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 Vout. We can also take as the output a current through any element of the network; we call it output current, Iout. 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, Iin. If the network is excited by a current source, J, then the voltage across the current source is the input voltage, Vin. Suppose that we analyze the network and keep the letter E or J in our derivations. Then, we can define
the following network functions:
Voltage transfer function, Tv ¼ VoutE Input admittance, Yin ¼ IinE Transfer admittance, Ytr ¼ IoutE Current transfer function, Ti ¼ IoutJ Input impedance, Zin ¼ VinJ Transfer impedance, Ztr ¼ VoutJ
(22:1)
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
V ¼ ZI, I ¼ YV
The impedance of a capacitor is ZC¼ 1=sC, the impedance of an inductor is ZL¼ sL, and the impedance of a resistor is R. The inverse of these values are admittances: YC¼ sC, YL¼ 1=sL, and the admittance of a resistor is G¼ 1=R. To demonstrate the derivations of the above functions two examples are used. Consider the network in
Figure 22.1, 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:
(G1 þ sC1 þ G2)V1 G2V2 EG1 ¼ 0
Similarly, the sum of currents flowing away from node 2 is
V1G2 þ (G2 þ sC2 þ G3)V2 ¼ 0
The independent source is denoted by the letter E, and is assumed to be known. Inmathematics, we transfer known quantities to the right. Doing so and collecting the equations into one matrix equation results in
G1 þ G2 þ sC1 G2 G2 G2 þ G3 þ sC2
V1 V2
¼ EG1
If numerical values from the figure are used, this system simplifies to
sþ 3 2 2 2sþ 5
V1 V2
¼ E
or
YV ¼ E
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,
D ¼ 2s2 þ 11sþ 11
To obtain the solution for the variable V1(V2), 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
N1 ¼ E 20 2sþ 5 ¼ (2sþ 5)E
Then,
V1 ¼ N1D ¼ 2sþ 5
2s2 þ 11sþ 11 E
Now, divide the equation by E, which results in the voltage transfer function
Tv ¼ V1E ¼ 2sþ 5
2s2 þ 11sþ 11
To find the nodal voltage V2, replace the second column by the elements of the vector on the righthand side:
N2 ¼ sþ 3 E2 0 ¼ 2E
The voltage is
V2 ¼ N2D ¼ 2
2s2 þ 11sþ 11 E
and another voltage transfer function of the same network is
Tv ¼ V2E ¼ 2
2s2 þ 11sþ 11
Note that many network functions can be defined for any network. For instance, we may wish to calculate the currents Iin or Iout, marked in Figure 22.1. Because the voltages are already known, they are used: For the output current Iout¼G3V2 and divided by E
Ytr ¼ IoutE ¼ 3V2 E
¼ 6 2s2 þ 11sþ 11
The input current Iin¼ EG1V1¼ EV1¼E(2s2þ 9sþ 6)=(2s2þ 11sþ 11) and dividing by E
Yin ¼ IinE ¼ 2s2 þ 9sþ 6 2s2 þ 11sþ 11
In order to define the other possible network functions, we must use a current source, J, as in Figure 22.2, 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
(G1 þ sC1)V1 þ IL J ¼ 0
from node 2 it is
G2V2 IL ¼ 0
and the properties of the inductor are expressed by the additional equation
V1 V2 sLIL ¼ 0
Inserting numerical values and collecting in matrix form:
sþ 1 0 1 0 2 1 1 1 s
24 35 V1V2 IL
24 35 ¼ J0 0
24 35 The determinant of the system is
D ¼ (2s2 þ 3sþ 3)
To solve for V1, we replace the first column by the right-hand side and evaluate the determinant
N1 ¼ J 0 1 0 2 1 0 1 s
¼ (2sþ 1)J
Then, V1¼N1=D and dividing by J we obtain the network function
Ztr ¼ V1J ¼ 2sþ 1
2s2 þ 3sþ 3
To obtain the inductor current, evaluate the determinant of a matrix in which the third column is replaced by the right-hand side: N3¼2J. Then, IL¼N3=D and
Ti ¼ ILJ ¼ 2
s2 þ 3sþ 3
In general,
F ¼ Output variable E or J
¼ Numerator polynomial Denominator polynomial
(22:2)
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 Figure 22.2, but use the admittance of the inductor, YL¼ 1=sL, and do not consider the current through the inductor. In such a case, the nodal equations are
1þ sþ 1 s
V1 1s V2 ¼ J
1 s V1 þ 1s þ 2
V2 ¼ 0
We can proceed in two ways:
1. We canmultiply each equation by s and thus remove the fractions. This provides the system equation
s2 þ sþ 1 1 1 2sþ 1
V1 V2
¼ sJ
The determinant of this matrix is D¼ 2s3þ 3s2þ 3s. To calculate V1, find N1¼ s(2sþ 1)J. Their ratio is the same as before because one s in the numerator can be canceled against the denominator.
