Mechatronics Engineering MT-144 NETWORK ANALYSIS Mechatronics Engineering (08)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 Basic RC and RL Circuits 8.2 Transients in First-order Networks 8.3 Step, Pulse and Pulse-Train Responses 8.4 First-order Op-Amp Circuits 8.5 Transient Analysis Using SPICE

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The The RL Circuit In the circuit of Figure 8.7 assume the inductance has zero initial stored energy so that, by Equation (7.19), i(0-) = 0. Flipping up the switch connects the source IS to the RL pair, causing current and, hence, energy to build up in the inductance. For t ≥ 0+, this process is governed by Equation (7.24) By the current continuity rule we must have i(0+)= i(0-)= 0. Consequently, the solution to Equation (8.9) is the forced response Figure 8.7 Flipping a switch to investigate the forced response of the RL Circuit

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The The RL Circuit The voltage is found using element law, v= L di/ dt . Differentiating Equations (8.10 a & b) and multiplying with L yields: These responses are shown in Figure 8.8. Comparison with the capacitive responses of Figure 8.2 further substantiates the duality principle (see next slide) Figure 8.7 Flipping a switch to investigate the forced response of the RL Circuit

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The The RL Circuit We also note that i is continuous at the origin, but v exhibits a spike. This does not violate Rule 1, which applies to inductive current, not to voltage. Figure 8.7 Flipping a switch to investigate the forced response of the RL Circuit

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit Both responses are governed by the time constant L/R. Increasing L slows down the responses because it takes longer to build up energy in a larger inductance. However, increasing R now speeds up the responses. We justify this by noting that the voltage response is scaled by R. Consequently, a larger R results in a larger v and, hence, in a faster rate of current buildup, since di/ dt = v/ L .

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit In the limit R  ∞ , current buildup in the inductance would be instantaneous and it would be accompanied by an infinitely large and infinitely brief voltage spike (a physical impossibility as we know). In a practical circuit this spike will be limited by the finite internal resistance of the source, confirming that current buildup in an inductance, however rapid, cannot be instantaneous. Home Work: Do Exercise 8.3 (page 337 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit If the switch is left in the up position for a sufficiently long time, all of IS will eventually be diverted through L and none through R, thus causing v to collapse, by Ohm's Law, to zero. As we know, this is the dc steady state for the inductance. The final stored energy in the inductance is wL = (1/2) L IS2.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit Suppose now the switch is flipped back down at an instant again chosen as t = 0, as depicted in Figure 8.9. Once more, we end up with a source-free circuit. By the continuity rule we have i(0+) =i(0-) = IS, indicating that after switch activation L will continue to draw current. With the source out of the circuit this current will have to come from R, so that v(0+) = - Ris. As a consequence of this voltage, the inductance current will start to decay, producing the natural response.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit The voltage is found as v = L di/dt. Differentiation of Equations 8.12 and multiplying with L yields Both responses are shown in Figure 8.10. The responses are governed by the time constant L/R. The smaller R, the slower the decay. In the limit R  0 the decay would become infinitely slow because L would have no means to dissipate its energy. Thus, the current IS would flow indefinitely inside the inductance, and we would have a memory effect.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit In practice, the nonzero resistance of the winding and connections will cause the inductance current to decay to zero eventually. This is similar to capacitor leakage, except that the parasitic resistance of a practical inductor is a far more serious limitation than is the parallel leakage of a practical capacitor. This explains why inductors are never used for long-term energy storage.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.1 BASIC RC AND RL CIRCUITS… DC Steady-State Behavior…The RL Circuit…Conversely, the larger R is, the quicker the decay, and the larger and briefer the voltage spike. If an attempt is made to open-circuit an inductor or a transformer carrying a nonzero current, an arc may develop across the switch whereby the stored energy is dissipated in ionizing the air along the path of the arc. This feature is exploited in automobile ignition systems, where the coil current is periodically interrupted by the distributor to develop an arc across a spark plug- Home Work: Do Exercise 8.4 & 8.5 (page 338 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS We now generalize the techniques of the previous section to the case in which an energy-storage element is part of a more complex circuit, or network. Subjecting the network to a sudden change, such as the activation of a switch, will cause the energy-storage element to respond with a transient. This, in turn, will force all other voltage and current variables in the circuit to readjust accordingly, since Kirchhoff’s laws must be satisfied at all times. If the circuit surrounding the energy-storage element consists of resistances and sources and is thus linear, all voltage and current transients in the circuit will be exponential and will be governed by the same time constant Ƭ. Assuming t = 0 as the instant of occurrence of the sudden change, each transient will take on the form of Equation (7.42), namely,

