exam in microwave engineering (mcc121)
TRANSCRIPT
MCC121 Page 1 of 4
Exam in Microwave Engineering (MCC121)
Friday, December 21, 2012
MCC121 Exam in Microwave Engineering
Friday, December 21, 2012, 1400 - 18:00, “V-salar” Teachers: Jan Stake phone: 031 -772 1836
Vincent Desmaris phone: 031 -772 1846 During the exam, teacher will visit around 1430 and 1600.
Examiner: Prof. Jan Stake. Terahertz and millimetre wave lab., Department of microtechnology and nanoscience (MC2), Chalmers University of Technology.
The inspection of the results can be done in my office (D615), MC2-building, Monday, January 7th, 13:00-14:30. The final results will be sent to registrar office on January 11th, 2013.
This is an open book exam. The following is allowed: - Calculator (approved by Chalmers) - ”Microwave Engineering” by Pozar - Mathematics handbook (Beta) - Smith charts
To pass this written examination, you need at least 24p out of 60p. Final grade of the course will also include results from assignment 1. That is: 3 (≥28p), 4 (≥42p) and 5 (≥56p).
Teamwork is not permitted on this examination. The university academic integrity policy will be strictly enforced. Failure to comply with the academic integrity policy will result in a zero for this examination. Make sure you have understood the question before you go ahead. Write shortly but make sure your way of thinking is clearly described. It is imperative to clearly explain how the results have been obtained. Solve the problem as far as you can – constructive, creative and valuable approaches are also rewarded. Assume realistic numbers/parameters when needed if data is missing in order to solve the problem.
v v v
MCC121 Page 2 of 4
Problem 1 Prove that a short (l < λ12 ) low impedance transmission line ( Z0 <
ZL3 ) can be
approximated as a shunt capacitor. Determine the value of this capacitor in terms of the line parameters.
(10p)
Problem 2 a. Design a three-port resistive power divider for an equal power split. The impedance
level is 75Ω. b. Calculate also the signal in port 3 if port 2 is mismatched and connected to 100Ω load.
Assume other ports matched. c. Has this leakage signal a significant influence on the performance of the power
divider?
1
3
2
R
R
R
(10p)
Problem 3 The circuit below can be used as a simple attenuator. Just by soldering two shunt resistors, separated by a quarter wavelength, it is possible to make a quick-fix-attenuator without major modifications in a planar microstrip circuit.
MCC121 Page 3 of 4
Determine a condition for Z1 and R so the attenuator is matched. Calculate R and Z1 to obtain 2dB attenuation. What’s the return loss if you choose Z1=Zc ? ( Zc=50Ω)
(10p)
Problem 4 A transistor requires a source reflection coefficient of Γs=0.5/166°for proving minimum noise figure in an amplifier circuit. Design the input matching circuit using distributed transmission lines, so embedding impedance corresponding to Γs is presented to the transistor as shown below. The amplifier will be used in a Zc = 50 ohm system impedance environment.
(10p)
Problem 5 Consider a rectangular waveguide:
Select the ratio a/b to obtain the largest possible frequency range at the lowest possible attenuation for the dominant mode.
(10p)
x
b
a
y
MCC121 Page 4 of 4
Problem 6 Show that near resonance for the circuit below, the following approximations can be made:
22
0
2
221
11
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−+
=
wwQ
S
or
2202
221
11
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠
⎞⎜⎝
⎛−+
=
ww
Q
S
Where Q is the loaded Q-factor and w0 is the angular frequency of resonance.
