In Fig. 3.6, V+ and V– represent incident and reflected phasors at load plane. Write voltage V1
using V+ and V–
. When the VSWR at the input is 2, find the magnitude of Γin and the corresponding
return loss. In above condition, setting Zo =50 ohms, find the value of resistor ZL.
Figure 3.6 Transmission line terminated by load.
Solution
V1 is composed of two voltage waves Vf and Vb. Vf is propagating to the right and resulting in V+ at
the load plane. The phase of the voltage Vf advances that of V+ by
θ. This can be expressed as
j
fV V e
θ+
=
The other propagating to the left from the load whose value is V– at the load plane. The phase of Vb
is delayed by
θ compared with V–
. Thus
j
bV V e
θ− −
=

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As a result,
1
j j
f bV V V V e V e
θ
θ+ − −
= + = +
Denote |Γin|=ρ, then
1 1 1
2 20 log 9.54(dB)
1 3 3
VSWR RL
ρ ρ
ρ
+  
= = → = → = − = 
−  
Since ρ=1/3,
1 1 1
2 or
1 3 2
L o L
L L
L o L
Z Z z z z
Z Z z
ρ − −
= = = → = =
+ +
3.8 Using the scale at the bottom of Smith chart, find VSWR and return loss corresponding to Γ=0.6
using compass.
Solution
The scale in the bottom of the Smith chart is shown in Fig. 3.7. From Fig. 3.7, the return loss and
VSWR corresponding to Γ=0.6 are about 4.4 dB and 4, respectively.
Figure 3.7 The scale in the bottom of the Smith chart.
3.9 For a transmission line with Zo and
θ, derive ABCD parameters defined as,
1 2
1 2
V VA B
I IC D
    
=    
    
Solution
The voltage and current of a transmission line at
θ can be expressed as
2
j j
V V e V e
θ
θ+ − − +
= +
( )2
1 j j
o
I V e V e
Z
θ
θ+ − − +
= − .(1)
From
1V V V+ −
= +

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( )1
1
o
I V V
Z
+ −
= − ,
V+ and V– can be obtained as
1 1 1 1
and
2 2
o oV Z I V Z I
V V+ −+ −
= = .
Substituting the above V+ and V– into (1),
1 1 1 1
2 1 1cos sin
2 2
j jo o
o
V Z I V Z I
V e e V jZ I
θ
θ
θ
θ− ++ −
= + = +
1 1 1 1 1
2 1
1 sin cos
2 2
j jo o
o o
V Z I V Z I jV
I e e I
Z Z
θ
θ
θ
θ− ++ − 
= + = + 
 
Thus
cos , sin , and sino oA D B jZ C jY
θ
θ
θ= = = = .
3.10 Figure 3.8 shows an equivalent circuit of a distribution amplifier similar to that in Fig. 3.1 in
the text. Find V1 and V2 .
Figure 3.8 Equivalent circuit of the output of a distribution amplifier.
Solution
The voltage V1 and V2 can be computed using superposition principle. The voltage Vc1 across the
current source I is computed to be
1 2
o
c
Z I
V =
This will propagates to both ends of the transmission lines and the resulting voltages Vo1 and Vo2 are
2
1, 1 2, 1and
2 2
j jo o
c c
Z I Z I
V e V e
θ
θ− −
= =
Similarly for the current source Ie -j
θ
2 3
1, 2 2, 2= and
2 2 2
j
j j jo o o
c c
Z Ie Z I Z I
V e e V e
θ θ
θ
θ
−
− − −
= =
Thus
2 2
1 1, 1 1, 2 2 =
2
j jo
c c o
Z I
V V V e Z Ie
θ
θ− −
= + = ×

