The driving-point functions have been presented in the previous page. Apart from the impedance and admittance of a network, we may be interested in the ratio of voltage across an element or the current through an element to the source voltage or current. Hence a versatile function, called the transfer function, is used to represent any of the ratios. It is better to define a transfer function with reference to a two-port network. The topic of two-port network is handled in detail in another chapter. In this page, a two-port network is introduced in order to define what a transfer function is.
TWO-PORT NETWORK & TRANSFER FUNCTIONS
The network shown in Fig. 36 has two ports, an input port and an output port. The source is connected to the input port and the load is connected to the output port. For this network, we can define many transfer functions. . Transfer functions may relate voltage at one port to voltage at another port, or it may be the ratio of two currents. It may also relate voltage at one port to current through an element or it may relate the source current to a voltage at some other port. By definition, a transfer function is defined as the ratio of Laplace transform of output function to the Laplace transform of input function. That is,
A transfer function is usually defined for a two-port network. A two-port network can be represented by a black box as shown in Fig. 36. The directions used for voltages and the currents at the two ports have been marked in Fig. 36. You need to remember that the convention that will be used when the two ports are discussed in detail is different from what is presented here.
For the network shown in Fig. 36, the following transfer functions can be defined.
Equation (3.2) specifies the voltage gain. The ratio of transform of voltage at input port to transform of voltage at the output port can also be called as the voltage transfer ratio. In fact, we can define the voltage transfer ratio in relation to any pair of ports. For the network in Fig. 36, the voltage across the load impedance ZL(s) is defined as the output voltage. The voltage transfer ratio can be defined for voltage across any other element too, and this is the reason why directions used for voltages and the currents differ from the convention followed for the two-port networks described later in another chapter.
Equation (3.3) specifies the current gain. The ratio of transform of current supplied by the source connected to the input port to transform of current delivered to the load connected to the output port can also be called as the current transfer ratio. As stated for voltage transfer ratio, the current transfer ratio can be defined for current through any other element in the network.
Apart from the transfer functions for currents and voltages, we can define transfer function for impedance and admittance.
As defined by equation (3.4), the ratio of transformed load voltage to the transformed source current is the transfer impedance. As defined by equation (3.5), the ratio of transformed load current to the transformed source voltage is the transfer admittance. Some worked examples are presented below to illustrate how we obtain the transfer functions.
WORKED EXAMPLES
Example 3.1:
For the RC network in Fig. 37(a), obtain the voltage transfer ratio Vo(s)/Vs(s).
Solution:
From the circuit in Fig. 37(a), we get that
The output voltage is the voltage across the capacitor. Hence the product of current and the reactance of capacitor yields the output voltage. The source current is the ratio of source voltage to the impedance of the circuit. Hence we get equation (3.6). The transfer obtained is expressed by equation (3.7).
We can obtain the same expression using the voltage division rule.
Equation (3.8) is obtained using the voltage division rule. The voltage is shared proportionate with the impedances. By replacing the corresponding expressions for the impedances, we get equation (3.9).
The transfer function expressed by equation (3.9) has a single pole at - (1/RC) and has no zero. A transfer function need not have the same number of poles and zeros.
Example 3.2:
For the RC network in Fig. 38(a), obtain the voltage transfer ratio Vo(s)/Vs(s).
Solution:
From the circuit in Fig. 38(a), we get the transfer function using the voltage division rule. The circuit in Fig. 38(b) is used to illustrate how the voltage division rule can be used.
Equations (3.10) and (3.11) illustrate how the transfer function can be obtained.
Example 3.3:
For the LC network in Fig. 39(a), obtain the voltage transfer ratio V2(s)/V1(s).
Solution:
From the circuit in Fig. 39(a), we get the transfer function using the voltage division rule. The circuit in Fig. 39(b) is used to illustrate how the voltage division rule can be used.
The transfer function expressed by equation (3.12) has a pair of conjugate poles on the imaginary axis and has no zero. A transfer function need not have the same number of poles and zeros. The order of the numerator polynomial and the denominator polynomial can differ by more than one. In equation (3.12), the order of denominator polynomial is two, whereas the order of numerator polynomial is zero.
