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CONTROL SYSTEMS ENGINEERING LECTURE SERIES CONTROL SYSTEMS ENGINEERING LECTURE SERIES PART PART VIII. III. Transient Response Transient Response Prepared by: Engr. Michael C. Pacis BSEE, MEP-EE, PhD EE (units) 1 Michael Calizo Pacis ECE 131

PART VIII. Transient Response

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Page 1: PART VIII. Transient Response

CONTROL SYSTEMS ENGINEERING LECTURE SERIESCONTROL SYSTEMS ENGINEERING LECTURE SERIES

PART PART VVIII. III. Transient ResponseTransient Response

Prepared by:

Engr. Michael C. Pacis

BSEE, MEP-EE, PhD EE (units)

1Michael Calizo PacisECE 131

Page 2: PART VIII. Transient Response

OUTLINEOUTLINE

1. Introduction to Poles, Zeros and System Response

2. Evaluating Responses using Poles and Zeros

3. Transient Response of First Order Systems

4. Transient Response of Second Order Systems

5. Analysis and Design of Feedback Control Systems5. Analysis and Design of Feedback Control Systems

ECE 131 2Michael Calizo Pacis

Page 3: PART VIII. Transient Response

Poles, Zeros, and System ResponsePoles, Zeros, and System Response

The output response of a system is the sum of two responses:

the forced response and the natural response. Although many

techniques, such as solving a differential equation or taking the inverse

Laplace transform, enable us to evaluate this output response, these

techniques are laborious and time-consuming. Productivity is aided by

analysis and design techniques that yield results in a minimum of time.

If the technique is so rapid that we feel we derive the desired result by

ECE 131 3Michael Calizo Pacis

If the technique is so rapid that we feel we derive the desired result by

inspection, we sometimes use the attribute qualitative to describe the

method. The use of poles and zeros and their relationship to the time

response of a system is such a technique. Learning this relationship

gives us a qualitative "handle" on problems. The concept of poles and

zeros, fundamental to the analysis and design of control systems,

simplifies the evaluation of a system's response.

Page 4: PART VIII. Transient Response

DefinitionsDefinitions

Poles of a Transfer FunctionThe poles of a transfer function are (1) the values of the

Laplace transform variable, s, that cause the transfer function

to become infinite or (2) any roots of the denominator of the

transfer function that are common to roots of the numerator.

Strictly speaking, the poles of a transfer function satisfy

ECE 131 4Michael Calizo Pacis

Strictly speaking, the poles of a transfer function satisfy

part (1) of the definition. For example, the roots of the

characteristic polynomial in the denominator are values of s

that make the transfer function infinite, so they are thus poles.

However, if a factor of the denominator can be cancelled by the

same factor in the numerator, the root of this factor no longer

causes the transfer function to become infinite.

Page 5: PART VIII. Transient Response

DefinitionsDefinitions

Zeros of a Transfer FunctionThe zeros of a transfer function are (1) the values of the

Laplace transform variable, s, that cause the transfer function to

become zero, or (2) any roots of the numerator of the transfer

function that are common to roots of the denominator.

Strictly speaking, the zeros of a transfer function satisfy

ECE 131 5Michael Calizo Pacis

Strictly speaking, the zeros of a transfer function satisfy

part (1) of this definition. For example, the roots of the numerator

are values of s that make the transfer function zero and are thus

zeros. However, if a factor of the numerator can be cancelled by

the same factor in the denominator, the root of this factor no

longer causes the transfer function to become zero.

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ExampleExample 1. 1. Poles and Zeros of a FirstPoles and Zeros of a First--Order Order SystemSystem

Given the transfer function G(s) in

Figure 1(a), a pole exists at s = - 5 ,

and a zero exists at -2.

These values are plotted on the Figure 1a

ECE 131 6Michael Calizo Pacis

These values are plotted on the

complex s-plane in Figure 1(b),

using an x for the pole and a O for

the zero.