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS ... This expression is uniquely defined once we know y(0+), y(∞), and Ƭ. For completeness, we also want to know y(0-) in order to tell whether there is any discontinuity at t= 0. In the following we shall assume that by the time the sudden change is applied, the circuit has had sufficient time to reach its dc steady state. Following is a systematic procedure for finding the above parameters: (1) Find y(0-), the steady-state value of the response preceding the sudden change. This is achieved by examining the circuit before the change, when the capacitance acts as an open circuit and the inductance as a short circuit. Also record the voltage vc(0-) across the capacitance, or the current iL(0-) through the inductance, which will be needed in the next step.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS … Procedure for finding the above parameters (continued): (2) Find y(0+), the value of the response just after the sudden change. This is achieved by replacing the capacitance with a voltage source of value vc(0+) = vc(0-), or the inductance with a current source of value iL(0+) = iL(0-). Note that these replacements hold only at the instant t= 0+, after which vc(t) or iL(t) will indeed change to reflect energy buildup or decay. Clearly, the equivalent circuit of the present step is only a snapshot of our network at t= 0+. (3) Find y(∞), the steady-state value of the response in the limit t  ∞. This is achieved by examining the circuit long after the sudden change, when the capacitance will again act as an open circuit and the inductance as a short circuit. (4) Find the….

