19.1 Introduction
• A two-port network is an electrical network
with two separate ports for input and output.
Two-Port Networks
Chapter 19
ch19 Two-Port Networks
19.2 Impedance Parameters, z
2
Impedance Parameters, z
z11 =
z 21 =
V1 = z11I1 + z12I 2
V2 = z 21I1 + z 22I 2
,
z12 =
2 =0
V2
I1 I
V1
I 2 I =0
1
, z 22 =
2 =0
V2
I 2 I =0
1
z11 = Open-circuit input impedance
z12 = Open-circuit transfer impedance from port 1 to port 2
⎡ V1 ⎤ ⎡ z11 z12 ⎤ ⎡ I1 ⎤
⎡ I1 ⎤
⎢V ⎥ = ⎢z
⎥ ⎢I ⎥ = [z ]⎢I ⎥
z
⎣ 2⎦
⎣ 2 ⎦ ⎣ 21 22 ⎦ ⎣ 2 ⎦
ch19 Two-Port Networks
V1
I1 I
z21 = Open-circuit transfer impedance from port 2 to port 1
z22 = Open-circuit output impedance
3
ch19 Two-Port Networks
4
Fig 19.3
Fig 19.4 & Fig 19.5
z11 =
V1
V
, z 21 = 2
I1
I1
z12 =
V1
V
, z 22 = 2
I2
I2
ch19 Two-Port Networks
5
ch19 Two-Port Networks
6
Fig 19.6
Example 19.1
• Determine the z parameters for the circuit in Fig. 19.7.
• Solution:
1
V1 = V2 , I1 = − nI 2
n
z11 =
z 21 =
z12 =
z 22 =
ch19 Two-Port Networks
7
V1
I1
V2
I1
V1
I2
V2
I2
=
(20 + 40)I1
= 60 Ω
I1
=
40I1
= 40 Ω
I1
=
40I 2
= 40 Ω
I2
=
(30 + 40)I 2
= 70 Ω
I2
I 2 =0
I 2 =0
I1 =0
I1 =0
⎡60 Ω 40 Ω⎤
Thus [z ] = ⎢
⎥
ch19 Two-Port Networks
⎣ 40 Ω 70 Ω ⎦
8
Practice Problem 19.1
Example 19.2
• Find the z parameters of the two-port network in Fig. 19.9.
• Find I1 and I2 in the circuit in Fig. 19.10.
• Solution:
V1 = 40I1 + j 20I 2
V2 = j 30I1 + 50I 2
Since V1 = 100∠0°, V2 = −10I 2
⇒ 100 = 40I1 + j 20I 2
⇒ −10I 2 = j 30I1 + 50I 2 ⇒ I1 = j 2I 2
⇒ 100 = j80I 2 + k 20I 2 ⇒ I 2 =
100
=−j
j100
Since I 2 = j 2(− j ) = 2, thus I1 = 2∠00 A, I 2 = 1∠ − 90° A
ch19 Two-Port Networks
9
Practice Problem 19.2
ch19 Two-Port Networks
10
19.3 Admittance Parameters, y
• Find I1 and I2 in the circuit in Fig. 19.11.
I1 = y11V1 + y12 V2
I 2 = y 21V1 + y 22 V2
⎡ I1 ⎤ ⎡ y11
⎢I ⎥ = ⎢ y
⎣ 2 ⎦ ⎣ 21
y12 ⎤ ⎡ V1 ⎤
⋅
y 22 ⎥⎦ ⎢⎣V2 ⎥⎦
⎡V ⎤
= [y ]⋅ ⎢ 1 ⎥
⎣V2 ⎦
ch19 Two-Port Networks
11
ch19 Two-Port Networks
12
Admittance Parameters, y
y11 =
I1
,
V1 V =0
y12 =
2
y 21 =
Fig 19.13
I1
V2 V =0
1
I
I2
, y 22 = 2
V2 V =0
V1 V =0
2
1
y11 =
V1
V
, y 21 = 2
I1
I1
y12 =
V1
V
, y 22 = 2
I2
I2
y11 = Short-circuit input admittance
y12 = Short-circuit transfer admittance from port 1 to port 2
y21 = Short-circuit transfer admittance from port 2 to port 1
y22 = Short-circuit output admittance
ch19 Two-Port Networks
13
ch19 Two-Port Networks
Example 19.3
Practice Problem 19.3
• Obtain the y parameters for the ∏ network shown in
Fig. 19.14.
