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Intro Introdu duct ctio ion n
A pair of terminals through which a current may enter or leave a network is known as a port. A port is an access to the network and consists of a pair of terminals; the current entering one terminal leaves through the other terminal so that the net current entering the port equals zero. There are several reasons why we should study two-ports and the parameters that describe them. For example, most circuits have two ports. We may apply an input signal in one port and obtain an output signal from the other port. The parameters parameters of a two-port network completely completely describes its behaviour in terms of the voltage and current at each port. Thus, knowing the parameters of a two port network permits us to describe its operation when it is connected into a larger network. Sign up to vote on this title Two-port networks are also important in modeling electronic devices and system components. Useful useful Notand For example, in electronics, two-port networks are employed to model transistors Op-amps. Other examples of electrical components modeled by two-ports are transformers and transmission lines.
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Network Theory
7.2 Admittan Admittance ce paramet parameters ers The network shown in Fig. 7.2 is assumed to be linear and contains no independent independent sources. Hence, Hence, principle of superposition can be applied to determine the current I1 , which can be written as the sum of two components, one due to V 1 and the other due to V 2 . Using this principle, we can write
I1 = y = y 11 V1 + y12 V2 Figure 7.2 A linear two-port network
where y11 and y12 are the constants of proportionality with units of Siemens. Siemens. In a similar way, we can write
I2 = y = y 21 V1 + y22 V2 Hence, the two equations that describe the two-port network are
I1 = y = y 11 V1 + y12 V2
(7.1
I2 = y = y 21 V1 + y22 V2
(7.2
Putting the above equations equations in matrix form, we get
I1 I2
=
y11 y12 y21 y22
Here Here the consta constants nts of proport proportiona ionality lity y11 y12 y21 and y22 are called y param paramete eters rs for a network network.. If y y and y are known, these parameters y
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Network Theory
7.2 Admittan Admittance ce paramet parameters ers The network shown in Fig. 7.2 is assumed to be linear and contains no independent independent sources. Hence, Hence, principle of superposition can be applied to determine the current I1 , which can be written as the sum of two components, one due to V 1 and the other due to V 2 . Using this principle, we can write
I1 = y = y 11 V1 + y12 V2 Figure 7.2 A linear two-port network
where y11 and y12 are the constants of proportionality with units of Siemens. Siemens. In a similar way, we can write
I2 = y = y 21 V1 + y22 V2 Hence, the two equations that describe the two-port network are
I1 = y = y 11 V1 + y12 V2
(7.1
I2 = y = y 21 V1 + y22 V2
(7.2
Putting the above equations equations in matrix form, we get
I1 I2
=
y11 y12 y21 y22
Here Here the consta constants nts of proport proportiona ionality lity y11 y12 y21 and y22 are called y param paramete eters rs for a network network.. If y y and y are known, these parameters y
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Two PortNetworks
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Similarly Similarly,, we obtain y12 and y22 by connecting a current source I2 to port 2 and shortcircuiting circuiting port 1 as in Fig. 7.4, finding I1 and V2 , and then calculating, calculating,
y12 =
I1 V2
y22 = V1 =0
I2 V2
V1 =0
called d the the short-circuit short-circuit transy12 is calle fer admittance and y22 is called the shortCollectively ly the circuit output admittance. Collective parameters are referred to as short-circuit short-circuit y parameters admittance parameters. Please note that y 12 = y21 only when there are no dependent sources or Op-amps within the two-port network. EXAMPLE
Figure 7.4 Determination of y12 and
y22
7.1
Determine Determine the admittance parameters parameters of the T network network shown in Fig. 7.5.
Figure 7.5
Sign up to vote on this title To find y11 and y 21 , we have to short the output terminals and connect a current source I 1 to the Useful Not useful input terminals. terminals. The circuit so obtained is shown in Fig. 7.6(a). 7.6(a). SOLUTION
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Network Theory
To find y12 and y22 , we have have to short-circui short-circuitt the input terminals terminals and connect connect a current current sourc terminals. The circuit so obtained is shown in Fig. 7.6(b). 7.6(b). I2 to the output terminals.
V2 4 2 2+ 4+2 V2 = 4 2+ 3 3V2 = 10
I2 =
Hence
y22 =
I2 V2
= V1 =0
3 S 10
Figure 7.6(b)
Employing the principle of current division, we have
I2 2 2+4 2I2 I1 = 6 1 3V2 I1 = 3 10 I1 1 y12 = = S V2 V =0 10 I1 =
Hence
1
It may be noted that, y12 = y = y 21 . Thus, in matrix form we have
I1 I2
I = YV = YV 1 5 = 1
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1 Useful V1 10 3 V2
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SOLUTION
To find y11 and y21 , short-circuit the output terminals and connect a current source I1 to the input terminals. The resulting circuit diagram is shown in Fig. 7.8(a).
V1 V1 = 1 2 1Ω 2Ω 1+2 3 I1 = V1 2 I1 3 y11 = = S V1 V =0 2 I1 =
Hence
2
Using the principle of current division,
Figure 7.8(a)
I1 1 You're Reading a Preview I2 = 1+2 1 Unlock full access with a free trial. I2 = I1 3 1 3 Download I2 = With V1Free Trial 3 2 I2 1 y21 = = S V1 2
Hence
To find y 12 and y 22 , short the input terminals and connect a current source I 2 to the output terminals. The resulting circuit diagram is shown in Fig. 7.8(b). Sign up to vote on this title
V2 I2 = 2Ω 3Ω
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Network Theory
3 5V2 5 6 1 I1 = V2 2 I1 1 y12 = = S V2 V =0 2 I1 =
Hence
1
Therefore, the equations that describe the two-port network are
3 1 I1 = V1 V2 2 2 1 5 I2 = V1 + V2 2 6
(7.3
(7.4
Putting the above equations (7.3) and (7.4) in matrix form, we get
3 2 1 2
1 2 5 6
You're Reading a Preview I1
V1 = V2
Unlock full access with a free trial.
I2 Download With Free Trial
Referring to Fig. 7.8(c), we find that I1 = 2A and V2 = 4I2 Substituting I 1 = 2A in equation (7.3), we get
3 2 = V1 4 Multiplying equation (7.4) by
4, we get
Figure 7.8(c)
Sign up to vote on this title 1 V2 2 Useful Not useful
(7.5
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It may be noted that the above equations are simply the nodal equations for the circuit shown in Fig. 7.8(c). Solving these equations, we get
3 V2 = V 2 1 3 I2 = V2 = A 4 8
and hence, EXAMPLE
7.3
Refer the network shown in the Fig. 7.9 containing a current-controlled current source. For this network, find the y parameters.
You're Reading a Preview Unlock full access with a free trial.
Figure 7.9
Download With Free Trial SOLUTION
To find y11 and y21 short the output terminals and connect a current source I1 to the input terminals. The resulting circuit diagram is as shown in Fig. 7.10(a) and it is further reduced to Fig. 7.10(b). Sign up to vote on this title
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Network Theory
Applying KCL at node A gives (Referring to Fig. 7.10(a)).
Hence
I3 + I2 = 3I1 V1 + I2 = 3I1 2 V1 + I2 = 3V1 2 5V1 = I2 2 I2 5 y21 = = S V1 2
To find y 22 and y 12 , short the input terminals and connect a current source I 2 at the outpu terminals. The resulting circuit diagram is shown in Fig. 7.10(c) and further reduced to 7.10(d).
