This document discusses Smith charts and impedance matching. It begins with an introduction to resonators, Q factor, and resonant bandwidth. It then covers basic impedance matching networks including L, T, and π networks. The document explains how to use Smith charts to represent LC circuits and perform impedance matching. It also discusses loaded Q versus unloaded Q and how to match impedances for different cases. Matching bandwidth is defined and conversions between series and parallel circuits are covered. The document provides an overview of important concepts regarding resonators, Q factor, impedance matching, and the use of Smith charts.
This Presentation is related to multistage amplifiers. different topics related to multistage amplifiers like two stage amplifiers. Two stage RC coupled amplifiers, cascading techniques, CE-CB cascod amplifiers, darlington pair, fet analysis
Use s parameters-determining_inductance_capacitancePei-Che Chang
1. Use s parameters-determining_inductance_capacitance
2. Relationship Between Common Circuits and the ABCD Parameters
3. Converts Z-parameters to S-parameters
4. Relationships Between Two-Port S and ABCD Parameters
5. Via and equivalent circuit
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
COLLEGE BUS MANAGEMENT SYSTEM PROJECT REPORT.pdfKamal Acharya
The College Bus Management system is completely developed by Visual Basic .NET Version. The application is connect with most secured database language MS SQL Server. The application is develop by using best combination of front-end and back-end languages. The application is totally design like flat user interface. This flat user interface is more attractive user interface in 2017. The application is gives more important to the system functionality. The application is to manage the student’s details, driver’s details, bus details, bus route details, bus fees details and more. The application has only one unit for admin. The admin can manage the entire application. The admin can login into the application by using username and password of the admin. The application is develop for big and small colleges. It is more user friendly for non-computer person. Even they can easily learn how to manage the application within hours. The application is more secure by the admin. The system will give an effective output for the VB.Net and SQL Server given as input to the system. The compiled java program given as input to the system, after scanning the program will generate different reports. The application generates the report for users. The admin can view and download the report of the data. The application deliver the excel format reports. Because, excel formatted reports is very easy to understand the income and expense of the college bus. This application is mainly develop for windows operating system users. In 2017, 73% of people enterprises are using windows operating system. So the application will easily install for all the windows operating system users. The application-developed size is very low. The application consumes very low space in disk. Therefore, the user can allocate very minimum local disk space for this application.
Automobile Management System Project Report.pdfKamal Acharya
The proposed project is developed to manage the automobile in the automobile dealer company. The main module in this project is login, automobile management, customer management, sales, complaints and reports. The first module is the login. The automobile showroom owner should login to the project for usage. The username and password are verified and if it is correct, next form opens. If the username and password are not correct, it shows the error message.
When a customer search for a automobile, if the automobile is available, they will be taken to a page that shows the details of the automobile including automobile name, automobile ID, quantity, price etc. “Automobile Management System” is useful for maintaining automobiles, customers effectively and hence helps for establishing good relation between customer and automobile organization. It contains various customized modules for effectively maintaining automobiles and stock information accurately and safely.
When the automobile is sold to the customer, stock will be reduced automatically. When a new purchase is made, stock will be increased automatically. While selecting automobiles for sale, the proposed software will automatically check for total number of available stock of that particular item, if the total stock of that particular item is less than 5, software will notify the user to purchase the particular item.
Also when the user tries to sale items which are not in stock, the system will prompt the user that the stock is not enough. Customers of this system can search for a automobile; can purchase a automobile easily by selecting fast. On the other hand the stock of automobiles can be maintained perfectly by the automobile shop manager overcoming the drawbacks of existing system.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Vaccine management system project report documentation..pdfKamal Acharya
The Division of Vaccine and Immunization is facing increasing difficulty monitoring vaccines and other commodities distribution once they have been distributed from the national stores. With the introduction of new vaccines, more challenges have been anticipated with this additions posing serious threat to the already over strained vaccine supply chain system in Kenya.