• Solution:
• Obtain the y parameters for the T network shown in
Fig. 19.16.
4
I
V1 = I1 (4 || 2) = I1 ⇒ y11 = 1
3
V1
=
8
I
V2 = I 2 (8 || 2) = I 2 ⇒ y 22 = 2
5
V2
=
14
I1
= 0.75S
4
V2 =0
I1
3
2
− I1
4
2
I2
− I2 =
= 3 = −0.5S
I1 = I1 ⇒ y12 =
4
4+2
3
V1 V =0
I1
2
3
I2
= 0.625S
8
V1 =0
I2
5
4
− I1
8
4
I1
5
− I1 =
=
= −0.5S
I 2 = I 2 ⇒ y 21 =
8
8+ 2
5
V2 V =0
I1
1
5
ch19 Two-Port Networks
15
ch19 Two-Port Networks
16
Example 19.4
Example 19.4
• Determine the y parameters for the T network shown
in Fig. 19.17.
V1 − Vo
V V −0
= 2I 1 + o + o
8
2
4
V − Vo
V − Vo 3Vo
→0= 1
+
But I1 = 1
8
8
4
At node 1,
0 = V1 − Vo + 6Vo → V1 = −5Vo → I1 =
⇒ y11 =
− 5Vo − Vo
= −0.75 Vo
8
I1 − 0.75Vo
= 0.15 S
=
V1
− 5Vo
Vo − 0
+ 2I1 + I 2 = 0 → − I 2 = 0.25Vo − 1.5Vo = −1.25Vo
4
1.25Vo
I
⇒ y 21 = 2 =
= −0.25 S
V1 − 5Vo
At node 2,
ch19 Two-Port Networks
17
Example 19.4
18
Practice Problem 19.4
Similarly, we get y12 and y 21 using Fig.19.18(b).
• Determine the y parameters for the T network shown
in Fig. 19.19.
0 − Vo
V V − V2
At node 1,
= 2I 1 + o + o
8
2
4
0 − Vo
V V V − V2
But I1 =
→0=− o + o + o
8
8
2
4
→ 0 = − Vo + 4Vo + 2Vo − 2V2 → V2 = 2.5Vo
⇒ y12 =
ch19 Two-Port Networks
I1 − Vo /8
=
= −0.05 S
V2 2.5Vo
Vo − V2
+ 2I 1 + I 2 = 0
4
2V
1
→ −I 2 = 0.25Vo − (2.5)Vo − o = −0.625Vo
4
8
I 2 0.625Vo
⇒ y 22 =
=
= 0.25 S
V2
2.5Vo
At node 2,
Notice that y12 ≠ y 21 in this case, since the network isn' t reciprocal.
ch19 Two-Port Networks
19
ch19 Two-Port Networks
20
19.4 Hybrid Parameters, h
Hybrid Parameters, h
h11 =
V1
V
, h12 = 1
I1 V =0
V2 I =0
2
h 21 =
1
I2
I
, h 22 = 2
I1 V =0
V2 I =0
2
V1 = h11I1 + h12 V2
h11 = Short-circuit input impedance
I 2 = h 21I1 + h 22 V2
h12 = Open-circuit reverse voltage gain
⎡V1 ⎤ ⎡h11 h12 ⎤ ⎡ I1 ⎤
⎡I ⎤
= [h]⎢ 1 ⎥
⎢ I ⎥ = ⎢h
⎥
⎢
⎥
⎣ 2 ⎦ ⎣ 21 h 22 ⎦ ⎣V2 ⎦
⎣V2 ⎦
h21 = Short-circuit forward current gain
ch19 Two-Port Networks
h22 = Open-circuit output admittance
21
Inverse Hybrid Parameters, g
ch19 Two-Port Networks
22
Inverse Hybrid Parameters, g
g11 =
g 21 =
I1 = g11V1 + g12I 2
ch19 Two-Port Networks
1
I1
V1 I
,
g12 =
2 =0
V2
V1 I
1
, g 22 =
2 =0
I1
I 2 V =0
V2
I 2 V =0
1
g11 = Open-circuit input impedance
V2 = g 21V1 + g 22I 2
g12 = Short-circuit reverse voltage gain
⎡ I1 ⎤ ⎡ g11 g12 ⎤ ⎡V1 ⎤
⎡V ⎤
= [g ]⎢ 1 ⎥
⎢ V ⎥ = ⎢g
⎥
⎢
⎥
⎣ 2 ⎦ ⎣ 21 g 22 ⎦ ⎣ I 2 ⎦
⎣I2 ⎦
g21 = Open-circuit forward current gain
g22 = Short-circuit output admittance
23
ch19 Two-Port Networks
24
Example 19.5
Example 19.5
• Find the hybrid parameters for the two-port network
of Fig. 19.22.
