In this page, it shown how the transform impedance and the transform admittance of a single-port network can be obtained. Quite a few examples are presented to illustrate the process of obtaining the transform impedance and the transform admittance. We also study the commonly used network configurations. In this page, ladder networks are introduced. We also find out how mesh equations and nodal equations can be used to obtain the driving-point functions. A network function can be expressed in terms of its poles and zeros. We study what poles and zeros are.
A network with only two terminals is shown in Fig. 3. A 2-terminal network is called a single-port network. A single-port network is a load on the source that excites it. The ratio of source voltage to current drawn by the network is called as the driving-point impedance, and equation (2.10) expresses how the driving-point impedance is obtained. The ratio of source current to source voltage is called as the driving-point admittance. Both the driving-point impedance and the driving-point admittance are the driving-point functions. The term, the driving-point immittance, is used to either the the driving-point impedance or the driving-point admittance.
Given a general network function F(s), it can be expressed as the ratio of two polynomials, as shown below.
Both the numerator and the denominator polynomials can be expressed in terms of their factors. The values of s, for which F(s) has zero value, are called as the zeros of F(s). The values of s, for which F(s) has infinite value, are called as the poles of F(s). A finite zero or a finite pole has a finite value. It is possible for a pole or a zero to be located at infinity. If the value of a pole or a zero is zero, then it is located at the origin.
Poles and zeros of a network determine the time response and the frequency response. We will study how the poles and zeros determine the response of a system, when we take up the study of time response and frequency response. Now we find out how we can obtain the driving-point immittance functions of some networks and where their poles and zeros are.
Example 1:
Obtain the driving-point functions of the RC circuit in Fig. 4.
Solution:
From the circuit in Fig. 4, we get that
The driving-point impedance is expressed by equation (2.13). It is seen that Zin(s) has a zero at - (1/RC), and a pole at origin. The driving-point admittance is expressed by equation (2.14). It is seen that Yin(s) has a pole at - (1/RC), and a zero at origin. The driving point-admittance is the reciprocal of the driving-point impedance and vice versa. It is seen that the number of poles of a driving-point function equals its number of zeros. This property is true for most passive networks.
The plot of pole and zero of the RC circuit in Fig. 4 is shown in Fig. 5. A zero is represented by a small circle and a pole by a cross. Given that R and C are positive values, the finite-pole of the driving-point impedance of the RC circuit is always in the left-half of s-plane. It is a real, negative value. If it is a real value, it is always a negative value.
Example 2:
Obtain the driving-point functions of the RC circuit in Fig. 6.
Solution:
From the circuit in Fig. 6, we get that
The driving-point impedance is expressed by equation (2.15). It is seen that Zin(s) has a pole at - (1/RC). When s tends to infinity, the driving-point impedance tends to become zero and the driving-point impedance has a zero at infinity. The driving-point admittance is expressed by equation (2.16). It is seen that Yin(s) has a zero at - (1/RC). When s tends to infinity, the driving-point admittance tends to become infinite and the driving-point admittance has a pole at infinity. The driving point-admittance is the reciprocal of the driving-point impedance and vice versa.
The plot of pole and zero of the RC circuit in Fig. 6 is shown in Fig. 7. Given that R and C are positive values, the finite-pole of the driving-point impedance of the RC circuit is always in the left-half of s-plane.
Example 3:
Obtain the driving-point functions of the RL circuit in Fig. 8.
Solution:
From the circuit in Fig. 8, we get that
The driving-point impedance is expressed by equation (2.17). It is seen that Zin(s) has a zero at - (R/L). When s tends to infinity, the driving-point impedance tends to become infinite and the driving-point impedance has a pole at infinity. The driving-point admittance is expressed by equation (2.18). It is seen that Yin(s) has a pole at - (R/L). When s tends to infinity, the driving-point admittance tends to become zero and the driving-point admittance has a zero at infinity. The driving point-admittance is the reciprocal of the driving-point impedance and vice versa.
The plot of pole and zero of the RL circuit in Fig. 8 is shown in Fig. 9. Given that R and L are positive values, the finite-pole or zero of the driving-point immittance function of the RL circuit is always in the left-half of s-plane.
