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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 ittance 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
ittance 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 ittance y12 = Short-circuit transfer ittance from port 1 to port 2 y21 = Short-circuit transfer ittance from port 2 to port 1 y22 = Short-circuit output ittance 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 ittance 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 ittance 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 ittance
⎡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 ittance 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 adt 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 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 ratio
V (s) Z in = 1 I1 ( s ) Z out
h fe = Basic - collector current gain hoe = Output ittance
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
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 ittance 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
ittance 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 ittance y12 = Short-circuit transfer ittance from port 1 to port 2 y21 = Short-circuit transfer ittance from port 2 to port 1 y22 = Short-circuit output ittance 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 ittance 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 ittance 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 ittance
⎡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 ittance 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 adt 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 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 ratio
V (s) Z in = 1 I1 ( s ) Z out
h fe = Basic - collector current gain hoe = Output ittance
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
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