Chapter 6 Electrical and Electromechanical Systems PDF

320 CHAPTER 6 Electrical and Electromechanical Systems Section 6.1 introduces the basic physics, common elements, and terminology of electrical circuits and treats the two main physical laws needed to develop circuit models. These are Kirchhoff’s current and voltage laws. Section 6.2 is an extensive collection of circuit examples that emphasize resistance networks and circuits having one or two capacitors or inductors. Impedance, a generalization of the electrical resistance concept, is covered in Section 6.3. This concept enables you to derive circuit models more easily, especially for more complex circuits, and is especially useful for obtaining models of circuits containing operational ampliﬁers, the subject of Section 6.4. The principles of direct-current (dc) motors are established in Section 6.5, and these principles are used to develop transfer function and state-variable models of motors. Section 6.6 examines some practical considerations in motor modeling, including methods for assessing the performance of motors and ampliﬁers in motion control systems. In Section 6.7, these principles are extended to other electromechanical devices such as sensors and speakers. Many of the systems treated in this chapter are more easily designed with computer simulation, and Sections 6.8 and 6.9 show how to apply MATLAB and Simulink in electromechanical systems analysis. ■ 6.1 ELECTRICAL ELEMENTS Voltage and current are the primary variables used to describe a circuit’s behavior. Current is the ﬂow of electrons. It is the time rate of change of electrons passing through a deﬁned area, such as the cross section of a wire. Because electrons are negatively charged, the positive direction of current ﬂow is opposite to that of the electron ﬂow. The mathematical description of the relation between the number of electrons (called charge Q) and current i is dQ or Q(t) = i dt i= dt The unit of charge is the coulomb (C), and the unit of current is the ampere (A), which is one coulomb per second. These units, and the others we will use for electrical systems, are the same in both the SI and FPS systems. Energy is required to move a charge between two points in a circuit. The work per unit charge required to do this is called voltage. The unit of voltage is the volt (V), which is deﬁned to be one joule per coulomb. The voltage difference between two points in a circuit is a measure of the energy required to move charge from one point to the other. The sign of voltage difference is important. Figure 6.1.1a shows a battery connected to a lightbulb. The electrons in the wire are attracted to the battery’s positive terminal; thus the positive direction of current is clockwise, as indicated by the arrow. Because the battery supplies energy to move electrons and the lightbulb dissipates energy (through light and heat), the sign of voltage difference across the battery is the opposite of the sign of the voltage difference across the lightbulb. This is indicated by the + and − signs Figure 6.1.1 (a) A battery-lightbulb circuit. (b) Circuit diagram representation of the battery-lightbulb circuit 1 i vs 1 2 2 (a) 1 vs 2 i (b) 1 R 2 6.1 Electrical Elements in the diagram. Although the charge ﬂows counterclockwise, we can think of a positive current ﬂowing clockwise. This current picks up energy as it passes through the battery from the negative to the positive terminal. As it ﬂows through the lightbulb, the current loses energy; this is indicated by the + and − signs above and below the bulb. The bulb current depends on the voltage difference and the bulb’s material properties, which resist the current and cause the energy loss. When a current ﬂows through wire or other circuit elements, it encounters resistance. Sometimes this resistance is desirable and intentionally introduced; sometimes not. A resistor is an element designed to provide resistance. Most resistors are designed to have a linear relation between the current passing through them and the voltage difference across them. This linear relation is Ohm’s law. It states that v = iR where i is the current, v is the voltage difference, and R is the resistance. The unit of resistance is the ohm (), which is one volt per ampere. Figure 6.1.1b shows a voltage source, such as a battery, connected to a resistor. Because of conservation of energy, the voltage increase vs supplied by the source must equal the voltage drop i R across the resistor. Thus the model of this circuit is vs = iR. If we know the voltage and the resistance, we can calculate the current that must be supplied by the source as follows: i = vs /R. 6.1.1 ACTIVE AND PASSIVE ELEMENTS Circuit elements may be classiﬁed as active or passive. Passive elements such as resistors, capacitors, and inductors are not sources of energy, although the latter two can store it temporarily. Elements that provide energy are sources, and elements that dissipate energy are loads. The active elements are energy sources that drive the system. There are several types available; for example, chemical (batteries), mechanical (generators), thermal (thermocouples), and optical (solar cells). Active elements are modeled as either ideal voltage sources or ideal current sources. Their circuit symbols are given in Table 6.1.1. An ideal voltage source supplies the desired voltage no matter how much current is drawn by the loading circuit. An ideal current source supplies whatever current is needed by the loading circuit. Obviously no real source behaves exactly this way. For example, battery voltage drops because of heat produced as current is drawn from the battery. If the current is small the battery can be treated as an ideal voltage source. If not, the battery is frequently modeled as an ideal voltage source plus an internal resistance whose value can be obtained from the voltage-current curve for the battery. Power P is work done per unit time, so the power generated by an active element, or the power dissipated or stored by a passive element, can be calculated as follows: work charge work = = voltage × current power = time unit charge time Thus the power generated, dissipated, or stored by a circuit element equals the product of the voltage difference across the element and the current ﬂowing through the element. That is, P = iv. The unit of power in the SI system is one joule per second, which is deﬁned to be a watt (W). If the element is a linear resistor, the power dissipated is given by P = iv = i 2 R = v2 R 321 322 CHAPTER 6 Electrical and Electromechanical Systems Table 6.1.1 Electrical quantities. Quantity Voltage Units Circuit symbol volt (V) Voltage Source ⫹ ⫺ Charge Current coulomb (C) = N · m/V ampere (A) = C/s Resistance ohm () = V/A Capacitance farad (F) = C/V Inductance henry (H) = V · s/A Battery — Ground — Terminals (input or output) — Current Source v i R C L The appropriate form to use depends on which two of the three quantities i, v, and R are known. 