Chapter 17 - Two-Port and Three-Port Networks

Esta é uma pré-visualização de arquivo. Entre para ver o arquivo original

Problems 
 
Section 17-3: T-to-T1 Transformations 
 
P17.2-1 
 
 
 
 
 
P17.2-2 
 
 
 
 
1 
P17.2-3 
 
 
21 1 2 21
2 i
22 L 1 22 L
 (forward current gain)
 
z I I zI A
z R I z R
− −= ⇒ = =+ + 
( )12 i 11 11 1 12 2 12 21in 11 111 1 1 22 L
 (input resistance)
z A IV z I z I z zR z z
I I I z R
+= = = − = − + 
 
i L2
2 2 L i L 1 1 in 1 v
1 in
 and (forward voltage gain)
A RVV I R A R I V R I A
V R
= − = = ⇒ = = 
L2
p i v i
in
 
R
A A A A
R
∴ = = 
 
P17.2-4 
First, simplify the circuit using a Δ-Y transformation: 
 
 
 
1
eq || 5 || 20 4 3
R
R R= = = Ω 
Mesh equations: 
2
2
30 18 10
50 10 20
I I
I I
= −
= − 
Solving for the required current: 
30 10
50 20 100 0.385 A
18( 20) ( 10)10 260
I
−
− −= = =− − − − 
2 
P17.2-5 
 
 
 
 
3 
Section 17-3: Equations of Two-Port Networks 
 
P17.3-1 
 
 
18 6
 
6 9
⎡ ⎤= Ω⎢ ⎥⎣ ⎦Z 
 
12
11 12 11
22 21 22
 6 
12 18
3 9
Z
Z Z Z
Z Z Z
 
 
= Ω
− = Ω ⇒ = Ω
− = Ω ⇒ = Ω
 
 
 
 
 
 
2
1
1
1
11
1 0
1 2
12 21
2 20
2 2
22
2 20
1 S
14
6 1 S 
(6 12) 21
/ 7 1 S
7
V
V
V
IY
V
I IY
V V
I VY
V V
=
=
=
= =
−= = = − = Υ+
= = =
 
 
 
 
 
1 1 14 21= S
1 1 21 7
⎡ ⎤−⎢ ⎥⎢ ⎥−⎣ ⎦
Υ 
 
 
P17.3-2 
 
 
12 21
11 12 11
22 21 22
4 
2 2 4 
2 2 4
z z j
z z z j
z z j z j j j
= = − Ω
− = Ω ⇒ = − Ω
2 − = Ω ⇒ = − = − Ω
 
 
2 4 4
 
4 2 
j j
j j
− −⎡ ⎤ = Ω⎢ ⎥− −⎣ ⎦Z 
 
 
1 
P17.3-3 
 
 
 
22
1 2
11 21
1 1 00
and 
 VV
I IY Y
V V ==
= = 
 
1 2 1 2 2
1 1
1 1 2
 and I I I I IV b
G G G
V+ += + =
1
 
so 
1 1 2 1 1 2 2 1 ( ( 1) ) 1 and ( 1) 3 I G b G V V I b G V V= − − = − = − = 
Finally 
11 211 S and 3 SY Y= − = 
Next 
 
 
 
1 2 2
2 2
3 2
 and 
I I I
V V
G G
+ −= = 
 
1
1
1
12 2
2 0
2
22 2 3
2 0
 1 S
 4 S
V
V
I
Y G
V
I
Y G G
V
=
=
= = − = −
= = + =
 
 
 
P17.3-4 
 
Using Fig. 17.3-2 as shown: 
 
12 21 12 21
11 12
22 21
0.1 S or 0.1 S
 0.2 0.3 S
 0.05 0.15 S
Y Y Y Y
Y Y
Y Y
− = − = = = −
= − =
= − =
 
 
 
P17.3-5 
 
12 21
11 12
11
10 S
13.33 S
23.33 S
Y Y
Y Y
Y
μ
μ
μ
= − =
+ =
=
 
 
22 21 2220 S 30 SY Y Yμ μ+ = ⇒ = 
2 
 
P17.3-6 
 
 
2
2
1
11
1 0
2 1
21
1 1 0
 3 3 2 (3 )
2 2
I
I
VZ j j
I
V j IZ j
I I
=
=
j= = + − = + Ω
−= = = − Ω
 
 
 
1
1
1
12
2 0
2
22
2 0
 2
 2
I
I
VZ j
I
VZ j
I
=
=
= = −
= = −
Ω
Ω
 
 
 
 
P17.3-7 
 
11 21
11
21 12
2
22 21 22
Z 4
1 4 1 41Z 
1 2 1 2 2
Z
sZ
ssZ
s
sZ Z s Z s
ss
− = ⎫ +⎪ ⇒ = + =⎬− = ⎪⎭
+− = ⇒ = + =
 
 
 
 
 
 
P17.3-8 
Given: 
1 1
 = 
1 1
s
s
s
+⎡ ⎤−⎢ ⎥⎢ ⎥− +⎣ ⎦
Y 
 
Try a π circuit as shown at the right. 
 
