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  • Slide 1
  • ES250: Electrical Science HW8: Complete Response of RL and RC Circuits
  • Slide 2
  • 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, t 0, typically t 0 = 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 Introduction
  • Slide 3
  • Circuits that contain only one inductor or capacitor can be represented by a first-order differential equation these circuits are called first-order circuits Thvenin 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: First-Order Circuits
  • Slide 4
  • Slide 5
  • Consider the first-order circuit with input voltage v s (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 First-Order Circuits
  • Slide 6
  • Slide 7
  • 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: First-Order Circuits
  • Slide 8
  • 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/ 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., M cos(1000t + ) The complete response of a first-order circuit can be represented in several ways, e.g.: First-Order Circuits
  • Slide 9
  • 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 First-Order Circuits
  • Slide 10
  • 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 = t 0 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 = Ket/ to the forced response to get the complete response; use the initial condition to evaluate the constant K First-Order Circuits
  • Slide 11
  • Questions?
  • Slide 12
  • Find the complete response of a first-order circuit shown below for time t 0 > 0 when the input is constant: Complete Response to a Constant Input
  • Slide 13
  • 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 t 0 = 0 disturbing the steady state condition for t 0 < 0 Closing the switch at t 0 = 0 removes resistor R 1 from the circuit; after the switch closes the circuit can be represented with all elements except the capacitor replaced by its Thvenin equivalent circuit, as shown: Complete Response to a Constant Input
  • Slide 14
  • The capacitor current is given by: The same current, i(t), passes through the resistor R t, Appling KVL to the circuit yields: Combining these results yields the first-order diff. eqn.: What is v(0 - )=v(0 + )? Complete Response to a Constant Input
  • Slide 15
  • Find the complete response of a first-order circuit shown below for time t 0 > 0 when the input is constant: What is i(0 - )=i(0 + )? Complete Response to a Constant Input
  • Slide 16
  • Closing the switch at t 0 = 0 removes resistor R 1 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: Complete Response to a Constant Input
  • Slide 17
  • Both of these circuits have eqns. of the form: where the parameter 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 = e D, which is determined from the IC x(0) To find A, let t = 0, then: Complete Response to a Constant Input
  • Slide 18
  • Therefore, we obtain: where the parameter is called the time constant Since the solution can be written as: Complete Response to a Constant Input
  • Slide 19
  • The circuit time constant can be measured from a plot of x(t) versus t, as shown: Complete Response to a Constant Input
  • Slide 20
  • Applying these results to the RC circuit yields the solution: Complete Response to a Constant Input
  • Slide 21
  • Applying these results to the RL circuit yields the solution: Complete Response to a Constant Input
  • Slide 22
  • The circuit below is at steady state before the switch opens; find the current i(t) for t > 0: Note: Example 8.3-5: First-Order Circuit
  • Slide 23
  • The figures below show circuit after the switch opens (left) and its the Thvenin equivalent circuit (right): The parameters of the Thvenin equivalent circuit are: Solution
  • Slide 24
  • 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: Solution
  • Slide 25
  • Solving for v a (t) yields: Finally, we calculating i(t) using Ohm's law yields: Solution
  • Slide 26
  • >> 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)')
  • Slide 27
  • Questions?
  • Slide 28
  • 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 t 0 and equal to the voltage V 0 thereafter, as shown below: We can represent voltage v(t) using the unit step, as shown: The Unit Step Source
  • Slide 29
  • Where the unit step forcing function is defined as: The Unit Step Source Note: That value of u(t 0 ) is undefined
  • Slide 30
  • Consider the pulse source v(t) = V 0 u(t t 0 )V 0 u(t t 1 ) defined: We can represent voltage v(t) using the unit step, as shown: The Unit Step Source
  • Slide 31
  • 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 = t 0 ; 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. The Unit Step Source
  • Slide 32
  • Consider the application of a pulse source to an RL circuit as shown below with t 0 = 0, implying a pulse duration of t 1 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 = i 1 + i 2 where i 1 is the response to V 0 u(t) and i 2 is the response to V 0 u(t t 1 ) Ex: Pulse Source Driving an RL Circuit
  • Slide 33
  • We recall that the response of an RL circuit to a constant forcing function applied at t = t n with i(0) = 0, I sc = V 0 /R, and where = L/R is given by: Consequently, we may add the two solutions to the two- step sources, carefully noting t 0 = 0 and t 1 as the start of each response, respectively, as shown: Ex: Pulse Source Driving an RL Circuit
  • Slide 34
  • Adding the responses provides the complete response of the RL circuit shown: Ex: Pulse Source Driving an RL Circuit
  • Slide 35
  • Questions?
  • Slide 36
  • Slide 37
  • Slide 38
  • Slide 39

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