Figure 1a

Figure 1b

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ExampleExample 1. 1. Poles and Zeros of a FirstPoles and Zeros of a First--Order Order SystemSystem

To show the properties of the poles and zeros, let us find the

unit step response of the system. Multiplying the transfer

function of Figure 1(a) by a step function yields

Solution:

ECE 131 7Michael Calizo Pacis

Page 8: PART VIII. Transient Response

ExampleExample 1. 1. Poles and Zeros of a FirstPoles and Zeros of a First--Order Order SystemSystem

ECE 131 8Michael Calizo Pacis

Figure 1c

Page 9: PART VIII. Transient Response

ExampleExample 1. 1. Poles and Zeros of a FirstPoles and Zeros of a First--Order Order SystemSystem

CONCLUSIONS:

1. A pole of the input function generates the form of the forced

response (that is, the pole at the origin generated a step function

at the output).

2. A pole of the transfer function generates the form of the natural

response (that is, the pole at - 5 generated e-5t).

ECE 131 9Michael Calizo Pacis

response (that is, the pole at - 5 generated e-5t).

3. A pole on the real axis generates an exponential response of the

form e-αt where -α is the pole location on the real axis. Thus, the

farther to the left a pole is on the negative real axis, the faster the

exponential transient response will decay to zero (again, the pole

at -5 generated e-5t; see Figure 2 for the general case).

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ExampleExample 1. 1. Poles and Zeros of a FirstPoles and Zeros of a First--Order Order SystemSystem

CONCLUSIONS:

4. The zeros and poles generate the amplitudes for both the

forced and natural responses

ECE 131 10Michael Calizo Pacis

Figure 2. Effect of a real-axis pole upon transient

response

Page 11: PART VIII. Transient Response

ExampleExample 2. 2. Evaluating Response Using PolesEvaluating Response Using Poles

PROBLEM: Given the system of Figure 3, write the output,

c(t), in general terms. Specify the forced and natural parts of

the solution.

SOLUTION: By inspection, each system pole generates an

ECE 131 11Michael Calizo Pacis

SOLUTION: By inspection, each system pole generates an

exponential as part of the natural response. The input's pole

generates the forced response. Thus,

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ExampleExample 2. 2. Evaluating Response Using PolesEvaluating Response Using Poles

Taking the inverse Laplace transform, we get

ECE 131 12Michael Calizo Pacis

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FirstFirst Order SystemsOrder Systems

We now discuss first-order systems without zeros to define a

performance specification for such a system. A first-order system

without zeros can be described by the transfer function shown in

Figure 4.4(a). If the input is a unit step, where R(s) = 1/s, the Laplace

transform of the step response is C(s), where

ECE 131 13Michael Calizo Pacis

Figure 3.a First Order System Figure 3.b Pole Plot

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FirstFirst Order SystemsOrder Systems

If the input is a unit step, where R(s) = 1/s, the Laplace transform of

the step response is C(s), where

Taking the inverse transform, the step response is given by

ECE 131 14Michael Calizo Pacis

Taking the inverse transform, the step response is given by

1

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FirstFirst Order SystemsOrder Systems

where the input pole at the origin generated the forced response cf(t) = 1, and

the system pole at -a, as shown in Figure 3(b), generated the natural response

cn(t) = -eat. Equation (1) is plotted in Figure 5.

ECE 131 15Michael Calizo Pacis

Figure 4 First Order System Response to a Unit Step

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FirstFirst Order SystemsOrder Systems

Examine the significance of parameter a, the only parameter

needed to describe the transient response. When t = 1 /a,

2

ECE 131 16Michael Calizo Pacis

3

Now use Eqs. (1), (2), and (3) to define three transient response

performance specifications.

Page 17: PART VIII. Transient Response

Time ConstantTime Constant

We call 1/a the time constant of the response. From Eq. (2),

the time constant can be described as the time for e-at to decay to

37% of its initial value. Alternately, from Eq. (3) the time constant is

the time it takes for the step response to rise to 63% of its final

value.