What is S21 at resonance? (10p)
Good Luck! / JS Enclosures:
1) Smith Charts
RADIALLY SCALED PARAMETERS
ROTARENEG DRAWOT —<>— DAOL DRAWOT1.11.21.41.61.822.5345102040100
SWR 1∞
12345681015203040dBS 1∞
1234571015 ATTEN. [dB]
1.1 1.2 1.3 1.4 1.6 1.8 2 3 4 5 10 20 S.W. L
OSS COEFF
1 ∞
0 1 2 3 4 5 6 7 8 9 10 12 14 20 30
RTN. LOSS [dB] ∞
0.010.050.10.20.30.40.50.60.70.80.91
RFL. COEFF, P0
0.1 0.2 0.4 0.6 0.8 1 1.5 2 3 4 5 6 10 15 RFL. LOSS
[dB]
∞0
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 10 S.W. P
EAK (CONST
. P)
0 ∞
0.10.20.30.40.50.60.70.80.91
RFL. COEFF, E or I 0 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TRANSM. C
OEFF, P
1
CENTER1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TRANSM
. COEFF, E
or I
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
ORIGIN
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.4
0.4
0.4
0.50.5
0.5
0.60.6
0.6
0.70.7
0.70.8
0.8
0.80.9
0.9
0.9
1.01.0
1.0
1.21.2
1.2
1.41.4
1.4
1.61.6
1.6
1.81.8
1.8
2.02.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.01.0
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.50.5
0.5
0.60.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.61.6
1.6
1.81.8
1.8
2.02.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.01.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
90-9
085
-85
80-8
0
75-7
5
70-7
0
65-6
5
60-6
0
55-55
50-50
45
-45
40
-40
35
-35
30
-30
25
-25
20
-20
15
-15
10
-10
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
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
0.20.2
0.21
0.210.22
0.220.23
0.230.24
0.240.25
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
00
AN
GLE
OF
TRAN
SMISS
ON
CO
EFFICIEN
TIN
DEG
REES
AN
GLE
OF
REFLECTIO
NC
OEFF
CIEN
TIN
DEG
REES
—>
WA
VEL
ENG
THS
TOW
ARD
GEN
ERA
TOR
—>
<—W
AVE
LEN
GTH
STO
WA
RDLO
AD
<—
IND
UCT
IVE
REA
CTAN
CECOM
PONENT (+
jX/Zo), OR IN
D UCTIVE SUSCEPTANCE (- jB/Yo)
EACTANCECOMPONENT (-jX/Zo),
ORCAPA
CITIV
ESUSC
EPTA
NCE
(+jB
Yo)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
SMITH CHART FORM ZY-01-N
ELTITEMAN
COLOR BY J. COLVIN, UNIVERSITY OF FLORIDA, 1997
DWG. NO.
DATE
NORMALIZED IMPEDANCE AND ADMITTANCE COORDINATES
R VE I TI CAPAC
0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.50.5
0.5
0.6
0.6
0.6
0.7
0.7
0.7
0.8
0.8
0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.02.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.01.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140
-140
150
-150
160
-160
170
-170
180
±
90-9
085
-85
80-8
0
75-7
5
70-7
0
65-6
5
60-6
0
55-5
5
50-5
0
45
-45
40
-40
35
-35
30
-30
25
-25
20
-20
15
-15
10
-10
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
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
0.20.2
0.210.21
0.22
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
GLE O
F TRA
NSM
ISSION
CO
EFFIC
IENT IN
DEG
REES
AN
GLE O
F REFL
EC
TIO
N C
OE
FFICIE
NT
IN D
EGR
EES
—>
WA
VEL
ENG
THS
TOW
AR
D G
ENER
ATO
R —
><—
WA
VEL
ENG
THS
TOW
AR
D L
OA
D <
—
IND
UC
TIV
E R
EAC
TAN
CE
COM
PON
ENT (+
jX/Zo), O
R CAPACITIVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COMPONENT (-jX
/Zo), O
R IND
UCTI
VE
SUSC
EPTA
NC
E (-
jB/Y
o)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
RADIALLY SCALED PARAMETERS
TOWARD LOAD —> <— TOWARD GENERATOR1.11.21.41.61.822.5345102040100
SWR 1∞
12345681015203040dBS
1∞
1234571015 ATTEN. [dB]
1.1 1.2 1.3 1.4 1.6 1.8 2 3 4 5 10 20 S.W. L
OSS C
OEFF
1 ∞0 1 2 3 4 5 6 7 8 9 10 12 14 20 30
RTN. LOSS [dB] ∞
0.010.050.10.20.30.40.50.60.70.80.91
RFL. COEFF, P0
0.1 0.2 0.4 0.6 0.8 1 1.5 2 3 4 5 6 10 15 RFL. LOSS
[dB]
∞0
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 10 S.W. P
EAK (CONST
. P)
0 ∞0.10.20.30.40.50.60.70.80.91
RFL. COEFF, E or I 0 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TRANSM. C
OEFF, P
1
CENTER1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TRANSM
. COEFF, E
or I
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
ORIGIN
Black Magic Design
The Complete Smith Chart
MCC121 Page 1 of 8
Solutions to Exam in Microwave Engineering (MCC121)
Friday, December 21, 2012
MCC121 Exam in Microwave Engineering
Friday, December 21, 2012, 1400 - 18:00, “V-salar” Teachers: Jan Stake phone: 031 -772 1836
Vincent Desmaris phone: 031 -772 1846 During the exam, teacher will visit around 1430 and 1600.