3. 串聯諧振器(Series Resonators)
• 串聯諧振電路
I R L
C
( )
1
inZ R j L
j C
ω ω
ω
= + +
CV
+
−
21
2
RP I R=
21
4
mW I L=
2 2 2
2 2 2
1 1 1 1
4 4 4
e C
C
W V C I I
C Cω ω
= = =
在諧振頻率時, :0 02 fω π= m eW W= 0
1
LC
ω =
( )0inZ Rω =
當電感與電容平均儲能相等時,串聯電路
的輸入阻抗將只剩下純電阻性的成分。
• 品質因素 (Quality Factor, Q):
average energy stored
energy loss/sec
m e
R
W W
Q
P
ω ω
+
=≜
當諧振發生時, 0ω ω=
0
0 0
0
2 2 1m e
R R
W W L
Q
P P R CR
ω
ω ω
ω
= = = =
X
Q
R
= 其中 0
0
1
orX L
C
ω
ω
=
此時,
Q是一種度量諧振器損耗多寡的參數,
電路損耗越低代表Q值越高。
Department of Electronic Engineering, NTUT3/70
4. 在諧振頻率時, :
此時,
並聯諧振器
2
2
R
V
P
R
=
21
4
eW V C=
2 2
2
2 2 2
1 1 1
4 4 4
m L
V V
W I L L
L Lω ω
= = =
0 02 fω π= m eW W= 0
1
LC
ω =
( )0
1
inY
R
ω =
LI
LR C
( )
1 1
inY j C
R j L
ω ω
ω
= + +
V
+
−
• 並聯諧振電路
當電感與電容平均儲能相等時,並聯電路
的輸入導納將只剩下純電導性的成分。
• 品質因素 (Quality Factor, Q):
average energy stored
energy loss/sec
m e
R
W W
Q
P
ω ω
+
=≜
B
Q
G
=
0
0
1
orB C
L
ω
ω
=
1
G
R
=
當諧振發生時, 0ω ω=
其中
Department of Electronic Engineering, NTUT4/70
5. 與串聯RL與RC電路比較
L R
C R
1X
Q
R CRω
=≜
雖然Q可以被定義,但是無法定義出
諧振頻率,因RC與RL電路並不會發
生諧振(沒有能量互丟的現象),因此
不屬於諧振電路。
• 串聯RL與RC電路可沿用Q之定義
• 並聯RL與RC電路
B R
Q
G Lω
=≜
B
Q CR
G
ω=≜
L
R
C
R
X L
Q
R R
ω
=≜
Department of Electronic Engineering, NTUT5/70
6. 定義諧振頻寬
• 以串聯諧振器為例,當 0ω ω→
( )
1
inZ R j L
j C
ω ω
ω
= + +
0ω ω ω= + ∆令 0ω∆ → 0
1
LC
ω =
( )
2 2 2
0 0
2 2 2
1
1 1inZ R j L R j L R j L
LC
ω ω ω
ω ω ω ω
ω ω ω
−
= + − = + − = +
( )( )0 0 0
2
0 0
2
2 2
QR
R j L R jL R j L R j
ω ω ω ω ω ω
ω ω ω
ω ω ω
+ − ⋅∆
= + + = + ∆ = + ∆
≃ 0L
Q
R
ω
=其中
其中 而
I L R
C
( )inZ ω
CV
+
−
• 定義頻寬為
0
2
BW
ω
ω
∆
≜
0
0
2
2
in
BW QR
Z R j R jBW QR
ω
ω
⋅
= + = + ⋅
inZ
0ω
ω
R
2R
2 ω∆
當 1
BW
Q
= 2inZ R jR R= + =
0
03-dB Bandwidth BW
Q
ω
ω= ⋅ =
1
3-dB Bandwidth in % BW
Q
= =
Department of Electronic Engineering, NTUT6/70
7. 並聯諧振器的3-dB頻寬
• 以並聯諧振電路為例,當 0ω ω→
( )
1 1 1
2inY j C jC
R j L R
ω ω ω
ω
= + + + ∆≃
0
2
BW
ω
ω
∆
≜定義
( ) 0
0
1 1
2
2
in
BW Q Q
Y j jBW
R R R R
ω
ω
ω
⋅
= + = +
當 1
BW
Q
=
1 1 2
inY j
R R R
= + =
0
03-dB Bandwidth BW
Q
ω
ω= ⋅ =
1
3-dB Bandwidth in % BW
Q