Example 4:
Obtain the driving-point functions of the RL circuit in Fig. 10.
Solution:
From the circuit in Fig. 10, we get that
The driving-point impedance is expressed by equation (2.19). It is seen that Zin(s) has a pole at - (R/L), and a zero at origin. The driving-point admittance is expressed by equation (2.120). It is seen that Yin(s) has a zero at - (R/L), and a pole at origin.
The plot of pole and zero of the RL circuit in Fig. 10 is shown in Fig. 11.
Example 5:
Obtain the driving-point impedance of the LC circuits in Fig. 12.
Solution:
For the circuit with L and C in series, the driving-point impedance obtained is expressed by equation (2.21). For the circuit with L and C in parallel, the driving-point impedance obtained is expressed by equation (2.22).
The driving-point impedance function of the circuit with L and C in series contains a pair of conjugate zeros on the imaginary axis. The values of the zeros can be obtained from equation (2.21). Then it should contain two poles. One pole is at the origin, due to the presence of s in the denominator. As s tends to infinity, the value of Zin(s) tends to infinity, and hence Zin(s) has a pole at infinity. The pole-zero plot for Zin(s) is shown in Fig. 13.
The driving-point impedance function of the circuit with L and C in parallel contains a pair of conjugate poles on the imaginary axis. The values of the poles can be obtained from equation (2.22). Then it should contain two zeros. One zero is at the origin, due to the presence of s in the numerator. As s tends to infinity, the value of Zin(s) tends to zero, and hence Zin(s) has a zero at infinity. The pole-zero plot for Zin(s) is shown in Fig. 14.
Example 6:
Obtain the driving-point impedance of the RLC circuit in Fig. 14.
Solution:
For the circuit with R, L and C in series, the driving-point impedance obtained is expressed by equation (2.23).
Since the numerator of equation (2.23) is a second-order polynomial, it has two roots. This means that Zin(s) has two zeros, as stated below.
If L and C are fixed, the position of zeros varies as a function of R. The trajectory of zeros is displayed in red colour in Fig. 15.
Since Zin(s) has two zeros, it should have two poles. One pole is the origin and the other is at infinity. If the circuit has a dc input, it means that s is zero, the circuit should have infinite impedance. It is indeed so, since the capacitor acts like an open-circuit for a dc circuit in steady-state. If the circuit has an ac input and if the frequency is very high, it means that s is approaching infinity and the circuit should have infinite impedance. It is indeed so, since the inductor acts like an open-circuit when the applied frequency is very high. The circuits in Fig. 16 illustrate what has been stated. For a dc input, the inductor can be replaced by a short-circuit and the capacitor by an open circuit. At a very high frequency, the inductor can be replaced by an open circuit and the capacitor by a short-circuit.
Example 7:
Obtain the driving-point impedance of the RLC circuit in Fig. 17.
Solution:
For the circuit with R, L and C in parallel, the driving-point impedance obtained is expressed by equation (2.24).
Since the numerator of equation (2.24) is a second-order polynomial, it has two roots. This means that Zin(s) has two poles, as stated below.
If L and C are fixed, the position of poles varies as a function of R. The trajectory of poles is displayed in red colour in Fig. 18.
Since Zin(s) has two poles, it should have two zeros. One zero is the origin and the other is at infinity. If the circuit has a dc input, it means that s is zero, the circuit should have zero impedance. It is indeed so, since the inductor can be replaced by a short-circuit and the capacitor by an open circuit for a circuit in steady-state when the input is a dc signal. If the circuit has an ac input and if the frequency is very high, it means that s is approaching infinity and the circuit should have zero impedance. It is indeed so, since the capacitor can be replaced by a short-circuit. The inductor acts like an open-circuit when the applied frequency is very high. In the case of the circuit in Fig. 16, there are three elements in parallel. When any of them is replaced by a short-circuit, the circuit has zero impedance. This is the case when the frequency of source is either zero or very high.