From Fig.19.23(a),
V1 = I1 (2 + 3 6) = 4I1
⇒ h11 =
V1
I1
= 4Ω
V2 = 0
6
2
I1 = I1
6+3
3
I
2
⇒ h 21 = 2
=−
I1 V = 0
3
− I2 =
2
ch19 Two-Port Networks
25
Example 19.5
ch19 Two-Port Networks
26
Practice Problem 19.5
• Find the hybrid parameters for the two-port network
of Fig. 19.24.
From Fig.19.23(b),
6
2
V2 = V2
6+3
3
2
V
⇒ h12 = 1
=
V2 I =0 3
V1 =
1
Also, V2 = (3 + 6)I 2 = 9I 2
⇒ h 22 =
ch19 Two-Port Networks
I2
V2
=
I1 =0
1
S
9
27
ch19 Two-Port Networks
28
Example 19.6
Example 19.6
• Determine the Thevenin equivalent at the output port
of the circuit in Fig. 19.25.
V1 = h11I1 + h12 V2
I 2 = h 21I1 + h 22 V2
But V2 = 1, and V1 = −40I1 ,
we get
− 40I1 = h11I1 + h12
⇒ I1 =
h12
40 + h11
⇒ I 2 = h 21I1 + h 22
ch19 Two-Port Networks
29
ch19 Two-Port Networks
Example 19.6
I 2 = h 22 −
Example 19.6
h 21h12
h h − h 21h12 + h 22 40
= 11 22
h11 + 40
h11 + 40
From Fig.19.26(b), at the intput
− 60 + 40I1 + V1 = 0 ⇒ V1 = 60 − 40I1
Therefore,
Z TH
At the output, I 2 = 0
V
h11 + 40
1
= 2= =
I 2 I 2 h11h 22 − h 21h12 + h 22 40
⇒ 60 − 40I1 = h11I1 + h12 V2
or 60 = (h11 + 40)I1 + h12 V2
1000 + 40
10 × 200 × 10 + 20 + 40 × 200 × 10−6
1040
=
= 51.46 Ω
20.21
=
ch19 Two-Port Networks
3
30
and 0 = h 21I1 + h 22 V2 ⇒ I1 = −
−6
h 22
V2
h 21
⎡
⎤
h
⇒ 60 = ⎢− (h11 + 40) 22 + h12 ⎥ V2
h 21
⎣
⎦
31
ch19 Two-Port Networks
32
Example 19.6
Practice Problem 19.6
• Determine the Thevenin equivalent at the output port
of the circuit in Fig. 19.25.
⇒ VTH = V2 =
60
− (h11 + 40)h 22 / h 21 + h12
=
60h 21
h12h 21 − h11h 22 − 40h 22
=
60 × 10
= −29.69 V
− 20.21
ch19 Two-Port Networks
33
ch19 Two-Port Networks
Example 19.7
34
Example 19.7
• Find the g parameters as functions of s for the circuit
in Fig. 19.28.
In the s domain,
1 H ⇒ sL = s, 1 F ⇒
1 1
=
sC s
From Fig.19.28(a),
I1 =
V1
I
or g11 = 1
s +1
V1
=
I 2 =0
1
s +1
By voltage division,
V2 =
ch19 Two-Port Networks
35
V1
V
or g 21 = 1
s +1
V1
ch19 Two-Port Networks
=
I 2 =0
1
s +1
36
Example 19.7
Practice Problem 19.7
• Find the g parameters as functions of s for the circuit
in Fig. 19.30.