= =
inY
0ω
ω
1
R
2
R
2 ω∆
關於關於關於關於Q的重要觀念的重要觀念的重要觀念的重要觀念::::
1. Q值越高、損耗越低、頻寬越窄
2. Q值越低、損耗越高、頻寬越寬
Department of Electronic Engineering, NTUT7/70
8. Loaded-Q與Unloaded Q
LR
Resonant
Circuit
Unloaded Q
• 串聯RLC電路
• 並聯RLC電路
( )
0
0
1
L
L L
X L
Q
R R R C R R
ω
ω
= = =
+ +
( )0
0
//
//L
L L
B R R
Q C R R
G L
ω
ω
= = =
• 定義Qe為外部Q (external Q)
0
0
1
e
L L
L
Q
R R C
ω
ω
= = for series RLC
0
0
L
e L
R
Q R C
L
ω
ω
= =
1 1 1
L eQ Q Q
= +
for parallel RLC
for both cases
負載效應
負載Q是無負載Q
與外部Q的並聯。
Department of Electronic Engineering, NTUT8/70
9. 阻抗匹配
Matching
Network
in sZ R=
+
−
sV
sR
LZo sZ R=
0inΓ =
Goal:
• 假設匹配網路為理想無損耗的情況下,為了達到最大功率傳輸的目
的,匹配網路是要設計來將ZL轉換為Z0 (matched with the transmission line)
或 Rs (matched with the source impedance when no line connected)。
• 當負載與傳輸線阻抗匹配時,將有最大功率傳輸至負載(assuming the
generator is matched)。
Department of Electronic Engineering, NTUT9/70
11. L型匹配 – Case (a) Rs < 1/GL
+
−
sV
sR
jX
jB LY
L L LY G jB= +
inZ
目標在於求匹配元件之電抗X與電受B,能使
Case (a) 1s LR G< in sZ R= 0Γ =
+
−
sV
sR
jX
( )Lj B B+ LG
L
L
B B
Q
G
+
=l
Series RC or RL
Parallel RC or RL
s
s
X
Q
R
=
( )
1
in s
L L
Z jX R
G jB jB
= + =
+ +
實部: ( ) 1s L LR G X B B+ + =
虛部: ( ) 0s L LR B B XG+ − =
阻抗匹配時,由匹配
網路往負載端視入的
輸入阻抗虛部為0。
L
s
s L
B BX
Q Q Q
R G
+
= = = =l
( )2
1 1s LR G Q + =
1
1
s L
Q
R G
= ± −
選擇
1
1s
s L
Q Q Q
R G
= = = + −l
1
1s s
s L
X R Q R
R G
= = −
1
1L L L L
s L
B G Q B G B
R G
= − = − − (>0, 電容)
(<0, 電感)
(>0, 電感)
當Q被決定了,X與B也就決定了。
或
Department of Electronic Engineering, NTUT11/70
12. L型匹配 – Case (b) Rs > RL
+
−
sV
sR
jX
jB LZ
L L LZ R jX= +
inY
Case (b) s LR R>
目標在於求匹配元件之電抗X與電受B,能使
1in sY R= 0Γ =或
s sQ BR= L
L
X X
Q
R
+
=l
+
−
sV
sR
( )Lj X X+
jB LR
Parallel
RC or RL
Series RC or RL
( )
1 1
in
L L s
Y jB
R jX jX R
= + =
+ +
( )s L s LBR X X R R+ = −
( ) 0L s LX X BR R+ − =虛部:
實部:
阻抗匹配時,由匹配
網路往負載端視入的
輸入導納虛部為0。
L
s s
L
X X
Q BR Q Q
R
+
= = = =l
2
L s LQ R R R= − 1s
L
R
Q
R
= ± −
選擇 1s
s
L
R
Q Q Q
R
= = = + −l
1s
L L L L
L
R
X R Q X R X
R
= − = − −
1
1s
s s L
RQ
B
R R R
= = −
(>0, 電感)