Example 8A:
Obtain the driving-point impedance of the series-parallel circuit in Fig. 19A. Find the value of R such the driving-point impedance equals 1 W .
Solution:
For the circuit in Fig. 19A, the driving-point impedance is obtained as shown below.
The driving-point impedance equals 1 W when the imaginary part of the numerator equals the numerator part of the denominator.
The condition that needs to be satisfied for driving-point impedance to be equal to 1 W is stated by equation (2.27).
Example 8B:
Obtain the driving-point impedance of the series-parallel circuit in Fig. 19B. When is the driving-point impedance of this circuit is equal to R.
Solution:
For the circuit in Fig. 19B, the driving-point impedance is obtained as shown below.
Since Zin(s) = R, we get the following equations.
Equation (2.27e) states the condition for Zin(s) to be equal to R. If Z1 = sL and Z2 = 1/( sC), then we get that Zin(s) = 1 W .
Example 9A:
Obtain the driving-point admittance of the series-parallel circuit in Fig. 20.
Solution:
For the circuit in Fig. 20, the driving-point admittance is obtained as shown below.
The admittance function has two finite poles, one finite zero and another zero at infinity.
Example 9B:
Obtain the driving-point impedance of the network in Fig. 20B. Determine the poles and zeros of this network’s input impedance.
Solution:
For the circuit in Fig. 20B, the driving-point impedance is obtained as shown below.
The poles of Zin(s) are located at s = -1 and = -3. When s equals either - 1 or -3, value of Zin(s) becomes infinite. The zeros of Zin(s) are located at s = -2 and s = -4. At these values of s, Zin(s) equals zero value. The pole-zero plot of Zin(s) is shown in Fig. 20c.
Ladder networks form a special class of network structures. The driving-point immittance functions can be obtained in the form of ‘continued fraction’. Often it is possible to represent the various loads connected to a source in the form of a ladder network. In electronics, a ladder network is used in analog-to-digital converters. Here the ladder network is used to obtain binary fraction of a reference voltage. A few examples are presented here to show how the driving-point function of a ladder network can be obtained.
Example 10:
Obtain the driving-point impedance of the circuit in Fig. 21(a).
Solution:
For the circuit in Fig. 21(a), the driving-point impedance is obtained as shown below.
It is seen from Fig. 20(a) that the driving-point impedance can be expressed as the sum of R1 and Z2(s), where Z2(s) is yet to be determined. This step is shown in Fig. 20(b). We have that Z2(s) and Y2(s) are reciprocals of each other. From the circuit in Fig. 20(a), we can extract the circuit that represents Z2(s). This circuit is shown in Fig. 20(c). We can express Y2(s) as the sum of the admittance of the capacitor and the admittance of the rest of the circuit, represented as Y3(s). The admittance of the capacitor is sC. From equation (2.30), equation (2.31) can be obtained. From equation (2.31), equation (2.32) can be obtained.
For a ladder circuit that is not as simple as the circuit shown in Fig. 20(a), the approach outlined above turns out to be the easier way. Since the circuit in Fig. 20(a) is relatively simple, its driving-point impedance can be obtained more easily, as shown by equation (2.33).
Example 11:
Obtain the driving-point impedance of the circuit in Fig. 22(a).
Solution:
Note that the circuit in Fig. 22(a) specifies the transform admittance of the capacitor, and the value of capacitance is 1 F. For the circuit in Fig. 22(a), the driving-point impedance is obtained as shown below.
To start with, the driving-point impedance can be expressed as shown by equation (2.34), where the rest of the circuit apart form the 1 W resistor can be represented as Z2(s), as shown in Fig. 22(b). As expressed by equation (2.35), Y2(s) can be represented as the sum of the admittance of the capacitor and the admittance of the rest of the circuit, represented as Y3(s). This stage is represented by Fig. 22(c). From equations (2.34) and (2.35), we obtain equation (2.35) for the driving-point impedance of the circuit in Fig. 22(a). The circuit that shows the first resistor, the first capacitor and the rest of the circuit as Z3(s) is presented by Fig. 22(d).