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( )3
2 2, 1 2, 2 2
j jo
c c
Z I
V V V e e
θ
θ− −
= + = +
3.11 If a 50 ohm transmission line has a 1/4 wavelength length at 1 GHz and its end is short-
circuited, can this be considered a parallel resonant circuit near the 1 GHz frequency? What then is
the capacitance of the parallel resonant circuit?
Solution
The transmission line of a quarter wavelength whose end is short-circuited can be approximated by
parallel resonant circuit according to section 3.5.2 and the capacitance of the parallel resonant
circuit is
9
1 1 1
= = (nF)
8 8 10 50 400o o
C f Z
= × ×
3.12 In Fig. 3.9, answer the following. Find the characteristic impedance Zx for maximum power
transfer to load 2Zo. For this, find the time domain voltage waveform v1(t). Find also the time
domain voltage waveform v2(t).
Figure 3.9 Problem 3.12 circuit.
Solution
For the maximum power transfer, the impedance Zin should be
2 2
2
2
x x
in o x o
L o
Z Z
Z Z Z Z
Z Z
= = = → =
Since Zin=Zo,
1
1
( ) cos
2 o ov t E t
ω=
This voltage propagates the transmission line and is delivered to the load 2Zo .whichis equal to the
power delivered to Zo. Thus
2 21
2 4
2
1
2 2
o
o
o o
V E V E
Z Z
= → =
The phase of V 2 should be –90° because the voltage propagates the quarter wavelength transmission
line. Thus

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2
2
1 1
2 2
j
o oV E e j E
π
−
= = −
And
2
1 1
( ) Re sin
2 2
oj t
o o ov t jE e E t
ω
ω
 
= − = 
 
3,13 (ADS Problem) The characteristic impedance of a transmission line can be found applying a
pulse signal to the transmission line. Fig. 3.10 shows a simulation set up to compute the
characteristic impedance of the transmission line. By computing the current and voltage pulses and
the ratio of the pulse, show that the ratio is the characteristic impedance given by 50 ohms.
Figure 3.10 Simulation setup to compute the characteristic impedance using a pulse
Solution
In Fig. 3.10, the time of arrival t d is
8
150 0.5 sec
3 10 [ / sec]
d
l m
t c m
m= = =
×
Fig. 3.11 shows the simulated pulse for Vin and Iin.i. From Fig. 3.11, it can be found that the pulse
appears after 0.5msec. Also from the values of current and voltage, the characteristic impedance is
50 Ω.

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Figure 3.11 The simulated voltage and current pulses.
3,14 (ADS Problem) The standing wave can be simulated by sweeping the length of the
transmission line as shown in Fig. 3.12. Plot of mag(v1) vs l1 results in the standing wave pattern.
Also plot the standing waveform with time as a parameter. It can be found that the minimum point
does not move despite of time change, which is why it is called “standing wave.”
Figure 3.12 Standing waveform simulation.
Solution
After the simulation given in Fig. 3.12, v1 is plot for the length variable l1. Fig. 3.13 shows the
simulated standing waveform. Fig. 3.14 shows the time variation of the standing waveform. This
can be obtained using ts() function in ADS. As can be seen from Fig. 3.14, the minimum point
called “node” does not change with respect to time. This is the reason why we naming “standing
wave.”

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Figure 3.13 Standing waveform
Figure 3.14 Standing waveform change for time
3.15 (ADS Problem) Using ADS, extract the equivalent circuit of corner-discontinuity MCORN
which has a width of 50 ohm line microstrip width. Note that it can be represented by Tee
equivalent circuit because the corner is a passive circuit. Here the substrate is a 10 mil thick alumina
with
εr=9.6.
Solution
Fig. 3.15 shows the schematic setup for the simulation of the corner discontinuity. Also the
microstrip line length which has the corner discontinuity along the center line is also included for
the comparison. The variable w50 represents the 50 ohm microstrip line width which is computed
using LineCalc in ADS.

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Figure 3.15 Schematic setup for the corner discontinuity simulation
After the simulation, the transmission characteristics are compared in Fig. 3.16. From Fig. 3.16, one
can find that the length of the corner discontinuity has a shorter than that of the microstrip line.
Figure 3.16 The S21 of the corner discontinuity and microstrip line
Generally, the corner discontinuity shows a shorter length than the microstrip line with the length of
the corner discontinuity along the center line. Since the corner discontinuity can be represented by a
Tee-equivalent circuit shown in Fig. 3.12.