From Fig.19.29(b),
I1 = −
I
1
I 2 or g12 = 1
s +1
I2
=−
V1 = 0
1
s +1
V
⎛1
⎞
Also, V2 = I 2 ⎜ + s 1⎟ or g 22 = 2
I2
⎝s
⎠
V1 = 0
1
1
s2 + s +1
= +
=
s s + 1 s ( s + 1)
Thus,
⎡ 1
⎢
[g] = ⎢ s 1+ 1
⎢
⎢⎣ s + 1
1 ⎤
s +1 ⎥
s 2 + s + 1⎥⎥
s ( s + 1) ⎥⎦
−
ch19 Two-Port Networks
37
19.5 Transmission Parameters, T
ch19 Two-Port Networks
38
Transmission Parameters, T
A=
C=
I1
V2
, B=−
I 2 =0
,
D=−
I 2 =0
V1
I2
V2 = 0
Ι1
I2
V2 = 0
V1 = AV2 − BI 2
A = Open-circuit voltage ratio
I1 = CV2 − DI 2
B = Negative short-circuit transfer impedance
C = Open-circuit transfer admittance
⎡V1 ⎤ ⎡ A B ⎤ ⎡ V2 ⎤
⎡ V2 ⎤
⎢ I ⎥ = ⎢ C D⎥ ⎢− I ⎥ = [T]⎢− I ⎥
⎦⎣ 2 ⎦
⎣ 1⎦ ⎣
⎣ 2⎦
ch19 Two-Port Networks
V1
V2
D = Negative short-circuit current ratio
39
ch19 Two-Port Networks
40
Example 19.8
Inverse Transmission Parameters, t
V
a= 2
V1
c=
I2
V1
V
, b=− 2
I1
I =0
1
,
I1 = 0
d=−
Ι2
I1
• Find the transmission parameters for the two-port network
in Fig. 19.32
V1 = 0
V1 = 0
a = Open-circuit voltage gain
b = Negative short-circuit transfer impedance
c = Open-circuit transfer admittance
d = Negative short-circuit current gain
AD − BC = 1, ad − bc = 1
ch19 Two-Port Networks
41
ch19 Two-Port Networks
42
Example 19.8
Example 19.8
From Fig.19.33(a),
V1 = (10 + 20)I1 = 30I1 and V2 = 20I1 − 3I1 = 17I1
Thus
A=
V1
V2
=
I 2 =0
I
30I1
= 1.765, C = 1
V2
17I1
=
I 2 =0
I1
= 0.0588 S
17I1
From Fig.19.33(b),
V1 − Va Va
−
+ I2 = 0
10
20
But Va = 3I1 and I1 = (V1 − Va ) / 10,
⇒ Va = 3I1 , V1 = 13I1
17
3I1
I1 = − I 2
+ I2 = 0 ⇒
20
20
Therefore,
⇒ I1 −
D=−
ch19 Two-Port Networks
43
I1
I2
=
V2 = 0
ch19 Two-Port Networks
V
20
= 1.176, B = − 1
V2
17
=
V2 = 0
− 13I1
= 15.29 Ω
(17 / 20)I1
44
Example 19.9
Example 19.9
• The ABCD parameters of the two-port network in Fig.
20 Ω⎤
⎡ 4
19.34 are
⎢0.1 S
2 ⎥⎦
⎣
The output port is connected to a variable load for
maximum power transfer. Find RL and the maximum
power transferred.
V1 = 4V2 − 20I 2
I1 = 0.1V2 − 2I 2
At the input port, V1 = −10I1
− 10I1 = 4V2 − 20I 2
or I1 = −0.4V2 + 2I 2
⇒ 0.1V2 − 2I 2 = −0.4V2 + 2I 2 ⇒ 0.5V2 = 4I 2
Hence,
Z TH =
ch19 Two-Port Networks
45
Example 19.9
ch19 Two-Port Networks
46
19.6 Relationships Between Parameters
To find VTH , we use the circuit in Fig.19.35(b).