6.1.2 MODELING CIRCUITS The dynamics of physical systems result from the transfer, loss, and storage of mass or energy. A basic law used to model electrical systems is conservation of charge, also known as Kirchhoff’s current law. Another basic law is conservation of energy. In electrical systems, conservation of energy is commonly known as Kirchhoff’s voltage law, which states that the algebraic sum of the voltages around a closed circuit or loop must be zero. These physical laws alone do not provide enough information to write the equations that describe the system. Three more types of information must be provided; the four requirements are 1. The appropriate physical laws, such as conservation of charge and energy. 2. Empirically based descriptions, called constitutive relations, for some or all of the system elements. 3. The speciﬁc way the system elements are arranged or connected. 4. Any relationships due to integral causality, such as the relation between charge and current, Q = i dt. The voltage-current relation for a resistor, v = iR, is an example of an empirically based description of a system element, the third type of required information. This type of description is an algebraic relation not derivable from a basic physical law. Rather, the relation is obtained from a series of measurements. For example, if we apply a range of currents to a resistor, then measure the resulting voltage difference for each current, we would ﬁnd that the voltage is directly proportional to the applied current. The constant of proportionality is the resistance R, and we can determine its value from the test results. 6.1 Electrical Elements 323 Knowledge of the elements’ arrangement is important because it is possible to connect two elements in more than one way. For example, two resistors can be connected differently to form two different circuits (Figures 6.1.2 and 6.1.3); the models are different for each circuit. We can use the voltage-current relation for a resistor along with conservation of charge and conservation of energy to obtain the models. Figure 6.1.2 Series resistors. 6.1.3 SERIES RESISTANCES For Figure 6.1.2, conservation of charge implies that the current is the same through each resistor. When traversing the loop clockwise, a voltage increase occurs when we traverse the source vs , and a voltage decrease occurs when we traverse each resistor. Assign a positive sign to a voltage increase, and a negative sign to a voltage decrease. Then Kirchhoff’s voltage law gives vs − v1 − v2 = vs − iR1 − iR2 = 0 i 1 vs 2 R2 1 v1 2 1 v2 2 Figure 6.1.3 Parallel resistors. i or vs = (R1 + R2 )i R1 (6.1.1) Thus, the supply voltage vs must equal the sum of the voltages across the two resistors. Equation (6.1.1) is an illustration of the series resistance law. If the same current passes through two or more electrical elements, those elements are said to be in series, and the series resistance law states that they are equivalent to a single resistance R that is the sum of the individual resistances. Thus the circuit in Figure 6.1.2 can be modeled as the simpler circuit in Figure 6.1.1b. Because i = vs /(R1 + R2 ) in Figure 6.1.2, R1 vs (6.1.2) v1 = R1 i = R1 + R2 R2 v2 = R 2 i = vs (6.1.3) R1 + R2 These equations express the voltage-divider rule. They show that R1 v1 = (6.1.4) v2 R2 This rule is useful for reducing a circuit with many resistors to an equivalent circuit containing one resistor. 6.1.4 PARALLEL RESISTANCES For Figure 6.1.3, conservation of charge gives i = i 1 + i 2 . From Kirchhoff’s voltage law we see that the voltage across each resistor must be vs , and thus, i 1 = vs /R1 and i 2 = vs /R2 . Combining these three relations gives vs vs i = i1 + i2 = + R1 R2 Solve for i to obtain 1 1 + vs (6.1.5) i= R1 R2 Equation (6.1.5) is an illustration of the parallel resistance law. If the same voltage difference exists across two or more electrical elements, those elements are said to be 1 vs 2 i1 i2 R1 R2 324 CHAPTER 6 Electrical and Electromechanical Systems in parallel. For two resistors in parallel, the parallel resistance law states that they are equivalent to a single resistance R given by 1 1 1 = + R R1 R2 (6.1.6) Thus, the circuit in Figure 6.1.3 can be modeled as the simpler circuit in Figure 6.1.1b. This formula can be extended to more than two resistors. Note that R2 i (6.1.7) i1 = R1 + R2 R1 i2 = i (6.1.8) R1 + R2 These equations express the current-divider rule. They show that i1 R2 = i2 R1 (6.1.9) The current-divider rule can be used to ﬁnd the equivalent resistance of a circuit with many resistors. 6.1.5 NONLINEAR RESISTANCES Not all resistance elements have the linear voltage-current relation v = iR. An example of a speciﬁc diode’s voltage-current relationship found from experiments is i = 0.16(e0.12v − 1) For low voltages, we can approximate this curve with a straight line whose slope equals the derivative di/dv at v = 0. di = 0.16 0.12e0.12v v=0 = 0.0192 dv v=0 Thus, for small voltages, i = 0.0192v, and the resistance is R = 1/0.0192 = 52 . 6.1.6 CAPACITANCE A capacitor is designed to store charge. The capacitance C of a capacitor is a measure of how much charge can be stored for a given voltage difference across the element. Capacitance thus has the units of charge per volt. This unit is named the farad (F). For a capacitor, Q = i dt, where Q is the charge on the capacitor, and i is the current passing through the capacitor. The constitutive relation for a capacitor is v = Q/C, where v is the voltage across the capacitor. Combining these two relations gives 1 t Q0 1 i dt = i dt + v= C C 0 C where Q 0 is the initial charge on the capacitor at time t = 0. In derivative form, this relation is expressed as i =C dv dt 6.1 Electrical Elements 6.1.7 INDUCTANCE A magnetic ﬁeld (a ﬂux) surrounds a moving charge or current. If the conductor of the current is coiled, the ﬂux created by the current in one loop affects the adjacent loops. This ﬂux is proportional to the time integral of the applied voltage, and the current is proportional to the ﬂux. The constitutive relation for an inductor is φ = Li, where L is the inductance and φ is the ﬂux across the inductor. The integral causality relation between ﬂux and voltage is φ = v dt Combining the two preceding expressions for φ gives the voltage-current relation for the inductor 1 v dt i= L which is equivalent to v=L di dt The unit of inductance is the henry (H), which is one volt-second per ampere. 6.1.8 POWER AND ENERGY The power dissipated by or stored by an electrical element is the product of its current and the voltage across it: P = iv. Capacitors and inductors store electrical energy as stored charge and in a magnetic ﬁeld, respectively. The energy E stored in a capacitor can be found as follows: dv 1 C v dt = C v dv = Cv 2 E = P dt = iv dt = dt 2 Similarly, the energy E stored in an inductor is E= P dt = iv dt = i di L dt dt = L i di = 1 2 Li 2 Table 6.1.2 summarizes the voltage-current relations and the energy expressions for resistance, capacitance, and inductance elements. Table 6.1.2 Voltage-current and energy relations for circuit elements. Resistance: v = iR Capacitance: v = Inductance: 1 C di v=L dt P = Ri 2 = t i dt + 0 Q0 C 1 2 Cv 2 1 E = Li 2 2 E= v2 R 325 326 CHAPTER 6 Electrical and Electromechanical Systems 6.2 CIRCUIT EXAMPLES The examples in this section illustrate how to apply the basic circuit principles introduced in Section 6.1. E X A M P L E 6.2.1 Current and Power in a Resistance Circuit Figure 6.2.1 A simple resistance circuit. ■ Problem i 1 vs 2 For the circuit shown in Figure 6.2.1, the applied voltage is vs = 6 V and the resistance is R = 10 . Determine the current and the power that must be produced by the power supply. 