(<0, 電容)
(>0, 電容)
當Q被決定了,X與B也就決定了。
Department of Electronic Engineering, NTUT12/70
13. (1) 已知串聯,要轉並聯:
串聯與並聯轉換
sR
sjX
pR
pjX
s
s
s
X
Q
R
=
s s sZ R jX= +
p
p
p
R
Q
X
=
p p
p
p p
R jX
Z
R jX
⋅
=
+
ps
s p
s p
RX
Q Q Q
R X
= = = =
p p
s p s s
p p
R jX
Z Z R jX
R jX
⋅
= = + =
+
( )2
1s pR Q R+ =
等效
彼此互轉
( )2
1p sR R Q= +
p
p
p
R
X
Q
=
(2) 已知並聯,要轉串聯:
2
1
p
s
R
R
Q
=
+
s s sX R Q=
Department of Electronic Engineering, NTUT13/70
14. 匹配頻寬 (I)
+
−
sV
sR
jX
jB LY
L L LY G jB= +
inZ
in s
in s
Z R
Z R
−
Γ =
+
+
−
sV
sR
jX
( )Lj B B+ LG
+
−
sV
sR
jX eqjB
eqR
2
1 1
1
eq
L
R
G Q
=
+
( )2
11 L
eq
eq
G Q
B
QR Q
+
= =
Case (a) 1s LR G<
把匹配後的完整網路
想辦法轉成「串聯」
或「並聯」RLC電路,
就可以直接使用諧振
器頻寬的定義來計算
匹配頻寬。
並聯準備轉成串聯
Department of Electronic Engineering, NTUT14/70
15. 匹配頻寬 (II)
阻抗匹配時的中心頻率 0
1
LC
ω ω= =
( )0
1
in eq s
eq
Z jX R R
jB
ω ω= = + + =
虛部虛部虛部虛部:
1
0
eq
jX
jB
+ = 1eqXB =
令 0X Lω= 0eqB Cω=
實部實部實部實部: eq sR R= 2
1 1
1
s
L
R
G Q
=
+
1
1
L s
Q
G R
= ± −
與
與
+
−
sV
sR
jX eqjB
eq sR R=
inQ1
2
L inQ Q=
• 定義RLC諧振器的 QL 與 Qin: ( )2
1in s L
eq
X
Q R Q G Q Q
R
= = ⋅ ⋅ + =
1
2
L inQ Q=而
• 找出 的 3-dB頻寬 :Γ
令 X Lω= eqB Cω=
當 0ω ω→ ,令 0ω ω ω= + ∆ 0
1
LC
ω =0ω∆ →
1
2
1 2 22
eqin s
in s s
s
eq
jX
jBZ R jL
Z R jL RjX R
jB
ω
ω
+
− ∆
Γ = =
+ ∆ ++ +
≃
與
其中 及
Department of Electronic Engineering, NTUT15/70
16. 匹配頻寬 (III)
( )2 ω= ∆
2 2 sL Rω∆ = 02 2
2 s sR R
L X
ω
ω∆ = =
3-dB Bandwidth in % BW=
0
22 2 1 2
1
1
s
L
s L
R
BW
X Q Q
R G
ω
ω
∆
= = = = =
−
1 2 2
1
1L
s L
BW
Q Q
R G
= = =
−
Γ
0ω
ω
1
2
1
2 ω∆
1
1
s L
Q
R G
= −
Case (a)
1
s
L
R
G
<
1s
L
R
Q
R
= −
Case (b)
s LR R>
• 找出 的 3-dB頻寬 :Γ
當
Case (b) s LR R>
同理可求得:
Department of Electronic Engineering, NTUT16/70
17. 範例 – L型匹配 (I)
• 下圖電路請以L型網路匹配之,並求出3-dB匹配頻寬。
1 2000
1 1
50s L
Q
R G
= ± − = ± −
1
, choose case (a)s
L
R
G
<
6.245Q = +
6
6
50 6.245 2 100 10
6.245 / 2000 0 2 100 10
s
L L
X R Q L
B G Q B C
π
π
= = ⋅ = ⋅ × ⋅
= − = − = ⋅ × ⋅
1st Solution: choose
Matching
Network
( )100in sZ f MHz R= =
+
−
sV
50sR = Ω 2000LR = Ω
( )100 0in f MHzΓ = =
Goal:
LR
LR4.9696 pFC =