⎧ ⎡ V1 ⎤ ⎡ z 11 z 12 ⎤ ⎡ I1 ⎤
⎡I ⎤
= [z ]⎢ 1 ⎥
⎪⎢ ⎥ = ⎢
⎥
⎢
⎥
⎪ ⎣ V2 ⎦ ⎣ z 21 z 22 ⎦ ⎣I 2 ⎦
⎣I 2 ⎦
−1
⇒ [y ] = [z ]
⎨
⎪ ⎡ I1 ⎤ = ⎡ y 11 y 12 ⎤ ⎡ V1 ⎤ = [y ]⎡ I1 ⎤ = [z ]−1 ⎡ I1 ⎤
⎢I ⎥
⎢I ⎥
⎥⎢ ⎥
⎪ ⎢I ⎥ ⎢ y
⎣ 2⎦
⎣ 2⎦
⎩ ⎣ 2 ⎦ ⎣ 21 y 22 ⎦ ⎣ V2 ⎦
The adjoint of the [z] matrix and its determinan t are
At the output port I 2 = 0,
and the input port V1 = 50 − 10I1
⇒ 50 − 10I1 = 4V2 ⇒ I1 = 0.1V2
⇒ 50 − V2 = 4V2 ⇒ V2 = 10
⎡ z 22
⎢− z
⎣ 21
Thus, VTH = V2 = 10 V
− z 12 ⎤
, Δ z = z11z 22 - z 12 z 21
z11 ⎥⎦
⎡ z 22 − z 12 ⎤
y 12 ⎤ ⎢⎣ − z 21 z11 ⎥⎦
⎡y
⇒ [y ] = ⎢ 11
⎥=
Δz
⎣ y 21 y 22 ⎦
z
z
z
z
⇒ y 11 = 22 , y 12 = − 12 , y 21 = − 21 , y 22 = 11 ,
Δz
Δz
Δz
Δz
The equivalent circuit is shown in Fig.19.35(c).
RL = Z TH = 8 Ω
2
⎛V ⎞
100
V2
⇒ P = I RL = ⎜⎜ TH ⎟⎟ RL = TH =
= 3.125 W
4 RL 4 × 8
⎝ 2 RL ⎠
2
ch19 Two-Port Networks
V2
4
=
=8Ω
I 2 0.5
47
ch19 Two-Port Networks
48
19.6 Relationships Between
Parameters
V1 = z 11I1 + z 12 I 2
V2 = z 21I1 + z 22 I 2 → I 2 = −
1
z 21
I1 +
V2
z 22
z 22
z z −z z
z
→ V1 = 11 22 12 21 I1 + 12 V2
z 22
z 22
⎡ z 11z 22 − z 12 z 21
⎡V ⎤ ⎢
z 22
→ ⎢ 1⎥ = ⎢
z
⎣ I2 ⎦ ⎢
− 21
⎢⎣
z 22
z
Δ
⇒ h11 = z , h12 = 12 , h 21
z 22
z 22
[g ] = [h]−1
[t ] ≠ [T]−1
z 12 ⎤
z 22 ⎥ ⎡ I1 ⎤ ⎡h11 h12 ⎤ ⎡ I1 ⎤
⎥
=
1 ⎥ ⎢⎣V2 ⎥⎦ ⎢⎣h 21 h 22 ⎥⎦ ⎢⎣V2 ⎥⎦
z 22 ⎥⎦
z
1
= − 21 , h 22 =
,
z 22
z 22
ch19 Two-Port Networks
49
ch19 Two-Port Networks
Example 19.10
⎡ 10 1.5 Ω⎤
• Find [z] and [g] of a two-port network if
[T] = ⎢
4 ⎥⎦
⎣2 S
• Solution:
If A = 10, B = 1.5, C = 2, D = 4, the determinant of the matrix is
Δ T = AD − BC = 40 − 3 = 37. From Table 19.1,
37
A 10
Δ
z11 = =
= 5, z12 = T =
= 18.5
2
C 2
C
1 1
D 4
z 21 = = = 0.5, z 22 = = = 2
C 2
C 2
37
C 2
Δ
g11 = =
= 0.2, g12 = − T = −
= −3.7
10
A 10
A
1 1
B 1 .5
g 21 = =
= 0.1, g 22 = =
= 0.15
A 10
A 10
⎡ 5 18.5⎤
⎡0.2 S − 3.7 ⎤
Thus, [z ] = ⎢
Ω, [g ] = ⎢
⎥
⎥
ch19 Two-Port Networks
51
⎣ 0 .5 2 ⎦
⎣ 0.1 0.15 Ω⎦
50
Example 19.11
• Obtain the y parameters of the op amp circuit in Fig. 19.37.