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Figure 3.17 T-shape equivalent circuit
The values of ZA and ZB can be obtained from Z-parameters as
Eqn ZA=Z11-Z12
Eqn ZB=Z12
Eqn YB=1/ZB
Writing the above equations in the Display window, their imaginary parts can be plotted as shown
in Fig. 3.18. Here Y B=1/ZB is plotted.
(a) (b)
Figure 3.18 Imaginary parts of Z A and Z B vs frequency.(a) YB (=1/ Z B) and (b) Z A
From Fig. 3.18(a), ZB is close to a capacitor. In Fig. 3.18(b) ZA is plotted with the microstrip line
ZA . Since the microstrip line ZA is
tan 2
A oZ jZ
θ 
=  
  ,
it will increase linearly with frequency. On the contrary, ZA of the corner discontinuity decrease
with frequency and its value is negative. This phenomenon occurs because the length of ZA is
negative. Using the marker values, the inductance and capacitance corresponding to ZA and ZB can
be computed inserting equations as

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Eqn LA=(m4-m3)/(indep(m4)-indep(m3))/(2*pi)
Eqn CB=(m2-m1)/(indep(m2)-indep(m1))/(2*pi)
The value of CB and LA are
33 fF and 7.3 pHB AC L= = − .
CHAPTER 4 PROBLEMS
4.1 For a circuit in Fig. 4.1, calculate 2-port S-parameters normalized by the reference impedance
Zo.
Figure 4.1 A two-port circuit
Using the previous results, calculate two-port S-parameters for the circuit where two transmission
lines of electrical lengths
θ1 and
θ2 are added to the circuit in Fig. 4.1 as shown in Fig. 4.2.
Figure 4.2 A two-port circuit with transmission lines
Solution
Due to symmetry, S12=S21 and S11 =S22 .
1 1
2 2
11 211 1
2 2
1 2
, and
3 3
o
o o o o
o o
Z
E
Z Z Z Z
S S
Z Z E
− +
= = − = =
+
Therefore
1 2
3 3
2 1
3 3
− 
=  
− 
S
For the circuit in Fig. 4.2, S can be computed as

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1 1 2
1 2 2
2 ( )1 2
3 3
( ) 22 1
3 3
j j
j j
e e
e e
θ
θ θ
θ θ
θ
− − +
− + −
 −
=  
− 
S ,
using the shift of reference plane explained in Section 4.1.5.
4.2 Calculate the S-parameters of the following parallel and serial resonance circuits.
(a) (b)
Figure 4.3 (a) Parallel resonator and (b) the serial resonator
Solution
In the case of Fig. 4.3(b), the S-parameters are presented in Example 4.5 in Section 4.1.5 and are
11 22
1
1 2 a
S S jQ
δ
−
= = + and 21 12
2
1 2
a
a
jQ
S S jQ
δ
δ
= = + .
Here Qa and
δ are defined as
2 o
a
o
L
Q Z
ω
= and = o
o
ω ω
δ ω
−
In the case of Fig. 4.3(a), the S-parameters are
( )
21 12
2 ||
||
p o
o o p
Z Z
S S Z Z Z
= = + and 11 22
||
||
p o o
o o p
Z Z Z
S S Z Z Z
−
= = + .
Zo|| Zp is computed as
1
|| and , =
1 4 2
o o
o p b o o
b o
Z
Z Z Q Z C
jQ
ω ω
ω δ
δ
ω
−
≅ =
+
Substituting Zp into S21
( ) ( )
21 11
2 || ||1 2
, and
|| 1 2 || 1 2
p o p o o b
o o p b o p o b
Z Z Z Z Z jQ
S S
Z Z Z jQ Z Z Z jQ
δ
δ
δ
− −
= = = =
+ + + +
4.3 Explain how to obtain the circuit values of the above parallel resonant or series resonant circuits
using |S21 | and |S11 | (Refer Example 4.5).
Solution