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial
Figure 7.10(c)
I2 = I1 =
I1 = V2 2 I1
V2 Sign up to vote on this title 2
1
S
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Short-cut method : Referring to Fig. 7.9, we have KCL at node V1 :
V1 V 1 V2 + 2 2 = V1 0 5V2
I1 =
Comparing with
I1 = y 11 V1 + y12 V2 we get
y11 = 1S and y12 =
0 5S
KCL at node V2 :
V2 V2 V1 + 2 2 V 2 V 2 V1 You're Reading a Preview = 3 [V1 0 5 V2 ] + + 2 2 Unlock 5 full access with a free trial. I2 = V1 0 5 V2 2 Download With Free Trial Comparing with I2 = y 21 V1 + y22 V2 I2 = 3I1 +
we get
y21 = 2 5S and y22 = EXAMPLE
0 5S
7.4
Sign up to vote on this title Find the y parameters for the two-port network shown in Fig. 7.11. Useful Not useful
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Network Theory
Figure 7.12(a)
KCL at node V1 :
V1 V1 V I1 = a Preview + You're Reading 1 1 2 Unlock full access with a free trial. 3V1
KCL at node V :
2V = I1
(7.7
Download With Free Trial V
V1 1 2 2V
+
0
V 1
+ 2V1 = 0
2V1 + V + 2V1 = 0 V Sign= up0to vote on this title
Making use of equation (7.8) in (7.7), we get
3V
I
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(7.8
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Figure 7.12(b)
KCL at node V2 :
V2 V2 V + =aI2Preview 1You're Reading 1 2Unlock full access with a free trial. 3V2 V = I2
(7.9)
Download With Free Trial
KCL at node V :
V
V2 1
+
V 1 2 3V
or Substituting equation (7.10) in (7.9), we get
0
+0=0 V2 = 0 Sign up to vote on this title 1 Not useful (7.10) V = V2 Useful 3
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Network Theory
Making use of equation (7.12) in (7.11), we get
I1 1 = V2 2 3 I1 y12 = V2 V
Hence
1
EXAMPLE
= =0
2 S 3
7.5
Find the y parameters for the resistive network shown in Fig. 7.13.
You're Reading a Preview Unlock full access with a free trial.
Figure 7.13 Download With Free Trial SOLUTION
Converting the voltage source into an equivalent current source, we get the circuit diagram show in Fig. 7.14(a). a curren To find y 11 and y 21 , the output terminals of Fig. 7.14(a) are shorted and connect Sign up to vote on this title source I1 to the input terminals. This results in a circuit diagram as shown in Fig. 7.14(b). Useful Not useful
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Since V2 = 0, we get
Hence
V1 + V1 = I1 + 3V1 2 3 I1 = V1 2 I1 3 y11 = = S V1 V =0 2 2
KCL at node V2 :
V2 V2 V1 + 3V1 + = I 2 2 1
Since V2 = 0, we get
0 + 3V1 Hence
V1 = I2 I2 = 2V1 I1 y21 = Reading = 2Sa Preview You're V2
To find y21 and y22 , the input terminals aretrial. shorted and connect a current Unlockof fullFig. access7.14(a) with a free source I2 to the output terminals. This results in a circuit diagram as shown in Fig. 7.14(c).
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Network Theory
KCL at node V2 :
V2 V 2 0 + = I2 2 1 3 V2 = I2 2 I2 3 y22 = = S V2 2
Hence KCL at node V1 :
I1 =
V1 V 1 V2 + =0 2 1
Since V1 = 0, we get
You're Reading I1 = aVPreview 2 I1 y12 =with a = 1S Unlock full access free trial. V2
Hence EXAMPLE
7.6
Download With Free Trial
The network of Fig. 7.15 contains both a dependent current source and a dependent voltag source. Find the y parameters.
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Figure 7.16(a)
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Figure 7.16(b)
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Since V2 = 0, we get
V1 + V1 + 2V1 = I1 4V1 = I1
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Network Theory
To find y22 and y12 , refer the circuit diagram shown in Fig. 7.16(c). KCL at node V1 :
V1 V1 V2 + + 2V1 = 2V2 + I1 1 1
Since V1 = 0 , we get
V2 = 2V2 + I1 3V2 = I1 I1 y12 = V2
Hence
=
3S
V1 =0
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Download With Free Trial
Figure 7.16(c)
KCL at node V2 :
V2 V2 V1 + = 2V1 + ISign up to vote on this title 2 1 1
Since V1 = 0, we get
V2 + V2 = 0 + I2
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511
Putting the above equations in matrix from, we get
V1 V2
z11 z12 z21 z22
=
I1 I2
The z parameters are defined as follows:
z11 =
V1 I1
z12 = I2 =0
V1 I2
z21 = I1 =0
V2 I1
z22 = I2 =0
V2 I2
I1 =0
In the preceeding equations, letting I 1 or I 2 = 0 is equivalent to open-circuiting the input or output port. Hence, the z parameters are called open-circuit impedance parameters. z 11 is defined as the open-circuit input impedance, z22 is called the open-circuit output impedance, and z12 and z21 are called the open-circuit transfer impedances. If z 12 = z21 , the network is said to be reciprocal network. Also, if all the z-parameter are identical, then it is called a symmetrical network.
You're Reading a Preview
EXAMPLE
7.7
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Refer the circuit shown in Fig. 7.18. Find the z parameters of this circuit. Then compute the current in a 4Ω load if a 24 0 V sourceDownload is connected at the input port. With Free Trial
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Figure 7.18
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Network Theory
Applying KVL to the left-mesh, we get
12I1 + 6I1 = V1 V1 = 18I1 V1 z11 = I1 I
Hence
= 18Ω
2 =0
Applying KVL to the right-mesh, we get
V2 + 3 Hence
0 + 6I1 = 0 V2 = 6I1 V2 z21 = = 6Ω I1
To find z22 and z12 , the input terminals are open circuited. Also connect a voltage source to the output terminals. This results in a network as shown in Fig. 7.19(b).
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Figure 7.19(b)
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Applying KVL to the left-mesh, we get
V1 = 12 V1 = 6I2
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The terminal voltages for the network shown in Fig. 7.19(c) are
V1 = 24 0
(7.15)
V2 =
(7.16)
4I2
Figure 7.19(c)
Combining equations (7.15) and (7.16) with equations (7.13) and (7.14) yields
24 0 Reading = 18I1 +a6IPreview You're 2 0 = 6I1 + 13I2
Solving, we get EXAMPLE
7.8
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I2 =
0 73 0 A
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Determine the z parameters for the two port network shown in Fig. 7.20.
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Network Theory
By inspection, we find that I3 = Applying KVL to mesh 1, we get
V1
1
(I1
I3 ) = V1 I = V1 V1 = V1
I
1 1
1 3
I
1 1
1
(1 + Hence
z11 =
Applying KVL to the path V1
V1 I1
I2 =0
3
I
2 3
1
1+
1
V2 , we get I + V2 = 0
3 2
You're get Reading a Preview V1 , we
V1 +
Hence
I
1 1
=
2
V1 + Since I2 = 0 and I3 =
) V1 =
1
Unlock full access with a free trial. 2
V1
0 + V2 = 0 V2 = V1 (1 Download With Free Trial = (1 2) z21 =
V2 I1
2
) I
1 1
1+
= I2 =0
1
(1 1+
1
2
)
1
The circuit used for finding z12 and z22 is shown in Fig. Sign7.21(b). up to vote on this title
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Two PortNetworks
Applying KVL to the path
3
2
I + (
3 2
I + (
3 2
I2
2 2
1
) I3
I2 1+ 2 + 3 + 1+
+
1
515
V2 , we get
1
+
V2 = 0
)
Hence
= V2 1 1
= V2
1
z22 =
V2 I2
=
3
I1 =0
+ 1+ 2
+
1
Ω
1
Applying KCL at node a, we get
V1 + I3 = 2 V 1 a Preview You're Reading V1 + = I2 1
Unlock full access 1 with a free trial.
V1
+
= I2
1
Download V1 With Free 1 Trial z12 = = 1 I2 I =0 + 1
1
= EXAMPLE
1
1+
1
Sign up to vote on this title
7.9
Construct a circuit that realizes the following z parameters.
12 4
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Network Theory
Comparing the preceeding equation with
V1 = z 11 I1 + z12 I2 we get
z11 =
1
+
z12 =
3
= 4Ω
1
3
= 12
3
= 12Ω Figure 7.22(a)
= 8Ω
Applying KVL to the output loop, we get
V2 =
2 2
I +
V2 =
3 1
I + (
(I1 + I2 )
3
2
+
3
) I2
Comparing the immediate preceeding You're Reading a Preview equation with Unlock full access with a free trial.