Show that the circuit has no z parameters.
ch19 Two-Port Networks
52
Example 19.11
Practice Problem 19.11
Since no current can enter the input terminals of the op ams, I1 = 0,
• Obtain the z parameters of the op amp circuit in Fig. 19.38.
Show that the circuit has no y parameters.
which can be expressed in terms of V1 and V2 as
I1 = 0V1 + 0V2 , y11 = 0 = y12
Also, V2 = R3I 2 + I o ( R1 + R2 ),
But I o = V1/R1. Hence,
⇒ V2 = R3I 2 +
⇒ y 21 = −
V1 ( R1 + R2 )
( R + R2 )
V
V1 + 2
⇒ I2 = − 1
R1
R1 R3
R3
( R1 + R2 )
1
, y 22 =
R1 R3
R3
The determinant of the [y ] matrix is
Δ y = y11y 22 − y12 y 21 = 0
Since Δ y = 0, the [y ] matrix has no inverse.
ch19 Two-Port Networks
53
ch19 Two-Port Networks
Fig 19.40
19.7 Interconnection of Networks
⎧I 1a = y 11 V1a + y 12 V2a
⎨
⎩I 2a = y 21 V1a + y 22 V2a
• The series connection
⎧V1a = z 11I 1a + z 12 I 2a
⎨
⎩V2a = z 21I 1a + z 22 I 2a
⎧I 1b = y 11V1b + y 12 V2b
⎨
⎩I 2b = y 21 V1b + y 22 V2b
⎧V1b = z 11I 1b + z 12 I 2b
⎨
⎩V2b = z 21I 1b + z 22 I 2b
I 1 = I 1a = I 1b , I 2 = I 2a = I 2b
⎧V1 = V1a + V1b
⎪= ( z + z ) I + ( z + z ) I
⎪
11a
11b
1
12 a
12 b
2
⇒⎨
V
V
V
=
+
2a
2b
⎪ 2
⎪⎩= (z 21a + z 21b )I 1 + (z 22a + z 22b )I 2
z 12 ⎤ ⎡ z 11a + z 11b z 12 a + z 12b ⎤
⎡z
⇒ ⎢ 11
⎥
⎥=⎢
⎣z 21 z 22 ⎦ ⎣z 21a + z 21b z 22 a + z 22b ⎦
ch19 Two-Port Networks
54
V1 = V1a = V1b , V2 = V2a = V2b
⎧I 1 = I 1a + I 1b
⎪= (y + y )V + (y + y )V
⎪
11a
11b
1
12a
12b
2
⇒⎨
⎪I 2 = I 2a + I 2b
⎪⎩= (y 21a + y 21b )V1 + (y 22a + y 22b )V2
y 12 ⎤ ⎡ y 11a + y 11b y 12a + y 12b ⎤
⎡y
⇒ ⎢ 11
⎥
⎥=⎢
⎣y 21 y 22 ⎦ ⎣y 21a + y 21b y 22a + y 22b ⎦
[z ] = [z a ] + [z b ]
55
ch19 Two-Port Networks
[y ] = [y a ] + [y b ]
56
Fig 19.41
Example 19.12
• Evaluate V2/Vs in the circuit in Fig. 19.42.
This may be regarded as two - ports in series.