The given circuit is presented again in Fig. 23(a). It can be seen that Z3(s) can be presented as the sum of 1 W and Z4(s). As shown in Fig. 23(b), Z4(s) represents the remaining part of the circuit. As shown in Fig. 23(c), Y4(s) can be represented as the sum of the admittance of the capacitor and the admittance of the rest of the circuit, represented as Y5(s). This stage is represented by Fig. 23(c). From Figs. 23(b) and 23(c), we get the circuit in Fig. 23(d).
As expressed by equation (2.37), Z3(s) can be represented as the sum of 1 W and Z4(s). Next, as expressed by equation (2.38), Y4(s) can be represented as the sum of the admittance of the capacitor and the admittance of the rest of the circuit, represented as Y5(s). This stage is represented by Fig. 23(c). From equations (2.36), (2.37) and (2.38), we obtain equation (2.39) for the driving-point impedance of the circuit in Fig. 23(a). The circuit that shows the two resistors, the two capacitors and the rest of the circuit as Z5(s) is presented by Fig. 23(d). From Fig. 23(a), it can be seen that Z5(s) is the sum of 1 W and the transform reactance of the capacitor.
Equation (2.40) represents Y5(s). We get equation (2.41) from equations (2.39) and (2.40). Now equation (2.41) can be simplified. The first step in simplification is expressed by the second part of equation (2.41). The way simplification is carried out has been illustrated above.
Development of equations (2.42) and (2.43) has been shown above. The technique is to backtrack and simplify. Equation (2.43) is obtained from equation (2.42) by moving (s2 + 3s) from denominator to numerator.
From equation (2.43), we get equation (2.44). From equation (2.44), we find that Zin(s) has three zeros and three poles. The zeros are at - 0.198, - 1.555 and - 3.247. The poles are at origin, - 1 and - 3. The pole-zero plot is shown in Fig. 24. We can draw one conclusion from the pole-zero plot. When all the poles and the zeros of a driving-point function lie on the real axis, the poles and zeros occur alternately. This property is true for most driving-point functions. Another property worth mentioning is that the driving-point function is a rational function. This statement can be proved, but it is left out.
Example 12:
Obtain the driving-point admittance of the circuit in Fig. 23(a).
Solution:
Note that the circuit in Fig. 25(a) specifies the transform admittance of the capacitor, , and the value of capacitance is 1 F. For the circuit in Fig. 25(a), the driving-point admittance is obtained as shown below.
To start with, the driving-point admittance can be expressed as shown by equation (2.45), where the rest of the circuit apart form the 1 F capacitor can be represented as Y2(s), as shown in Fig. 25(b). As expressed by equation (2.46), Z2(s) can be represented as the sum of the resistance of 1 W resistor and the impedance of the rest of the circuit, represented as Z3(s). This stage is represented by Fig. 25(c). From equations (2.45) and (2.46), we obtain equation (2.47) for the driving-point admittance of the circuit in Fig. 25(a). The circuit that shows the first capacitor, the first resistor and the rest of the circuit as Z3(s) is presented by Fig. 25(d).
The given circuit is presented again in Fig. 26(a). As shown in Fig. 26(c), Y3(s) can be presented as the sum of 1 F capacitor and Y4(s). As shown in Fig. 26(c), Z4(s) represents the remaining part of the circuit. As shown in Fig. 26(d), Z4(s) can be represented as the sum of the resistance of 1 W resistor and the impedance of the rest of the circuit, represented as Z5(s). This stage is represented by Fig. 26(d). From Figs. 26(a), 26(b), and 26(c), we get the circuit in Fig. 26(d).
As expressed by equation (2.48), Y3(s) can be represented as the sum of admittance of capacitor and Y4(s). Next, as expressed by equation (2.49), Z4(s) can be represented as the sum of the resistance of 1 W resistor and the impedance of the rest of the circuit, represented as Z5(s). This stage is represented by Fig. 26(d). From equations (2.47), (2.48) and (2.49), we obtain equation (2.50) for the driving-point admittance of the circuit in Fig. 26(a). The circuit that shows the two capacitors, the two resistors and the rest of the circuit as Y5(s) is presented by Fig. 26(d). From Fig. 26(a), it can be seen that Y5(s) is the sum of conductance 1 Siemens and the transform admittance of the capacitor. We can now get an expression for the driving-point admittance of the circuit in Fig. 26(a) by backtracking steps.
