Mechatronics Engineering MT-144 NETWORK ANALYSIS Mechatronics Engineering (09)

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) 8.3 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.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits Referring to Figure 8.22, we note that in dc steady state both circuits yield vo = vi. This is so because in this state C acts as an open, so that vo = vi - Ri = vi - R x 0 = vi in the R-C circuit; In L-R Circuit, L acts as a short, so that vo = vi - vL = vi - 0 = vi

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits If we apply a positive-going step of amplitude Vm each circuit will yield vo (0-) = 0 and vo (∞) = Vm . Moreover, by the continuity rules, vo (0+) = 0 in both cases. Substituting into Equation (8.14), we have: … (8.17 a) … (8.17 b) where the two circuits have, respectively, Ƭ = RC … (8.18a), Ƭ = L/ R … (8.18b). This response is shown in Figure 8.23(a). Figure 8.23(a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits A measure of how rapidly a circuit responds to a positive-going step is the rise time tr defined as the amount of time it takes for the response to rise from 10% to 90% of its final value. Using Equation (8.16) we get, tr = Ƭ ln [(0.1Vm - Vm) / (0.9Vm- Vm)] = Ƭ ln9, or tr = 2.2Ƭ … (8.19) Figure 8.23(a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits By similar reasoning, the response to a negative-going step is, … … (18.20 a) … (18.20 b) and is shown in Figure 8.23 (b). This response is characterized by the fall time tf, the time it takes for vo to fall from 90% to 10% of its initial value. Using Equation (8.16) as before yields: tf = 2.2Ƭ … (8.21 ) Figure 8.23(b)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits Example 8.10: Home Work: Do Exercises 8.14 (page 350 of text book) Figure 8.22(a)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse Response of R-C and L-R Circuits

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse Response of R-C and L-R Circuits…

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse Response of R-C and L-R Circuits… Example 8.11: An R-C circuit with R = 10 kΩ and C = 1 nF is driven with a 5-V, 7.5-μs pulse. Find vo for (a) t = 4 μs, (b) t = 7.5 μS, and (c) t = 12 μs. Solution (a) We have Vm = 5 V, W = 7.5 μs, and Ƭ = l04 x 10-9 = 10 μs. Using Equation (8.22a), vo(4μs) = 5(1 - e-4/10) = 1.648 V. (b) By Equation (8.22a), vo (7.5μS) = 5(1 - e-7.5/10)= 2.638 V. (c) By Equation (8.22b). vo(12μs)= 5(e7.5/10 - 1)e-12/10 = 1.682 V. … 8.22(a) … 8.22(b) Home Work: Do Exercises 8.15 (page 351 of text book) Figure 8.24: Pulse response of R-C Circuit

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits If an R-C or an L-R circuit is subjected to a pulse train with duty cycle d, the response consists of exponential buildups alternated with exponential decays. After a sufficient number of pulses, the steady-state situation of Figure 8.25 is reached, where the response alternates between VL and VH. To find VL and VH, we use the following:

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… To find VL and VH, we use the following: Calculation of the first expression at t = dT and the second at t= T yields, respectively, VH= (VL-Vm) e -d T/ Ƭ +Vm ... (8.23a) VL = VH e -(T-dT) / Ƭ ... (8.23b)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… To find VL and VH, we use the following: Calculation of the first expression at t = dT and the second at t= T yields, respectively, VH= (VL-Vm) e -d T/ Ƭ +Vm ... (8.23a) VL = VH e -(T-dT) / Ƭ ... (8.23b) This system of equations is readily solved for the two unknowns, Depending on the magnitude of Ƭ relative to T, we have three different possibilities, depicted in Figure 8.26 for the case d = 50%:

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… Depending on the magnitude of Ƭ relative to T, we have three different possibilities, depicted in Figure 8.26 for the case d = 50%: (1) Ƭ << T. One can readily verify that Equations (8.24) predict VL  0 and VH  Vm indicating that vo has sufficient time to essentially reach the extremes of its range. Physically, the pulses and their spacings are long enough to allow for the energy in the capacitance or inductance to fully build up or decay within each subcycle. Figure 8.26(a) reveals that, aside from a slight rounding of the edges, the shape of vo is similar to that of vI . In the limit Ƭ << T the input pulses are thus transmitted to the output with negligible distortion.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… Depending on the magnitude of Ƭ relative to T, we have three different possibilities, depicted in Figure 8.26 for the case d = 50%: (2) Ƭ= T/2 . Now vo has only a limited amount of time to perform its exponential transitions. Letting T/ Ƭ= 2 and d= 0.5 in Equations (8.24) yield VL = 0.269 Vm and VH = 0.731 Vm. As shown in Figure 8.26(b), vo is a fairly distorted version of v I .

