29. Network Functions for Circuits Containing Op AmpsIntroductionWorked ExamplesExample 1:Example 2:29. Network Functions for Circuits Containing Op Amps Introduction Each of the circuits in this problem set contains at least one op amp. Also each circuit is represented by a given network function. These problems can be solved by analyzing the circuit to determine its network function and then comparing the network function of the circuit to the given network function. Network functions are described in Section 13.3 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. Op amps are discussed in Chapter 6. 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 10 kΩ resistor, vo(t). The network function that represents this circuit is given to be ()()()1250002012500jjωωωωω+==−+oiVHV (1) Determine the values of the capacitance, C1, and of the resistances, R1 and R2. 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 capacitance and resistances, C1, R1 and R2, appear 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 and resistances. We will determine the value of the capacitance and resistances 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 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. To analyze the circuit in Figure 2, we write a node equation at the node labeled as node a. (The node voltage at node a is zero volts because the voltages at the input nodes of an ideal op amp are equal. The current entering the inverting input of the op amp is zero, so there are four currents in this node equation, the currents in the impedances corresponding to R1, R2, C1 and C2.) ()()()()1212011RRjC jCωωωωωω+++ooiiVVV V= Doing a little algebra gives () ()121211jC jCRRωωω +=−+   oiVVω Then, () ()11 2 21211jCR jCRRRωωωω ++=−   oiVV 2Finally, the network function of the circuit is ()()()1112122221111jCR112RRjCRRjCRjCRRωωωωωωω++== =−++−oiVHV (2) The network functions given in Equations 1 and 2 must be equal. That is ()2112112500020112500j12RjCRRjCRjωωωωω++−==−++H Equating coefficients gives: 2120RR=, 11125000CR= , and 2212500CR= Now using C gives 922 nF 2 10 F−==× ()929110200 k50002500 2 10R−===××Ω, 2110 k20RR==Ω and ()()81441110.4 10 4 nF250002.5 10 1 10CR−== =×=×× 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 node voltage at the output terminal of the op amp, 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 (3) 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. 3Figure 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 3. 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. To analyze the circuit in Figure 4, we write a node equation at the node labeled as node a. (The current entering the non-inverting input of the op amp is zero, so there are two currents in this node equation, the currents in the impedances corresponding to 20 kΩ resistor and 0.2 µF capacitor.) 4()()()63aa51020 10jωωωω−=××iVV V Doing a little algebra gives ()()336a120 10 20 10 5 10jωωω=+×××iVV Then ()()() () ()()36aaa12010 15 10 2501250jjjωωωωωωωω=+ × =+ ⇒ =×+iiVVVVV The op amp together with the 10 kΩ and 40 kΩ resistors comprise a noninverting amplifier so () () ()oa40 51101250jωωωω=+ =+iVV V Finally, the network function is ()()()o51250jωωωω==+iVHV (4) Comparing the network functions given by Equations 3 and 4 gives k = 5 V/V and p = 250 rad/s