( )
12 21
11 12 11
22 21 22
1 S
1 1 1 
 1 1
Y Y
sY Y Y
s s
Y Y s Y s s
= = −
++ = ⇒ = − =
+ = ⇒ = − − = +
1
s
 
 
 
 
 
 
 
 
 
3 
 
P17.3-9 
Given: 
2
2 2
2
2 2
2 2 1 
1 1 = 
1 1 
1 1
s s
s s s s
s
s s s s
⎡ ⎤+ +⎢ ⎥+ + + +⎢ ⎥+⎢ ⎥⎢ ⎥+ + + +⎣ ⎦
Z 
Try : 
 
 
From the circuit, we calculate: 
( ) ( )22 1 1 2 12
11 1 1 2 2
2 2
2
1
1 1 1
R L s LC R s R R C L s R RR L sC sz R R
R C s LC s LC s R C sR L s
C s
+ + + + ++= + = + =+ + + ++ +
2 
 
Comparing to the given yields: 11z
1
2
2
1
1 2
1 2
1
1 
1
1 
1
1 H
2
1 F
2
LC
R
R C
R
LC R
L
R R C L
C
R R
= ⎫ = Ω⎧⎪= ⎪⎪ = Ω⎪ ⎪= ⇒⎬ ⎨ =⎪ ⎪+ = ⎪ ⎪ =⎩+ = ⎪⎭
 
 
Then check . The are all okay. If they were not, we would have to try a different 
circuit structure.. 
12 21 22, and z z z
 
 
P17.3-10 
It is sufficient to require that the input resistance of each section of the circuit is equal to Ro, that 
is 
 
Then 
2 2(2 ) 2 3 ( 3
3
o
o o
o
R R R 1)R R R R R R R
R R
+= ⇒ = − ± + = − ± =+ R− 
 
 
 
4 
 
5 
Section 17-4: Z and Y Parameters 
 
P17.4-1 
 
1 1
2 1
1 1
1 2
1 2
1 2
( )
 and 
 
V b
i I
R R
b R R
2
1
R
V
R
I I i V
R R
+= − = −
⎛ ⎞+ += − − = ⎜ ⎟⎜ ⎟⎝ ⎠
 
 
 
2 2
1 2 11 2
11 21
1 11 2 0 
 ( )
 and 
V V
b R R b RI IY Y
V VR R= =
+ + += = = = −
1 20 R R
 
 
 
 
2 1
2
2 2 2 2
2
 
I I
VV R I I
R
= −
= ⇒ = 
 
 
1 1
2 1
22 12
2 2 2 0 
1 1 and 
V V
I IY Y
V R V= =
= = = = −
20 R
 
Finally 
1 2
1 2 2
1
1 2 2
1
1
b R R
R R R
b R
R R R
+ +⎡ ⎤−⎢ ⎥⎢ ⎥= ⎢ ⎥+⎢ ⎥−⎢ ⎥⎣ ⎦
Y 
 
 
 
P17.4-2 
 
( )1 1
2
2 1
1 3 4
0
3
v i
i
v i
⎧ = + =⎪= ⇒ ⎨ =⎪⎩
1i 
therefore 
11 214 and 3 Z Z= Ω = Ω 
 
1 
 
( )
( )
1 2 2
1
1 2 2
3
0
3 2
v i i
i
v i i
α
α
⎧ = +⎪= ⇒ ⎨ = + +⎪⎩ 2i
 
therefore 
( )12 213 1 and 5 3Z Zα α= + = + 
Finally, 
4 3(1 )
 = 
3 5 3
α
α
+⎡ ⎤⎢ ⎥+⎣ ⎦Z 
 
 
 
P17.4-3 
Treat the circuit as the parallel connection of two 2-port networks: 
 
 
The admittance matrix of the entire network 
can be obtained as the sum of the admittance 
matrices of these two 2-port networks 
 