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… Depending on the magnitude of Ƭ relative to T, we have three different possibilities, depicted in Figure 8.26 for the case d = 50%: (3) Ƭ >> T . Because of the long time constant, vo can now accomplish only the initial portions of its exponential transitions. Since these portions are essentially straight segments, vo now approximates a triangle wave with a small peak-to-peak amplitude. In Chapter 7 we learned that integrating a square wave yields a triangle wave. For this reason the R-C and L-R circuits are, in the limit Ƭ >> T, referred to as integrators. Figure 8.26(c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… Depending on the magnitude of Ƭ relative to T, we have three different possibilities, depicted in Figure 8.26 for the case d = 50%: We also observe that in the limit (T/Ƭ)  0 Equations (8.24) predict VL  VH  d Vm indicating that vo now approximates a dc signal of value d Vm. Since this coincides with the full-wave average of vI, the R-C and L-R circuits are, in the limit Ƭ >> T, also referred to as averaging filters. Figure 8.26(c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… Depending on the magnitude of Ƭ relative to T, we have three different possibilities, depicted in Figure 8.26 for the case d = 50%: We also observe that in the limit (T/Ƭ)  0 Equations (8.24) predict VL  VH  d Vm indicating that vo now approximates a dc signal of value d Vm. Since this coincides with the full-wave average of vI, the R-C and L-R circuits are, in the limit Ƭ >> T, also referred to as averaging filters. Figure 8.26(c)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of R-C and L-R Circuits… Example 8.12: In the circuit of Figure 8.22(a). let R= 10 kΩ and C= 0.1 μF, and let vI be a 5-V pulse train with a 50% duty cycle. Find the pulse train frequency for which vo will approach a triangular wave with a peak-to-peak amplitude of 10 mV? Figure 8.22 (a) Solution: We have Ƭ= RC= 104 x 10-7 = 1 ms. The average value of vo is 2.5 V, and the peak values of vo are 2.5 V ± 5 mV, that is, 2.495 V and 2.505 V. Use Equation (8.23b) , VL = VH e -(T-dT) / Ƭ to impose Solving for T yields T = 0.008Ƭ = 0.008 x 10-3 = 8 μs. Thus, f = 1/T = 1/(8 x 10-6) = 125 kHz. Home Work: Do Exercises 8.16 (page 353 of text book)

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits Referring to Figure 8.27, we observe that in steady state both circuits yield vo = 0, so that we must have vo(0-) = 0 and vo (∞) = 0. Using the continuity rules, it is also seen that, at t = 0+, vo must experience the same jump as vl.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits Referring to Figure 8.27, we observe that in steady state both circuits yield vo = 0, so that we must have vo(0-) = 0 and vo (∞) = 0. Using the continuity rules, it is also seen that, at t = 0+, vo must experience the same jump as vl.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Step Response of R-C and L-R Circuits

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of CR and RL Circuits Figure 8.29 shows the steady-state response of the C-R and R-L circuits to a 50% duty-cycle pulse train. We observe that the current through C and the voltage across L must average to zero, for otherwise there would be unlimited energy buildup in these elements. Consequently, vo will also average to zero, this being why the C-R and R-L circuits are also known as dc blocking circuits. We again identify three cases of interest: (1) Ƭ >> T. With a long time constant there is insufficient time for appreciable energy buildup or decay in the capacitance or inductance within each subcycle. The input pulses are thus transmitted to the output with negligible distortion. Figure 8.29(a): Input Pulse Train vI

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of CR and RL Circuits… We again identify three cases of interest: (1) Ƭ >> T. With a long time constant there is insufficient time for appreciable energy buildup or decay in the capacitance or inductance within each sub-cycle. The input pulses are thus transmitted to the output with negligible distortion. (2) Ƭ = T/2 . The pulses undergo greater distortion because there is more time for energy buildup or decay within each sub-cycle. Figure 8.29(a): Input Pulse Train vI Fig. 8.29(b) Response for Ƭ ≫ T Fig. 8.29(c) Response for Ƭ = T/2

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.3 STEP, PULSE AND PULSE-TRAIN RESPONSES… Pulse-Train Response of CR and RL Circuits… We again identify three cases of interest: (3) Ƭ << T. Now there is sufficient time for essentially complete energy buildup or decay within each subcycle. Consequently, vo consists of a series of spikes, the polarity of each spike reflecting the direction of the corresponding transition of vI. Since the response now approximates the derivative of the applied signal, the C-R and R-L circuits are, in the limit t << T, referred to as differentiators. . Figure 8.29(a): Input Pulse Train vI Home Work: Do Exercises 8.17 (page 356 of text book) Fig. 8.29(d) Response for Ƭ << T

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.4 First Order Op Amp Circuits. RC circuits provide a fertile area of application for op amps. Thanks to its ability to draw energy from its own power supplies and inject it into the surrounding circuitry, the op amp can be used in ingenious ways to create effects that cannot be achieved with purely passive RLC components, let alone with resistors and capacitors but no inductors. The Differentiator. Figure 8.30 Differentiator and input / output waveform example.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.4 First Order Op Amp Circuits. The Integrator (Inverting). Figure 8.31 Integrator and input / output waveform example.

TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS: (Chapter 8) 8.4 First Order Op Amp Circuits. The Integrator (Non-inverting). 0 T/2 T 3T/2 t Vm vO -Vm Figure 8.32 Integrator and input / output waveform example.