Mechatronics Engineering

Mechatronics Engineering
MT-144 NETWORK ANALYSIS Mechatronics Engineering (11)

TRANSIENT RESPONSE OF SECOND-ORDER CIRCUITS: (Chapter 9)
9.1 Natural Response of Series & Parallel RLC Circuits 9.2 Transient Response of Second Order Circuits 9.3 Step Response of Second Order Circuits 9.4 Second Order Op Amp Circuits 9.5 Transient Analysis Using Spice

Having studied the natural or source-free response of RLC circuits, we now investigate the transient response, that is, the response to a dc forcing function x(t) = XS The equation governing the circuit now takes on the general form : The response will generally consist of a transient component and a dc steady state component: y(t) = yxsient + yss … (9.33) The transient component takes on the same functional form as the natural response. Depending on the damping conditions of the circuit, this component will be over-damped if ζ > I, critically damped if ζ = I , and under-damped if 0 < ζ < 1.

As we know, the corresponding functional forms are, respectively, For ζ > 1 : yxsient = B1 e-t/Ƭ1 + B2 e-t/Ƭ … (9.34a) For ζ = 1 : yxsient = (B1 + B2 t ) e -t/Ƭ … (9.34b) For 0 < ζ < 1 : yxsient = B e-αt cos (ωdt + θ ) … (9.34c) where B1and B2, or B and θ, are determined on the basis of the initial value y(0+) of the function and the initial value dy(0+)/dt of its derivative. Moreover,

The steady-state component represents what is left of y(t) after all transients have died out, that is: yss = y(∞) … (9.37) The procedure to find how a second-order circuit, assumed to be initially in its dc steady state, responds to a sudden change such as the activation of a switch or the application of a step function is as follows: (1) Find y(0+) and dy(0+)/dt using the continuity rules for capacitance and inductance. (2) Find y(∞) using dc steady-state circuit analysis techniques. Recall that in steady state C acts as an open circuit and L as a short circuit.

(3) Find ζ and ω0. If, with all independent sources suppressed, L and C appear either in series or in parallel with each other, then ω0 = 1/ √(LC) and ζ = (½)Req √(C/L) for the series case, and ζ = (½) / (Req √(C/L) for the parallel case, where Req is the equivalent resistance seen by the LC pair. If L and C are neither in series nor in parallel, then we must derive the differential equation governing the circuit and put it in the standard form of Equation (9.3) to find ζ and ω0. (4) Based on the value of ζ select the proper form of Equation (9.34), and find B1 and B2, or B and θ, so as to make Equation (9.33) satisfy the initial conditions of Step 1. Compared with first-order circuits, second-order circuits require the additional initial condition dy(0+)/dt.

Moreover, instead of the single circuit parameter Ƭ, we now need the parameter pair ζ and ω0, which may require the tedious task of formulating a differential equation. Later (In Chapters 14 & 16) network functions and Laplace transforms are used to find transient responses, much more efficiently.

Example 9.8: Assuming the circuit of Figure 9.9 is in steady state prior to switch activation, find v(t) ? Solution. By the voltage continuity rule, v(0+) = v(0-) = 25 V. By the capacitance law and the current continuity rule, dv(0+)/dt= ic(0+)/C = iL(0+)/C = iL(0-)/C = 0. In summary, v(0+) = 25 V, dv(0+)/dt = 0 By inspection, v(∞)= [4k/ (2.25k+4k) ] 25, or v(∞)= 16 V, Since the resistance and capacitance are in series with each other, we have ω0 = I/ √(LC) = 1 / √{9x10-3 (1/90)10-6} = 100 k rad/ s, and ζ = (1/2) Req √(C/L) ,where Req is the resistance seen by the LC pair

Example 9.8: Assuming the circuit of Figure 9.9 is in steady state prior to switch activation, find v(t) ? Solution… with the source suppressed, or Reg= 2.25k // 4k = 1.44 kΩ. Hence, ζ = (½ ) x 1.44 x 103 x √{(1/90)10-6/(9 x 10-3)} = 0.8 < 1 . We thus have an under-damped response with α = ζω0= 0.8 x 100 = 80 k Np/ s, and ωd = √(1-ζ2) x ω0 = √(1-0.82) x 100 = 60 k rad/ s. Its functional form, as per (9.34c) is : V(t ≥0+) = B e - 80 x 1000t cos( 60 x 103t + θ) + 16 To find B and θ we observe that: v(0+) = B cos θ + 16 = 25 V : dv(0+)/dt = B(-80 x 103 cos θ - 60 x 103 sinθ) = 0 The second equation yields tan θ = - 80/60, or θ = - tan-1 (4/3) = o ; the first equation yields B = ( ) / [cos (-53.13o)] = 15 V. Consequently, v(t ≥ 0+) = 15e-80 x 1000t cos (60 x l03 t o) + 16 V

Example 9.9: Assuming the circuit of Figure 9.10 is in steady state prior to switch activation, find v(t) ? Solution… By the voltage continuity rule, v(0+) = vc(0+) = vc(0-) = vL(0-) = 0. Applying the capacitance law, KCL, the current continuity rule, and the current divider formula, Cdv(0+)/dt = ic(0+) = iL(0+) = iL(0+)= [12/(12+13)]6.5 = 3.38 A. Hence, dv(0+)/dt = 3.38/ (10-3 / 3) = V/s. To summarize, V(0+) = 0, dv(0+)/ dt = V/s, by inspection v(∞) =0