For N b ,
⎡V1a ⎤ ⎡ A a B a ⎤ ⎡ V2a ⎤ ⎡V1b ⎤ ⎡ A b B b ⎤ ⎡ V2b ⎤
⎢ I ⎥ = ⎢ C D ⎥ ⋅ ⎢ − I ⎥ , ⎢I ⎥ = ⎢ C D ⎥ ⋅ ⎢ − I ⎥
a⎦ ⎣
b⎦ ⎣
2a ⎦
2b ⎦
⎣ 1b ⎦ ⎣ b
⎣ 1a ⎦ ⎣ a
⎡V1 ⎤ ⎡V1a ⎤ ⎡ V2a ⎤ ⎡V1b ⎤ ⎡ V2b ⎤ ⎡ V2b ⎤
⇒ ⎢ ⎥ = ⎢ ⎥, ⎢
⎥=⎢
⎥ = ⎢ ⎥, ⎢
⎥
⎣ I 1 ⎦ ⎣ I 1a ⎦ ⎣− I 2b ⎦ ⎣ I 1b ⎦ ⎣− I 2b ⎦ ⎣− I 2 ⎦
⎡V ⎤ ⎡ A B a ⎤ ⎡ A b B b ⎤ ⎡ V2 ⎤
⇒ ⎢ 1⎥ = ⎢ a
⎥
⎥⋅⎢
⎥⋅⎢
⎣ I 1 ⎦ ⎣ Ca Da ⎦ ⎣ Cb Db ⎦ ⎣ − I 2 ⎦
⎡A B ⎤ ⎡A a
⇒⎢
⎥=⎢
⎣ C D⎦ ⎣ C a
Ba ⎤ ⎡A b
⋅
Da ⎥⎦ ⎢⎣ Cb
z12b = z 21b = 10 = z11b = z 22b
Thus,
[z ] = [z a ] + [z b ]
⎡12 8 ⎤ ⎡10 10⎤ ⎡22 18 ⎤
=⎢
⎥
⎥=⎢
⎥+⎢
⎣ 8 20⎦ ⎣10 10⎦ ⎣18 30⎦
But
V1 = z11I1 + z12 I 2 = 22I1 + 18I 2
Bb ⎤
⇒ [T] = [Ta ] ⋅ [Tb ]
Db ⎥⎦
ch19 Two-Port Networks
V2 = z 32 I1 + z 22 I 2 = 18I1 + 30I 2
57
Example 19.12
58
Practice Problem 19.12
• Find V2/Vs in the circuit in Fig. 19.43.
Also, at the input port V1 = Vs − 5I1
and at the output port V2 = −20I 2 ⇒ I 2 = −
ch19 Two-Port Networks
V2
20
18
V2 ⇒ Vs = 27I1 − 0.9V2
20
30
2.5
⇒ V2 = 18I1 − V2 ⇒ I1 =
V2
20
18
2.5
⇒ Vs = 27 ×
V2 − 0.9V2 = 2.85V2
18
1
V
And also, 2 =
= 0.3509
Vs 2.85
⇒ Vs − 5I1 = 22I1 −
ch19 Two-Port Networks
59
ch19 Two-Port Networks
60
Example 19.13
Example 19.13
• Find the y parameters of the two-port in Fig. 19.44.
y12 a = − j 4 = y 21a , y11a = 2 + j 4, y 22 a = 3 + j 4
⎡2 + j 4 − j 4 ⎤
or [y a ] = ⎢
⎥S
⎣ − j 4 3 + j 4⎦
and
y12b = −4 = y 21a , y11a = 4 − j 2, y 22b = 4 − j 6
−4 ⎤
⎡4 − j 2
or [y b ] = ⎢
⎥S
4
4
6
j
−
−
⎦
⎣
⎡ 6 + j 2 − 4 − j 4⎤
⇒ [y ] = [y a ] + [y b ] = ⎢
⎥S
⎣− 4 − j 4 7 − j 2 ⎦
ch19 Two-Port Networks
61
ch19 Two-Port Networks
Practice Problem 19.13
62
Example 19.14
• Find the transmission parameters for the circuit in Fig. 19.46.
• Solution:
• Find the y parameters of the two-port in Fig. 19.45.
A = 1+
R1
R ( R + R3 )
R
1
, B = R3 + 1 2
, C = , D = 1+ 3
R2
R2
R2
R2
and A a = 1 + 4 = 5, B a = 8 + 4 × 9 = 44 Ω,
Ca = 1 S, Da = 1 + 8 = 9
⎡ 5 44 Ω ⎤
⇒ [Ta ] = ⎢
9 ⎥⎦
⎣1 S
and A b = 1 + 4 = 5, Bb = 8 + 4 × 9 = 44 Ω,
Cb = 1 S, Db = 1 + 8 = 9
6 Ω⎤
⎡ 1
⇒ [Tb ] = ⎢
⎥
⎣0.5 S 4 ⎦
ch19 Two-Port Networks
63
ch19 Two-Port Networks
64
Example 19.14
Practice Problem 19.14
• Find the transmission parameters for the circuit in Fig. 19.48.
