ES250: Electrical Science

1. ES250:Electrical Science HW8: Complete Response of RL and RC Circuits

2. Introduction • RL and RC circuits are called first-order circuits. In this chapter we will do the following: • develop vocabulary that will help us talk about the response of a first-order circuit • analyze first-order circuits with inputs that are constant after some particular time, t0, typically t0 = 0 • analyze first-order circuits that experience more than one abrupt change, e.g., when a switch opens or closes • introduce the step function and use it to determine the step response of a first-order circuit

3.   First-Order Circuits • Circuits that contain only one inductor or capacitor can be represented by a first-order differential equation • these circuits are called first-order circuits • Thévenin and Norton equivalent circuits simplify the analysis of first-order circuits by permitting us to represent all first-order circuits as one of two possible simple equivalent first-order circuits, as shown:

4.   First-Order Circuits

5.   First-Order Circuits • Consider the first-order circuit with input voltage vs(t); the output, or response, is the voltage across the capacitor: • Assume the circuit is in steady state before the switch is closed at time t = 0, then closing the switch disturbs the circuit; eventually, the disturbance dies out and the resulting circuit assumes a new steady state condition, as shown on the next slide

6.   First-Order Circuits

7.   First-Order Circuits • When the input to a circuit is sinusoidal, the steady-state response is also sinusoidal; furthermore, the frequency of the response sinusoid must be the same as the frequency of the input sinusoid • If the prior circuit is at steady state before the switch is closed, the capacitor voltage will be of the form: • The switch closes at time t = 0, the capacitor voltage is: • After the switch closes, the response will consist of two parts: a transient part that eventually dies out and a steady-state part, as shown:

8.   First-Order Circuits • The steady-state part of the circuit response to a sinusoidal input will also be sinusoidal at the same frequency as the input, while the transient part of the response of a first-order circuit is exponential of the form Ke−t/τ • Note, the transient part of the response goes to zero as t becomes large; when this part of the response “dies out,” the steady-state response remains, e.g., Mcos(1000t + δ) • The complete response of a first-order circuit can be represented in several ways, e.g.:

9.   First-Order Circuits • Alternatively, the complete response can be written as: • The natural response is the part of the circuit response solely due to initial conditions, such as a capacitor voltage or inductor current, when the input is zero; while the forced response is the part of the circuit response due to a particular input, with zero initial conditions, e.g.: • In the case when the input is a constant or a sinusoid, the forced response is the same as the steady-state response and the natural response is the same as the transient response

10.   First-Order Circuits • Steps to find the complete response of first-order circuits: • Step 1: Find the forced response before the disturbance, e.g., a switch change; evaluate this response at time t = t0 to obtain the initial condition of the energy storage element • Step 2: Find the forced response after the disturbance • Step 3: Add the natural response = Ke−t/τ to the forced response to get the complete response; use the initial condition to evaluate the constant K

11. Questions?

12.   Complete Response to a Constant Input • Find the complete response of a first-order circuit shown below for time t0 > 0 when the input is constant:

13.   Complete Response to a Constant Input • Note, the circuit contains a single capacitor and no inductors, so its response is first order in nature • Assume the circuit is at steady state before the switch closes at t0 = 0 disturbing the steady state condition for t0 < 0 • Closing the switch at t0 = 0 removes resistor R1 from the circuit; after the switch closes the circuit can be represented with all elements except the capacitor replaced by its Thévenin equivalent circuit, as shown:

14.   Complete Response to a Constant Input • The capacitor current is given by: • The same current, i(t), passes through the resistor Rt, Appling KVL to the circuit yields: • Combining these results yields the first-order diff. eqn.: • What is v(0-)=v(0+)?

15.   Complete Response to a Constant Input • Find the complete response of a first-order circuit shown below for time t0 > 0 when the input is constant: • What is i(0-)=i(0+)?

16.   Complete Response to a Constant Input • Closing the switch at t0 = 0 removes resistor R1 from the circuit; after the switch closes the circuit can be represented with all elements except the capacitor replaced by its Norton equivalent circuit, as shown:

17.   Complete Response to a Constant Input • Both of these circuits have eqns. of the form: • where the parameter τ is called the time constant • Separating the variables and forming an indefinite integral, we have: • where D is a constant of integration • Performing the integration and solving for x yields: • where A = eD, which is determined from the IC x(0) • To find A, let t = 0, then:

18.   Complete Response to a Constant Input • Therefore, we obtain: • where the parameter τ is called the time constant • Since the solution can be written as:

19.   Complete Response to a Constant Input • The circuit time constant can be measured from a plot of x(t) versus t, as shown:

20.   Complete Response to a Constant Input • Applying these results to the RC circuit yields the solution:

21.   Complete Response to a Constant Input • Applying these results to the RL circuit yields the solution:

22. Example 8.3-5: First-Order Circuit • The circuit below is at steady state before the switch opens; find the current i(t) for t > 0: • Note:

23. Solution • The figures below show circuit after the switch opens (left) and its the Thévenin equivalent circuit (right): • The parameters of the Thévenin equivalent circuit are:

24. Solution • The time constant is: • Substituting these values into the standard RC solution: • where t is expressed in units of ms • Now that the capacitor voltage is known, node voltage applied to node “a” at the top of the circuit yields: • Substituting the expression for the capacitor voltage yields:

25. Solution • Solving for va(t) yields: • Finally, we calculating i(t) using Ohm's law yields:

26. Solution >> tau=120e-3; >> t=0:(5*tau)/100:5*tau; >> i=66.7e-6-16.7e-6*exp(-t/120e-3); >> plot(t,i,'LineWidth',4) >> xlabel('t [s]') >> ylabel('i [A]') >> title('Plot of i(t)')

27. Questions?

28.  The Unit Step Source • The application of a constant source, e.g., a battery, by means of switches may be considered equivalent to a source that is zero up to t0 and equal to the voltage V0 thereafter, as shown below: • We can represent voltage v(t) using the unit step, as shown:

29.  The Unit Step Source • Where the unit step forcing function is defined as: Note: That value of u(t0) is undefined

30.  The Unit Step Source • Consider the pulse source v(t) = V0u(t − t0)−V0u(t − t1) defined: • We can represent voltage v(t) using the unit step, as shown:

31.  The Unit Step Source • The pulse source v(t) can schematically as: • Recognize that the unit step function is an ideal model. No real element can switch instantaneously at t = t0; However, if it switches in a very short time (say, 1 ns), we can consider the switching as instantaneous for medium-speed circuits • As long as the switching time is small compared to the time constant of the circuit, it can be ignored.

32. Ex: Pulse Source Driving an RL Circuit • Consider the application of a pulse source to an RL circuit as shown below with t0 = 0, implying a pulse duration of t1 sec: • Assume the pulse is applied to the RL circuit when i(0) = 0; since the circuit is linear, we may use the principle of superposition, so that i=i1 + i2 where i1 is the response to V0u(t) and i2 is the response to V0u(t − t1)

33. Ex: Pulse Source Driving an RL Circuit • We recall that the response of an RL circuit to a constant forcing function applied at t = tn with i(0) = 0, Isc = V0/R, andwhere τ = L/R is given by: • Consequently, we may add the two solutions to the two-step sources, carefully noting t0 = 0 and t1 as the start of each response, respectively, as shown:

34. Ex: Pulse Source Driving an RL Circuit • Adding the responses provides the complete response of the RL circuit shown:

35. Questions?