Example 3.4:
For the p network in Fig. 40, obtain the transfer functions and the driving-point impedance. The expressions for the transfer functions and the driving-point impedance are stated below.
Solution:
From the circuit in Fig. 40, the voltage transfer ratio is obtained as shown below.
Next an expression for the transfer impedance is obtained.
Next expressions for the current transfer ratio and the transfer admittance are obtained.
Next an expression for the driving-point impedance is obtained. Two ways of obtaining the expression are displayed.
We can use nodal analysis to get the transfer ratios and the driving-point impedance. The equation obtained using nodal analysis is presented below.
Using Cramer's rule, we get the following equation.
Then an expression for the driving-point impedance is obtained.
Using Cramer's rule, we can express V2(s) as shown below.
Then expressions for the the voltage transfer ratio and the transfer impedance are obtained.
The expressions for the current transfer ratio and the transfer admittance are obtained.
Example 3.5:
Find the current transfer ratio a21(s) for the circuit shown in Fig. 3.5
Solution:
The current transfer ratio a21(s) is the ratio of I2(s) to I1(s). The first suffix refers to the port related to the variable present in the numerator. The second suffix refers to the port related to the variable present in the denominator. We can form the following equations for the circuit in Fig. 3.5.
Equation (3.27) expresses the current transfer ratio a21(s).
Example 3.6:
For the circuit shown in Fig. 42, obtain the transfer functions G21(s), a21(s), Z21(s) and Y21(s) and the driving-point impedance function Zin(s).
Solution:
At first, the voltage transfer ratio G21(s) is obtained. We get this ratio by starting with the output voltage. Then we work our way through to the input voltage, which is expressed as a function the output voltage and the components of the circuit.
Equation (3.28) expresses I2(s) as the product of V2(s) and the transform susceptance. Based on V2(s) and I2(s), equation (3.29) expresses V2(s) as the sum of V2(s) and the drop across the inductance. Then I2(s) can be expressed as the product of Vb(s) and the transform susceptance.
Next express Va(s) as the sum of Vb(s) and the drop across the inductance, as shown by equation (3.31). The drop across inductance is the product of transform reactance and the current flowing through it, which is the sum of I2(s) and Ib(s). These currents can be replaced by the corresponding expressions in equations (3.28) and (3.30) and the resultant equation is equation (3.32). Group terms and get equation (3.33). Use equation (3.29) to replace Vb(s) and get equation (3.34). Group the coefficients and get equation (3.35).
Next express V1(s) as the sum of Va(s) and the drop across the inductance, as shown by equation (3.36). Replace Va(s) and Vb(s) in terms of V2(s).
Then we obtain G21(s) as expressed by equation (3.39). It is possible to obtain G21(s) using nodal analysis.
Equation (3.39) expresses the matrix equation obtained for the circuit in Fig. 42. The nodal admittance matrix can be formed by inspection. Use Cramer's rule to get G21(s).
Equation (3.40) expresses the determinant of the nodal admittance matrix.
Cofactor D13 is obtained, as expressed by equation (3.41). Using Cramer's rule, we get equation (3.42) and G21(s) can then be obtained as shown by equation (3.43). The transfer admittance ratio Y21(s) is obtained as shown by equation (3.44).
We can use mesh analysis to obtain the other transfer ratios and the driving-point impedance.
Equation (3.45) expresses the matrix equation obtained for the circuit in Fig. 42. The loop impedance matrix can be formed by inspection.
Equation (3.46) expresses the determinant of the loop impedance matrix.
Cofactor D11 is obtained, as expressed by equation (3.47). Then the driving-point impedance can be obtained, as shown by equation (3.48).
Currents I1(s) and I2(s) can be expressed, as shown by equation (3.49). Then the current transfer ratio a21(s) can be obtained, as shown by equation (3.50).
The transfer impedance ratio Z21(s) can be obtained, as shown by equation (3.51).
Example 3.7:
Find the voltage transfer ratio of the lattice network in Fig. 43a.
