Network Functions for Simple CircuitsIntroductionWorked ExamplesExample 1:Example 2:Example 3:Example 4:Example 5:Example 6:Example 7:Network Functions for Simple Circuits Introduction Each of the circuits in this problem set is represented by a network function. Network functions are defined, in the frequency-domain, to be quotient obtained by dividing the phasor corresponding to the circuit output by the phasor corresponding to the circuit input. We calculate the network function of a circuit by representing and analyzing the circuit in the frequency-domain. Network functions are described in Section 13.3 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. Also, Table 10.7-1 summarizes the correspondence between the time-domain and the frequency domain. Worked Examples Example 1: Consider the circuit shown in Figure 1. The input to the circuit is the voltage of the voltage source, vi(t). The output is the voltage across the 8 W resistor, vo(t). The network function that represents this circuit is ()()()0.66130jωωωω==+oiVHV (1) Determine the value of the inductance, L. Figure 1 The circuit considered in Example 1. Solution: The circuit has been represented twice, by a circuit diagram and also by a network function. The unknown inductance, L, appears in the circuit diagram, but not in the given network function. We can analyze the circuit to determine its network function. This second network function will depend on the unknown inductance. We will determine the value of the inductance by equating the two network functions. 1A network function is the ratio of the output phasor to the input phasor. Phasors exist in the frequency domain. Consequently, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 2 shows the frequency domain representation of the circuit from Figure 1. Figure 2 The circuit from Figure 1, represented in the frequency domain, using impedances and phasors. The impedances of the inductor and the two resistors are connected in series in Figure 2. Vi(w) is the voltage across these three series impedances and Vo(w) is the voltage across one of the impedances. Apply the voltage division principle to get () ()848 jLωωω=++oiVV Divide both sides of this equation by Vi(w) to obtain the network function of the circuit ()()()812jLωωωω==+oiVHV (2) The network functions given in Equations 1 and 2 must be equal. That is ()()()()8 0.66121308 1 0.66 123088 8 0.663080.6630830 0.660.4 HjLjjjjjLLLωωωωωω=+++= ++=+===LL 2We can simply the algebra required to find L by putting the network function in Equation 2 into the same form as the network function in Equation 1 before equating the two network functions. Notice that the real part of the denominator of the network function is 1 in Equation 1. Let’s make the real part of the denominator be 1 in the network function given by Equation 2. Divide the numerator and denominator by 12 to get ()()()80.661212112 12jLjωωωωω== =++oiVHVL (3) Equating the network functions given by Equations 1 and 3 gives: 0.66 0.66 10.4 H12 301112 30LLLjjωω=⇒=⇒=++ The same result is obtained with less algebra. Example 2: Consider the circuit shown in Figure 3. The input to the circuit is the voltage of the voltage source, vi(t). The output is the voltage across the 4 W resistor, vo(t). This circuit is an example of a “first order low-pass filter”. The network function that represents a first order low-pass filter has the form ()()()1kjpωωωω==+oiVHV (4) This network function depends on two parameters, k and p. The parameter k is called the “dc gain” of the first order low-pass filter and p is the pole of the first order low-pass filter. Determine the values of k and of p for the first order low-pass filter in Figure 3. Figure 3 The circuit considered in Example 2. 3Solution: We will analyze the circuit to determine its network function and then put the network function into the form given in Equation 4. A network function is the ratio of the output phasor to the input phasor. Phasors exist in the frequency domain. Consequently, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 4 shows the frequency domain representation of the circuit from Figure 3. Figure 4 The circuit from Figure 3, represented in the frequency domain, using impedances and phasors. The impedances of the inductor and the two resistors are connected in series in Figure 4. Vi(w) is the voltage across these three series impedances and Vo(w) is the voltage across one of the impedances. Apply the voltage division principle to get ()()()49 4 0.25jωωω=++oiVV Divide both sides of this equation by Vi(w) to obtain the network function of the circuit ()()()()413 0.25jωωωω==+oiVHV (5) Next, we put the network function into the form specified by Equation 4. Notice that the real part of the denominator is 1 in Equation 4. Divide the numerator and denominator by 13 in Equation 5 to get ()()()()40.308130.251315213 13jjωωωωω== =++oiVHV (6) Comparing the network functions given by Equations 4 and 6 gives k = 0.308 V/V and p = 52 rad/s 4Example 3: Consider the circuit shown in Figure 5. The input to the circuit is the voltage of the voltage source, vi(t). The output is the voltage across the 5 W resistor, vo(t). The network function that represents this circuit is ()()()0.20813jjωωωωω==+oiVHV (7) Determine the value of the capacitance, C. Figure 5 The circuit considered in Example 3. Solution: The circuit has been represented twice, by a circuit diagram and also by a network function. The unknown capacitance, C, appears in the circuit diagram, but not in the given network function. We can analyze the circuit to determine its network function. This second network function will depend on the unknown capacitance. We will determine the value of the capacitance by equating the two network functions. A network function is the ratio of the output phasor to the input phasor. Phasors exist in the frequency domain. Consequently, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 6 shows the frequency domain representation of the circuit from Figure 5. Figure 6 The circuit from Figure 5, represented in the frequency domain, using impedances and phasors. 5The impedances of the capacitor