ABSTRACT
This chapter demonstrates how the various functions can be derived, but first we introduce some explanations and definitions. If we analyze any linear network, we can take as output any nodal voltage, or a difference of any two nodal voltages; denote such as output voltage by
V
. We can also take as the output a current through any element of the network; we call it output current,
I
. If the network is excited by a voltage source,
E
, then we can also calculate the current delivered into the network by this source; this is the input current,
I
. If the network is excited by a current source,
J
, then the voltage across the current source is the input voltage,
V
. Suppose that we analyze the network and keep the letter
E
or
J
in our derivations. Then, we can define the following network functions:
(21.1)
Voltage transfer function,
Input admittance,
Transfer admittance,
T V
E
Y I
E
Y I
E
=
=
=
Output impedance or output admittance are also used, but the concept is equivalent to the input impedance or admittance. The only difference is that, for calculations, the source is placed temporarily at a point from which the output normally will be taken. In the Laplace transform, it is common to use capital letters,
V
for voltages and
I
for currents. We also deal with impedances,
Z
, and admittances,
Y
. Their relationships are
The impedance of a capacitor is
Z
= 1/
sC
, the impedance of an inductor is
Z
=
sL,
and the impedance of a resistor is
R
. The inverse of these values are admittances:
Y
=
sC
,
Y
= 1/
sL,
and the admittance of a resistor is
G
= 1/
R
. To demonstrate the derivations of the above func-
tions two examples are used. Consider the network in Fig. 21.1, with input delivered by the voltage source,
E
. By Kirchhoff ’s current law (KCL), the sum of currents flowing
away
from node 1 must be zero:
Similarly, the sum of currents flowing away from node 2 is
The independent source is denoted by the letter
E
, and is assumed to be known. In mathematics, we transfer known quantities to the right. Doing so and collecting the equations into one matrix equation results in
If numerical values from the figure are used, this system simplifies to
or
Any method can be used to solve this system, but for the sake of explanation it is advantageous to use Cramer’s rule. First, find the determinant of the matrix,
Current transfer function,
Input impedance,
Tran
T I
J
Z V
J
Z V
J
=
=
=
V ZI I YV= =
G sC G V G V EG1 1 2 1 2 2 1 0+ + − − =( )
− + + +( ) =V G G sC G V1 2 2 2 3 2 0
G G sC G
G G G sC
V
V
+ + −
− + +
=
s
s
V
V
E+ −
− +
=
3 2
2 2 5 0
YV E=
To obtain the solution for the variable
V
(
V
), replace the first (second) column of
Y
by the right-hand side and calculate the determinant of such a modified matrix. Denoting such a determinant by the letter
N
with an appropriate subscript, evaluate
Then,
Now, divide the equation by
E
, which results in the voltage transfer function
To find the nodal voltage
V
, replace the second column by the elements of the vector on the right-hand side:
The voltage is
and another voltage transfer function of the same network is
Note that many network functions can be defined for any network. For instance, we may wish to calculate the currents
I
or
I
, marked in Fig. 21.1. Because the voltages are already known, they are used: For the output current
I
=
G
V
and divided by
E
The input current
I
=
E
–
G
V
=
E
–
V
=
E
(2
s
+ 9
s
+ 6)/(2
s
+ 11
s
+ 11) and dividing by
E
In order to define the other possible network functions, we must use a current source,
J
, as in Fig. 21.2, where we also take the current through the inductor as an output variable. This method of formulating the network equations is called
modified nodal
. The sum of currents flowing away from node 1 is
D s s= + +2 11 112
N E
s s E1
0 2 5 2 5=
−
+ = +( )
V N
D
s
s s E1
2 5
2 11 11 = =
+
+ +
T V
E
s
s sv = =
+
+ + 1
2 5
2 11 11
N s E
E2 3
2 0 2=
+
−
=
V N
D s s E2
2 11 11 = =
+ +
T V
E s sv = =
+ + 2
2 11 11
Y I
E
V
E s str out
= = =
+ +
3 6
2 11 11 2
Y I
E
s s
s sin = =
+ +
+ + in 2 9 6
2 11 11
from node 2 it is
and the properties of the inductor are expressed by the additional equation
Inserting numerical values and collecting in matrix form:
The determinant of the system is
To solve for
V
, we replace the first column by the right-hand side and evaluate the determinant
Then,
V
=
N
/
D
and dividing by
J
we obtain the network function
To obtain the inductor current, evaluate the determinant of a matrix in which the third column is replaced by the right-hand side:
N
= –2
J
. Then,
I
=
N
/
D
and
In general,
(21.2)
G sC V I JL1 1 1 0+( ) + − =
G V IL2 2 0− =
V V sLIL1 2 0− − =
s
s
V
V
I
J
+
−
− −
=
1 0 1
0 2 1
1 1
D s s= − + +( )2 3 32
N
J
s
s J1
0 1
0 2 1
0 1
2 1= −
− −
= − +( )
Z V
J
s
s str = =
+
+ + 1
2 1
2 3 3
T I
J s si L
= =
+ +
3 32
F E J
= = Output variable
or
Numerator polynomial
Denominator polynomial
Any method that may be used to formulate the equations will lead to the same result. One example shows this is true. Reconsider the network in Fig. 21.2, but use the admittance of the inductor,
YL = 1/sL, and do not consider the current through the inductor. In such a case, the nodal equations are
We can proceed in two ways:
1. We can multiply each equation by s and thus remove the fractions. This provides the system equation
The determinant of this matrix is D = 2s3 + 3s2 + 3s. To calculate V1, find N1 = s(2s + 1)J. Their ratio is the same as before because one s in the numerator can be canceled against the denominator.