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Procedure for finding the above parameters (continued): (4) Find the equivalent resistance Req seen by the energy-storage element during the transient. Then use, respectively, Req, is found using Method 2 of Section 5.2, though in particular cases simple inspection may suffice. (Study / revise section 5.2) In general, the procedure uses three equivalent circuits, one for each of the first three steps, and possibly a fourth circuit to find Req.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Procedure for finding the above parameters (continued): Once we have all parameters in hand, we substitute them into Equation (8.14) to obtain the response as a function of time. It is often of interest to know how long it takes for a response to swing from a given value y(t1) to a value y(t2). By Equation (8.14) we can write y(t1) - y (∞) = [y(0+)- y (∞)]e-t1/Ƭ and y(t2) - y(∞) = [y(0+)- y(∞)] e-t2/Ƭ2 Dividing the two equations pair-wise yields [y(t1) - y(∞)] / [y(t2) - y(∞)] = e(t2-t1)/ Ƭ. Taking the natural logarithm of both sides and simplifying we obtain: …(8.16) Let us now apply the procedure to actual examples. Capacitive Examples (next slide Example 8.4)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples Example 8.4: Sketch and label v(t) for the circuit of Figure 8.11(a) ? Solution We need to find v(0-), v(0+), v(∞), and Ƭ. The sequence is illustrated in Figure 8.12, where the units are [V], [mA], and [kΩ]. Figure 8.12(a) depicts the steady state prior to switch activation. With C acting as an open circuit, we can use the voltage divider formula and write v(0-) = [1/(3 + 1)]12 = 3 V. Moreover, using KVL, vc(0-) = 12- 3 = 9 v. Figure 8.11 (a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples Example 8.4: Sketch and label v(t) for the circuit of Figure 8.11(a) ? Solution… Figure 8.12(b) provides a snapshot at t = 0+. By the continuity rule, the voltage across the capacitance just after switch activation must still be 9 V regardless of the external circuit conditions. We model this behavior with a 9-V voltage source, as shown. Then, by KVL, v(0+) = -9 V. Figure 8.11 (a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples Example 8.4: Sketch and label v(t) for the circuit of Figure 8.11(a) ? Solution Figure 8.12(c) depicts the steady state after the transient has died out. The voltage divider formula again yields v(∞) = 3 V. From Figure 8.12(c) we also note that the resistance seen by the capacitance is Req = 3K // 1K = 750 Ω. Thus, Ƭ = ReqC = 750 x 10-7 = 75 μs. Substituting into Equation (8.14) yields Figure 8.11 (a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples Example 8.4: Sketch and label v(t) for the circuit of Figure 8.11(a) ? Solution… Substituting into Equation (8.14) yields , The sketch of v versus t is shown below in Figure 8.11(b). Figure 8.11 (a) Home Work: Do Exercise 8.6 (page 341 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Example 8.5: Find v(t) in the circuit of Figure 8.13 ? Solution… The sequence is illustrated in Figure 8.14, where the units are [V], [mA], and [kΩ]. In Figure 8.14(a) we have, by Ohm's Law, v(0-) = -1K x 4 ix and ix = 5/10K = 0.5 mA. Hence, v(0-) = - 2 V. Moreover, KVL yields vc(0-) = 5 - (-2) = 7 v. Figure 8.13 Figure 8.14(a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Example 8.5: Find v(t) in the circuit of Figure 8.13 ? Solution… Opening the switch removes the 5-V source from the picture, as shown in the remaining figures. KCL at the super-node surrounding the 7-V source in the snapshot circuit of Figure 8.14(b) yields 0= ix + 4ix + v(0+)/1K , or v(0+)= -(K)5ix. KVL around the outer loop clockwise yields 10Kix = 7 + v(0+) or ix = [7+ v(0+)]/10K. Eliminating ix and solving for v(0+) yields v(0+) = -7/3 V. Figure 8.13 Figure 8.14(b)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Example 8.5: Find v(t) in the circuit of Figure 8.13 ? Solution… For the steady State: in Figure 8.14(c) it is apparent that ix = 0, so 4ix = 0 and, hence, v(∞) = 0 Figure 8.13 Figure 8.14(c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Example 8.5: Find v(t) in the circuit of Figure 8.13 ? Solution… To find the resistance Req, seen by C during the transient, replace C with a test source as in Figure 8.14(d). KVL around the outer loop clockwise yields 10ix= v +1i1. KCL at the super-node surrounding the test source yields 0= ix + 4ix + i1 . Eliminating i1 yields ix = v/15. But, i = ix , so that i = v/15. Hence, Req = v / i = v/ (v/15) = 15 kΩ. Finally, Ƭ = ReqC = 15 x l03 x 0.2 x 10-6 = 3 ms. Substituting the calculated data into Equation (8.14), yields  Figure 8.13 Figure 8.14(d)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Home Work: Do Exercise 8.7 (page 342 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Example 8.6: (a) Sketch and label i(t) in the circuit of Figure 8.15(a). (b) Find the time at which i = 0. ? Solution… (a) After gaining sufficient experience we can visualize the intermediate steps mentally. At t= 0- the switch is open and C acts as an open. KCL at the upper center node yields i(0-)+1mA= 0, or i(0-)= -1 mA Figure 8.15 (a) Since the capacitance voltage does not change during switch closure, the voltage across the 3-k Ω resistance also remains unchanged, thus yielding i(0+)= i(0-) = - 1mA. For t  ∞, C acts again as an open. Considering that the switch is now closed, we can apply the superposition principle, along with Ohm's Law and the current divider formula, and write (current passing through 3KΩ) i(∞) = 12/ (3+2)k - [2/ (3 + 2)] 1 = 2 mA.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Capacitive Examples… Example 8.6: (a) Sketch and label i(t) in the circuit of Figure 8.15(a). (b) Find the time at which i = 0. ? Solution… (a) The value of Req during the transient is found by suppressing the sources. Considering that the switch is closed, we have Req = 3k // 2k = 1.2 kΩ. Hence, Ƭ = RC = 1.2 x 103 x 0.1 x 10-6 = 120 μs. The sketch of i versus t is shown in Figure 8.15(b). (b) The time it takes for i(t) to reach 0A is, by Equation (8.16), t = 120 ln [(-1 - 2) / (0 - 2)] = 48.66 μs. Figure 8.15 (a) Figure 8.15(b) Home Work: Do Exercise 8.8 (page 343 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.7: (a) Sketch and label v(t) in the circuit of Figure 8.16(a). ? Solution… We need to find v(0-), v(0+), v(∞) and Ƭ. The sequence is illustrated in Figure 8.17, where the units are [V], [mA], and [KΩ]. Figure 8.16 (a) (a) t= 0- , (b) t=0+ and (c) t  ∞ Figure 8.17: Circuits of Figure 8.16(a) for (a) t= 0- , (b) t=0+ and (c) t  ∞.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.7: (a) Sketch and label v(t) in the circuit of Figure 8.16(a). ? Solution… We need to find v(0-), v(0+), v(∞) and Ƭ. The sequence is illustrated in Figure 8.17, where the units are [V], [mA], and [KΩ]. Figure 8.16 (a) (a) t= 0- , (b) t=0+ and (c) t  ∞ Figure 8.17(a) depicts the steady state prior to switch activation. With L acting as a short circuit, v(0-) = 5 V. Moreover, iL(0-) = 5/1 + 5/2 = 7.5 mA.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.7: (a) Sketch and label v(t) in the circuit of Figure 8.16(a). ? Solution… We need to find v(0-), v(0+), v(∞) and Ƭ. The sequence is illustrated in Figure 8.17, where the units are [V], [mA], and [KΩ]. Figure 8.16 (a) (a) t= 0- , (b) t=0+ and (c) t  ∞ Figure 8.17(b) provides a snapshot at t= 0+, with L replaced by a 7.5-rnA source to reflect current continuity. By Ohm's Law, v(0+) = 2 x 7.5 = 15 V.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.7: (a) Sketch and label v(t) in the circuit of Figure 8.16(a). ? Solution… We need to find v(0-), v(0+), v(∞) and Ƭ. The sequence is illustrated in Figure 8.17, where the units are [V], [mA], and [KΩ]. Figure 8.16 (a) (a) t= 0- , (b) t=0+ and (c) t  ∞ Figure 8.17(c) depicts the steady state after the transient has died out. By inspection, v(∞)= 5 V.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.7: (a) Sketch and label v(t) in the circuit of Figure 8.16(a). ? Solution… We need to find v(0-), v(0+), v(∞) and Ƭ. The sequence is illustrated in Figure 8.17, where the units are [V], [mA], and [KΩ]. Figure 8.16 (a) (a) t= 0- , (b) t=0+ and (c) t  ∞ From Figure 8.17(c) we also note that the resistance seen by L during the transient is Req = 2 kΩ. Then, Ƭ = L/Req = 10 x 10-3 / (2 x 103) = 5 μs. Substituting into Equation (8.14) : v(0-) = 5 v , v(0+) = 15 V , v(∞) = 5 V : yields : v(t ≥ 0+) = 5 + 10 e – t/ (5 μs) V