496.96 nHL =
+
−
sV
50sR = Ω
-9
12
496.96 10 (H) 496.96 (nH)
4.9696 10 (F) 4.9696 (pF)
L
C −
= × =
= × =
6.245= ±
Department of Electronic Engineering, NTUT17/70
18. 範例 – L型匹配 (II)
6
6
50 ( 6.245) 1/ (2 100 10 )
( 6.245) / 2000 0 1/ (2 100 10 )
s
L L
X R Q C
B G Q B L
π
π
= = ⋅ − = − ⋅ × ⋅
= − = − − = − ⋅ × ⋅
6.245Q = −2nd Solution: choose
12
9
5.097 10 (F) 5.097 (pF)
509.7 10 (H) 509.7 (nH)
C
L
−
−
= × =
= × =
LR
+
−
sV
50sR = Ω
509.7 nHL =
5.097 pFC =
3dB
1 2 2
32%
| | 6.245L
BW
Q Q
= = = =
Department of Electronic Engineering, NTUT18/70
19. 三元件匹配(High Q匹配、窄頻匹配)
Case (a) π型匹配
+
−
sV
sR
2jX
3jB L L LY G jB= +
,inZ Γ
1jB
+
−
sV
sR
1jX
LZ L L LZ R jX= +2jB
3jX
Case (b) T型匹配
LY
,inZ Γ
目標在於求匹配元件之電抗X2與電受B1、B3,
能使 in sZ R= 0Γ =或
目標在於求匹配元件之電抗X1、X3 與電受B2,
能使 in sZ R= 0Γ =或
Department of Electronic Engineering, NTUT19/70
20. π型匹配 – 兩個L型的結合
+
−
sV
sR
2ajX
3jB LY L L LY G jB= +
VR
1jB
2bjX
VR
2 2 2a bX X X= +
1
V
L
R
G
<V sR R<
• π型網路可拆分成兩個“L型”網路來進行分析:
+
−
sV
sR
2ajX
1jB VR
in sZ R= 0Γ =
3jB LY
L L LY G jB= +
2bjX
+
−
sV
sR
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
Department of Electronic Engineering, NTUT20/70
21. π型匹配 – Q值與匹配頻寬
1 1s
V
R
Q
R
= ± − 2
1
1
V L
Q
R G
= ± −1
1
2
BW
Q
= 2
2
2
BW
Q
=
( )1 2min ,BW BW BW≃
28/51
+
−
sV
sR
2ajX
1jB VR
in sZ R= 0Γ =
3jB LY
L L LY G jB= +
2bjX
+
−
sV
sR
in VZ R= 0Γ =
• 因RV同時小於Rs與(1/GL),因此我們可以透過設計RV值來得到高Q匹配(窄頻)。
• 我們也可以反過來說,如果要求匹配頻寬為BW,我們便可以反推虛擬電阻值RV
應該要設計為多少。
Department of Electronic Engineering, NTUT21/70
22. T型匹配 – 兩個L型的結合
+
−
sV
sR
1jX
LZ L L LZ R jX= +
VR
2ajB
3jX
2bjB
VR V LR R>V sR R>
sR
+
−
sV
1jX
in sZ R= 0Γ =
2ajB VR LZ
L L LZ R jX= +
3jX
2bjB
+
−
sV
VR
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
• T型網路可拆分成兩個“L型”網路來進行分析:
Department of Electronic Engineering, NTUT22/70
23. T型匹配 – Q值與匹配頻寬
1 1V
s
R
Q
R
= ± − 2 1V
L
R
Q
R
= ± −1
1
2
BW
Q
= 2
2
2
BW
Q
=
( )1 2min ,BW BW BW≃
sR
+
−
sV
1jX
in sZ R= 0Γ =
2ajB VR LZ
L L LZ R jX= +
3jX
2bjB
+
−
sV
VR
in VZ R= 0Γ =
• 因RV同時大於Rs與(1/RL),因此我們可以透過設計RV值來得到高Q匹配(窄頻)。