1 R 2 ■ Solution The current is found from i = vs /R = 6/10 = 0.6 A. The power is computed from P = vs2 /R = 62 /10 = 3.6 W. Note that we can also compute the power from P = ivs . A Summing Circuit E X A M P L E 6.2.2 ■ Problem Figure 6.2.2 shows a circuit for summing the voltages v1 and v2 to produce v3 . Derive the expression for v3 as a function of v1 and v2 , for the case where R1 = R2 = 10 , and R3 = 20 . Figure 6.2.2 A summing circuit. i2 1 v2 2 R2 ■ Solution R1 i 1 1 v1 2 i3 Deﬁne the currents shown in the diagram. The voltage-current relation for each resistor gives v1 − v 3 R1 v2 − v 3 i2 = R2 v3 i3 = R3 v3 i1 = R3 There is no capacitance in the circuit. Therefore, no charge can be stored anywhere in the circuit. Thus conservation of charge gives i 3 = i 1 + i 2 . Substituting the expressions for the currents into this equation, we obtain v3 v1 − v 3 v2 − v3 = + R3 R1 R2 which gives v3 = 0.4(v1 + v2 ) Thus, v3 is proportional to the sum of v1 and v2 . This is true in general only if R1 = R2 . E X A M P L E 6.2.3 Application of the Voltage-Divider Rule ■ Problem Consider the circuit shown in Figure 6.2.3. Obtain the voltage vo as a function of the applied voltage vs by applying the voltage-divider rule. Use the values R1 = 5 , R2 = 10 , R3 = 6 , and R4 = 2 . 6.2 R1 R3 A i1 2 i2 vs R2 5V vA vo R4 2 6V 5V 1 vs 327 Figure 6.2.3 A resistance network with two loops. 1 i3 1 Circuit Examples 10 V 2V 5V vA 1 vo vs 2 1 10 V 8V 2 40 V 9 vs 2 (b) (a) Figure 6.2.4 Application of the voltage-divider rule. vA (c) ■ Solution Let v A be the voltage at the node shown in Figure 6.2.4a. The voltage-divider rule applied to resistors R3 and R4 gives R4 2 1 vA = (1) vA = vA R3 + R4 6+2 4 Because resistors R3 and R4 are in series, we can add their values to obtain their equivalent resistance Rs = 2 + 6 = 8 . The equivalent circuit is shown in Figure 6.2.4b. Resistors Rs and R2 are parallel, so we can combine their values to obtain their equivalent resistance R p as follows: vo = 1 1 9 1 = + = Rp 10 8 40 Thus, R p = 40/9. The equivalent circuit is shown in Figure 6.2.4c. Finally, we apply the voltage-divider rule again to obtain vA = 40/9 8 vs = vs 5 + 40/9 17 Using (1) and (2) we obtain 1 1 vo = v A = 4 4 8 17 vs = (2) 2 vs 17 A potentiometer is a resistance with a sliding electrical pick-off (Figure 6.2.5a). Thus the resistance R1 between the sliding contact and ground is a function of the distance x of the contact from the end of the potentiometer. Potentiometers, commonly Figure 6.2.5 A translational (linear) potentiometer. L V x 1 V 2 R2 R1 vo (a) (b) vo 328 CHAPTER 6 Electrical and Electromechanical Systems called pots, are used as linear and angular position sensors. The volume control knob on some radios is a potentiometer that is used to adjust the voltage to the speakers. Potentiometers E X A M P L E 6.2.4 ■ Problem Assuming the potentiometer resistance R1 is proportional to x, derive the expression for the output voltage vo as a function of x. ■ Solution The length of the pot is L and its total resistance is R1 + R2 . Figure 6.2.5b shows the circuit diagram of the system. From the voltage-divider rule, R1 V (1) vo = R1 + R2 Because the resistance R1 proportional to x, x R1 = (R1 + R2 ) L Substituting this into equation (1) gives x vo = V = K x L where K = V /L is the gain of the pot. Figure 6.2.6 A rotational potentiometer. ␪ The voltage source for the pot can be a battery or it can be a power supply. A rotational potentiometer is shown in Figure 6.2.6. Following a similar procedure we can show that if the resistance is proportional to θ, then vo = ␪max vo θ θmax V = Kθ where K = V /θmax . V E X A M P L E 6.2.5 Maximum Power Transfer in a Speaker-Ampliﬁer System ■ Problem A common example of an electrical system is an ampliﬁer and a speaker. The load is the speaker, which requires current from the ampliﬁer in order to produce sound. In Figure 6.2.7a the resistance R L is that of the load. Part (b) of the ﬁgure shows the circuit representation of the system. The source supplies a voltage v S and a current i S , and has its own internal resistance R S . For optimum efﬁciency, we want to maximize the power supplied to the speaker, for given values of v S and R S . Determine the value of R L to maximize the power transfer to the load. ■ Solution The required model should describe the power supplied to the speaker in terms of v S , R S , and R L . From Kirchhoff’s voltage law, vs − i S R S − i S R L = 0 6.2 Power Source (e.g., an amplifier) RL Circuit Examples 329 Figure 6.2.7 (a) An ampliﬁer-speaker system. (b) Circuit representation with a voltage source and a resistive load. Load (e.g., a speaker) (a) Load Source iS 1 2 1 RS RL vS vL 2 (b) We want to ﬁnd v L in terms of v S . From the voltage-divider rule, vL = RL vS RS + RL The power consumed by the load is PL = i S2 R L = v L2 /R L . Using the relation between v L and v S we can express PL in terms of v S as PL = RL v2 (R S + R L )2 S To maximize PL for a ﬁxed value of v S , we must choose R L to maximize the ratio r= RL (R S + R L )2 The maximum of r occurs where dr/d R L = 0. dr (R S + R L )2 − 2R L (R S + R L ) = =0 d RL (R S + R L )4 This is true if R L = R S . Thus to maximize the power to the load we should choose the load resistance R L to be equal to the source resistance R S . This result for a resistance circuit is a special case of the more general result known as impedance matching. A Feedback Ampliﬁer ■ Problem Early in the twentieth century engineers struggled to design vacuum-tube ampliﬁers whose gain remained constant at a predictable value. The gain is the ratio of the output voltage to the input voltage. The vacuum-tube gain G can be made large but is somewhat unpredictable and unreliable due to heat effects and manufacturing variations. A solution to the problem is shown in Figure 6.2.8. Derive the input-output relation for vo as a function of vi . Investigate the case where the gain G is very large. E X A M P L E 6.2.6 330 CHAPTER 6 ■ Solution Figure 6.2.8 A feedback ampliﬁer. vi Electrical and Electromechanical Systems Part of the voltage drop across the resistors is used to raise the ground level at the ampliﬁer input, so the input voltage to the ampliﬁer is vi − R2 vo /(R1 + R2 ). Thus the ampliﬁer’s output is vo G vo = G vi − R2 vo R1 + R2 Solve for vo : R1 R2 vo = G vi 1 + G R2 /(R1 + R2 ) If G R2 /(R1 + R2 ) 1, then vo ≈ R1 + R2 vi R2 Presumably, the resistor values are sufﬁciently accurate and constant enough to allow the gain (R1 + R2 )/R2 to be predictable and reliable. 