ECE 131 17Michael Calizo Pacis

Figure 4 First Order System Response to a Unit Step

Page 18: PART VIII. Transient Response

Time ConstantTime Constant

The reciprocal of the time constant has the units

(1/seconds), or frequency. Thus, we can call the parameter a the

exponential frequency. Since the derivative of e-at is -a when t = 0,

a is the initial rate of change of the exponential at t = 0. Thus, the

time constant can be considered a transient response

specification for a first-order system, since it is related to the

speed at which the system responds to a step input.

ECE 131 18Michael Calizo Pacis

speed at which the system responds to a step input.

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Rise Time, Rise Time, TrTr

Rise time is defined as the time for the waveform to go

from 0.1 to 0.9 of its final value. Rise time is found by solving Eq.

(1) for the difference in time at c(t) = 0.9 and c(t) = 0.1. Hence:

4

ECE 131 19Michael Calizo Pacis

4

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Settling Time, Settling Time, TsTs

Settling time is defined as the time for the response to

reach, and stay within, 2% of its final value.2 Letting c(t) = 0.98 in

Eq. (1) and solving for time, t, we find the settling time to be

5

ECE 131 20Michael Calizo Pacis

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ExampleExample 3. Evaluating Time Constants3. Evaluating Time Constants

PROBLEM: A system has a transfer function, G(s) = 50 / s + 50. Find

the time constant, Tc, settling time, Ts, and rise time, Tr.

ECE 131 21Michael Calizo Pacis

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SeconSecond Order Systemsd Order Systems

Compared to the simplicity of a first-order system, a

second-order system exhibits a wide range of responses that must

be analyzed and described. Whereas varying a first-order system's

parameter simply changes the speed of the response, changes in

the parameters of a second-order system can change the form of

the response. For example, a second-order system can display

ECE 131 22Michael Calizo Pacis

the response. For example, a second-order system can display

characteristics much like a first-order system, or, depending on

component values, display damped or pure oscillations for its

transient response.

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Numerical Examples: Numerical Examples: SeconSecond Order Systemsd Order Systems

All the next examples are derived from Figure 5(a), the

general case, which has two finite poles and no zeros. The term

in the numerator is simply a scale or input multiplying factor that

can take on any value without affecting the form of the derived

results. By assigning appropriate values to parameters a and b,

ECE 131 23Michael Calizo Pacis

results. By assigning appropriate values to parameters a and b,

we can show all possible second-order transient responses. The

unit step response then can be found using C(s) = R(s)G(s), where

R(s) = 1/s, followed by a partial-fraction expansion and the

inverse Laplace transform.

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Numerical Examples: Numerical Examples: SeconSecond Order Systemsd Order Systems

ECE 131 24Michael Calizo Pacis

Figure 5

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Numerical Examples: Numerical Examples: SeconSecond Order Systemsd Order Systems

ECE 131 25Michael Calizo Pacis

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OverdampedOverdamped Response, Figure 5(b)Response, Figure 5(b)

For this response,

ECE 131 26Michael Calizo Pacis

This function has a pole at the origin that comes from the unit step

input and two real poles that come from the system. The input pole at the

origin generates the constant forced response; each of the two system poles

on the real axis generates an exponential natural response whose exponential

frequency is equal to the pole location.

Page 27: PART VIII. Transient Response

OverdampedOverdamped Response, Figure 5(b)Response, Figure 5(b)

Hence, the output initially could have been written as

c(t) = K1 +K2e-7.854t + K3e-1.146t. This response, shown in Figure 5(b),

is called overdamped. We see that the poles tell us the form of the

response without the tedious calculation of the inverse Laplace

transform.

ECE 131 27Michael Calizo Pacis

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UnderdampedUnderdamped Response, Figure 5(c)Response, Figure 5(c)

For this response,

ECE 131 28Michael Calizo Pacis

This function has a pole at the origin that comes from the unit step

input and two complex poles that come from the system. We now compare the

response of the second-order system to the poles that generated it. First we

will compare the pole location to the time function, and then we will compare

the pole location to the plot.