V2 = z 21 I1 + z22 I2
Download With Free Trial
we get
z21 =
3
= 4Ω
z22 =
2
+
2
= 8
3 3
= 8Ω
Figure 7.22(b)
= 4Ω
Sign up to vote on this title Hence, the network to meet the given z parameter set is shown in Fig. 7.22(b). Useful Not useful EXAMPLE 7.10
40 10
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517
SOLUTION
We are given z11 = 40Ω z12 = 10Ω z21 = 20Ω z22 = 30Ω Since z 12 = z21 , this is not a reciprocal network. Hence, it cannot be represented only by passive elements. We shall draw a network to satisfy the following two KVL equations:
V1 = 40I1 + 10I2 V2 = 20I1 + 30I2 One possible way of representing a network that is non-reciprocal is as shown in Fig. 7.24(a).
You're Reading a Preview Unlock full access with a free trial. Figure 7.24(a)
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Now connecting the source and the load to the two-port network, we get the network as shown in Fig. 7.24(b).
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Network Theory
Solving the above mesh equations, we get
24 24II2 + I2 = 10 23 23II2 = 10 10 I2 = 23 Power Power supplied to the load
= I2
2
100 50 (23)2 = 9 45 W =
EXAMPLE 7.11
Refer the network network shown in Fig. 7.25. Find the z the z parameters parameters for the network. Take
=
4 3
Figure 7.25
SOLUTION
Sign up to vote on this title
Usefulin Fig. useful Not To find z 11 and z 21 , open-circuit open-circuit the output terminals terminals as shown 7.26(a). 7.26(a ). Also connec voltage source V1 to the input terminals.
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519
KCL at node b leads to
I1
V2
I3 = 0
(7.19)
Substituting Substituting equation (7.18) in (7.19), (7.19), we get
I1 = V2 +
V2 = 3
+
1 V2 3
4 1 + V2 3 3 V2 3 z21 = = Ω I1 I =0 5 =
Hence
2
Substituting Substituting equation (7.18) in (7.17), (7.17), we get
V1 = 4I1 + 5 = 4I 4 I1 + Therefore
z11 =
V2 3
5 3
I1
3 5
Since
V2 3 = I1 5
V1 = 5Ω I1
To obtain z22 and z12 , open-circuit the input terminals as shown in Fig. 7.26(b). Also, connect a voltage source V2 to the output terminals. terminals. Sign up to vote on this title
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Network Theory
Substituting Substituting equation (7.22) in (7.21), (7.21), we get
Hence
V2 + 3 V2 3I2 = 0 V2 (1 + 3 ) = 3I2 V2 z22 = I2
= I1 =0
3
=
1+3
4 3
3 1+3 3 = Ω 5
Substituting Substituting equation (7.22) in (7.20), (7.20), we get
V1 + 5 V2 = 3I 3 I2 Substituting V2 =
3 I2 , we get 5 3 V1 + 5 5
I2
= 3I 3 I2 V1 I2 I =3 3
Hence
z12 =
1
=3
=0
4 3 = 3
1Ω
Finally, in the matrix form, we can write
z11 z12 z =
= z21 z22
5 1 5 3 Sign up to vote on this title 3 5
Useful Not useful Please note that z12 = z 21 , since a dependent dependent source is present present in the circuit. EXAMPLE 7.12
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SOLUTION
The two port network is defined by
V1 = z = z 11 I1 + z12 I2 ; V2 = z = z 21 I1 + z22 I2 ; here, V1 = V = V I1 and V2 = I = I = I2 To find Thevenin equivalent circuit as seen from the output terminals, we have to remove the load resistance . The resulti resulting ng circuit circuit diagram diagram is shown in Fig. 7.28(a). 7.28(a).
= V2
Figure 7.28(a) I2 =0
= z21 I1
(7.23)
With I2 = 0, we get
V1 = z11 I1 V1 I1 = = z11
I1 z11
Solving for I1 , we get
I1 =
Substituting Substituting equation (7.24) into equation equation (7.23), we get
=
z11 +
z21 z11 +
(7.24)
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Network Theory
We know that,
Thus
V2 = z21 I1 + Z22 I2 z12 I2 = z21 + z22 I2 z11 + V2 z21 z12 = = z 22 I2 z11 +
The Thevenin equivalent circuit with respect to the output terminals along with load impedance is as shown in Fig. 7.28(c).
Figure 7.28(c)
EXAMPLE 7.13 (a) Find the z parameters for the two-port network shown in Fig. 7.29.
(b) Find V2 ( ) for
0 where
( ) = 50 ( )V.
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Figure 7.29
Sign up to vote on this title SOLUTION
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The Laplace transformed network with all initial conditions set to zero is as shown in Fig. 7.30(a
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(a) To find z 11 and z 21 , open-circuit the output terminals and then connect a voltage source V1 across the input terminals as shown in Fig. 7.30(b). Applying KVL to the left mesh, we get
V1 = Hence
z11 =
+ V1 I1
= + Also Hence
V2 = I1
1
I1
I2 =0
2
1
=
+1
1
V2 1 = I1 I =0 Reading aand Preview and z22 , open-circuit theYou're input terminals then connect a voltage source V 2 z21 =
2
To find z 21 across the output terminals as shown in Fig. 7.30(c). Unlock full access with a free trial.
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Figure 7.30(b)
Figure 7.30(c)
Applying KVL to the right mesh, we get
1
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Network Theory
(b)
Figure 7.30(d)
Refer the two port network shown in Fig. 7.30(d).
V1 =
I1You're = zReading 11 I1 + z12 2 aIPreview = (z11 +
) I1 + z12 I2
Unlock full access with a free trial.
= (z11 + ) I1 + z12 Download With Free Trial V2 = z21 I1 + z22 I2 V2 V2 = z21 I1 z22
and
I1 =
V2
1 z 22 V2 1+ z21
(7.25
Sign up to vote on this title Substituting equation (7.26) in equation (7.25) and simplifying, we get Useful Not useful
V2
=
(
z21 z + z22 ) (z11 +
)
z12 z21
(7.26
(7.27
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Two PortNetworks
The equation
2
+ + 2 = 0 gives 1 2
This means that,
=
7 2
1 2
( )
V2 ( ) = ( + 1)
Given
Hence
525
+
1 2
7 2
+
1 + 2
7 2
( ) = 50 ( ) 50 ( )= 50
V2 ( ) = ( + 1)
+
1 2
7 2
1 + + 2
1 2 3 = You're + Reading + a Preview + +1 1 7 + a free trial. Unlock full access with 2 2
7 2 3
1 + + 2
7 2
By performing partial fraction expansion, we get
With Trial = 25Download 25 Free 2 = 3 = 9 45 /90 25 25 9 45 /90 9 45 / 90 V2 ( ) = + + +1 1 7 1 7 + + + 2 2 2 2 1
Hence
Taking inverse Laplace transform of the above equation, we get Sign up to vote on this title
V2 ( ) = 25
Verification:
25
+ 18 9
0 5
cos(1 32 + 90 )
Useful
( )V
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Network Theory
EXAMPLE 7.14
The following measurements were made on a resistive two-port network:
Measurement 1: With port 2 open and 100V applied to port 1, the port 1 current is 1.125A an port 2 voltage is 104V.
Measurement 2: With port 1 open and 50V applied to port 2, the port 2 current is 0.3A, and th port 1 voltage is 30 V. Find the maximum power that can be delivered by this two-port network to a resistive load port 2 when port 1 is driven by an ideal voltage source of 100 Vdc. SOLUTION
z11 = z21 =
V1 I1
= I2 =0
100 = 88 89Ω 1 125
V2 104 You're Reading a Preview = = 92 44Ω I1 I =0 1 125 2
Unlock full access with a free trial.
V1 30 = = 100Ω I2 I =0 0 3 Download With Free Trial V2 50 z22 = = = 166 67Ω I2 I =0 0 3 z12 =
1
1
We know from the previous example 7.12 that,
= z22
z12 z21 Sign up to vote on this title z11 +
= 166 67
Useful
92 44 100 88 89 + 0
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527
The Thevenin equivalent circuit with respect to the output terminals with load resistance is as shown in Fig. 7.31. max
= I2 104 = 62 68 2
2
62 68
= 43 14 W
Figure 7.31
EXAMPLE 7.15
Refer the network shown in Fig. 7.32(a). Find the impedance parameters of the network.