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.7: (a) Sketch and label v(t) in the circuit of Figure 8.16(a). ? Solution… v(t ≥ 0+) = 5 + 10 e – t/ (5 μs) V The sketch of v versus t is shown in Figure 8.16(b). Figure 8.16 (a) Figure 8.16 (b) Home Work: Do Exercise 8.9 (page 344 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.8: Sketch and label v ( t ) in the circuit of Figure 8.18(a); hence, find the instant when v(t) = 10 V ? Solution… Refer to Figure 8.19, where the units are [V], [A], and [Ω]. In Figure 8.19(a) we have, by the voltage divider formula, v(0-) = [1/(3+1)]12= 3 V. Also, iL(0-)=12/(3+1) = 3 A. Figure 8.18 (a) Fig. 8.19 (a) (b) (c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.8: Sketch and label v ( t ) in the circuit of Figure 8.18(a); hence, find the instant when v(t) = 10 V ? Solution… In Figure 8.19(b) we apply the superposition principle, along with the voltage divider formula and Ohm's Law, to write v(0+) = [1/(2 + 1)]12+ 3 x (2 // 1) = 6 V. Figure 8.18 (a) Fig. 8.19 (a) (b) (c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.8: Sketch and label v ( t ) in the circuit of Figure 8.18(a); hence, find the instant when v(t) = 10 V ? Solution… In Figure 8.19(c) we have v(∞) = 12 V . Moreover, the resistance seen by L is Req = 1 // 2 = 2/ 3 Ω. Consequently, Ƭ = L/ Req = 4 x 10-3 / (2/3) = 6 ms. Substituting into Equation (8.14), Figure 8.18 (a) Fig. 8.19 (a) (b) (c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.8: Sketch and label v ( t ) in the circuit of Figure 8.18(a); hence, find the instant when v(t) = 10 V ? Solution… The sketch of v versus t is shown in Figure 8.18(b). Figure 8.18 (a) Figure 8.18(b) Home Work: Do Exercise 8.10 (page 346 of text book) The amount of time it takes for v(t) to reach 10 V is, by Equation (8.16),