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In the case of Fig. 4.3(b),
11 21
0 0 and
1 o
S S
ω
ω
ω ω
→ → ∞
= =  =
Thus both |S21 | and |S11 | have the bell-shaped frequency response. From the peak value of |S21 | and
|S11 |, one can find the resonance frequencies, which are equal to
ωo=2πf o as
2 1
o
LC
ω =
In addition, Qa and Qb of the circuits in Fig. 4.3(a) and (b) can be found from the 3-dB bandwidth
BW as
2
o o o
a
f CZ
Q
BW
ω
= = and 2o o
b
o
f L
Q
BW Z
ω
= =
Using Qs and
ωo , the values of L and C in Fig 4.3 can be determined.
4.4 Find the maximum value of |S21 | for the circuit shown in Fig. 4.4, and sketch the frequency
response (ωo=2πf o=2π×10 GHz).
Figure 4.4 A series resonant circuit
Solution
The transmission lines at the input and output have no effects on |S21 |. Excluding the transmission
lines, one can form a new two port which yield the same |S21 |. Since S21 can be expressed as
2
21
2V
S E
=
|S21 | approaches to 0 as frequency away from the resonance frequency 10 GHz. Thus, |S21 | has a
bell-shaped frequency response centered at 10 GHz. The maximum value of |S21 | appears at the
resonance and the value is
2
21
2 100
2 0.953
2 105
o
o
ZV
S E Z r
= = = =
+
4.5 Find voltage Vout for the circuit in Fig. 4.5.

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Figure 4.5 Circuit for problem 4.5
Solution
P=polar(dbmtow(0),0) in P_1Tone port means that, when a conjugate matched load is connected
to this port, the power of 0 dBm is delivered the conjugate matched load. Thus the power 0 dBm is
delivered to 50 ohm load. The voltage Vout is
2
3 21
1mW 10 W 10 0.1(V)
2
out
out out
o
V V V
Z
− −
= = → = → =
Note that the value Vout represents the peak value.
4.6 Given that the insertion loss in a passive lossless 2-port network is 3 dB, calculate the return
loss.
Solution
Since lossless,
2 2
11 21
1 1
1 1 2 2
S S= − = − = .
Thus the return loss= 3dB.
4.7 Calculate the S-parameters of the attenuator circuit shown below. Determine whether |S21 |2
+
|S11 |2 =1 is satisfied.
Figure 4.6 T-type 3 dB attenuator
Solution
S11 and S22 can be manipulated as
11 22
{(50 8.56) ||141.8 8.56} 50 0
{(50 8.56) ||141.8 8.56} 50
S S + + −
= = =
+ + +

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21 12
141.8 || (50 8.56) 50
2 50 8.56 141.8 || (50 8.56) 50 8.56
2
S S +
= = ×
+ + + +
= 41.44
2
×
100
50
× 1
0.707
58.56 2
= =
2 2
11 21
1 1
2
S S+ = <
Since the circuit is lossy, |S11 |2
+|S21 |2 <1.
4.8 Prove equation (4.83)
Solution
The noise figure in (4.82) is
( ) ( ){ }2 22
1 1
u n u n
s s s
s s s s
G R G R
F Y Y G G B B
G G G G
γ
γ
γ= + + + = + + + + +
For Bs change, minimum occurs at
optB B
γ=
The resulting noise figure is
( )
( ) ( )
2
2
2
1
1 2 1 2 2
u n
s
s s
u n
n n s n u n n
s
G R
F G G
G G
G R G R G R G R G G R R G
G
γ
γ γ
γ
γ
= + + +
+
= + + + ≥ + + +
The minimum occurs at
( ) ( )
2
2u n u
n s opt
s n
G R G G
R G G G
G R
γ
γ
+ = → = +
Then Fmin is
( ) ( )
2
min 1 2 2 1 2n u n n n optF R G G R R G R G G
γ
γ
γ= + + + = + +
Using Fmin, Gopt, and Bopt, the noise figure relation can be re-arranged as
( ) ( ){ }2 2
min
n
s opt s opt
s
R
F F G G B B
G
= + − + −
4.9 Prove equation (4.84)
Solution
From problem 4.8, the relation can be transformed into the equation using normalized admittances
as
( ) ( ){ }2 2 2
min min2
1n o n
s opt s opt s opt
s o s
o
R Y r
F F G G B B F y y
G Y g
Y
= + − + − = + −
Substituting the following