You're Reading a Preview Unlock full access with a free trial.
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Figure 7.32(a)
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SOLUTION
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Network Theory
Referring to Fig. 7.32(b), we can write
V3 = 2 (I1 + I2 ) KVL for mesh 1:
2I1 + 2I2 + 2 (I1 + I2 ) = V1 4I1 + 4I2 = V1 KVL for mesh 2:
2 (I2 2I2
4
2V3 ) + 2 (I1 + I2 ) = V2
2 (I1 + I2 ) + 2 (I1 + I2 ) = V2 2I2
6 (I1 + I2 ) = V2 6I1 4I2 = V2 You're Reading a Preview V1 4I1 + 4I2 z11 = Unlock full= access with a free trial.= 4Ω I1 I =0 I1 I =0 2
2
V2 6I1 4I2 z21 = Download = With Free Trial = I1 I =0 I1 I =0 2
z12 = z22 =
V1 I2 V2 I2
= I1 =0
= I1 =0
4I1 + 4I2 I2
= 4Ω I1 =0
6I1 4I2 I2
=
4Ω
I1 =0 Sign up to vote on this title
EXAMPLE 7.16
6Ω
2
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Figure 7.33
The defining equations for the two-port network shown in Fig. 7.33 are:
1 V1 = V2 I1 = I2 You're Reading a Preview It is not possible to express the voltages in terms of the currents, and viceversa. Thus, the ideal Unlock full access with a free trial. transformer has no z parameters and no y parameters.
Download With Free Trial 7.4 z and y parameters by matrix partitioning For z parameters, the mesh equations are
V1 = z11 I1 + z12 I2 + V2 = z21 I1 + z22 I2 + 0= 0 = z 1 I1 + z 2 I2 +
+ z1 I + z2 I Sign up to vote on this title
+z I Useful
By matrix partitioning, the above equations can be written as
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Network Theory
The above equation can be simplified as (exact analysis not required)
V1 V2 M
NQ
1
=
M
N Q
1
I1 I2
P
P gives z parameters.
Similarly for y parameters,
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial I1 I2 M NQ
1
=
M
N Q
1
P
V1 V2
P gives y parameters. Sign up to vote on this title
EXAMPLE 7.17
Find y and z parameters for the resistive network shown in 7.34(a). the result by usin Useful Not useful Fig. Verify
Y
∆ transformation.
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V1 V2
Then
1 3 5
3 0 0 0 5
=
1 8571 0 2857 0 2857 0 4285 1
y=z
0 6 0 4
=
2 0 5
2 0 5
531
I1 I2
= [z]
0 4 2 5
Verification
You're Reading a Preview Unlock full access with a free trial. Figure 7.34(b)
Refer Fig 7.34(b), converting T of 1, Download 1’, 2 into equation, With Free Trial 1
=
1
1+1
2+1 1
2
=5
= 5 5 3 = 2 2
Therefore,
2
1 5 2 = 5 = 5 5 11 3 y = ;
Sign up to vote on this title
y = y
=
2
;
y =
Useful
13
Figure 7.34(c)
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Network Theory
SOLUTION
At node 1,
1 5V1
0 5V2 = I 1
At node 2,
0 5 V1 + V2 = I 2
3I1
In matrix form,
1 5 0 5
0 5 1
V1 V2 V1 V2
=
1 0 3 1
=
1 5 0 5
I1 I2 0 5 1
1
1 0 3 1
I1 I2
0 4 0 4 I1 3 2 1 2 I2 You're Reading0 4 a Preview 0 4 [z] = 3 2 1 2 =
Therefore
Unlock full access with a free trial.
1 5 0 5 [y] = [z] 1 = 4 0 5 Download With Free Trial
7.5 Hybrid parameters
The z and y parameters of a two-port network do not always exist. Hence, we define a third se of parameters known as hybrid parameters. In the pair of equations that define these parame terms V1 and I2 are the dependent variables. Hence, the two-port Sign upequations to vote onin this title of the hybri parameters are Useful Not useful
V1 = h11 I1 + h12 V2
(7.30
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EXAMPLE 7.19
Refer the network shown in Fig. 7.36(a). For this network, determine the h parameters.
Figure 7.36(a)
You're Reading a Preview SOLUTION
Unlockterminals full access so withthat a free To find h 11 and h 21 short-circuit the output Vtrial. 2 = 0. Also connect a current source I1 to the input port as in Fig. 7.36(b).
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Network Theory
Hence
h11 =
V1 I1
+ I1 I1
= = =
V2 =0
(1 ( (1
V2 =0
)I1 + + ) I1 ) + +
I1
KCL at node y:
I1 + I2 + I3 = 0 V 0 I1 + I2 + =0 You're Reading a Preview 1 (1 )I1 I1 + I2 + =0 Unlock full access with a + free trial. Hence
I2 h = 21 Download With Free Trial I 1 V2 =0
(1
= =
) +
(
+ +
)
vote on this title I1 =up0to To find h22 and h12 open-circuit the input port so thatSign . Also, connect a voltage sourc Useful Not useful V2 between the output terminals as shown in Fig. 7.36(c).
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Two PortNetworks
535
Since I1 = 0, we get
V1
+
V1
V1
V2
1
1
h12 =
+ V1 V2
=0 =
V2
= I1 =0
+
Applying KVL to the output mesh, we get
V2 +
( I1 + I2 ) +
I2 = 0
Since I1 = 0, we get
I2 + I2 = V2 You're Reading a 1Preview I2 h22 = = V2 full I =0 Unlock access with+a free trial. 1
EXAMPLE 7.20
Download With Free Trial
Find the hybrid parameters for the two-port network shown in Fig. 7.37(a).
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Figure 7.37(a)
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Network Theory
Referring to Fig. 7.37(b), we find that
V1 = I1 [2Ω + (8Ω 4Ω)] = I1 4 67 V1 h11 = = 4 67Ω I1 V =0
Hence
2
By using the principle of current division, we find that
I1 8 2 = I1 8+4 3 I2 2 h21 = = I1 V =0 3 I2 =
Hence
2
To obtain h12 and h22 , open-circuit the input port and connect a voltage source V2 to th output port as in Fig. 7.37(c). You're Reading a Preview Unlock full access with a free trial. Using the principle of voltage division,
8 2 V2Download = V2 With Free Trial 8+4 3 V1 2 h12 = = V2 3 V2 = (8 + 4)I2 V1 =
Hence Also
= 12I2 I2 h22 = V2 I
1
Figure 7.37(c)
=0
1 = S 12
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537
SOLUTION
Performing ∆ to Y transformation, the network shown in Fig. 7.38(a) takes the form as shown in
=
Fig. 7.38(b). Please note that since all the resistors are of same value,
1 3
∆.
Figure 7.38(b)
To find h 11 and h 21 , short-circuit the output port and connect a current source I1 to the input port as in Fig. 7.38(c).
You're Reading a Preview V1 = I1 [4Ω + (4Ω 4Ω)]
Hence
= 6I1 V1 h11 = I1
Unlock full access with a free trial.
= 6Ω Download With Free Trial V2 =0
Using the principle of current division,
I1 4 4+4 I1 I2 = 2 I2 1 h21 = = I1 V =0 2 I2 =
Hence
2
Figure 7.38(c)
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Network Theory
EXAMPLE 7.22
Determine the Thevenin equivalent circuit at the output of the circuit in Fig. 7.39(a).
Figure 7.39(a)
SOLUTION
You're Reading a Preview
To find , deactivate the voltage source and apply a 1 V voltage source at the output port, Unlock full access with a free trial. shown in Fig. 7.39(b).
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Figure 7.39(b)
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Two PortNetworks
539
Substituting equation (7.34) into equation (7.35), we get
h21 h12 h11 + h11 h22 h21 h12 + h22 = h11 + h11 + V2 1 = = = I2 I2 h11 h22 h12 h21 + h22
I2 = h22
Therefore To get
, we find open circuit voltage V2 with I2 = 0. To find
, refer the Fig. 7.39(c).