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.9: Find v(t) in the circuit of Figure 8.20 ? Solution… Refer to the sequence of Figure 8.21, where the units are [V], [A], and [Ω]. In Figure 8.21(a) we have v(0-) = 4ix, where ix= 1 A. Hence, v(0-) = 4 V . Moreover, KCL at the upper rightmost node yields iL(0-) + iX = v(0-)/3, or iL(0-) = v(0-)/ 3 - i x = 4/ 3 - 1= 1/ 3 A. Figure 8.20 Figure 8.21 By the continuity rule, the inductance current at t = 0+ is still 1/3 A, as shown in Figure 8.21(b). KCL at the upper rightmost node yields 1/3 + ix = v(0+)/3, where ix = [4ix – v(0+) ] / 2 . Eliminating ix yields v(0+) = - 2 v.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.9: Find v(t) in the circuit of Figure 8.20 ? Solution… In Figure 8.21(c) we have v(∞) = 4ix, where ix = 0 because the 2 Ω resistance is now short-circuited. Hence, v ( ∞ ) = 0. Suppressing the independent source and applying a test voltage as in Figure 8.21(d), Figure 8.20 Figure 8.21 we have, by KCL at the upper rightmost node, i + ix = (4ix + v ) / 3 . But, by Ohm's Law, ix = - v/ 2. Eliminating ix

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.2 TRANSIENTS IN FIRST-ORDER NETWORKS… Let us now apply the procedure to actual examples. Inductive Examples… Example 8.9: Find v(t) in the circuit of Figure 8.20 ? Solution… Eliminating ix yields i = v / 6. Thus, Req = v / i = 6 Ω, Ƭ = 3 x 10-3/ 6 = 0.5 ms = 500μS, and v(0-) = 4 v v(t ≥ 0+) = - 2e -t/(500μS) V Figure 8.20 Figure 8.21 Home Work: Do Exercises 8.11, 8.12 & 8.13 (page 348 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES The activation of a switch is not the only form of effecting a sudden change in a circuit. Another common form is by means of a step signal, that is, a signal that changes abruptly from one value to another. As we proceed we shall see that the step response provides precious information about the dynamic characteristics of a circuit. If a positive-going step is followed at some later instant by a negative-going step of equal amplitude, the resulting signal is the pulse. The pulse response is of great interest in computer electronics, where information is represented by means of pulses.