• 我們也可以反過來說,如果要求匹配頻寬為BW,我們便可以反推虛擬電阻值RV
應該要設計為多少。
Department of Electronic Engineering, NTUT23/70
24. 範例 – π與T型匹配網路
• 使用π型與T型網路進行匹配,並要求匹配頻寬 BW < 5%。
Matching
Network
( )100in sZ f MHz R= =
+
−
sV
50sR = Ω 2000LR = Ω
( )100 0in f MHzΓ = =
Goal:
LR
Department of Electronic Engineering, NTUT24/70
25. 範例 – π型匹配網路 (I)
+
−
sV
50sR = Ω
2ajX
1jB 1.249Ω
( )100 50inZ f MHz= = Ω ( )100 0f MHzΓ = =
VR
1
50
1 1 6.247
1.2492
s
V
R
Q
R
= ± − = ± − = ± 2
1 2000
1 1 40
1.2492V L
Q
R G
= ± − = ± − = ±
3jB
2bjX
+
−
sV
1.249VR = Ω
( )100 1.2492inZ f MHz= = Ω ( )100 0f MHzΓ = =
LG
1
2000
LG =
Ω
1 2max(| |,| |) max( 50 / 1, 2000 / 1) 2000 / 1v v vQ Q Q R R R= = − − = −
2
5% 1.249
(2000 / ) 1
v
v
BW R
R
= ≤ ⇒ ≤ Ω
−
Case (a) π型匹配
1 6.247Q = +
1 1 0sB Q R Cω= = 198.85 pFC =
2 1 0a VX R Q Lω= = 12.42 nHL =
1 6.247Q = −
( )1 1 01sB Q R Lω= = −
203.95 pFC =( )2 1 01a VX R Q Cω= = −
12.74 nHL =
2 40Q = +
3 2 0L LB G Q B Cω= − = 31.83 pFC =
2 2 0b VX R Q Lω= = 79.53 nHL =
2 40Q = −
( )3 2 01L LB G Q B Lω= − = − 79.58 nHL =
( )2 2 01b VX R Q Cω= = − 31.85 pFC =
Department of Electronic Engineering, NTUT25/70
30. 範例 – T型匹配網路 (III)
Department of Electronic Engineering, NTUT30/70
31. 串接L型匹配 (Low Q匹配、寬頻匹配)
Case (a) 1s V LR R G< <
+
−
sV
sR
1jX
1jB LY L L LY G jB= +
in sZ R=
2jX
2jB
0Γ =
VRVR
+
−
sV
sR
1jX
1jB VR
in sZ R= 0Γ =
+
−
sV
VR
2jX
LY
L L LY G jB= +
2jB
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
• 串接L型網路當然可以拆成兩個“L型”網路來進行分析:
Department of Electronic Engineering, NTUT31/70
32. 串接L型匹配 (Low Q)
1 1V
s
R
Q
R
= ± − 2
1
1
L V
Q
G R
= ± −
令 1 2Q Q Q= = s
V
L
R
R
G
=
2 2 2
2 1
1
s L
BW
QQ
R G
= =
−
≃
+
−
sV
sR
1jX
1jB VR
in sZ R= 0Γ =
+
−
sV
VR
2jX
LY
L L LY G jB= +
2jB
in VZ R= 0Γ =
• 我們想要最大頻寬的匹配,也就是要找到最小Q匹配(寬頻)。
Department of Electronic Engineering, NTUT32/70
33. 串接L型匹配 (Low Q)
Case (b) s V LR R R> >
+
−
sV
sR
1jX
1jB LZ L L LZ R jX= +
in sZ R=
2jX
2jB
VRVR
0Γ =
+
−
sV
sR
1jX
1jB
in sZ R= 0Γ =
VR LZ
L LR jX+
2jX
2jB
+
−
sV
VR
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