Examiner: Prof. Jan Stake. Terahertz and millimetre wave lab., Department of microtechnology and nanoscience (MC2), Chalmers University of Technology.
This is an open book exam. The following is allowed: - Calculator (approved by Chalmers) - ”Microwave Engineering” by Pozar - Mathematics handbook (Beta) - Smith charts
To pass this written examination, you need at least 24p out of 60p. Final grade of the course will also include results from assignment 1. That is: 3 (≥28p), 4 (≥42p) and 5 (≥56p). Teamwork is not permitted on this examination. The university academic integrity policy will be strictly enforced. Failure to comply with the academic integrity policy will result in a zero for this examination. Make sure you have understood the question before you go ahead. Write shortly but make sure your way of thinking is clearly described. It is imperative to clearly explain how the results have been obtained. Solve the problem as far as you can – constructive, creative and valuable approaches are also rewarded. Assume realistic numbers/parameters when needed if data is missing in order to solve the problem.
v v v
MCC121 Page 2 of 8
Problem 1 Prove that a short (l < λ12 ) low impedance transmission line ( Z0 <
ZL3 ) can be
approximated as a shunt capacitor. Determine the value of this capacitor in terms of the line parameters.
(10p)
Solution
Alt 1) Study the ABCD matrix for a short transmission line, 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽𝛽𝛽 𝑗𝑗𝑍𝑍 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽𝛽𝛽𝑗𝑗𝑗𝑗 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽𝛽𝛽 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽𝛽𝛽 ≈
1 𝑗𝑗𝑍𝑍𝑗𝑗𝑗𝑗 𝛽𝛽𝛽𝛽 1 . For low impedance line (high admittance), the ABCD matrix becomes
identical to the matrix representation for a shunt capacitor. I.e. 1 0𝑗𝑗𝑗𝑗 𝛽𝛽𝛽𝛽 1 ≅ 1 0
𝑗𝑗𝑗𝑗𝑗𝑗 1 so
𝐶𝐶 = .
Alt 2) Study the input impedance for a short low impedance transmission line, loaded with ZL. 𝑍𝑍 = 𝑍𝑍 ( )
( )≈ 𝑍𝑍 ≈ 𝑍𝑍 = . The last expression
corresponds to the impedance of a capacitor, C, in parallel to a load, ZL. So 𝐶𝐶 = =
Answer: 𝐶𝐶 = .
MCC121 Page 3 of 8
Problem 2 a. Design a three-port resistive power divider for an equal power split. The impedance
level is 75Ω. b. Calculate also the signal in port 3 if port 2 is mismatched and connected to 100Ω load.
Assume other ports matched. c. Has this leakage signal a significant influence on the performance of the power
divider?
1
3
2
R
R
R
(10p)
Solution a) For equal split, and matched -> R = Zo/3. So R =25 ohm.
b) 𝑆𝑆 =0 1 11 0 11 1 0
for the equal power split divider. For matched ports: 𝑉𝑉 = 𝑉𝑉 . A
100 ohm termination in port 2 will result in a reflection coefficient of Γ = = . So
additional signal at port 3 due to a mismatch at port 2: 𝛿𝛿𝑉𝑉 = 𝑉𝑉 = 𝑉𝑉 . Hence, the
total signal at port 3: 𝑉𝑉 = 𝑉𝑉 + 𝛿𝛿𝑉𝑉 = 𝑉𝑉 (1+ ) = 𝑉𝑉
c) The leak signal will be, 20 log = 22.9𝑑𝑑𝑑𝑑 , below the main signal, so in many cases negligible.
MCC121 Page 4 of 8
Problem 3 The circuit below can be used as a simple attenuator. Just by soldering two shunt resistors, separated by a quarter wavelength, it is possible to make a quick-fix-attenuator without major modifications in a planar microstrip circuit.