Equation (2.51) represents Y5(s). We get equation (2.52) from equations (2.50) and (2.51). Now equation (2.52) can be simplified. The first step in simplification is expressed by the second part of equation (2.52). The way simplification is carried out has been illustrated above.
Development of equations (2.53) and (2.54) has been shown above. The technique is to backtrack and simplify. Equation (2.54) is obtained from equation (2.53) by moving (s2 + 3s + 1 ) from denominator to numerator. From equation (2.54), we find that Yin(s) has three zeros and three poles. The zeros are at - 0.198, - 1.555 and - 3.247. The poles are at origin, - 1 and - 3.
A General Ladder Circuit
The ladder circuits presented in the examples shown above are called as simple ladder networks. A network is presented below in Fig. 27. If each impedance and each admittance in circuit in Fig. 27 represents a single passive element such as a resistor, an inductor or a capacitor, then the circuit in Fig. 27 is a simple ladder network. If the impedances and admittances contain combination of passive elements, then the circuit in Fig. 27 is not a simple ladder circuit.
A general ladder circuit is shown in Fig. 27. The technique of obtaining the driving-point impedance is illustrated in Fig. 27. Continuing along these lines, the driving-point impedance is expressed by equation shown below.
For example, a circuit is shown below.
Example 13:
Its driving-point admittance is expressed by equation shown below.
The mesh and nodal analysis techniques can be used to obtain the driving-point immittance functions.
When a network is excited by a voltage source as shown in Fig. 29(a), it is easier to use mesh equations. In such a case, we get that
In equation (2.55), I and V are column vectors and Z is a square matrix, usually representing the loop impedances. Using Cramer's rule, the driving-point functions can then be expressed by equations (2.56) and (2.57), where D11 is the co-factor of the element at row 1 and column 1 of the loop impedance matrix and DZ is the determinant of the loop impedance matrix.When a network is excited by a voltage source as shown in Fig. 29(b), it is easier to use nodal equations. In such a case, we get that
In equation (2.58), Y is a square matrix, usually representing the nodal admittances. Using Cramer's rule, the driving-point functions can then be expressed by equations (2.58) and (2.59), where D11 is the co-factor of the element at row 1 and column 1 of the nodal admittance matrix and DY is the determinant of the nodal admittance matrix. The examples presented below illustrate how mesh and nodal analysis can be used to obtain the driving-point immittance functions.
Example 14:
Obtain the driving-point impedance of the circuit in Fig. 30.
Solution:
For the circuit in Fig. 30, we get the following equations.
Equation (2.61) presents the matrix equation for the circuit in Fig. 30. This equation has been obtained using mesh analysis. By the use of Cramer's rule, we get equation (2.62).
The cofactor D11 is presented by equation (2.63). The determinant of the loop impedance matrix is also presented by equation (2.63). Equation (2.64) presents the driving-point impedance of the circuit in Fig. 30. The same expression has been obtained earlier in example 2 on this page.
Example 15:
Obtain the driving-point impedance of the circuit in Fig. 31.
Solution:
For the circuit in Fig. 31, we get the following equations.
Equation (2.66) presents the matrix equation for the circuit in Fig. 31. This equation has been obtained using mesh analysis. By the use of Cramer's rule, we get equation (2.66).
The cofactor D11 is presented by equation (2.67). The determinant of the loop impedance matrix is also presented by equation (2.67). Equation (2.68) presents the driving-point impedance of the circuit in Fig. 31. The same expression has been obtained earlier in example 4 on this page.
Example 16:
Obtain the driving-point admittance of the circuit in Fig. 32.
Solution:
For the circuit in Fig. 32, we get the following equations.
Equation (2.69) presents the matrix equation for the circuit in Fig. 32. This equation has been obtained using nodal analysis. By the use of Cramer's rule, we get equation (2.70).