6.2.1 LOOP CURRENTS Sometimes the circuit equations can be simpliﬁed by using the concept of a loop current, which is a current identiﬁed with a speciﬁc loop in the circuit. A loop current is not necessarily an actual current. If an element is part of two or more loops, the actual current through the element is the algebraic sum of the loop currents. Use of loop currents usually reduces the number of unknowns to be found, although when deriving the circuit equations you must be careful to use the proper algebraic sum for each element. E X A M P L E 6.2.7 Analysis with Loop Currents ■ Problem We are given the values of the voltages and the resistances for the circuit in Figure 6.2.9a. (a) Solve for the currents i 1 , i 2 , and i 3 passing through the three resistors. (b) Use the loop-current method to solve for the currents. ■ Solution a. Applying Kirchhoff’s voltage law to the left-hand loop gives v1 − R 1 i 1 − R 2 i 2 = 0 For the right-hand loop, v2 − R 2 i 2 + i 3 R 3 = 0 Figure 6.2.9 Example of loop analysis. R3 R1 1 i1 v1 2 i3 1 1 2 2 v2 R2 i2 (a) v1 R1 R3 iA R2 iB 1 v2 2 (b) 6.2 Circuit Examples 331 From conservation of charge, i 1 = i 2 + i 3 . These are three equations in three unknowns. Their solution is b. i1 = (R2 + R3 )v1 − R2 v2 R1 R2 + R1 R3 + R2 R3 i2 = R3 v1 + R1 v2 R1 R2 + R1 R3 + R2 R3 i3 = R2 v1 − (R1 + R2 )v2 R1 R2 + R1 R3 + R2 R3 (1) Deﬁne the loop currents i A and i B positive clockwise, as shown in Figure 6.2.9b. Note that there is a voltage drop R2 i A across R2 due to i A and a voltage increase R2 i B due to i B . Applying Kirchhoff’s voltage law to the left-hand loop gives v1 − R1 i A − R2 i A + R2 i B = 0 For the right-hand loop, v2 + R 3 i B + R 2 i B − R 2 i A = 0 Now we have only two equations to solve. Their solution is iA = (R2 + R3 )v1 − R2 v2 R1 R2 + R1 R3 + R2 R3 (2) iB = R2 v1 − (R1 + R2 )v2 R1 R2 + R1 R3 + R2 R3 (3) The current i 1 through R1 is the same as i A , and the current i 3 through R3 is the same as i B . The current i 2 through R2 is i 2 = i A − i B , which gives expression (1) when expressions (2) and (3) are substituted. 6.2.2 CAPACITANCE AND INDUCTANCE IN CIRCUITS Examples 6.2.8 through 6.2.13 illustrate how models are developed for circuits containing capacitors or inductors. Series RC Circuit E X A M P L E 6.2.8 ■ Problem The resistor and capacitor in the circuit shown in Figure 6.2.10 are said to be in series because the same current ﬂows through them. Obtain the model of the capacitor voltage v1 . Assume that the supply voltage vs is known. ■ Solution For the capacitor, v1 = 1 C t i dt + 0 vs (1) Q0 C R 1 From Kirchhoff’s voltage law, vs − Ri − v1 = 0. This gives 1 i = (vs − v1 ) R Figure 6.2.10 A series RC Circuit. 2 i C v1 332 CHAPTER 6 Electrical and Electromechanical Systems Differentiate this with respect to t to obtain 1 dv1 = i dt C Then substitute for i from (1): dv1 1 = (vs − v1 ) dt RC This the required model. It is often expressed in the following rearranged form: RC E X A M P L E 6.2.9 dv1 + v1 = vs dt (2) Pulse Response of a Series RC Circuit ■ Problem A rectangular pulse input is a positive step function that lasts a duration D. One way of producing a step voltage input is to use a switch like that shown in Figure 6.2.11a. The battery voltage V is constant and the switch is initially closed at point B. At t = 0 the switch is suddenly moved from point B to point A. Then at t = D the switch is suddenly moved back to point B. Obtain the expression for the capacitor voltage v1 (t) assuming that v1 (0) = 0. ■ Solution When at t = 0 the switch is suddenly moved from point B to point A, the circuit is identical to that shown in Figure 6.2.10 with vs = V , and its model is dv1 (1) + v1 = V dt The input voltage is a step function of magnitude V. Using the methods of Chapter 3, we can obtain the following solution for v1 as a function of time. Since v1 (0) = 0, the solution is the forced response. RC v1 (t) = V 1 − e−t/RC (2) When the switch is moved back to point B at time t = D, the circuit is equivalent to that shown in Figure 6.2.11b, whose model is equation (1) with V = 0. The solution of this equation for t ≥ D is simply the free response with the initial condition v1 (D), whose expression can be obtained from equation (2). v1 (t) = v1 (D)e−(t−D)/RC = V 1 − e−D/RC e−(t−D)/RC The solution is sketched in Figure 6.2.12. Figure 6.2.11 (a) Series RC circuit with a switch. (b) Circuit with switch in position B. R A 1 V B R C v1 C 2 (a) (b) v1 6.2 Circuit Examples 333 Figure 6.2.12 Pulse response of a series RC circuit. v1(D) v1(t) 0 0 D 0.02v1(D) D 1 4RC t Series RCL Circuit E X A M P L E 6.2.10 ■ Problem The resistor, inductor, and capacitor in the circuit shown in Figure 6.2.13 are in series because the same current ﬂows through them. Obtain the model of the capacitor voltage v1 with the supply voltage vs as the input. ■ Solution Figure 6.2.13 A series RCL circuit. R i 1 From Kirchhoff’s voltage law, vs di vs − Ri − L − v1 = 0 dt (1) L For the capacitor, v1 = 1 C t i dt 0 Differentiate this with respect to t to obtain i =C dv1 dt and substitute this for i in (1): vs − RC dv1 d 2 v1 − LC 2 − v1 = 0 dt dt This is the required model. It can be expressed in the following form: LC dv1 d 2 v1 + RC + v1 = vs dt 2 dt C 2 (2) v1 334 CHAPTER 6 Electrical and Electromechanical Systems E X A M P L E 6.2.11 Parallel RL Circuit ■ Problem The resistor and inductor in the circuit shown in Figure 6.2.14 are said to be in parallel because they have the same voltage v1 across them. Obtain the model of the current i 2 passing through the inductor. Assume that the supply current i s is known. Figure 6.2.14 A parallel RL circuit. i1 is i2 R L ■ Solution v1 The currents i 1 and i 2 are deﬁned in the ﬁgure. Then, v1 = L di 2 = Ri 1 dt (1) From conservation of charge, i 1 + i 2 = i s . Thus, i 1 = i s − i 2 . Substitute this expression into (1) to obtain di 2 L = R(i s − i 2 ) dt This is the required model. It can be rearranged as follows: L di 2 + i2 = is R dt (2) Analysis of a Telegraph Line E X A M P L E 6.2.12 ■ Problem Figure 6.2.15 shows a circuit representation of a telegraph line. The resistance R is the line resistance and L is the inductance of the solenoid that activates the receiver’s clicker. The switch represents the operator’s key. Assume that when sending a “dot,” the key is closed for 0.1 s. Using the values R = 20 and L = 4 H, obtain the expression for the current i(t) passing through the solenoid. Figure 6.2.15 Circuit representation of a telegraph line. R 12v 1 2 L i ■ Solution From the voltage law we have L di + Ri = vi (t) dt (1) where vi (t) represents the input voltage due to the switch and the 12-V supply. We could model vi (t) as a rectangular pulse of height 12 V and duration 0.1 s, but the differential equation (1) is easier to solve if we model vi (t) as an impulsive input of strength 12(0.1) = 1.2 V · s. This model can be justiﬁed by the fact that the circuit time constant, L/R = 4/20 = 0.2, is greater than the duration of vi (t). Thus we model vi (t) as vi (t) = 1.2δ(t). The Laplace transform of equation (1) with i(0) = 0 gives (4s + 20)I (s) = 1.2 or I (s) = 0.3 1.2 = 4s + 20 s+5 This gives the solution i(t) = 0.3e−5t A Note that this solution gives i(0+) = 0.3, whereas i(0) = 0. The difference is due to the impulsive input. 