Determine a condition for Z1 and R so the attenuator is matched. Calculate R and Z1 to obtain 2dB attenuation. What’s the return loss if you choose Z1=Zc ? ( Zc=50Ω)
(10p) Solution ABCD representation: 𝐴𝐴 𝐵𝐵𝐶𝐶 𝐷𝐷 =
1 01𝑅𝑅 1
0 𝑗𝑗𝑍𝑍𝑗𝑗𝑌𝑌 0
1 01𝑅𝑅 1 =
1 01𝑅𝑅 1
𝑗𝑗𝑍𝑍 𝑅𝑅 𝑗𝑗𝑍𝑍𝑗𝑗𝑌𝑌 0 =
𝑗𝑗𝑍𝑍 𝑅𝑅 𝑗𝑗𝑍𝑍𝑗𝑗𝑍𝑍 𝑅𝑅 + 𝑗𝑗𝑌𝑌 𝑗𝑗𝑍𝑍 𝑅𝑅
(Reciprocal and AD-BC=1) S11=0 => 𝐴𝐴 + 𝐵𝐵 𝑍𝑍 − 𝐶𝐶𝑍𝑍 − 𝐷𝐷 = 0 , hence 𝑗𝑗𝑍𝑍 𝑍𝑍 − 𝑍𝑍 𝑗𝑗𝑍𝑍 𝑅𝑅 + 𝑗𝑗𝑌𝑌 = 𝑗𝑗𝑍𝑍 1 𝑍𝑍 −
𝑍𝑍 𝑅𝑅 − 𝑍𝑍 𝑍𝑍 = 0 so 1/𝑍𝑍 = 1 𝑍𝑍 − 1 𝑅𝑅 → 𝑍𝑍 = 𝑍𝑍 .
For 2-dB attenuation, we need an expression for S21 in order to find correct value for R:
𝑆𝑆21 =2 𝐴𝐴𝐴𝐴 − 𝐵𝐵𝐵𝐵
𝐴𝐴 + 𝐵𝐵 𝑍𝑍 + 𝐶𝐶𝑍𝑍 + 𝐷𝐷 =2
𝑗𝑗𝑍𝑍 𝑅𝑅 + 𝑗𝑗𝑍𝑍 𝑍𝑍 + 𝑗𝑗𝑍𝑍 𝑅𝑅 + 𝑗𝑗𝑌𝑌 𝑍𝑍 + 𝑗𝑗𝑍𝑍 𝑅𝑅
=2𝑗𝑗
12𝑍𝑍 𝑅𝑅 + 𝑍𝑍 𝑍𝑍 + 𝑍𝑍 𝑍𝑍 𝑅𝑅 + 𝑍𝑍 𝑍𝑍 =
2𝑗𝑗𝑍𝑍
12 𝑅𝑅 + 1 𝑍𝑍 + 𝑍𝑍 𝑅𝑅 + 𝑍𝑍 𝑍𝑍
= 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑓𝑓𝑓𝑓𝑓𝑓 𝑆𝑆11 = 0 =2 𝑅𝑅 − 𝑍𝑍
𝑗𝑗𝑗𝑗𝑍𝑍1
2 𝑅𝑅 + 1 𝑍𝑍 + 𝑍𝑍 1 𝑍𝑍
=𝑅𝑅 − 𝑍𝑍𝑗𝑗𝑗𝑗𝑍𝑍
11 𝑅𝑅 + 1 𝑍𝑍 =
𝑅𝑅 − 𝑍𝑍𝑗𝑗
1𝑅𝑅 + 𝑍𝑍 = −𝑗𝑗
𝑅𝑅 − 𝑍𝑍𝑅𝑅 + 𝑍𝑍
2dB-> 𝑆𝑆21 = 10 . = 𝑘𝑘 = → 𝑅𝑅 = 𝑍𝑍 = 50.
. = 221Ω
MCC121 Page 5 of 8
and Z1=51 ohm. Very close to Zc so lets etstimate return loss if we stick to same characteristic impedance. I.e. RL for Z1=Zc=50. We can use our ABCD representation above to find S11, or use the following approach to calculate the input impedance presented by the two-port. The quarter wave transformer will transform the impedance of a shunt connection between Zc and R, which after transformation is shunted with R. ZL=Zc//R which after the quarter wave transformer (Z1=Zc) become: 𝑍𝑍 = = 𝑅𝑅 + 𝑍𝑍 = 𝑅𝑅 + 𝑍𝑍 in shunt with R
1/𝑍𝑍 =𝑅𝑅
𝑍𝑍 𝑅𝑅 + 𝑍𝑍 +1𝑅𝑅 =
1𝑍𝑍
𝑅𝑅𝑅𝑅 + 𝑍𝑍 +
𝑍𝑍𝑅𝑅 → 𝑍𝑍 = 48𝑜𝑜ℎ𝑚𝑚
Γ = = 0.02 Hence, return loss is about RL=34dB.