Example 9.9 … Assuming the circuit of Figure 9.10 is in steady state prior to switch activation, find v(t) ? Solution… Since the resistance and capacitance are in parallel with each other, we have ω0= 1/ √(LC)= 100 rad/ s and ζ= 1/ (2 Req √(C/L)) , where ), Req is the resistance seen by the LC pair with the source suppressed, or Req= = 12 Ω. Hence, ζ= 1.25 > 1, indicating an overdamped response. Using Equation (9.35a) we find Ƭ1 = 20 ms and Ƭ2 = 5 ms, so

Example 9.9 … Assuming the circuit of Figure 9.10 is in steady state prior to switch activation, find v(t) ? Solution…

Example 9.10: Assuming the circuit of Figure 9.11 is in steady state prior to switch activation, find v(t ≥ 0+). Solution…

Example 9.10 …Assuming the circuit of Figure 9.11 is in steady state prior to switch activation, find v(t ≥ 0+). Solution…

9.3 Step Response of Second-Order Circuits The step response of second-order circuits plays a central role in electrical engineering. Its importance stems from the fact that many circuits of practical interest are of the second order. Even higher-order systems often exhibit a predominantly second-order behavior, this being the reason why design and performance specifications are often given in terms of second-order step-response parameters. In this section we investigate the step response using the RLC circuit of Figure 9.12 as a vehicle. We assume the circuit, after having had sufficient time to reach its dc steady state, is subjected to a voltage step vI of amplitude Vm That is: Figure 9.12 Investigating the step response of a series RLC Circuit

9.3 Step Response of Second-Order Circuits …

9.3 Step Response of Second-Order Circuits… Example 9.11 In the circuit of Figure 9.12 let R = 3 Ω, L = 1 H, and C = 1 F. Find the response to a 1-V input step ? Solution: Figure 9.12

9.3 Step Response of Second-Order Circuits… Example 9.11 In the circuit of Figure 9.12 let R = 3 Ω, L = 1 H, and C = 1 F. Find the response to a 1-V input step ? Solution:

9.3 Step Response of Second-Order Circuits… Example 9.12 Find the value of R that makes the circuit of Example 9.11 critically damped. Hence, find the response to a 1-V input step ? Solution: R= 2 ζ / √(C/L) = 2 x 1/ √(1/1) = 2 Ω. Using Equation (9.45), Ƭ= 1/1= 1 s. substituting into Equation (9.46) gives: vo (t ≥ 0+) = 1- (1+ t ) e –t V This response is shown in figure (down-right) as the curve labeled ζ =1

9.3 Step Response of Second-Order Circuits… It is interesting to compare the step response of a critically damped RLC circuit with that of a plain RC circuit having the same time constant Ƭ . Recall that the step response of the RC circuit is : vo(t ≥ 0+) = Vm (1 - e-t/Ƭ). Since its initial slope is dvo(0+)/ dt = Vm/ Ƭ ≠ 0, the RC response starts to rise immediately. By contrast, the RLC response exhibits some initial delay because its initial slope is zero, by Equation (9.39b). Physically, this stems from the current-choking action by the inductance. We also note that because of the presence of the (t/Ƭ) term in Equation (9.46), the RLC response is slower than the RC response. For instance, at t= Ƭ the RC response rises to vo(Ƭ) = Vm (1 - e-1 ) ≈ 0.63 Vm , while the RLC response is still at vo(Ƭ) = Vm[1 - ( 1 + t/Ƭ) e-1] ≈ 0.26Vm. Physically, this stems again from the current-choking action by the inductance, which slows down the charging of the capacitance.

9.3 Step Response of Second-Order Circuits… For 0 < ζ < 1 the transient component is underdamped, and it consists of a decaying sinusoid with α = ζω (9.47a) , ωd = ω0 √(1- ζ2) (9.47 b) Substituting Equation (9.34c) into Equation (9.42) and imposing the initial conditions of Equation (9.39), we obtain Equation (9.48):

9.3 Step Response of Second-Order Circuits… Example 9.13 Repeat Example 9.11 , but for R = 1 Ω ? Solution: We now have ζ= (1/2) √(1/1) = 0.5 < 1 , so α = 0.5 x 1 = 0.5 Np / s, and ωd = √( ) x 1 = rad / s. Substituting into Equation (9.48) yields : vo(t ≥ 0+) = e-0.5t cos (0.8660t – 30o) V This response is shown in Figure 9.13 as the curve labeled ζ = 0.5. Figure 9.12

9.3 Step Response of Second-Order Circuits… For convenience, also shown in Figure 9.13 are the responses corresponding to ζ= 1/ √2, ζ = 0.35, and ζ = 0.2. We immediately note that the smaller the value of ζ , the more rapidly the response rises. Moreover, for ζ < 1 we observe two distinctive features: (1) The underdamped responses rise above Vm , a phenomenon referred to as overshoot. (2) After the overshoot, the underdamped responses decay toward the steady-state value Vm in an oscillatory manner, a phenomenon referred to as ringing. It is readily seen that the smaller the value of ζ , the more pronounced the overshoot and the longer the time it takes for the ringing effect to die out.

9.3 Step Response of Second-Order Circuits… Overshoot. The maxima and minima of an underdamped response are found by calculating the time derivative of Equation (9.48) and setting the result to zero. The derivative is

9.3 Step Response of Second-Order Circuits …Overshoot …

9.3 Step Response of Second-Order Circuits …Overshoot … Multiplying by 100 yields the overshoot (OS) in percentage form. We note that the overshoot depends only on the damping ratio ζ . There is no overshoot for ζ ≥ 1. However, for 0 < ζ < 1, the smaller the value of ζ , the larger the overshoot. In the limit ζ  0 the overshoot approaches 100%.

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