The cofactor D11 is presented by equation (2.71). The determinant of the nodal admittance matrix is presented by equation (2.72). Equation (2.53) presents the driving-point admittance of the circuit in Fig. 32. The same expression has been obtained earlier in example 9A on this page.
Example 17:
Obtain the driving-point impedance of the circuit in Fig. 33.
Solution:
For the circuit in Fig. 33, we can get the driving-point impedance using either mesh analysis or nodal analysis. Both techniques are illustrated. The solution that uses mesh analysis is presented first. For the circuit in Fig. 33, we get that
Equation (2.74) presents the matrix equation for the circuit in Fig. 33. The loop impedance matrix can be obtained by inspection.
The cofactor D11 is presented by equation (2.75). The determinant of the loop impedance matrix is presented by equation (2.76). Equation (2.77) presents the driving-point impedance of the circuit in Fig. 31. The same expression has been obtained earlier in example 11 on this page. It is probably easier to use mesh analysis to get the driving-point impedance.
The use of nodal analysis is somewhat trickier, since the source is a voltage source. More importantly, there is an element in series with the source and this makes the use of nodal analysis somewhat trickier.
Equation (2.78) presents the matrix equation for the circuit in Fig. 33. The nodal admittance matrix can be obtained by inspection.
Using Cramer's rule, we get equation (2.79) from equation (2.78). It is seen that we can express node voltage V1(s) as a function of the source voltage and the circuit elements. Hence an expression for Is(s) can be obtained as shown below.
Equation (2.80) presents an expression for Is(s). Then Zin(s) can be obtained, as shown by equation (2.81).
Example 18:
Obtain the driving-point impedance of the circuit in Fig. 34.
Solution:
For the circuit in Fig. 34, we can get the driving-point impedance using either mesh analysis or nodal analysis. Both techniques are illustrated. The solution that uses nodal analysis is presented first. For the circuit in Fig. 34, we get that
Equation (2.82) presents the matrix equation for the circuit in Fig. 34. The nodal admittance matrix can be obtained by inspection.
Using Cramer's rule, we get equation (2.83) from equation (2.82).
Then Zin(s) can be obtained, as shown by equation (2.84). The same expression has been obtained earlier in example 12.
The solution obtained using mesh analysis is presented below.
Equation (2.85) presents the matrix equation for the circuit in Fig. 34. The loop impedance matrix can be obtained by inspection.
Using Cramer's rule, we get equation (2.86) from equation (2.85). It is seen that we can express node voltage I1(s) as a function of the source current and the circuit elements. Hence an expression for Vs(s) can be obtained as shown below.
Equation (2.87) presents an expression for Vs(s). Then Zin(s) can be obtained, as shown by equation (2.88).
Example 19:
Obtain the driving-point impedance of the circuit in Fig. 35.
Solution:
For the circuit in Fig. 35, we can get the driving-point impedance using either mesh analysis or nodal analysis. Both techniques are illustrated. The solution that uses mesh analysis is presented first. For the circuit in Fig. 35, the loop impedance matrix can be augmented and we get the equation shown below. We can adopt the technique described for the previous problem. In fact, the technique outlined in the previous problem is the preferred technique since the order of the square matrix is less than that the order of square matrix presented below. Increase in order of matrix increases the complexity of obtaining its determinant.
The first line in equation (2.89) expresses the source voltage as a function of Is(s) and I1(s). It does not describe a loop equation.
Using Cramer's rule, we get equation (2.90) from equation (2.89). The expression for Yin(s) is presented in equation (2.90). Using nodal analysis, the solution is obtained as shown below.
Equation (2.91) presents the matrix equation obtained using the nodal admittance matrix. The expression for Zin(s) is presented in equation (2.92).
Now that we have presented some examples, the properties of driving-point functions can be stated. Let a driving-point function F(s) be expressed as follows:
This page has explained what a driving-point function is. Numerous examples have been presented to illustrate how the driving-point function of a network can be obtained. The properties of a driving-point function have also been explained. In the next page, we take up transfer functions.