1 0 2 1 2 
2 1 2 2 1 2
s s s s
s s s s
− + −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦Y 
 
When ( ) ( )1 :i t u t=
1 1 1
2
2 2
2 1 
1 1( ) ( ) 2 2 1
 
( ) ( ) 3 6 10 0
s s
V s V s s s
s s
V s V s s s
−
+⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤ − +⎣ ⎦⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⇒ = =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Y Y
1
0
s 
so 
2 2
( 2) 1 6 1.25 7.25( ) 
(3 6 1) 3 1.82 0.184
S
S S S S S
V s
s
⎡ ⎤− − −= = + +⎢ ⎥+ + + +⎣ ⎦ 
 
Taking the inverse Laplace transform 
 
1.82 0.184
2
1(t) = 6 1.25 7.25 t 0 
3
t tv e e− −⎡ ⎤− − + ≥⎣ ⎦ 
 
2 
 
 
P17.4-4 
 
KVL: ( )1 1 2 2 11 2 02 i v v v v− + + − = 
 
KCL: 1 1 14 2i v v v− = + 2
 
 
1 1 1
1 1 11
11 1 2
1 1 2 1 2 2
1 2 21
1
5 23 6 
2 93 6
5 2 6 15 2
3 1
i v v
i v z
ii v v
i v v i v v
i v z
i
−⎧
 
8
= − ⇒ = =⎪= − ⎫ ⎪⇒⎬ ⎨= + +⎭ ⎪
Ω
= + ⇒ = = −⎪⎩
Ω
 
 
 
 
 
KVL: 2 1 1 1
1 11 5
2 2
v v v v= + + = 13 v 
 
KCL: 22 1 25 2 51 2
v
i v v= + = + 1v 
 
2 1 1 1 12
13 12 5 18
2 1
i v v v z⎛ ⎞= + = ⇒ =⎜ ⎟⎝ ⎠ 8 Ω 
and 
 2 2 2 2 22
22 5 2.769 0.361 
13
i v v v z⎛ ⎞= + = ⇒ =⎜ ⎟⎝ ⎠ Ω 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
3 
 
 
 
P17.4-5 
 
KCL: 11 2
1
v
i i
R
+ = 
KVL: 2 2 1 10 0R i bv v− − + − =
 
Then 
( )2 1
2 1 1 1
2 1 2 1
11 1 1 R R bb bi v and i v v
R R R R
⎛ ⎞ + ++ += − = + =⎜ ⎟⎜ ⎟⎝ ⎠ 12R
 
so 
( )2 1 2
21 11
1 2 1 1 2
11 and
i i Rby y
v R v R R
1R b+ ++= = − = = 
 
 
 
KVL: 2 1 2 1 2
2
10R i v i v
R
+ = ⇒ = − 
 
KCL: ( )2 3 1 2 3 2 3
2
1v R i i R v R i
R
⎛ ⎞= + = − +⎜ ⎟⎜ ⎟⎝ ⎠ 2
 
 
Then 
1 3 2
12 2 3 2 22
2 2 2 2 3
1 1and 1
i R i
y v R i y
v R R v R R
⎛ ⎞= = − + = ⇒ = = +⎜ ⎟⎜ ⎟⎝ ⎠ 2
1 
 
Finally 
( )2 1
1 2 2
2 3
2 2
1 1
1
R R b
R R R
3
R Rb
R R R
⎡ + + − ⎤⎢ ⎥⎢ ⎥= ⎢ ⎥++⎢ ⎥−⎢ ⎥⎣ ⎦
Y 
 
4 
Section 17-5: Hybrid Transmission Parameters 
 
P17.5-1
2
1 1
2 1
2 0
34 56.8 since
5 2V
V V
4||10 34
B I V
I =
= = = Ω − = =− +
 
2
1
2 1
2 0
10 4 101.4 since 
10 10 4V
ID I
I =
+= = = = −− + I 
 
 
2
1
2 1
2 0
12 10 1.2 since
10 10 2I
VA V V
V =
= = = = + 
 
2
1
2 0
1 0.1 S
10I
IC
V =
= = = 
 
 
 
P17.5-2 
2 0V = 
so 
1 i 1 2( ||V R R R= + 1) I 
therefore 
 
2
1
11 i 1 2
1 0
|| 600 k
V
V
h R R R
I =
= = + = Ω 
KVL: 
1 i
2 1
1 2 o
R R
1I I AR R R
+ = −+ I 
therefore 
2
2 i 1 6
21
1 o 1 20
( )
V
I R R
h A
I R R R=
= = − + = −+ 10 
 