• 串接L型網路當然可以拆成兩個“L型”網路來進行分析:
Department of Electronic Engineering, NTUT33/70
34. 串接L型匹配 (Low Q)
1 1s
V
R
Q
R
= ± − 2 1V
L
R
Q
R
= ± −
1 2Q Q Q= = V s LR R R=
2 2 2
2 1
1
s L
BW
QQ
R G
= =
−
≃
+
−
sV
sR
1jX
1jB
in sZ R= 0Γ =
VR LZ
L LR jX+
2jX
2jB
+
−
sV
VR
in VZ R= 0Γ =
• 串接L型網路當然可以拆成兩個“L型”網路來進行分析:
令
Department of Electronic Engineering, NTUT34/70
35. 範例 – 串接L型匹配 (I)
• 使用串接L型匹配使匹配頻寬能夠達到BW > 60%。
Matching
Network
( )100in sZ f MHz R= =
+
−
sV
50sR = Ω 2000LR = Ω
( )100 0in f MHzΓ = =
Goal:
LR
選擇RV : 50 2000 316.23V s LR R R= = ⋅ =
2 2
61.29%
1 2000 50 11
s L
BW
R G
= = =
−−
s V LR R R< <
Department of Electronic Engineering, NTUT35/70
36. 範例 – 串接L型匹配 (II)
+
−
sV
50sR = Ω
1jX
1jB
( )100 50inZ f MHz= = Ω
2jX
2jB
( )100 0f MHzΓ = =
VR
+
−
sV
316.23VR = Ω
( )100 0f MHzΓ = =( )100 316.23inZ f MHz= = Ω
LR
1
316.23
1 1 2.3075
50
V
s
R
Q
R
= ± − = ± − = ± 2
2000
1 1 2.3075
316.23
L
V
R
Q
R
= ± − = ± − = ±
1 2.3075Q = +
1 1 0sX R Q Lω= = 183.63 nHL =
1 1 0VB Q R Cω= = 11.61 pFC =
1 2.3075Q = −
( )1 1 01sX R Q Cω= = − 13.79 pFC =
( )1 1 01VB Q R Lω= = − 218.11 nHL =
2 2.3075Q = +
2 2 0VX R Q Lω= = 1.161 µHL =
( )2 2 0L LB Q R B Cω= − = 1.836 pFC =
2 2.3075Q = −
( )2 2 01VX R Q Cω= = − 2.181 pFC =
( ) ( )2 2 01L LB Q R B Lω= − = − 1.379 µHL =
316.23VR = Ω 2 kLR =
Department of Electronic Engineering, NTUT36/70
41. 史密斯圖的建立
( ) o
o
Z Z
Z
Z Z
−
Γ =
+
• 史密斯圖(Smith chart)也稱為反射係數圖( plane),某一阻抗所代表
的反射係數與其阻抗具有以下關係:
Γ
對所有的正實數Z都成立,而其中Zo是傳輸線特徵阻抗或系統參考阻
抗,一般為50 。
• 定義正規化阻抗 z 為
o o
Z R jX
z r jx
Z Z
+
= = = +
( )
( )
11
1 1
r jxz
U jV
z r jx
− +−
Γ = = = +
+ + + ( )
2 2
2 2
1
1
r x
U
r x
− +
=
+ + ( )
2 2
2
1
x
V
r x
=
+ +
其中 及
• 反射係數
Department of Electronic Engineering, NTUT41/70
42. 史密斯圖 (Smith Chart)
r
x
( )U jVΓ = +Γ-plane
U
V
1z j=
1z =
0z =
1
1
z
z
−
Γ =
+
1 1 1 90z j j= ⇒ = ∠
0 1 1 180z = ⇒ Γ = − = ∠
1 0z = ⇒ Γ =
1 90Γ = ∠
0Γ =
1Γ = −
( )z r jx= +z-plane
1 1 1 90z j j= − ⇒ Γ = − = ∠ −
1z j= −
Short Load Open
1z = ∞ ⇒ Γ =
1Γ =
Pure Imaginary: inductive
1 90Γ = ∠ −
