Network Functions of Circuits Containing Dependents SourcesIntroductionWorked ExamplesExample 1:Example 2:Example 3:Example 4:Network Functions of Circuits Containing Dependents Sources 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 capacitor, vo(t). The network function that represents this circuit is ()()()31125jjωωωωω==++oiVHV (1) Determine the value of the inductance, L, and of the gain, A, of the Voltage Controlled Voltage Source (VCVS). Figure 1 The circuit considered in Example 1. 1Solution: The circuit has been represented twice, by a circuit diagram and also by the given network function. The unknown parameters, L and A, appear in the circuit diagram, but not in the given network function. We can analyze the circuit to determine its network function. This version of the network function will depend on the unknown parameters. We will determine the value of these parameters by equating the two version of the network function. 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 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 circuit in Figure 2 consists of two meshes. The mesh current of the left-hand mesh is labeled as I1(w) and the mesh current of the right-hand mesh is labeled as I2(w). Apply Kirchhoff’s Voltage Law (KVL) to the left-hand mesh to get ()()()40jLωω ω ω+−=11iIIV Solve for I1(w) to get ()()()0.25414LjLjωωωωω==++1iiVIV Next use Ohm’s Law to obtain represent Va(w) as () () ()1414Ljωωω==+1aiVI Vω (2) Apply Kirchhoff’s Voltage Law (KVL) to the right-hand mesh to get () () ()2040Ajωωωω+−=22aIIV Solve for I2(w) to get 2() () () ()2020420415jAAjAjjjωωωωωωωω== =+++2 aaIV VωaV The output voltage is obtained by multiplying the mesh current I2(w) by the impedance of the capacitor () () () ()20 20201155jAAjjjjωωωωωωωω==× =++2oaVI V Vωa (3) Substituting the expression for Va(w) from Equation 2 into Equation 3 gives () () ()111114545AALLjjjjωωωωωωω=× =++++oiiVV V Divide both sides of this equation by Vi(w) to obtain the network function of the circuit ()()()1145ALjjωωωωω==++oiVHV (4) Comparing the network functions given by Equations 1 and 4 gives A = 3 V/V and L = 2 H. 3Example 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 capacitor, vo(t). This circuit is an example of a “second order low-pass filter”. The network function that represents a second order low-pass filter has the form ()()()1211kjjppωωωωω==++oiVHV (5) This network function depends on three parameters, k, p1 and p2. The parameter k is called the “dc gain” of the second order low-pass filter. Both p1 and p2 are poles of the second order low-pass filter. Determine the values of k, p1 and p2 for the second order low-pass filter in Figure 3. Figure 3 The circuit considered in Example 2. Solution: 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. 4The circuit in Figure 4 consists of two meshes. The mesh current of the left-hand mesh is labeled as I1(w) and the mesh current of the right-hand mesh is labeled as I2(w). Apply Kirchhoff’s Voltage Law (KVL) to the left-hand mesh to get ()()()()0.66 4 0jωωωω+−=11iIIV Solve for I1(w) to get ()()()()()0.250.66 4 1 0.165jjωωωωω==++1iiVIV Next use Ohm’s Law to obtain represent Va(w) as () ()()()141 0.165jωωω==+1aiVI Vω (6) Apply Kirchhoff’s Voltage Law (KVL) to the right-hand mesh to get ()()() ()10006112.82jωωω50ω+−=22aIIV Solve for I2(w) to get ()()()()()()12.82 1515100012.82 6 1000612.82jjjωωωωωω==++2 aaIV V The output voltage is obtained by multiplying the mesh current I2(w) by the impedance of the capacitor ()()()()()()()()()()12.82 151000 100012.82 12.82 12.82 6 10001512.82 61100015113jjjjjjωωωωωωωωωωω==×+=+=+2o aaaVIVVV (7) Substituting the expression for Va(w) from Equation 6 into Equation 7 gives ()()()15 11 0.165113jjωωωω=×++ioVV 5Divide both sides of this equation by Vi(w) to obtain the network function of the circuit ()()()()()151 1 0.16513jjωωωωω==++oiVHV (8) Comparing the network functions given by Equations 4 and 8 gives k = 15 V/V, p1 = 13 rad/s and p2 = 6.06 rad/s (Or perhaps p1 = 6.06 rad/s and p2 = 13 rad/s. One pole is named p1 and the other pole is named p2.) Example 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 capacitor, vo(t). The network function that represents this circuit is ()()()5110jjωωωωω==+oiVHV (9) Determine the value of the inductance, L, and of the gain, A, of the Current Controlled Current Source (CCCS). Figure 5