 
1 
 
1 i0 0I v A= ⇒ = ⇒ =i 0v 
so 
2
2
o 1 2( )
V
I
R R R
= + 
therefore 
1
2 o 1 2 3
22
2 o 1 20
10
( )
I
I R R R
h
V R R R
−
=
+ += = =+ 
Next, 
1
1 2
1 2
R
V V
R R
= + 
therefore 
1
1 1
12
2 1 20
1
2
I
V R
h
V R R=
= = + = 
 
 
 
 
P17.5-3 
 
Compare : 
2 1
1 2
V nV
I n I
=
=−
to 
1 11 1 12
2 21 1 22 2
 
 
V h I h V2
I h I h V
= +
= + 
 
11 22 12 21
1 1Then 0, 0, and h h h hn n= = = = − 
 
 
P17.5-4 
 
( ) 2 31 1 2 3 1 11 1
2 3
2 2
2 1 21
2 3 2 3
|| + 
 
R R
V R R R I h R
R R
R R
I I h
R R R R
= + ⇒ = +
= − ⇒ = −+ +
 
 
 
2
2 22
2 3 2 3
2 2
1 2 12
2 3 2 3
1 
V
I h
R R R R
R R
V V h
R R R
= ⇒ =+ +
= ⇒ =+ +R
 
 
2 
P17.5-5 
 
2 1
2 1
0.1 and 950 
so 95 
I v v
I I
I= =
= 
 
2
2
1
11
1 0
2
21
1 0
50 950 1000 
95
V
V
V
h
I
I
h
I
=
=
= = + = Ω
= =
 
1 0 0I v= ⇒ = 
 
1
1
1
12
2 0
2 4
22
2 0
0
10 S
I
I
V
h
V
I
h
V
=
−
=
= =
= =
 
 
 
3 
Section 17.6: Relationships between Two-Port Parameters 
 
P17.6-1 
Start with 
1 11 1 121 11 1 12 2
2 21 1 22 2 2 21 1 22 2
 
Y parameters: and H parameters: 
 
V h I h VI Y V Y V 2
I Y V Y V I h I h V
= +⎧= +⎧ ⎪⎨ ⎨= + = +⎪⎩ ⎩
 
Solve the Y parameter equations for V to put them in the same form as the H parameter 
equations. 
1 and 2I
I
 
1
11 1 12 1 1 11 12 111 1 1 12 2
21 1 2 22 2 22 2221 2 2 2 21 2
Y 0 01 1 
 
 Y 1 0 1 0 
V I V YY YY V I Y V
Y V I Y V I Y V I Y Y V
−− −− −− + = ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⇒ = ⇒ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− + = − −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
12
1
11 12 11 12 11 11
21 22 21 22 21 12 21
22
11 11 11
1 1 0 
 0 1 1 
 = 
 1 0 0 1 
Y
Y Y Y Y Y Y
Y Y Y Y Y Y
Y Y
−
Y Y
Y
⎡ ⎤ ⎡− − ⎤⎢ ⎥ ⎢− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
 ⎥⎢ ⎥ ⎢ ⎥∴ = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ ⎦− − ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣
H
 ⎦ 
 
P17.6-2 
First Δ = . Then (3)(6) (2)(2) 14− =Z
22 12
21 11
6 2 
14 14= =
2 3 14 14
Z Z
Z Z
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥Δ Δ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ Δ Δ ⎦
Z ZY
Z Z
. 
 
 
P17.6-3 
First Δ = . Then (0.1)(0.5) (0.4)(0.1) .01 S− =Y
12
11 11
21
11 11
1 
10 1
 =
4 0.1 
Y
Y Y
Y
Y Y
⎡ ⎤−⎢ ⎥ −⎡ ⎤⎢ ⎥= ⎢ ⎥Δ⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦
H
Y
. 
 