You're Reading a Preview Unlock full access with a free trial. Figure 7.39(c)
At the input port, we can write
Download With Free Trial + I1
+ V1 = 0
V1 =
I1
(7.36)
Substituting equation (7.36) into equation (7.32), we get
I1
= h 11 I1 + h12Sign V2 up to vote on this title
= (h11 + and substituting I2 = 0 in equation (7.33), we get
) I1 + h12 V2
Useful
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Network Theory
EXAMPLE 7.23
Find the input impedance of the network shown in Fig. 7.40.
Figure 7.40 SOLUTION
You're Reading a Preview
For the two-port network, we can write
Unlock full access with a free trial.
But where
V1 = h11 I1 + h12 V2
(7.39
I2 = h21 I1 + h22 V2 Download With Free Trial V2 = = I2
(7.40
(7.41
= 75 kΩ
Substituting the value of V2 in equation (7.40), we get
I2 = h21 I1 h22 I2 Sign up to vote on this title h21 I1 I2 = 1 + h22 Useful Not useful Substituting equation (7.42) in equation (7.41), we get
(7.42
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EXAMPLE 7.24
Find the voltage gain,
V2
for the network shown in Fig. 7.41.
Figure 7.41
You're Reading a Preview Unlock full access with a free trial.
SOLUTION
For the two-port network we can write,
Hence
Also
Download With Free Trial V1 = h11 I1 + h12 V2 here V1 = I2 = h21 I1 + h22 V2 here V2 = I1 = h 11 I1 + h12 V2
I1 I2
= (h11 + ) I1 + h12 V2 h12 V2 Sign up to vote on this title I1 = h11 + Useful Not useful V2 I2 = = h21 I1 + h22 V2
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Network Theory
EXAMPLE 7.25
The following dc measurements were done on the resistive network shown in Fig. 7.42(a). Measurement 1 V1 = 20 V I1 = 0 8 A V2 = 0 V I2 = 0 4 A Find the value of
Measurement 2 V1 = 35 V I1 = 1 A V2 = 15 V I2 = 0 A
for maximum power transfer.
You're Reading a Preview Unlock full access with a free trial.
Figure 7.42(a)
Download With Free Trial SOLUTION
For the two-port network shown in Fig. 7.41, we can write:
V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2
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From measurement 1:
h11 =
V1 I1
=
Useful
20 = 25Ω 0 8
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|
543
For example (7.22), Vt =
= = Z t =
= =
V g h21
− h11 h22 − Z g h22 50 × (−0.5) 0.67 × (−0.5) − 25 × 0.033 − 20 × 0.033 −25 = 13.74 Volts −1.82 h11 + Z g h11 h22 − h12 h21 + h22 Z g 25 + 20 25 × 0.033 − 0.67 × (−0.5) + 0.033 × 20 45 = 24.72 Ω 1.82 h12 h21
You're Reading a Preview For maximum power transfer, Z L = Z t = 24.72 Ω (Please note that, Z L is purely resistive). The Thevenin equivalent circuit as seen from the output terminals along with Z L is shown in Unlock full access with a free trial. Fig. 7.42(b). 2
P max = I t
Download With Free Trial
× 24.72
13.74 = 24.72 + 24.72
2
× 24.72
(13.74)2 = = 1.9 Watts 4 × 24.72
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EXAMPLE 7.26
Useful Figure 7.42(b) Not useful
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Network Theory
SOLUTION
To find h 11 and h 21 , short-circuit the output terminals so that V 2 = 0. Also connect a cur source I1 to the input port as shown in Fig. 7.44(a).
You're Reading a Preview Figure 7.44(a) Unlock full access with a free trial.
Applying KVL to the mesh on the right side, we get
1 Trial Download With Free [ I1 + I2 ] = 0 2 [I1 + I2 ] + 2
[ +
I1 +
+
2
] I1 = I2 =
Hence
h21 =
[1 + [ + 1+ I2 I1
2
1
+
I2 = 0
] I2 ] up to vote on this title 2Sign I1 2
2
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Two PortNetworks
= =
1
( + 1+ )+
2
+
2
1 +
2 (1 1+
2
545
)
2 1
2
2
To find h 22 and h 12 , open-circuit the input terminals so that I 1 = 0. Also connect a voltage source V 2 to the output port as shown in Fig. 7.44(b). The dependent current source is open, because I1 = 0.
V1 = I2 R2 V2 = 1 2 + Hence
h12 = =
V1 V2
You're Reading a Preview
I1 =0
2
1+
Unlock full access with a free trial. 2
V2
I2 = 2
h22 =
2
+
I2 V2
=
1
2
Figure 7.44(b)
= I1 =0
1+
V2 Download With Free Trial
1+
2
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7.6 Transmission parameters The transmission parameters are defined by the equations:
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Network Theory
Putting the above equations in matrix form we get
V1 I1
=
A B C D
V2 I2
Please note that in computing the transmission parameters, I2 is used rather than I2 , becaus the current is considered to be leaving the network as shown in Fig. 7.45.
These parameters are very useful in the analysis of circuits in cascade like transmission line and cables. For this reason they are called Transmission Parameters. They are also known a ABCD parameters. The parameters are determined via the following equations:
A =
V1 V2
B = I2 =0
V1 I2
C = V2 =0
I1 V2
D = I2 =0
I1 I2
V2 =0
A, B, C and D represent the open-circuit voltage ratio, the negative short-circuit transfe impedance, the open-circuit transfer admittance, and the negative short-circuit current rati You're Reading a Preview respectively. When the two-port network does not contain dependent sources, the following rela tion holds good. Unlock full access with a free trial.
AD EXAMPLE 7.27
BC = 1
Download With Free Trial
Determine the transmission parameters in the
domain for the network shown in Fig. 7.46.
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547
To find the parameters A and C, open-circuit the output port and connect a voltage source V1 at the input port. The same is shown in Fig. 7.47(b).
I1 =
Then
V1 = 1 1+
V1 +1
1 V2 = I1 1 V1 V1 = +1 +1 V1 A= = +1 V2 I =0 1 V2 = I1
V2 =
Figure 7.47(b)
2
Also
C=
I1 V2
= I2 =0
You're Reading a Preview
To find the parameters B and D, short-circuit Unlock full access with a free trial. the output port and connect a voltage source V 1 to the input port as shown in Fig. 7.47(c). Download With Free Trial Figure 7.47(c) The total impedance as seen by the source V1 is
1 Z=1+
1
1 +1
1 + 2 Sign up to vote on this title =1+ = +1 + 1 Useful Not useful V1 V1 ( + 1) (7.44) I1 = = Z ( + 2)
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Network Theory
Verification We know that for a two port network without any dependent sources,
AD ( + 1)( + 1)
BC = 1 ( + 2) = 1
EXAMPLE 7.28
Determine the ABCD parameters for the two port network shown in Fig. 7.48.
You're Reading a Preview Unlock full access with a free trial.