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11 , and
1 1
opts
s opt
s opt
y y − Γ− Γ
= =
+ Γ + Γ ,
Into F, the following equation can be obtained after manipulation.
( )
2
min 2 2
4
1 1
n s opt
s opt
r
F F Γ − Γ
= +
− Γ + Γ
4.10 For the circuit in Fig. 4.7, find Fmin and Γopt. Use the definition in equation (4.68) after
transforming the circuit into Norton equivalent circuit at port 2
Figure 4.7 A two-port resistor circuit.
Solution
From Fig. 4.8, the Norton current i T at the port2
2 1
o s
T s
s o s o
Y G
i i i i
G Y G Y
= + −
+ +
Figure 4.8 Equivalent Norton current computation.
The contribution of the source noise to Norton current is
,
o
T s s
s o
Y
i i G Y
= +
Here

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( ) ( ) ( )2 2 2
1 24 , and 4s o s o oE i kT G E i E i kT Y= = =
Thus
( )
( )
2 22 2 2
2
2
1 1
1 2 2
T o s o s o s s s o o
s o s o o o ss
s o
o s
E i Y G Y G Y G G G Y Y
F G Y G Y Y Y GE i
G Y
Y G
   + + +
= = + + = + +   
   
= + + +
Hence
min 3 2 2F = + and 2 1
50 2 0.172
2 1
opt optR s −
= → = =
+
4.11 The noise temperature Te is defined in terms of the noise factor as T e=(F–1)To. Using this
relationship, with the noise temperature of each stage known, derive the expression below by
expressing the noise temperature in terms of Frii’s formula.
2 3
1
1 1 2
e e
e e
T T
T T G G G
= + + + ⋅⋅⋅
Solution
The noise temperature of a linear two-port network is often used in place of noise figure. Given that
the linear two-port network has the transducer power gain G (which is defined as the ratio of the
available power from the source to delivered power to a load and will be explained in detail in
Chapter 8). When the input termination with the temperature T s. is connected to the linear two port
network, the delivered output power PL can be expressed as
( )L s eP kG T T f= + ∆ .
In the above equation, kTe
∆f represents the added equivalent thermal noise power density at the
input due to the noises of the two-port network. Thus the noisy linear two-port network can be
modeled as shown in Fig. 4.9.
Figure 4.9 Two port’s noise temperature concept
One can find that from section 4.2.3,
( 1) ( 1)e o e okT f k F T f T F T∆ = − ∆ → = −
Now consider the following cascaded chains shown in Fig. 4.10.

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.
(a)
(b)
Figure 4.10 Cascaded block diagram (a) cascaded block and (b) its equivalent representation
Excluding thermal noise source contribution to PL, the total output noise power excluding the
contribution of the source noise power can be related as
{ }1 2 1 2 1 2 2
L
T e n e n e n e n en
P G T G G G T G G G T G G T G T
k f = = = + + +
∆    
Dividing both sides with total gain GT
2 3
1
1 1 2
e e
e e
T T
T T G G G
= + + + ⋅⋅⋅
4.12 The attenuation with an attenuator is L (L is not in decibel scale), given that its temperature is
T o, determine the noise factor of the attenuator.
Solution
When one end (source port) of the attenuator is terminated by 50 ohm termination, the impedance
seen from the other end (load port) is 50 ohm. Since the available power of a 50 ohm resistor is kTo,
the delivered power to the 50 ohm load connected to the load port is kTo.
On the other hand, the delivered power from 50 ohm termination connected to the source port of the
attenuator is kTo/L. Thus, from the definition of the noise figure
1
o
o
kT
F L
kT
L
= =
4.13 When the attenuator in Problem 4.12 is connected next to a noise source having ENR1 , what is
the resulting ENR2? Assume the cold temperature of the noise source is To.