6.2 Circuit Examples 335 An RLC Circuit with Two Input Voltages E X A M P L E 6.2.13 The RLC circuit shown in Figure 6.2.16 has two input voltages. Obtain the differential equation model for the current i 3 . Figure 6.2.16 An RLC circuit with two voltage sources. ■ Problem i1 ■ Solution Applying Kirchhoff’s voltage law to the left-hand loop gives di 3 v1 − Ri 1 − L =0 dt For the right-hand loop, (1) 1 v1 2 C i2 R L i3 1 v2 2 1 di 3 i 2 dt − L =0 C dt Differentiate this equation with respect to t: v2 − dv2 d 2i3 1 − i2 − L 2 = 0 dt C dt (2) i3 = i1 + i2 (3) From conservation of charge, These are three equations in the three unknowns i 1 , i 2 , and i 3 . To eliminate i 1 and i 2 , solve equation (1) for i 1 1 i1 = R di 3 v1 − L dt (4) In equation (2), substitute for i 2 from equation (3): dv2 d 2i3 1 − (i 3 − i 1 ) − L 2 = 0 dt C dt Now substitute for i 1 from equation (4): dv2 1 1 − i3 + dt C C 1 R di 3 v1 − L dt −L d 2i3 =0 dt 2 Rearrange this equation to obtain the answer: LRC d 2i3 di 3 dv2 +L + Ri 3 = v1 + RC dt 2 dt dt (5) 6.2.3 STATE-VARIABLE MODELS OF CIRCUITS The presence of several current and voltage variables in a circuit can sometimes lead to difﬁculty in identifying the appropriate variables to use for expressing the circuit model. Use of state variables can often reduce this confusion. To choose a proper set of state variables, identify where the energy is stored in the system. The variables that describe the stored energy are appropriate choices for state variables. State-Variable Model of a Series RLC Circuit ■ Problem Consider the series RLC circuit shown in Figure 6.2.17. Choose a suitable set of state variables, and obtain the state variable model of the circuit in matrix form. The input is the voltage vs . E X A M P L E 6.2.14 336 Figure 6.2.17 Series RLC circuit. CHAPTER 6 R 1 vs Electrical and Electromechanical Systems L i C v1 2 ■ Solution In this circuit the energy is stored in the capacitor and in the inductor. The energy stored in the capacitor is Cv12 /2 and the energy stored in the inductor is Li 2 /2. Thus a suitable choice of state variables is v1 and i. From Kirchhoff’s voltage law, di vs − Ri − L − v1 = 0 dt Solve this for di/dt: di 1 R 1 = vs − v1 − i dt L L L This is the ﬁrst state equation. Now ﬁnd an equation for dv1 /dt. From the capacitor relation, 1 i dt v1 = C Differentiating gives the second state equation. 1 dv1 = i dt C The two state equations can be expressed in matrix form as follows. ⎡ di ⎤ ⎡ ⎤ ⎡ ⎤ R 1 1 − − ⎢ dt ⎥ ⎢ L L⎥ i + ⎣L⎦v = ⎣ ⎦ ⎣ 1 ⎦ v s dv1 1 0 0 C dt 6.3 TRANSFER FUNCTIONS AND IMPEDANCE The Laplace transform and the transfer function concept enable us to deal with algebraic equations rather than differential equations, and thus ease the task of analyzing circuit models. This section illustrates why this is so, and introduces a related concept, impedance, which is simply the transfer function between voltage and current. 6.3.1 USE OF TRANSFORMED EQUATIONS Applying the Laplace transform to the circuit equations often helps in eliminating intermediate variables, because the transformed equations are algebraic and therefore are easier to solve. This is especially true when the circuit contains dynamic elements (capacitances or inductances). E X A M P L E 6.3.1 An RLC Circuit with Two Inputs ■ Problem Consider the RLC circuit treated in Example 6.2.14 and shown in Figure 6.2.17. Use the Laplace transform to obtain the differential equation model for the current i 3 . 6.3 Transfer Functions and Impedance ■ Solution Applying Kirchhoff’s voltage law to the left-hand loop gives di 3 =0 dt which, when transformed for zero initial conditions, gives v1 − Ri 1 − L V1 (s) − R I1 (s) − Ls I3 (s) = 0 Similarly for the right-hand loop, v2 − 1 C i 2 dt − L (1) di 3 =0 dt or 1 I2 (s) − Ls I3 (s) = 0 Cs From conservation of charge, i 3 = i 1 + i 2 , or V2 (s) − I3 (s) = I1 (s) + I2 (s) (2) (3) These are three equations in the three unknowns I1 (s), I2 (s), and I3 (s). To eliminate I1 (s) and I2 (s), solve equation (1) for I1 (s): 1 [V1 (s) − Ls I3 (s)] R (4) I2 (s) = Cs[V2 (s) − Ls I3 (s)] (5) I1 (s) = From equation (2), Substitute equations (4) and (5) into equation (3): I3 (s) = I1 (s) + I2 (s) = 1 [V1 (s) − Ls I3 (s)] + Cs[V2 (s) − Ls I3 (s)] R and collect the I3 (s) terms: (LRCs2 + Ls + R)I3 (s) = V1 (s) + RCsV2 (s) This transformed equation corresponds to the differential equation: LRC d 2i3 di 3 dv2 +L + Ri 3 = v1 + RC dt 2 dt dt (6) 6.3.2 TRANSFER FUNCTIONS We have seen that the complete response of a linear differential equation is the sum of the free and the forced responses. For zero initial conditions the free response is zero, and the complete response is the same as the forced response. Thus we can focus our analysis solely on the effects of the input by taking the initial conditions to be zero temporarily. When we have ﬁnished analyzing the effects of the input, we can add to the result the free response caused by any nonzero initial conditions. In Chapter 2 we introduced the concept of the transfer function, which is useful for analyzing the effects of the input. Consider the model τ ẋ + x = b f (t) (6.3.1) where τ is the time constant. Assume that x(0) = 0 and transform both sides of the equation to obtain τ s X (s) + X (s) = bF(s). Solve for the ratio X (s)/F(s) and denote 337 338 CHAPTER 6 Electrical and Electromechanical Systems it by T (s): T (s) = X (s) b = F(s) τs + 1 (6.3.2) The function T (s) is called the transfer function of the system whose model is (6.3.1). Note that we can obtain the time constant from a transfer function having a ﬁrstorder denominator by expressing it in the form of (6.3.2). For example, 3 1.5 X (s) = = F(s) 10s + 2 5s + 1 so the time constant is 5. The transfer function is the transform of the forced response divided by the transform of the input. It can be used as a multiplier to obtain the forced response transform from the input transform; that is, X (s) = T (s)F(s). It is important to realize that the transfer function is equivalent to the differential equation model. If we are given the transfer function, we can reconstruct the corresponding differential equation. For example, the transfer function 5 X (s) = 2 F(s) s + 7s + 10 corresponds to the equation ẍ + 7ẋ + 10x = 5 f (t). Relating the differential equation and the transfer function is easily done because the initial conditions are assumed to be zero when working with transfer functions. From the derivative property, this means that to work with a transfer function you can use the relations L(ẋ) = s X (s), L(ẍ) = s 2 X (s), and so forth. Note that the denominator of the transfer function is the characteristic polynomial, and thus the transfer function tells us something about the intrinsic behavior of the model, apart from the effects of the input and speciﬁc values of the initial conditions. In the previous transfer function X (s)/F(s), the characteristic polynomial is s 2 + 7s + 10 and the roots are −2 and −5. The roots are real, and this tells us that the free response does not oscillate and that the forced response does not oscillate unless the input is oscillatory. Because the roots are negative, the model is stable and its free response disappears with time. E X A M P L E 6.3.2 Coupled RC Loops ■ Problem (a) Determine the transfer function Vo (s)/Vs (s) of the circuit shown in Figure 6.3.1. (b) Use a block-diagram representation of the circuit to show how the output voltage vo affects the internal voltage v1 . i1 Figure 6.3.1 Coupled RC loops. R R v1 1 vs 2 C i2 C i3 vo 6.3 Transfer Functions and Impedance ■ Solution a. The energy in this circuit is stored in the two capacitors. Because the energy stored in a capacitor is expressed by Cv 2 /2, appropriate