Figure 7.48
Download With Free Trial SOLUTION
To find the parameters A and C, open-circuit the output port as shown in Fig. 7.49(a) and conne a voltage source V1 to the input port. Applying KVL to the output mesh, we get
V2 + mI1 + 0 V2 = I1 (m + I1
+ I1
=0
) 1
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Figure 7.49(b)
Applying KVL to the right-mesh, we get
mI1 + (m +
Hence
I2 +
(I1 + I2 ) = 0
) I1 =
( + ) I2 ( + ) I1 = You're Reading I2 a Preview (m + ) Unlock full access I1 ( with+a free )trial. D= = I2 V =0 (m + ) Download With Free Trial 2
Applying KVL to the left-mesh, we get
V1 +
I1 +
V1 = (
+
=(
+
=
(I1 + I2 ) = 0 ) I1 + I2 ( + Sign ) up to vote on this title ) I2 + I2 (m + ) Useful Not useful + + m I2 m+
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Network Theory
SOLUTION
For the two port network, we can write
V1 = AV2 From I1 = 0 (port 1 open):
1
From V1 = 0 (Port 1 short):
BI2
I1 = CV2 DI2 10 3 = A 10 B 0= A
5
B
200 80
10 10
6
6
Solving simultaneously yields,
4 10 4 B = 25Ω 0 = C 10 D 200 10 10 6 = C 5 D 80 10 6 A=
From I1 = 0: From V1 = 0:
0 5
6
Solving simultaneously yields,
You're Reading a Preview
C = 10 7 SwithD = 0 025 Unlock5full access a free trial. In summary,
A = 4 10 4 Download With Free Trial B = 25Ω C=
5
10
D=
0 025
7
S
EXAMPLE 7.30
Find the transmission parameters for the network shown in Fig. Sign up7.50. to vote on this title
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551
Figure 7.51(a)
Applying KVL to the input loop, we get
V1 = 1 5 Also KCL at node gives
103 I1 + 10
3
V2
You're Reading a Preview V2 40I =0 1 +full access with Unlock 40 103 a free trial. V2 I1 = = 6 25 10 6 V2 3 Download 160 10 With Free Trial
Substituting the value of I1 in the preceeding loop equation, we get
V1 = 1 5
103
V1 = Hence
6 25 10 6 V2 + 10 10 3 V2 + 10 3 V2
V2
9 375 = 8 375 10 3 V2 Sign up to vote on this title V1 3 A= = 8 375 10 Useful Not useful V2 I =0 2
Also
3
C=
I1
=
6 25
10
6
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Network Theory
Applying KCL at node b, we find
40I1 + 0 = I2 I2 = 40I1 I1 D= I2
Hence
= V2 =0
1 40
Applying KVL to the input loop, we get
Hence
V1 = 1 5
103 I1
V1 = 1 5
103
I2 40
V1 I2 V =0 1 5a Preview You're Reading = 103 40 Unlock full access with a free trial. = 37 5Ω B=
2
Download With Free Trial EXAMPLE 7.31
Find the Thevenin equivalent circuit as seen from the output port using the transmission param ters for the network shown in Fig. 7.52. Sign up to vote on this title
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Figure 7.53(a)
Refer the network shown in Fig. 7.53(a) to find At the input port,
.
You're I1 Reading = V1 a Preview Also
I2 = 0 with a free trial. Unlock full access
(7.48)
(7.49)
Making use of equations (7.48) and (7.49) in equations (7.46) and (7.47) we get,
Download With Free Trial I1
and
= AV2
(7.50)
I1 = CV2
(7.51)
Making use of equation (7.51) in (7.50), we get
CV2
= AV2
V2 =
=
A+C
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Network Theory
Referring Fig. 7.53(b), we can write
V1 =
I1
Substituting the value of V1 in equation (7.46), we get
I1
= AV2 BI2 A B V2 + I2 I1 =
(7.52
Equating equations (7.47) and (7.52) results
CV2 V2 C +
Hence
DI2 =
A
V2 +
B
I2
A
B = D+ I2 You're Reading a Preview B D+ Unlock full with a free trial. Vaccess 2 = = A I2 C+ Download With Free Trial B+D = A+C
Figure 7.54
Hence, the Thevenin equivalent circuit as seen from the output port is as shown in Fig. 7.54 EXAMPLE 7.32
For the network shown in Fig. 7.55(a), find power transferred.
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Two PortNetworks
555
SOLUTION
From the previous example 7.31,
= =
B+D A+C A+C
20 + 3 20 20 + 60 = = 6 .67Ω 4 + 0 4 20 4+8 100 100 = = = 8 .33V 4 + 0 4 20 12 =
For maximum power transfer, = = 6 67Ω (purely resistive) Hence, the Thevenin equivalent circuit as seen from output terminals along with is as shown in Fig. 7.55(b).
8 33 = 0 624A 6 67 + 6 67 You're Reading a Preview 2 )max = 6 67 =
(
= (0 624)2 6 67 = 2.6Watts
Unlock full access with a free trial.
Figure 7.55(b)
Download With Free Trial
EXAMPLE 7.33
Refer the bridge circuit shown in Fig. 7.56. Find the transmission parameters.
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Network Theory
Figure 7.57(a)
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial Figure 7.57(b)
To find the parameters A and D, open the output port and connect a voltage source V 1 at th input port as shown is Fig. 7.57(c). Sign up to vote on this title Applying KVL to the input loop we get
I1 + 4I1 = V 1
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Two PortNetworks
557
I1 4 4 = I1 4+1 5 I1 5 D= = I2 v =0 4 I2 =
Hence
2
Applying KVL to the input loop, we get Figure 7.57(d)
V1 + 1 Substituting I1 =
I1 + 4
(I1 + I2 ) = 0
5 I2 in the preceeding equation, we get 4 V1
Hence
5 5 I2 + 4 I2 + I2 = 0 4 4 5 V1 I2 5I2 + 4I2 = 0 You're Reading a Preview 4 4V1 = 9I2 Unlock full access with a free trial. V1 9 B = = Ω I2 v =0 4 Download With Free Trial 2
Verification: For a two port network which does not contain any dependent sources, we have
AD 5 4
7.7
5 4
1 4
BC = 1 9 25 = 4 16
9 = 1Sign up to vote on this title 16
Relations between two-port parameters
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Network Theory
Substituting equation (7.54) into equation (7.53), we get
V = zyV adj(y) ∆y ∆y = y y y y
Hence
z=y
where
1
11
= 22
21
12
This means that we can obtain z matrix by inverting y matrix. It is quite possible that two-port network has a y matrix or a z matrix, but not both. Next let us proceed to find z parameters in terms of ABCD parameters. The ABCD parameters of a two-port network are defined by
V1 = AV2
BI2
I1 = CV2 DI2 1 V2 = (I1+ DI2 ) C 1 D You're Reading a Preview V2 = I1 + I2 C C Unlock full access with a free trial. I1 DI 2 V1 = A + BI2 C C Download Free AIWith ADTrial 1 = + B I2 C C
(7.55
(7.56
Comparing equations (7.56) and (7.55) with
V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2
respectively, we find that
z11 =
A C
Sign up to vote on this title
z12 =
AD
BC C
z21 =
Useful
1 C
Not useful z22 =
D C
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Two PortNetworks
559
Comparing equations (7.59b) and (7.59a) with
V1 = h11 I1 + h12 V2 we get,
where
I2 = h21 I1 + h22 V2 z11 z22 z12 z21 ∆z h11 = = z22 Z22 z12 z21 h12 = h21 = z22 z22 ∆z = z11 z22 z12 z21
h22 =
1 z22
Finally, let us derive the relationship between y parameters and ABCD parameters.
I1 = y11 V1 + y12 V2
(7.60)
I2 = y21 V1 + y22 V2
(7.61)
From equation (7.61), we can write You're Reading a Preview
V1 =
I2 full y22access with a free trial. Unlock
V2 y21 y21 y22 1 = Download V2 + WithI2Free Trial y21 y21
(7.62)
Substituting equation (7.62) in equation (7.60), we get
y11 y22 y 11 V2 + y12 V2 + I2 y21 y21 Sign up to vote on this title y 11 ∆y (7.63) = V2 + I2 Useful Not useful y21 y21
I1 =
Comparing equations (7.62) and (7.63) with the following equations,
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Network Theory
Table 7.1 Parameter relationships z
y
z11 z12 z
z21 z22
y
z22 ∆z z21 ∆z
z12 ∆z z11 ∆z
T
z11 z21 1 z21
h
∆z z22 z21 z22
∆z = z 11 z22 EXAMPLE 7.34
∆z z21 z22 z21 z12 z22 1 z22
y22 ∆y y21 ∆y
T
y12 ∆y y11 ∆y
y11 y12 y21 y22
A C 1 C D B 1 B
h
∆h h22 h21 h22
∆T C D C ∆T B A B
y22 1 A B y21 Reading y21 a Preview You're ∆y y11 C D Unlock full access with a free trial. y21 y21 1 y12 B Download With Free Trial y11 y11 D y21 ∆y 1 y11 y11 D
z12 z21 ∆y = y 11 y22
∆h h21 h22 h21
∆T D C D
y12 y21 ∆h = h11 h22
1 h11 h21 h11
Useful
h22 h11 ∆h h11 h11 h21 1 h21
h11 h12 h21 h22
h12 h21 ∆T = AD
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h12 h22 1 h22
Not useful
Determine the y parameters for a two-port network if the z parameters are
BC
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Two PortNetworks
561
EXAMPLE 7.35
Following are the hybrid parameters for a network:
h11 h12 h21 h22
=
5 2 3 6
Determine the y parameters for the network. SOLUTION
∆h = 5
6 3 2 = 24 1 1 y11 = = S h11 5 h22 6 y12 = = S h11 5 h21 3 y21 = = S h11 5 ∆h Reading 24 a Preview y22 = You're = S h11 5 Unlock full access with a free trial.