 
P17.6-4 
First Δ = . Then (0.5)(0.6) ( 0.4)( 0.4) S − − −Y
12
11 11
21
11 11
1 Y -
Y Y 2 0.8
 
Y Δ 0.8 0.28 
Y Y
⎡ ⎤⎢ ⎥ ⎡ ⎤⎢ ⎥= = ⎢ ⎥−⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦
H
Y
 
 
 
1 
Section 17.7: Interconnection of Two-Port Networks 
 
P17.7-1 
 
12 21
22 21
11 12 11
1 S3
10 S3
41 S S3
Y Y
Y Y
Y Y Y
= = −
= − =
+ = ⇒ =
 
 
a
4 1 3 3
1 1 3 3
⎡ ⎤−⎢ ⎥= ⎢ ⎥−⎣ ⎦
Y 
 
 
 
12 21
11 12 11
21 22 22
1 S
31 S S2 2
1 4 S S3 3
Y Y
Y Y Y
Y Y Y
= = −
+ = ⇒ =
+ = ⇒ =
 
 
b
3 12 
41 3
⎡ ⎤−⎢ ⎥= ⎢ ⎥−⎣ ⎦
Y 
 
a b
3 174 4( ) 3 2 3 6 3
5 54 4 3 3 3
−4
3
⎡ ⎤ ⎡ ⎤+ −⎢ ⎥ ⎢ ⎥= + = =⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
Y Y Y 
 
 
P17.7-2 
 
Admittance parameters: 
 
10 6 
44 44
6 8 
44 44
−⎡ ⎤⎢ ⎥= ⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
Y 
Transmission parameters: 
 
8 44 
6 6
1 10 
6 6
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
T 
 
 
 
 
 
 
p
20 12 
44 44
12 16 
44 44
−⎡ ⎤⎢ ⎥= + = ⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
Y Y Y 
 
1 
 
 
C
108 792 
36 36
18 144 
36 36
⎡ ⎤⎢ ⎥= ⋅ = ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
T T T 
 
 
 
 
P17.7-3 
1 2 2
a b
2 2
1 
 
1 
sC sC G G Gs L
G G GsC sC
s L
⎡ ⎤+ −⎢ ⎥ + −
3
⎡ ⎤⎢ ⎥= + = + ⎢ ⎥− +⎢ ⎥ ⎢ ⎥⎣ ⎦− +⎢ ⎥⎣ ⎦
Y Y Y 
 
 
2 
Section 17.8 How Can We Check…? 
P17.8-1 
 
 
1 2
7550 15 
175 75
V I⎛ ⎞= =⎜ ⎟+⎝ ⎠ 2I 
 
1
1
12
2 0
15 
I
VZ
I =
= = Ω 
 
 
 
1 1
1 1 0.028
50 125 1
I V V⎛ ⎞= + =⎜ ⎟⎝ ⎠ 
 
2
1
11
1 0
28 mS
V
IY
V =
= = 
 
11 24 mS, so the report is not correct.Y ≠ 
 
P17.8-2 
 
 
 
111 1
212 1
2 0.2 0.2 ( 10) (2 0.2 ) 
0.1 (0.1 )
Z s sV s I
Z sV s I
= + = += + ⎫⇒⎬ == ⎭
 
 
 
 
22 122 0.2 and 0.1Z s Z s= + = 
 
2(2 0.2 )(2 0.2 ) (0.1 )(0.1 ) 0.01(3 80 400)s s s s s sΔ = + + − = + +Z 
 
 
 
 
1 
2
11
21 21
22
21 21
2( 10) 0.1(3 80 400)
= =
1 10 2( 10)
Z s s s
Z Z s s
Z s
Z Z s s
Δ⎡ ⎤ ⎡ ⎤+ + +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ +⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
Z
T 
 
This is the transmission matrix given in the report. 
 
2 
	Ch17Sec2
	Problems 
	Section 17-3: T-to-T1 Transformations 
	P17.2-1 
	P17.2-2 
	P17.2-3 
	P17.2-4 
	P17.2-5 
	Ch17Sec3
	Section 17-3: Equations of Two-Port Networks 
	P17.3-1
	P17.3-2
	P17.3-3
	P17.3-4
	P17.3-5
	P17.3-6
	P17.3-7
	P17.3-8
	P17.3-9
	P17.3-10 
	Ch17Sec4
	Section 17-4: Z and Y Parameters 
	P17.4-1
	P17.4-2
	P17.4-3 
	P17.4-4
	P17.4-5
	Ch17Sec5
	Section 17-5: Hybrid Transmission Parameters 
	P17.5-1
	P17.5-2
	P17.5-3
	P17.5-4
	Ch17Sec6
	Section 17.6: Relationships between Two-Port Parameters 
	P17.6-1 
	P17.6-2 
	P17.6-3 
	P17.6-4 
	Ch17Sec7
	Section 17.7: Interconnection of Two-Port Networks 
	P17.7-1
	P17.7-2
	P17.7-3 
	Ch17Sec8
	Section 17.8 How Can We Check…? 
	P17.8-1 
	P17.8-2