choices for the state variables are the voltages v1 and vo . The capacitance relations are 1 1 i3 v1 = i2 vo = C C which give the state equations i3 dvo = (1) dt C dv1 i2 = (2) dt C For the right-hand loop, v1 − Ri 3 − vo = 0 Thus i3 = v1 − v o R (3) and equation (1) becomes 1 dvo = (v1 − vo ) dt RC (4) For the left-hand loop, vs − Ri 1 − v1 = 0 which gives vs − v 1 R From conservation of charge and equations (3) and (5), vs − v 1 v1 − vo 1 − = (vs − 2v1 + vo ) i2 = i1 − i3 = R R R Substituting this into equation (2) gives 1 dv1 = (vs − 2v1 + vo ) dt RC Equations (4) and (7) are the state equations. To obtain the transfer function Vo (s)/Vs (s), transform these equations for zero initial conditions to obtain 1 [V1 (s) − Vo (s)] sVo (s) = RC 1 sV1 (s) = [Vs (s) − 2V1 (s) + Vo (s)] RC i1 = b. (5) (6) (7) Eliminating V1 (s) from these two equations gives 1 Vs (s) Vo (s) = 2 2 2 R C s + 3RCs + 1 So the transfer function is Vo (s) 1 = 2 2 2 (8) Vs (s) R C s + 3RCs + 1 In the block diagram in Figure 6.3.2a, the left-hand inner loop is based on equations (2) and (6). The right-hand inner loop is based on equations (1) and (3). The diagram can be simpliﬁed by reducing each inner loop to a single block, as shown in part (b) of the ﬁgure. 339 340 CHAPTER 6 Electrical and Electromechanical Systems Figure 6.3.2 Block diagrams of coupled RC loops. Vs(s) 1 1 1 2 1 R I2(s) 1 Cs V1(s) 1 1 R 2 I3(s) 1 Cs Vo(s) 2 (a) Vs(s) 1 1 RCs 1 2 1 V1(s) 1 RCs 1 1 Vo(s) (b) This simpler diagram shows how the output voltage vo affects the inner voltage v1 through the outer feedback loop. Some models have more than one input or more than one output. In such cases, there is one transfer function for each input-output pair. If a model has more than one input, a particular transfer function is the ratio of the output transform over the input transform, with all the remaining inputs ignored (set to zero temporarily). E X A M P L E 6.3.3 Transfer Functions for Two Inputs ■ Problem For the circuit of Example 6.3.1, the model is d 2i3 di 3 dv2 +L + Ri 3 = v1 + RC (1) 2 dt dt dt where the two inputs are v1 and v2 and the current i 3 has been selected as the output. Determine the transfer functions. LRC ■ Solution Transforming equation (1) for zero initial conditions and solving for I3 (s) gives V1 (s) + RCsV2 (s) RCs 1 = V1 (s) + V2 (s) LRCs2 + Ls + R LRCs2 + Ls + R LRCs2 + Ls + R Temporarily setting V2 (s) = 0, we have I3 (s) = (2) I3 (s) 1 = 2 V1 (s) LRCs + Ls + R This is the transfer function for the input v1 . To ﬁnd the transfer function for the input v2 , set V1 (s) = 0 in (2) to obtain RCs I3 (s) = V2 (s) L RCs 2 + Ls + R The characteristic polynomial for the system is L RCs 2 + Ls + R, which can be obtained from the denominator of either transfer function. The numerators are different, however. Note that the second transfer function has numerator dynamics, which indicates that i 3 responds to the derivative v̇ 2 . 6.3 Transfer Functions and Impedance 341 6.3.3 IMPEDANCE We have seen that a resistance resists or “impedes” the ﬂow of current. The corresponding relation is v/i = R. Capacitance and inductance elements also impede the ﬂow of current. In electrical systems an impedance is a generalization of the resistance concept and is deﬁned as the ratio of a voltage transform V (s) to a current transform I (s) and thus implies a current source. A standard symbol for impedance is Z (s). Thus V (s) (6.3.3) Z (s) = I (s) The impedance of a resistor is its resistance R. The impedances of the other two common electrical elements are found as follows. For a capacitor, 1 t i dt v(t) = C 0 or V (s) = I (s)/(Cs). The impedance of a capacitor is thus 1 Z (s) = (6.3.4) Cs For an inductor, di v(t) = L dt or V (s) = Ls I (s). Thus the impedance of an inductor is Z (s) = Ls (6.3.5) 6.3.4 SERIES AND PARALLEL IMPEDANCES The concept of impedance is useful because the impedances of individual elements can be combined with series and parallel laws to ﬁnd the impedance at any point in the system. The laws for combining series or parallel impedances are extensions to the dynamic case of the laws governing series and parallel resistance elements. Two impedances are in series if they have the same current. If so, the total impedance is the sum of the individual impedances. Z (s) = Z 1 (s) + Z 2 (s) For example, a resistor R and capacitor C in series, as shown in Figure 6.3.3a, have the equivalent impedance RCs + 1 1 = Z (s) = R + Cs Cs Thus the relation between the current i ﬂowing through them and the total voltage drop v across them is RCs + 1 V (s) = Z (s) = I (s) Cs i 1 v 2 (a) Figure 6.3.3 Series and parallel RC circuits. i R 1 C 2 v R (b) C 342 CHAPTER 6 Electrical and Electromechanical Systems and the differential equation model is di dv = RC + i(t) dt dt If the impedances have the same voltage difference across them, they are in parallel, and their impedances combine by the reciprocal rule 1 1 1 = + Z (s) Z 1 (s) Z 2 (s) C where Z (s) is the total equivalent impedance. If a resistor R and capacitor C are in parallel, as shown in Figure 6.3.3b, their equivalent total impedance Z (s) is found from 1 1 1 = + Z (s) 1/Cs R or R RCs + 1 Thus the relation between the total current i and the voltage drop v across them is Z (s) = V (s) R = Z (s) = I (s) RCs + 1 and the differential equation model is RC E X A M P L E 6.3.4 dv + v = Ri(t) dt Circuit Analysis Using Impedance ■ Problem For the circuit shown in Figure 6.3.4a, determine the transfer function between the input voltage vs and the output voltage vo . ■ Solution Note that R and C are in parallel. Therefore their equivalent impedance Z (s) is found from 1 1 1 = + Z (s) 1/Cs R or Z (s) = R RCs + 1 An impedance representation of the equivalent circuit is shown in Figure 6.3.4b. In this representation we may think of the impedance as a simple resistance, provided we express the relations i Figure 6.3.4 Circuit analysis using impedance. R1 i 1 vs R C 2 vo R1 1 vs Z(s) 2 (a) (b) vo 6.3 Transfer Functions and Impedance 343 in Laplace transform notation. Kirchhoff’s voltage law gives Vs (s) − R1 I (s) − Z (s)I (s) = 0 The output voltage is related to the current by Vo (s) = Z (s)I (s). Eliminating I (s) from these two relations gives Vs (s) − R1 Vo (s) − Vo (s) = 0 Z (s) which yields the desired transfer function: Z (s) R Vo (s) = = Vs (s) R1 + Z (s) R R1 Cs + R + R1 This network is a ﬁrst-order system whose time constant is τ= R R1 R + R1 If the voltage output is measured at the terminals to which the driving current is applied, the impedance so obtained is the driving-point or input impedance. If the voltage is measured at another place in the circuit, the impedance obtained is a transfer impedance (because the effect of the input current has been transferred to another point). Sometimes the term admittance is used. This is the reciprocal of impedance, and it is an indication of to what extent a circuit “admits” current ﬂow. 6.3.5 ISOLATION AMPLIFIERS A voltage-isolation ampliﬁer is designed to produce an output voltage that is proportional to the input voltage. It is intended to boost the electrical signal from a low-power source. Therefore, the ampliﬁer requires an external power source, which is not usually shown in the circuit diagrams. We will not be