Reinforcement problems R.P
7.1
Download With Free Trial
The network of Fig. R.P. 7.1 contains both a dependent current source and a dependent voltage source. Determine y and z parameters.
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Network Theory
At node ,
V1 = V 2
2V1
I2
V2 2
Simplifying, the nodal equations, we get
3 V2 2 3 3V1 + V2 2
I1 + I2 = V1 I2 = In matrix form,
3 1 1 I1 V1 2 = I2 V2 0 1 3 3 2 3 You're Reading a Preview 1 1 I1 1 1 2 = with a free trial. Unlock full access I2 0 1 3 3 2 Download With4Free Trial 3 3 y= 3 2 1
Therefore
and
R.P
Z=
7.2
1 3
V1 V2
3 0 5 1 3 4 2 = 1 3 Sign 4 up to vote on this3title
Useful
Compute y parameters for the network shown in Fig. R.P. 7.2.
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Two PortNetworks
563
SOLUTION
The circuit shall be transformed into -domain and then we shall use the matrix partitioning method to solve the problem. From Fig 7.2, Node equations in matrix form,
I1 I2
= =
=
y= R.P
P
Q N +3
1
M
V1 V2 1 5
V1 V2
2 2 1 1 +2 4 2 +3 V1 5 5 You're Reading a Preview V2 +2 2 1 5 5 Unlock full access with a free trial. + 2 2 ( + 0 4) ( + 0 4) + 1 8 Download With Free Trial
7.3
Determine for the circuit shown in Fig. R.P. 7.3(a): (a)
1,
2,
3 and
(b) Repeat the problem if the current source is connected across left.
3 with
in terms of y parameters. the arrow pointing to the
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Network Theory
Then from equations (7.64) and (7.65),
y11 =
1
; y12 = y22 = 3;
+
3
y21 =
3 2
+
3
Solving, 3
=
y12 ;
2
= y22 + y12 ;
1
= y 11 + y12 = y 21
y12
(b) Making the changes as given in the problem, we get the circuit shown in Fig R.P. 7.3(b).
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial Figure R.P. 7.3(b)
Node equations : At node 1
I1 = =(
1
V1 + (V1 1
+
3
V2 )
)V1 Sign3up V2to vote on this title
At node 2,
I2 = V2
2
+ (V2
V1
3
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V1 ) 3 +
V1
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(7.66
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Two PortNetworks
R.P
565
7.4
Complete the table given as part of Fig. R.P. 7.4. Also find the values for y parameters. Table Sl.no 1
V1 50
V2 100
I1 1
I2 27
2
100
50
7
24
3
200
0
4
20
0
5
10
30
50 100 100 50 Sign up to vote on this title
Figure R.P. 7.4
SOLUTION
From article 7.2,
I1 I2
y11 Reading y12 1 aVPreview = You're y21 y22 V2 Unlock full access with a free trial.
Substituting the values from rows 1 and 2,
1 7 27 24 Post multiplying by [V]
1
=
Download y11 y12With Free 50 Trial 100 y21 y22 100 50
,
y11 y12 y21 y22
=
1 7 27 24
=
0 1 0 14
1
0 06 0 2
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Network Theory
R.P
7.5
Find the condition on
and
for reciprocity for the network shown in Fig. R.P. 7.5.
Figure R.P. 7.5
SOLUTION
The loop equations are
You're Reading a Preview Unlock full access with a free trial. V V2 = 3(I1 + I3 ) 1
V2 = (I2 I3 ) I1 Download With Free Trial I3 = V2 (V1 V2 ) = (1 + )V2
(7.68
V1
(7.69
(7.70
Substituting equation (7.70) in equations (7.68) and (7.69),
V1
V2 = 3I1 + 3(1 + )V2
3V1
up to vote on this title 4V1 = (3 + 4 )VSign 2 + 3I1 Not useful )Useful V2 = I2 [(1 + V2 V I1 1]
V1 + (2 + ) V2 =
I1 + I2
(7.71
(7.72
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Two PortNetworks
R.P
567
7.6
For what value of is the circuit reciprocal? Also find h parameters.
Figure R.P. 7.6
You're Reading a Preview Unlock full access with a free trial. SOLUTION
Download With Free Trial
The node equations are
0 5 V1
V1
I1 = V 2
V2 = (I1 + I2 )2 + I1 0 5 0
h= =
1 ∆
1
0 2 2 0 0 0 5
2
2
1 1 2 + Sign up 1 to vote on this title 1 Useful 1 Not useful 2+ 1
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Network Theory
R.P
7.7
Find y 12 and y21 for the network shown in Fig. R.P. 7.7 for
= 10. What is the value of for
network to be reciprocal?
Figure R.P. 7.7 SOLUTION
Equations for I1 and I2 are
V1 0 01V2 You're I1 = Reading a Preview 5 = full 0 2V 0 002V 1 Unlock access with a free2trial. V2 V 2 0 01 V2 I2 = + I1 + 20 With Free Trial 50 Download
(7.73
(7.74
Substituting the value of I1 from equation (7.73) in equation (7.74), we get
I2 = (0 2 V1
0 002V2 ) +
Simplifying the above equation with
V2 V2 + 20
0 01V2 20
= 10,
I2 = 2V1 + 0 0498V2
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From equation (7.73), and from equation (7.75), For reciprocity
(7.75
y12 =
0 002 Useful
y21 = 0 2
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(7.76
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Two PortNetworks
569
SOLUTION
Network equations are
V1
10I1 = V 2 V2
I1
1 5 V1
1 5V1 + I2 25
(7.77)
V2 =0 20
(7.78)
Simplifying,
2 5V1
10I1 = V2
0 06 V1 + I1 = 0 09 V2
I2
In matrix form,
2 5 0 06
10 1
Unlock full access with a free trial.
Therefore
T = R.P
V1 V 1 0 You're Reading a Preview 2 = I1 0 09 1 I2
2 5 0 06
10 1
1
1 0 0 613 3 23 =Free Trial Download With 0 09 1 0 053 0 806
7.9
(a) Find T parameters for the active two port network shown in Fig. R.P. 7.9. (b) Find new T parameters if a 20 Ω resistor is connected across the output. Sign up to vote on this title
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Network Theory
Simplying the equations (7.79) and (7.80),we get
V1 14I1 = V2 5I2 V1 + 20I1 = 50I2
and
Therefore,
T =
1 1
14 20
1
1 5 0 50
3 33 133 33 0 167 9 17
=
(b) Treating 20 Ω across the output as a second T network for which
T = Then new T-parameters,
T =
1 1 20
0
1
You're Reading a Preview 1 a free 0 trial. full 3 33Unlock 133 33access with 10 133 33 = 1 0 167 9 17 0 625 9 17 1 20 Download With Free Trial
R.P 7.10
Obtain z parameters for the network shown in Fig. R.P. 7.10.
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Two PortNetworks
|
571
Putting in matrix form,
1 1
2 −8 3
V1
V2
=
I1
10
0
0
−10
I2
Therefore, z =
1 1
2 −8 3
−1
10
0
0
−10
=
5.71 2.143
−4.286 2.143
R.P 7.11
Obtain z and y parameters for the network shown in Fig. R.P. 7.11.