Pure Imaginary: capacitive
Department of Electronic Engineering, NTUT42/70
51. 考慮導納的史密斯圖
y g jb= +
U
V
U′
V′z r jx= +
1 1
1
y g jb
z
− Γ
= = = +
+ Γ
1
1
z
+ Γ
=
− Γ
Impedance Chart (Z-Chart) Admittance Chart (Y-Chart)
jx+
jx− jb+
jb−
Short Load Open Short Load Open
Department of Electronic Engineering, NTUT51/70
61. 匹配範例 (II)
( )10 10LZ j= + Ω
0.2 0.2Lz j= +
Goal
0.2j
0.4j−
0.6x j∆ = −
2j
0j
2y j∆ = −
0.2 0.4z j= −
L
C 0.2
0.2j
01@ 500 MHzinz f= =
0.6j−
( )
1
02 0.6 50 30f Cπ
−
= × Ω =
( )
1
0
1
2 2 0.04
50
f Lπ
−
= × =
Ω
10.6 pFC =
7.95 nHL =
L
C
10.6 pF
7.95 nH
10 Ω
3.18 nH
Department of Electronic Engineering, NTUT61/70
62. 匹配範例 (III)
1 L
C
( )8 12 mSoutY j= −
Goal
( )50 Ω
0.4 0.6outy j= −
Department of Electronic Engineering, NTUT62/70
63. 匹配至任意阻抗
LZC
L
50 20inZ j= + Ω
100 100LZ j= + Ω
Goal
100refZ = Ω
LZ
C
L
0.5 0.2inZ j= + Ω
1 1Lz j= + Ω
Department of Electronic Engineering, NTUT63/70
64. 頻率變高時的阻抗變化軌跡 (I)
L
R
C
R
L
R
C
L
R
C
( )1inZ R j Lω ω= +
( )
( )1
1
50
in
in
Z
z r jx
ω
ω = = +
Ω
( )1in aZ ω
( )1in bZ ω
( )2inZ ω
( )2in aZ ω
( )2in bZ ω
( )3inZ ω
( )3in aZ ω
( )1
1
inZ R j
C
ω
ω
= −
( )3in bZ ω ( )4inZ ω
( )4in bZ ω
( )4in aZ ω
Department of Electronic Engineering, NTUT64/70
65. 頻率變高時的阻抗變化軌跡 (II)
( )2inZ ω ( )1inZ ω
( )4inZ ω ( )3inZ ω
( )1in bZ ω
( )1in aZ ω
( )2in bZ ω
( )2in aZ ω
( )4in bZ ω
( )3in bZ ω
( )3in aZ ω
( )4in aZ ω
L R
C
L R
RCRC
L
Department of Electronic Engineering, NTUT65/70
66. 固定Q軌跡 (I)
n
X x
Q
R r
= =
1nQ =
2nQ =
Short Open
Department of Electronic Engineering, NTUT66/70
67. 固定Q軌跡 (II)
Short Open
very intensive
very intensive
intensive
Department of Electronic Engineering, NTUT67/70
68. 匹配頻寬與Q值的要求 (I)
• 前面我們已經知道,在阻抗匹配時:
2
n
L
Q
Q =
• 在特定的匹配頻寬BW要求下,QL 的值應設計為: 0
1
L
f BW
Q
=
• 範例:設計一個T型匹配網路,使其能將負載阻抗 轉到50LZ = Ω
10 15inZ j= − Ω
1
0.4
LQ
=
1
2.5
0.4
LQ = =
在阻抗匹配時: 2.5
2
n
L
Q
Q = =
5nQ =所以匹配網路本身的節點Q值應該要為
0
L
f
Q
BW
=
並且能達到匹配頻寬40%的要求。
在下一頁我們會看到如何利用Smith Chart與固定Q軌跡來完成
這個匹配條件。
Department of Electronic Engineering, NTUT68/70