concerned with the internal design details of ampliﬁers, but we need to understand their effects on any circuit in which they are used. Such an ampliﬁer may be considered to be a voltage source if it does not affect the behavior of the source circuit that is attached to the ampliﬁer input terminals, and if the ampliﬁer is capable of providing the voltage independently of the particular circuit (the “load”) attached to ampliﬁer output terminals. Consider the system in Figure 6.3.5. The internal impedances of the ampliﬁer at its input and output terminals are Z i (s) and Z o (s), respectively. The impedance of the source circuit is Z s (s) and the impedance of the load is Z L (s). A simple circuit analysis will reveal that Vs (s) − I1 (s)Z s (s) − Vi (s) = 0 1 i1 Zs(s) vs 2 vi 1 Zi(s) Gvi Zo(s) Figure 6.3.5 Input and output impedance in an ampliﬁer. io ZL(s) 2 Source Load Amplifier vL 344 CHAPTER 6 Electrical and Electromechanical Systems and I1 (s) = Vi (s)/Z i (s). Thus, if Z i (s) is large, the current i 1 drawn by the ampliﬁer will be small. Therefore, if the input impedance Z i (s) is large, the ampliﬁer does not affect the current i 1 , and thus the ampliﬁer does not affect the behavior of the source circuit. In addition, if Z i (s) is large, Vi (s) = Z i (s) Vs (s) ≈ Vs (s) Z i (s) + Z s (s) So we conclude that a voltage-isolation ampliﬁer must have a high input impedance. Denote the ampliﬁer’s voltage gain as G. This means that the ampliﬁer’s output voltage vo is vo = Gvi . For the load circuit, GVi (s) − Io (s)Z o (s) − Io (s)Z L (s) = 0 Thus Io (s) = GVi (s) Z o (s) + Z L (s) and VL (s) = Z L (s)Io (s) = Z L (s) GVo (s) Z o (s) + Z L (s) Thus, if Z o (s) is small, v L ≈ GVo . So if the ampliﬁer output impedance is small the voltage v L delivered to the load is independent of the load. A similar analysis applies to a current ampliﬁer, which provides a current proportional to its input signal regardless of the load. Thus such an ampliﬁer acts as a current source. 6.3.6 LOADING EFFECTS AND BLOCK DIAGRAMS One application of transfer functions is to provide a graphical representation of the system’s dynamics in the form of a block diagram, which illustrates the cause-and-effect relations operating in a particular system. The algebraic representation of the system’s equations in terms of transfer functions enables easier manipulation for analysis and design purposes, and the graphical representation allows the engineer to see the interaction between the system’s components. For example, the equation for the feedback ampliﬁer discussed in Example 6.2.6 and shown in Figure 6.2.8 is R2 vo vo = G vi − R1 The block diagram of this device is given in Figure 6.3.6. It shows the presence and action of the negative feedback loop, by which the output voltage vo is used to modify the behavior of the device. A block diagram example with two inputs is given in Figure 6.3.7, which shows the block diagram of the circuit in Figure 6.2.17, whose transfer functions were obtained in Example 6.3.3. Suppose two elements, whose individual transfer functions are T1 (s) and T2 (s), are physically connected end-to-end so that the output of the left-hand element becomes the input to the right-hand element. We can represent this connection by the block diagram shown in Figure 6.3.8, only if the output w of the right-hand element does not affect the inputs u and v or the behavior of the left-hand element. If it does, the right-hand element is said to “load” the left-hand element. Example 6.3.5 illustrates this point. 6.3 Figure 6.3.6 Block diagram of a feedback ampliﬁer. Vi(s) 1 2 V2(s) RCs R2 R1 V1(s) 1 1 U(s) V(s) T1(s) 345 Figure 6.3.7 Block diagram of an RLC circuit with two voltage sources. Vo(s) G Transfer Functions and Impedance I3(s) 1 LRCs2 1 Ls 1 R W(s) T2(s) Figure 6.3.8 Block diagram of two series elements. Loading Effects, Transfer Functions, and Block Diagrams E X A M P L E 6.3.5 ■ Problem The circuit shown in Figure 6.3.9a consists of two series RC circuits wired so that the output voltage of the ﬁrst circuit is the input voltage to an isolation ampliﬁer. The output voltage of the ampliﬁer is the input voltage to the second RC circuit. The ampliﬁer has a voltage gain G; that is, v2 (t) = Gv1 (t). Derive the transfer function Vo (s)/Vs (s) for this circuit, and for the case G = 1 compare it with the transfer function of the circuit shown in Figure 6.3.1. ■ Solution The ampliﬁer isolates the ﬁrst RC loop from the effects of the second loop; that is, the ampliﬁer prevents the voltage v1 from being affected by the second RC loop. This in effect creates two separate loops with an intermediate voltage source v2 = Gv1 , as shown in Figure 6.3.9b. Thus, for the left-hand loop we obtain V1 (s) 1 = Vs (s) RCs + 1 For the right-hand RC loop, Vo (s) 1 = V2 (s) RCs + 1 For the ampliﬁer with gain G, V2 (s) = GV1 (s) To obtain the transfer function Vo (s)/Vs (s), eliminate the variables V1 (s) and V2 (s) from these equations as follows: Vo (s) Vo (s) V2 (s) V1 (s) 1 1 G = = G = 2 2 2 Vs (s) V2 (s) V1 (s) Vs (s) RCs + 1 RCs + 1 R C s + 2RCs + 1 R 1 vs 2 R R C v1 G (a) v2 C vo 1 vs 2 Figure 6.3.9 Two RC loops with an isolation ampliﬁer. R C 1 Gv vs1 2 v1 (b) (1) C vo 346 CHAPTER 6 Electrical and Electromechanical Systems Figure 6.3.10 Block diagrams of two RC loops with an isolation ampliﬁer. Vs(s) 1 RCs 1 1 V1(s) G V2(s) Vo(s) 1 RCs 1 1 Vs(s) G (RCs 1 1)(RCs 1 1) Vo(s) (b) (a) This procedure is described graphically by the block diagram shown in Figure 6.3.10a. The three blocks can be combined into one block as shown in part (b) of the ﬁgure. The transfer function of the circuit shown in Figure 6.3.1 was derived in Example 6.3.2. It is Vo (s) 1 = 2 2 2 Vs (s) R C s + 3RCs + 1 (2) Note that it is not the same as the transfer function given by equation (1) with G = 1. A common mistake is to obtain the transfer function of loops connected end-to-end by multiplying their transfer functions. This is equivalent to treating them as independent loops, which they are not, because each loop “loads” the adjacent loops and thus changes the currents and voltages in those loops. An isolation ampliﬁer prevents a loop from loading an adjacent loop, and when such ampliﬁers are used, we can multiply the loop transfer functions to obtain the overall transfer function. This mistake is sometimes made when drawing block diagrams. The circuit of Figure 6.3.9 can be represented by the block diagram of Figure 6.3.10, where the transfer function of each block can be multiplied to obtain the overall transfer function Vo (s)/Vi (s). However, the circuit of Figure 6.3.1 cannot be represented by a simple series of blocks because the output voltage vo affects the voltage v1 . To show this effect requires a feedback loop, as shown previously in Figure 6.3.2. In general, even though elements are physically connected end-to-end, we cannot represent them by a series of blocks if the output of one element affects its input or the behavior of any preceding elements. 