Transcript Chapter_5_Lecture_PowerPoint

Chapter 5
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Transient Analysis
A steady-state region for 0 ≤ t ≤
0.2 s; a transient region for 0.2
≤ t ≤ 2 s (approximately); and a
new steady-state region for t >
2 s, where the voltage reaches
a steady DC or AC condition.
The objective of transient
analysis is to describe the
behavior of a voltage or a
current during the transition
between two distinct steadystate conditions.
Examples of transient response
A general model of the transient
analysis problem
Regardless of how many resistors the circuit
contains, it is a first-order circuit. The
response of a first-order circuit to a switched
DC source will appear in one of the two forms
shown below, which represent, in order, a
decaying exponential and a rising
exponential waveform.
Decaying and rising exponential
responses
Applying KVL around the loop, we may obtain
the following equation:
Circuit containing energy storage element
Any circuit containing a single energy storage
element can be described by a differential equation
of the form
Second-order circuit
DC Steady-State Solution
The term DC steady state refers to circuits that
have been connected to a DC (voltage or current)
source for a very long time, such that it is
reasonable to assume that all voltages and
currents in the circuits have become constant.
At DC steady state, all capacitors behave as
open circuits and all inductors behave as short
circuits.
Consider the defining equation for the capacitor
Abrupt change in capacitor voltage
The value of an inductor current or a capacitor
voltage just prior to the closing (or opening) of a
switch is equal to the value just after the switch
has been closed (or opened). Formally
Where the notation 0+ signified “just after
t = 0” and 0- means “just before t = 0.”
FOCUS ON METHODOLOGY
FIRST-ORDER TRANSIENT RESPONSE
1. Solve for the steady-state response of the circuit before the switch changes state
(t = 0−) and after the transient has died out (t→∞).We shall generally refer to these
responses as x(0−) and x(∞).
2. Identify the initial condition for the circuit x(0+), using continuity of capacitor
voltages and inductor currents [vC = vC(0−), iL(0+) = iL(0−)].
3. Write the differential equation of the circuit for t = 0+, that is, immediately after the
switch has changed position. The variable x(t) in the differential equation will be
either a capacitor voltage vC(t) or an inductor current iL(t). It is helpful at this time
to reduce the circuit to Thévenin or Norton equivalent form, with the energy
storage element (capacitor or inductor) treated as the load for the Thévenin
(Norton) equivalent circuit. Reduce this equation to standard form (equation 5.8).
4. Solve for the time constant of the circuit: τ = RTC for capacitive circuits, τ = L/RT for
inductive circuits.
5. Write the complete solution for the circuit in the form x(t) = x(∞) + [x(0) − x(∞)]e−t/τ
General Solution of First-Order Circuits
Let the initial condition of the system be x(t =
0) = x(0). Then we seek to solve the differential
equation
This solution consists of two parts: the natural response
(or homogeneous solution), with the forcing function set
equal to zero, and the forced response (or particular
solution), in which we consider the response to the forcing
function. The complete response then consists of the
sum of the natural and forced responses. Once the form
of the complete response is known, the initial condition can
be applied to obtain the final solution.
Natural Response
Forced Response
Complete Response
Energy Storage in Capacitors and Inductors
Decay through a resistor of energy stored
in a capacitor
Decay through a resistor of energy stored
in an inductor
Parallel case
Second-order circuits
Solution of Second-Order Circuits
Response of switched second-order system
1. The response asymptotically tends to a
final value of 1.
2. The response oscillates with a period
approximately equal to 6 s.
3. The oscillations decay (and eventually
disappear) as time progresses.
1. The final value of 1 is predicted by the DC gain KS = 1, which tells
us that in the steady state (when all the derivative terms are zero)
x(t) = f(t).
2. The period of oscillation of the response is related to the natural
frequency:
ωn = 1 leads to the calculation T = 2π/ωn = 2π ≈ 6.28 s. Thus, the
natural frequency parameter describes the natural frequency of
oscillation of the system.
3. Finally, the reduction in amplitude of the oscillations is governed
by the damping ratio ζ. You can see that as ζ increases, the
amplitude of the initial oscillation becomes increasingly smaller until,
when ζ = 1, the response no longer overshoots the final value of 1
and has a response that looks, qualitatively, like that of a first-order
system.
Response of switched second-order system
Elements of the Transient Response
The solution of a second-order differential equation
also requires that we consider the natural response
(or homogeneous solution), with the forcing
function set equal to zero, and the forced response
(or particular solution), in which we consider the
response to the forcing function. The complete
response then consists of the sum of the natural
and forced responses. Once the form of the
complete response is known, the initial condition can
be applied to obtain the final solution.
FOCUS ON METHODOLOGY
ROOTS OF SECOND-ORDER SYSTEMS
Case 1: Real and distinct roots. This case occurs when ζ >
1, since the term under the square root sign is positive in this
case, and the roots are s1,2 = −ζωn ± ωn √ ζ 2 − 1. This leads to
an overdamped response.
Case 2: Real and repeated roots. This case holds when ζ =
1, since the term under the square root is zero in this case,
and s1,2 = −ζωn = −ωn. This leads to a critically damped
response.
Case 3: Complex conjugate roots. This case holds when ζ <
1, since the term under the square root is negative in this case,
and s1,2 = −ζωn ± jωn √ 1 − ζ 2. This leads to an underdamped
response.
FOCUS ON METHODOLOGY
SECOND-ORDER TRANSIENT RESPONSE
1. Solve for the steady-state response of the circuit
before the switch changes state (t = 0−) and after the
transient has died out (t→∞).We shall generally refer
to these responses as x(0−) and x(∞).
2. Identify the initial conditions for the circuit x(0+),
and ˙x(0+), using the continuity of capacitor voltages
and inductor currents [vC(0+) = vC(0−), iL(0+) = i +
L(0−)] and circuits analysis. This will be illustrated by
examples.
3. Write the differential equation of the circuit for t
= 0+, that is, immediately after the switch has
changed position. The variable x(t) in the
differential equation will be either a capacitor
voltage vC(t) or an inductor current iL(t). Reduce
this equation to standard form.
4. Solve for the parameters of the second-order
circuit, ωn and ζ.
5. Write the complete solution for the circuit in one
of the three forms given below as appropriate:
Overdamped case (ζ > 1):
Critically damped case (ζ = 1):
Underdamped case (ζ < 1):
6. Apply the initial conditions to solve for the
constants α1 and α2.