Thud , for the total network in Fig.19.46,
⎡5 44⎤ ⎡ 1 6⎤
[T] = [Ta ] ⋅ [Tb ] = ⎢
⎥
⎥⎢
⎣1 9 ⎦ ⎣0.5 4⎦
⎡5 × 1 + 44 × 0.5 5 × 6 + 44 × 4⎤
=⎢
⎥
⎣ 1× 1 + 9 × 0.5 1× 6 + 9 × 4 ⎦
⎡ 27 206 Ω⎤
=⎢
42 ⎥⎦
⎣5.5 S
Notice that Δ Ta = Δ Tb = Δ T = 1.
ch19 Two-Port Networks
65
ch19 Two-Port Networks
19.9 Applications
19.9.1 Transistor Circuits
66
Transistor Circuits
Av =
V2 ( s )
V1 ( s )
hi = h11 , hr = h12 , h f = h21 , ho = h22
Ai =
I 2 ( s)
I1 ( s )
hie = Basic input imdepance
hre = Reverse voltage feedback ratio
V (s)
Z in = 1
I1 ( s )
Z out
h fe = Basic - collector current gain
hoe = Output admittance
V (s)
= 2
I 2 ( s ) V =0
s
ch19 Two-Port Networks
67
ch19 Two-Port Networks
68
Fig 19.56
Fig 19.57
Ai =
Vb = hieI b + hre Vc
I c = h feI b + hoe Vc
Av =
ch19 Two-Port Networks
69
Transistor Amplifiers
Z in =
h fe
Ιc
=
Ι b 1 + hoe RL
− h fe RL
Vc
=
Vb hie + (hie hoe − hre h fe ) RL
ch19 Two-Port Networks
70
19.9.2 Ladder Network Synthesis
h h R
Vb
= hie − re fe L
Ib
1 + hoe Rl
Z out =
Rs + hie
( Rs + hie )hoe − hre h fe
ch19 Two-Port Networks
71
ch19 Two-Port Networks
72
Fig 19.62
Ladder Network Synthesis
I1 = y11V1 + y12 V2
H(s) =
I 2 = y21V1 + y22 V2
H(s) = −
y 21 / YL
1 + y 22 / YL
N( s) N o + N e
=
D( s ) Do + De
⎧ N o / D e , ( N = 0)
⎧ N o , ( N = 0)
e
e
⎪⎪ D / D + 1
⎪⎪ D + D
e
H(s) = ⎨ o
⇒ H( s) = ⎨ o e
Ne
N /D
⎪ e o , ( N o = 0)
⎪
, ( N o = 0)
⎪⎩1 + De / Do
⎪⎩ Do + De
⎧ N o , ( N = 0)
e
y 21 ⎪⎪ De
=⎨
YL ⎪ N e , (N = 0)
o
⎪⎩ Do
ch19 Two-Port Networks
73
ch19 Two-Port Networks
Example 19.18
D( s ) = ( s 3 + 2 s ) + (2 s 2 + 1)
1
( s + 2 s ) + (2 s 2 + 1)
1
y
− 21
3
YL
= s + 22s =
2s + 1 1 + y22
1+ 3
YL
s + 2s
3
s3 + 2s
1
ZA =
=
= sL3 + Z B
y22 2 s 2 + 1
1.5s
⇒ Z A = 0.5s + 2
2s + 1
1.5s
⇒ L3 = 0.5 H, Z B = 2
2s + 1
1 2s 2 + 1
1
=
= 1.333s +
= sC2 + YC
YB =
1.5s
1.5s
ZB
from which C2 = 1.33 F and
− y21
1
2s 2 + 1
when RL = 1, YL = 1 → H ( s ) =
⇒ y21 = − 3
, y22 = 3
1 + y22
s + 2s
s + 2s
ch19 Two-Port Networks
74
Example 19.18
• Design the LC ladder network terminated with a 1-Ω
reistor that has the normalized transfer function.
• Solution:
H(s) =
⎧ D o , ( N = 0)
e
y 22 ⎪⎪ De
=⎨
and
YL ⎪ De , (N = 0)
o
⎪⎩ Do
75
YC =
1
1
=
⇒ L1 = 1.5 H
1.5s sL1
ch19 Two-Port Networks
76