Answer: R=221 ohm and Z1= 51ohm for 2dB attenuation (matched case). For Z1=50, we will have a low return loss of 34dB.
MCC121 Page 6 of 8
Problem 4 A transistor requires a source reflection coefficient of Γs=0.5/166°for proving minimum noise figure in an amplifier circuit. Design the input matching circuit using distributed transmission lines, so embedding impedance corresponding to Γs is presented to the transistor as shown below. The amplifier will be used in a Zc = 50 ohm system impedance environment.
(10p) Solution Mark desired source reflection coefficient in a SC. Using, distributed components, we will transform from origin (match) to Γs=0.5/166°. See moves in SC below. Answer: An open stub with an electrical length of 0.136λ followed by a line with length 0.1λ will result in desired Γs . See below.
MCC121 Page 7 of 8
Problem 5 Consider a rectangular waveguide:
Select the ratio a/b to obtain the largest possible frequency range at the lowest possible attenuation for the dominant mode.
(10p) Solution The dominant mode, TE10, in a rectangular hollow waveguide has a cut off frequency, which depends on the width, a. Corresponding to half wavelength equal to a. Next similar mode, TE20, exhibit a cut off when the wavelength is equal to a. Hence, twice the cut off frequency for TE10. The metallic loss (attenuation) depends on both width and height, and for all low order modes, a large waveguide is always preferred. (See fig 3.8 in Pozar or analytic expressions). Consequently, for a fixed width, a, determining the dominant mode, we would prefer a large height waveguide. However, corresponding TE01 mode might be excited if the height, b, is equal to half a wavelength. Since TE20 cannot be avoided, the maximum frequency span is a factor of two. So a natural design, which at the same time minimizes loss and provides the largest frequency band for the dominant TE10 mode, is when TE01 is designed to have the same cut off frequency as the TE20 mode. Hence, b=a/2. (This gives largest possible bandwidth for dominant mode and at the same time lowest loss.)
x
b
a
y
MCC121 Page 8 of 8
Problem 6 Show that near resonance for the circuit below, the following approximations can be made:
22
0
2
221
11
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−+
=
wwQ
S
or
2202
221
11
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠
⎞⎜⎝
⎛−+
=
ww
Q
S
Where Q is the loaded Q-factor and w0 is the angular frequency of resonance.
What is S21 at resonance? (10p) Solution
Start by defining: 𝐴𝐴 𝐵𝐵𝐶𝐶 𝐷𝐷 =
1 0𝑗𝑗𝑗𝑗𝑗𝑗 + 1 (Reciprocal and AD-BC=1)
𝑆𝑆21 =2 𝐴𝐴𝐴𝐴 − 𝐵𝐵𝐵𝐵
𝐴𝐴 + 𝐵𝐵 𝑍𝑍 + 𝐶𝐶𝑍𝑍 + 𝐷𝐷 =2
2+ 𝑍𝑍 𝑗𝑗𝑗𝑗𝑗𝑗 + 1𝑗𝑗𝑗𝑗𝐿𝐿
=1
1+ 12𝑍𝑍 𝑗𝑗𝑗𝑗𝑗𝑗 1− 1𝜔𝜔 𝐿𝐿𝐶𝐶
= 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ,𝑤𝑤𝑤𝑤 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝜔𝜔 =
1𝐿𝐿𝐿𝐿
𝑄𝑄 =12𝑍𝑍 𝜔𝜔 𝐶𝐶 =
𝑍𝑍2𝜔𝜔 𝐿𝐿
=1
1+ 𝑗𝑗𝑗𝑗 1− 𝜔𝜔𝜔𝜔
= 𝑜𝑜𝑜𝑜
=1
1+ 𝑄𝑄𝑗𝑗 1− 𝜔𝜔𝜔𝜔
𝑆𝑆21 = = V.S.V.
Answer: S21 is equal to 1 at resonance.