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 steady-

state 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

Circuit containing energy storage element

Applying KVL around the loop, we may obtain

the following equation:

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

Second-order circuits

Parallel case

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.