es250: electrical science

ES250:Electrical Science

HW8: Complete Response of RL and RC Circuits

• 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

Introduction

• 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:

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



6.28ms

6.28ms

2 rad1000 secT

T

T

• 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:


• 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., M cos(1000t + δ)• The complete response of a first-order circuit can be

represented in several ways, e.g.:


• 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


• 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


Questions?

• Find the complete response of a first-order circuit shown below for time t0 > 0 when the input is constant:

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:


• 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+)?


• 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+)?


• 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:


Element law:

KCL:

• 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 = e D, which is determined from the IC x(0)• To find A, let t = 0, then:


• Therefore, we obtain:

where the parameter τ is called the time constant• Since the solution can be written as:


• The circuit time constant can be measured from a plot of x(t) versus t, as shown:


Note: The circuit settles to

steady state within 5

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

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

• 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

• 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:

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:

Solution

• Solving for va(t) yields:

• Finally, we calculating i(t) using Ohm's law yields:

Solution

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)')

Questions?

• 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:

The Unit Step Source

• Where the unit step forcing function is defined as: The Unit Step Source

Note: That value of u(t0) is undefined

• 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:


• 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.


• 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)

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, and where 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:


• Adding the responses provides the complete response of the RL circuit shown:


Questions?

es250: electrical science

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