6.4 OPERATIONAL AMPLIFIERS A modern version of the feedback ampliﬁer discussed in Example 6.2.6 is the operational ampliﬁer (op amp), which is a voltage ampliﬁer with a very large gain G (greater than 105 ). The op amp has a large input impedance so it draws a negligible current. The op amp is an integrated circuit chip that contains many transistors, capacitors, and resistors and has several external terminals. We can attach two resistors in series with and parallel to the op amp, as shown in Figure 6.4.1a. This circuit diagram does not show all of the op amp’s external terminals; for example, some terminals are needed i2 Figure 6.4.1 An op-amp multiplier. Ri i3 i1 2 G 1 vi Rf 1 v1 2 i2 Ri vo i1 i3 1 1 vi v1 2Gv1 2 (a) Rf 2 (b) vo 6.4 Operational Amplifiers 347 to power the device and to provide constant bias voltages. Our diagram shows only two pairs of terminals: the input terminals intended for time-varying input signals and the output terminals. A plus sign or a minus sign on an input terminal denotes it as a noninverting terminal or an inverting terminal, respectively. Op amps are widely used in instruments and control systems for multiplying, integrating, and differentiating signals. Op-Amp Multiplier ■ Problem Determine the relation between the input voltage vi and the output voltage vo of the op-amp circuit shown in Figure 6.4.1a. Assume that the op amp has the following properties: 1. The op-amp gain G is very large, 2. vo = −Gv1 ; and 3. The op-amp input impedance is very large, and thus the current i 3 drawn by the op amp is very small. ■ Solution Because the current i 3 drawn by the op amp is very small, the input terminal pair can be represented as an open circuit, as in Figure 6.4.1b. The voltage-current relation for each resistor gives i1 = vi − v1 R1 i2 = v1 − vo R2 and From conservation of charge, i 1 = i 2 + i 3 . However, from property 3, i 3 ≈ 0, which implies that i 1 ≈ i 2 . Thus, vi − v1 v1 − v o = R1 R2 From property 1, v1 = −vo /G. Substitute this into the preceeding equation: vi vo vo vo + =− − R1 R1 G R1 G R2 Because G is very large, the terms containing G in the denominator are very small, and we obtain vi vo =− R1 R2 Solve for vo : vo = − R2 vi R1 (1) This circuit can be used to multiply a voltage by the factor R2 /R1 , and is called an op-amp multiplier. Note that this circuit inverts the sign of the input voltage. Resistors usually can be made so that their resistance values are known with sufﬁcient accuracy and are constant enough to allow the gain R2 /R1 to be predictable and reliable. E X A M P L E 6.4.1 348 CHAPTER 6 Electrical and Electromechanical Systems 6.4.1 GENERAL OP-AMP INPUT-OUTPUT RELATION We can use the impedance concept to simplify the process of obtaining a model of an op-amp circuit. A circuit diagram of the op amp with general feedback and input elements Z f (s) and Z i (s) is shown in Figure 6.4.2a. A similar but simpliﬁed form is given in part (b). The impedance Z i (s) of the input elements is deﬁned such that Vi (s) − V1 (s) = Z i (s)I1 (s) For the feedback elements, V1 (s) − Vo (s) = Z f (s)I2 (s) The high internal impedance of the op amp implies that i 1 ≈ 0, and thus i 1 ≈ i 2 . The ﬁnal relation we need is the ampliﬁer relation vo = −Gv1 or Vo (s) = −GV1 (s) (6.4.1) When the preceding relations are used to eliminate I1 (s) and I2 (s), the result is V1 (s) = Z f (s) Z i (s) Vi (s) + Vo (s) Z f (s) + Z i (s) Z f (s) + Z i (s) Using (6.4.1) to eliminate V1 (s), the transfer function between Vi (s) and Vo (s) is found to be Vo (s) Z f (s) G =− Vi (s) Z f (s) + Z i (s) 1 + G H (s) where H (s) = Z i (s) Z f (s) + Z i (s) Because G is a very large number, |G H (s)| 1, and we obtain Z f (s) Vo (s) ≈− (6.4.2) Vi (s) Z i (s) This is the transfer function model for op-amp circuits. An op-amp multiplier is created by using two resistors as shown in Figure 6.4.1a, where Z i (s) = Ri and Z f (s) = R f . Thus, Rf Vo (s) ≈− Vi (s) Ri The gain of this multiplier is R f /Ri , with a sign reversal. In some applications, we want to eliminate the sign reversal. We can do this by using an inverter, which is a multiplier i2 Figure 6.4.2 General circuit representation of an op-amp system. 1 vi Zi (s) i3 i1 Zf (s) 2 G 1 v1 Zf (s) vo vi vo Zi (s) 2 (a) (b) 6.4 Rf vi R Ri vo Inverter R3 v2 R R1 R Figure 6.4.4 An op-amp adder. (a) Circuit. (b) Block diagram. vo R2 (a) V1(s) V2(s) R3 R1 349 Figure 6.4.3 An op-amp multiplier with an inverter. R Multiplier vi Operational Amplifiers 1 21 1 21 Vo(s) Inverter R3 R2 Multiplier (b) having equal resistances. Using an inverter in series with the multiplier, as shown in Figure 6.4.3, eliminates the overall sign reversal. 6.4.2 OP-AMP COMPARATOR The multiplier circuit can be modiﬁed to act as an adder (Figure 6.4.4) or as a subtractor if another inverter is included (Figure 6.4.5). The output relation for the adder is vo = R3 R3 v1 + v2 R1 R2 The output relation for the subtractor is vo = R3 R3 v1 − v2 R1 R2 The block diagrams are shown in parts (b) of the ﬁgures. We can make a comparator by selecting R1 = R2 = R3 in Figure 6.4.5, in which case vo = v1 − v2 In Chapter 10 we will see how this is used in a control system. 6.4.3 INTEGRATION AND DIFFERENTIATION WITH OP AMPS Control systems and many types of instrumentation utilize op amps to integrate and differentiate electrical signals. 350 CHAPTER 6 Electrical and Electromechanical Systems R3 R Figure 6.4.5 An op-amp subtractor. (a) Circuit. (b) Block diagram. v1 R vo R1 v2 R2 (a) V1(s) 21 R3 R1 1 21 1 Inverter R3 R2 Vo(s) V2(s) Multiplier (b) Integration with Op Amps E X A M P L E 6.4.2 ■ Problem Determine the transfer function Vo (s)/Vi (s) of the circuit shown in Figure 6.4.6. Figure 6.4.6 An op-amp integrator. ■ Solution C vi R vo The impedance of a capacitor is 1/Cs. Thus, the transfer function of the circuit is found from (6.4.2) with Z i (s) = R and Z f (s) = 1/Cs. It is Vo (s) Z f (s) 1 =− =− Vi (s) Z i (s) RCs Thus in the time domain, the circuit model is 1 vo = − RC t vi dt (1) 0 Thus the circuit integrates the input voltage, reverses its sign, and divides it by RC. It is called an op-amp integrator, and is used in many devices for computing, signal generation, and control, some of which will be analyzed in later chapters. Differentiation with Op Amps E X A M P L E 6.4.3 ■ Problem Design an op-amp circuit that differentiates the input voltage. Figure 6.4.7 An op-amp differentiator. ■ Solution R vi C vo In theory, a differentiator can be created by interchanging the resistance and capacitance in the integrator circuit. The result is shown in Figure 6.4.7, where Z i (s) = 1/Cs and Z f (s) = R. The input-output relation for this ideal differentiator is Vo (s) Z f (s) =− = −RCs Vi (s) Z i (s) 6.5 Electric Motors 351 The model in the time domain is vo (t) = −RC dvi (t) dt The difﬁculty with this design is that no electrical signal is “pure.” Contamination always exists as a result of voltage spikes, ripple, and other transients generally categorized as “noise.” These high-frequency signals have large slopes compared with the more slowly varying primary signal, and thus they will dominate the output of the differentiator. Example 6.4.4 shows an improved differentiator design that does not have this limitation. Design of an Improved Differentiator E X A M P L E 6.4.4 ■ Problem The differentiator analyzed in Example 6.4.3 is susceptible to high-frequency noise. In practice, this problem is often solved by using a re

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