Transcript first-order circuits

ES250:
Electrical Science
HW8: Complete Response
of RL and RC Circuits
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
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:
First-Order Circuits
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
First-Order Circuits
T  6.28ms
2
 
T
 1000 rad
T  6.28ms
sec
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 steadystate part, as shown:
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 firstorder 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.:
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
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
Questions?
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:
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:
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+)?
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+)?
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:

Element law:
KCL:

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 = e D, which is determined from the IC x(0)
• To find A, let t = 0, then:

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:
Complete Response to a Constant Input
• 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
Complete Response to a Constant Input
• 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:


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:
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:
• The time constant is:
Solution
• 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:
• Solving for va(t) yields:
Solution
• Finally, we calculating i(t) using Ohm's law yields:
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 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:
The Unit Step Source
• Where the unit step forcing function is defined as:
Note: That value of u(t0) is undefined
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:

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.
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)
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 twostep sources, carefully noting t0 = 0 and t1 as the start of
each response, respectively, as shown:
Ex: Pulse Source Driving an RL Circuit
• Adding the responses provides the complete response of
the RL circuit shown:
Questions?