Solution:
A lattice network is shown in Fig. 43a. This network is known as an all-pass network since there is no attenuation either at low frequency or at high frequency. It has unity gain over the entire frequency range. This aspect will be explained later in the chapter on frequency response. The lattice network in Fig. 43a can be re-drawn as shown in Fig. 43b and it can be seen that the resulting network is a bridge network. A bridge network is used widely for measurement. From the circuit in Fig. 43b, we get that
Example 3.8:
For the bridged-T network shown in Fig 44, obtain the driving-point admittance and the voltage transfer ratio, G21(s).
Solution:
It is possible to get the the driving-point admittance and the voltage transfer ratio G21(s) in several ways. Here two ways are illustrated. The driving-point admittance is obtained using star-to-delta transformation and mesh analysis.
The circuit in Fig. 44 can be represented by the circuit in Fig. 45, where the star-connected network to be replaced comprises Z2, Z3and Z4.
The network obtained after transformation is shown in Fig. 46. Then
The driving-point impedance is obtained as follows.
The driving-point admittance is the reciprocal of the driving-point impedance, expressed by equation (3.59).
We can verify that the answer is correct by letting s = 0 and s = infinity, as shown in Fig. 47.
Now the solution using mesh analysis is illustrated. It is possible to use nodal analysis, in which case the number of unknowns is one less, but it is somewhat messy to get the driving-point admittance. Hence mesh analysis is used.
From the circuit in Fig. 48b, we get the following matrix equation.
Use Cramer's rule.
The required cofactor and the determinant can be obtained as shown above.
The expression obtained for the driving-point admittance is the same as that obtained earlier.
Using Cramer's rule, we can obtain the transfer admittance Y21(s), as illustrated by equations (3.65) and (3.66). Then the voltage transfer ratio G21(s) can be obtained.
Example 3.9:
For the network shown in Fig. 49, obtain the driving-point admittance and sketch its pole-zero plot.
Solution:
From the circuit in Fig. 49, we form the following equations.
The pole-zero plot is shown in Fig. 50.
Example 3.10:
For the op amp circuit in Fig. 51, obtain the gain of the circuit . If vS(t) is 3.cos(5t), obtain the sinusoidal response VO(j5) and then the steady-state response vO(t).
Solution:
Form the KCL equation at the node with voltage V1(s). It is expressed by equation (3.72). Then obtain the KCL equation at the non-inverting input terminal of the op amp. Because of unity feedback from the output to the inverting input terminal, the voltage at the non-inverting input terminal is equal to the output voltage. Then we obtain equation (3.73).
Eliminating V1(s), we get that
Example 3.11:
For the op amp circuit in Fig. 52, obtain the driving-point impedance and the voltage gain of the circuit . If vS(t) is 4.cos(2t - 30o), obtain the sinusoidal response VO(j2).
Solution:
Form the KCL equation at the node with voltage V1(s). It is expressed by equation (3.74). Then obtain the KCL equation at the inverting input terminal of the op amp. The inverting input terminal acts as the virtual ground and the voltage inverting input terminal is zero. Then we obtain equation (3.75).
The gain of the circuit is expressed by equation (3.76).
The sinusoidal response of the circuit is expressed by equation (3.78). The driving-point impedance is obtained as follows.
PROPERTIES OF TRANSFER FUNCTIONS
- The denominator polynomial and the numerator polynomial of a transfer functions have real coefficients.
- The coefficients of the numerator polynomial may be negative, as in the case of lattice network. But coefficients of the denominator polynomial are positive.
- If poles and zeros are either complex or imaginary, then occur in conjugate pairs.
- The poles have either zero or negative real parts. If there is a pole at the origin, then it must be simple. There cannot be multiple poles at the origin.
- If the denominator polynomial may be only or even. If it is not odd or even, then there should be no missing terms in the denominator polynomial.
- The degree of numerator polynomial may be zero. The degree of numerator polynomial may differ from the degree of denominator polynomial by more than one.
- For the transfer functions of voltage and current, the highest degree of numerator polynomial cannot exceed that of the denominator polynomial.
- For the transfer functions of impedance and admittance, the highest degree of numerator polynomial may exceed that of the denominator polynomial by one at the most.
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