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Solution
Defining the hot and cold state noise temperatures Th and Tc, the ENR of the noise source is
1
h c
c
T T
ENR T
−
=
Normally, Tc=To. After the attenuator, the corresponding output noise powers are
1c oN kT kT= =
2
1 1
1h h oN kT kT kT
L L
 
= = + − 
 
Thus
2 1
2 1
1
1 1
1 1h o o
h o
o o
T T T T TkT kT L L
ENR ENR
kT T LT L
 
+ − −  −−  = = = =
4.14 (ADS problem) It is explained that icor(1,1) and icor(2,2) appearing after S-parameter noise
analysis represent the short circuit noise current squares at ports 1 and 2 respectively. The short
circuit noise currents can be obtained through AC simulation in ADS. The following figure shows
AC noise simulation to obtain the short circuit noise currents. Since the noise currents cannot be
obtained directly, the short circuit noise currents are computed using the node voltages across small
valued resistors. Compute the short circuit noise currents through the AC noise simulation and
compare the results with the icor(1,1) and icor(2,2) obtained through the S-parameter noise
simulation.
Figure 4.11 AC noise simulation to compute the short circuit noise currents
Solution
After the AC noise simulation, the short circuit noise current squares are computed in the display
window using the equations as
Eqn Ii_sq=(Prob_15b..vi.noise/1e-3)**2

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Eqn Io_sq=(Prob_15b..vo.noise/1e-3)**2
Figure 4.12 AC noise simulation to compute the short circuit noise currents
In the above equations, Prob_15b..vi.noise and Prob_15b..vo.noise represent the simulated
noise voltages in the dataset name Prob_15b. To represent the noise voltage, Ads uses the
extension .noise after the node voltage name. The division factor 1e-3 appears tocompute the
current across 1mΩ resistors. In addition, the S-parameter noise simulation for the same device is
performed. The computed two values are compared in Fig. 4.12. As can be seen in Fig. 4.12, the
exact match can be found.
CHAPTER 5 PROBLEMS
5.1 Figure 5.1 shows a simplified equivalent circuit of GaAs MESFET at low frequency.
Figure 5.1 A simplified FET equivalent circuit

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Figure 5.2 A simplified FET equivalent circuit with a parallel feedback resistor
(1) Calculate S-parameters setting the reference impedance Zo.
Solution
Since input is open, S11=1. Also S12=0 by inspection. In the case of S21,
2
21 1 1
2 2
m o
m o
g Z EV
S g Z
E E
−
= = = −
When the gate is termed by Zo, and source is applied to the drain, gm Vgs=0. Thus the impedance seen
from the drain is open, which means that S22=1. Therefore
1 0
2 1m og Z
 
=  
− 
S
(2) Setting Zo =50 ohms, compute g m when S21 =5∠180°.
Solution
21 2 5 50 (mA/V)m o mS g Z g= − = − → =
(3) When resistor fR is connected in parallel as shown in Fig. 5P.2, find the value of Rf that makes
S11=0.
Solution
For S11=0, Zin =50. Applying KVL at the input
1 1 1 1( )f m oV R I I g V Z= + −
Thus the following condition can be obtained.
21
1 1
f o
in o f m o
m o
R ZV
Z Z R g Z
I g Z
+
= = = → =
+
5.2 Figure 5.3 shows a simplified equivalent circuit of FET with source resistance Rs at low
frequency. Due to Rs, the equivalent trans-conductance gme is lowered from g m as explained in
Example 5.3. Defining the equivalent trans-conductance g me as
Drain shorted
d
me
gs
i
g v
=

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Solution Manual for Microwave Circuit Design A Practical Approach Using ADS, 1st Edition

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