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial
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Figure R.P. 7.11
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Network Theory
By matrix partitioning,
+2 z=
2 +2
=
2 +2
=
=
2 +4 2 2 +4 2 2
1 1 2
2 3 +4
1
2 3 +4
1 2
1
1 2
1 2 1 4
2 3 +4
3 +4 +4 8 2 3 +4 3 +4 You're Reading a Preview 2 2 + 10 + 8 +6 +8 (3 + 4) full access (3 + 4)a free trial. Unlock with 2
2
2 +6 +8 +8 +8 Download With Free Trial (3 + 4) (3 + 4)
R.P 7.12
Find z and y parameters at shown in Fig. R.P. 7.12.
= 108 rad/sec for the transistor high frequency equivalent circu Sign up to vote on this title
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Sequential Circuits Analysis
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Two PortNetworks
|
573
Simplifying the above equations, I1 = 10
4
I2 = 10
4
−
[(0.1 + j 6)V1
2
[(100 + 1) + (1 + ) ] 0 1+ 6 − 1 × 10 = −
Therefore,
− j 1V ]
j
.
I1 I2
V1
j
j
4
100
− j 1 1 + j 1 × 5 × 10 = 5 × 10 × 10 = 10 12
−
ωC 2 = 108
12
∆ = 10
8
= 10
8
−
−
[(0.1 + j 6)(1 + j ) + (100 ◦
6/89 You're Reading − 1 a Preview × 10 = √ 100/−0 6 2/45 √ 2/45 Download With 1 Free Trial × 10 ÷ ∆ = 100/−180 6 6/89 √ 2/45 1 × 10 × 10610213 /92 64 = 100/−180 6 6/89 133 15 /−47 64 94 16 /−2 64 = j
z = y
4
−
Unlock . full access with a free trial. ◦
j
◦
1
− j 1)( j 1)]
× 106.213/92.64 ◦
−
4
−
◦
y
V2
4
−
−
Then,
V1
−
ωC 1 = 108
Therefore,
j V2
4
−
.
◦
◦
j
◦
.
.
◦
.
16/86
4
−
◦
◦
.
565/
8 . . Sign up to vote on this title −
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574
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Sequential Circuits Analysis
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Network Theory
SOLUTION
Using the equation for T-parameters for a -network
A =
1
+
3
3
We have for
For
For Overall T :
;
B =
1 3
3
;
C=
1
;
D =
2
+
3
3
3
7 2 5 T = 1 1 5 9 54 6 6 T = 1 10 6 6 1 0 T = 1 1 7 You're Reading a Preview 4 709 15 93 962 3 46
T= ][Twith ][T ] =trial. Unlock full[T access a free 0
This dervation is left as an exercise to the reader.
Download With Free Trial
Verification: Using T ∆ transformation, that is changing T (3, 4, 6 ) of Fig. R.P. 7.13, in to ∆,
= 13 5 =9 = 18
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Two PortNetworks
575
Reducing, the above circuit, we get the circuit shown in Fig. R.P. 7.13c.
Figure R.P. 7.13(c)
Converting the circuit into T, we get the circuit shown in Fig. R.P. 7.13(d). Now from Fig. R.P. 7.13(d),
3 8564 + 1 0396 = 4 709 1 0396 You're Reading a Preview 1 0396(3 8564 + 2 5644) + 3 8564 2 5644 B= Unlock full access with a free trial. 1 0396 = 15 93 Ω 1 1 = = 0 962 Download With Free Trial C= 1 0396 2 5644 + 1 0396 = 3 46 D= 1 0396 A=
Exercise Problems
E.P
7.1
Find the parameters for the network shown in Fig. E.P. 7.1.
Figure R.P. 7.13(d)
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576
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Network Theory
E.P
7.2
Find the z parameters for the network shown in Fig. E.P. 7.2.
Figure E.P. 7.2
Ans: E.P
z11 =
13 Ω 7
z12 =
2 Ω 7
z21 =
2 Ω 7
z22 =
3 Ω 7
7.3
Find the h parameters for the network shown in Fig. E.P. 7.3.
You're Reading a Preview
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Figure E.P. 7.3
Ans:
h11 =
+ R + (1 sC R + 1
sC R R sC R
m)R
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h21 =
Not useful
sC R
sC R sC
+m . +1
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Two PortNetworks
E.P
577
7.5
Find the y parameters for the network shown in Fig. E.P. 7.5. Give the result in domain.
Figure E.P. 7.5
Ans: E.P
y11
=
y22
=
2s(2s + 1) 4s + 1
y12
=
y21
=
4s2 . 4s + 1
7.6
You'reinReading Preview Obtain the y parameters for the network shown Fig. E.P.a7.6. Unlock full access with a free trial.
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Figure E.P. 7.6
Ans: E.P
y11
=
0 625 S
y12
=
0 125 S
y21
=
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0 375 S
Useful
y22
=
0 125 S
Not useful
7.7
Find the z parameters for the two-port network shown in Fig. E.P. 7.7. Keep the result in domain.
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578
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Network Theory
E.P
7.8
Find the h parameters for the network shown in Fig. E.P. 7.8. Keep the result in domain.
Figure E.P. 7.8 Ans:
h11
E.P
=
5s + 4 2(s + 2)
h12
s + 4 (s + 4) You're Reading a Preview h21 2(s + 2) 2(s + 2)
=
=
Unlock full access with a free trial.
h22
s =
2(s + 2)
7.9
With Free Trial Find the transmission parametersDownload for the network shown in Fig. E.P. 7.9. Keep the result in domain.
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Two PortNetworks
579
E.P 7.11
Select the values of
,
in the circuit shown in Fig. E.P. 7.11 so that A =1, B = 34 Ω,
, and
C = 20 mS and D = 1.4.
Figure E.P. 7.11
Ans:
R
= 10Ω
R
E.P 7.12
Find the
= 20Ω
R
= 50Ω
You're Reading a Preview
domain expression for the h parameters of the circuit in E.P. 7.12. Unlock full access with a free trial.
Download With Free Trial
Figure E.P. 7.12
1 Ans:
C
h11 = s2
+
1
s
1 LC
LC
h12 = h 21 = s2
+
1 LC
2 2 Sign upCs to vote +this title s on
h22Useful =
useful Not LC
s2
+
1
LC
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Network Theory
E.P 7.14
Find the two-port parameters h 12 and y12 for the network shown in Fig. E.P. 7.14.
Figure E.P. 7.14
Ans:
h12 = 1 2
E.P 7.15
y12 = 0 24S
You're Reading a Preview
Find the ABCD parameters for the 4 Ω resistor of Fig. E.P. 7.15. Also show that the ABCD 4 parameters for a single 16Ω resistor canfull beaccess obtained Unlock withby (ABCD) a free trial.
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Figure E.P. 7.15 Sign up to vote on this title
Useful Not useful parameters. Ans: Verify your answer using the relation between the E.P 7.16
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Two PortNetworks
581
E.P 7.17
Find y21 for the network shown in Fig. E.P. 7.17.
Ans:
y21 =
s
Figure E.P. 7.17
4s + 1
E.P 7.18
Determine the y-parameters for the network shown in Fig. E.P. 7.18.
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial Figure E.P. 7.18
Ans:
y11 =
s
3
+ s2 + 2s + 1 s(s2 + 2)
1 y12 = y 21 = s(s2 + 2)
E.P 7.19
Obtain the h-parameters for the network shown in Fig. 7.19.
y22 =
s
3
+ s2 + 2s + 1 s(s2 + 2)
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Network Theory
E.P 7.21
Find the network shown in figure, determine the z and y parameters.
Figure E.P. 7.21
Ans:
y11 = 4 z11 = 1Ω
y22 = 3 4 z22 = Ω 3
y12 = y 21 =
3,
z12 = z 21 = 1Ω
E.P 7.22
You're Reading Preview Determine the z, y and Transmission parameters ofathe network shown in Fig. 7.22. Unlock full access with a free trial.
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Figure E.P. 7.22
Ans:
3 , 55 z11 = 20Ω, y11 =
A = 55Ω
1 , 55 z12 = z 21 = 5Ω y12 = y 21 = B = 55Ω
4 y22 =Sign up , 55 to vote on this title z22 = 15Ω Useful Not useful C = 0 .2,
D = 3 .
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