Page 29: PART VIII. Transient Response

UnderdampedUnderdamped Response, Figure 5(c)Response, Figure 5(c)

From Figure 5(c), the poles that generate the natural

response are at s = -1 ± j√8. Comparing these values to c(t) in the

same figure, we see that the real part of the pole matches the

exponential decay frequency of the sinusoid's amplitude, while

the imaginary part of the pole matches the frequency of the

sinusoidal oscillation.

ECE 131 29Michael Calizo Pacis

sinusoidal oscillation.

Page 30: PART VIII. Transient Response

The General Sinusoidal Response : 2The General Sinusoidal Response : 2ndnd Order Order SystemsSystems

.

ECE 131 30Michael Calizo Pacis

Figure 6

The transient response consists of an exponentially decaying amplitude

generated by the real part of the system pole times a sinusoidal waveform

generated by the imaginary part of the system pole

Page 31: PART VIII. Transient Response

The General Sinusoidal Response : 2The General Sinusoidal Response : 2ndnd Order Order SystemsSystems

.

The time constant of the exponential decay is equal to the

reciprocal of the real part of the system pole. The value of the

imaginary part is the actual frequency of the sinusoid. This

sinusoidal frequency is given the name damped frequency of

oscillation, �d.

ECE 131 31Michael Calizo Pacis

oscillation, �d.

The steady-state response (unit step) was generated by the

input pole located at the origin. This is the type of response shown

in Figure 5 an underdamped response one which approaches a

steady-state value via a transient response that is a damped

oscillation.

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Example 3. Example 3. Form of Form of UnderdampedUnderdamped Response Response Using PolesUsing Poles

.

PROBLEM: By inspection, write the form of the step response of

the system

ECE 131 32Michael Calizo Pacis

SOLUTION:

First we determine that the form of the forced response is a step.

Next we find the form of the natural response. Factoring the

denominator of the transfer function the poles to be s = -5 ±j

13.23.

Page 33: PART VIII. Transient Response

Example 3. Example 3. Form of Form of UnderdampedUnderdamped Response Response Using PolesUsing Poles

.

The real part, -5, is the exponential frequency for the

damping. It is also the reciprocal of the time constant of the

decay of the oscillations. The imaginary part, 13.23, is the radian

frequency for the sinusoidal oscillations.

Using lecture series 1 and 2 and Figure 5(c) as a guide,

ECE 131 33Michael Calizo Pacis

Using lecture series 1 and 2 and Figure 5(c) as a guide,

and c(t) is a constant plus an exponentially damped sinusoid.

Page 34: PART VIII. Transient Response

UndampedUndamped Response, Figure 5(d)Response, Figure 5(d)For this response,

ECE 131 34Michael Calizo Pacis

This function has a pole at the origin that comes from the unit step

input and two imaginary poles that come from the system. The input pole at

the origin generates the constant forced response, and the two system poles

on the imaginary axis at ±j3 generate a sinusoidal natural response whose

frequency is equal to the location of the imaginary poles. Hence, the output

can be estimated as c(t) = K1 + K4 cos(3t - Ө). This type of response, shown in

Figure 5(d), is called undamped. Note that the absence of a real part in the

pole pair corresponds to an exponential that does not decay. Mathematically,

the exponential is e-0t = 1.

Page 35: PART VIII. Transient Response

Critically Damped Response, Figure 5(e)For this response,

This function has a pole at the origin that comes from the unit step

ECE 131 35Michael Calizo Pacis

This function has a pole at the origin that comes from the unit step

input and two multiple real poles that come from the system. The input pole at

the origin generates the constant forced response, and the two poles on the real

axis at -3 generate a natural response consisting of an exponential and an

exponential multiplied by time, where the exponential frequency is equal to the

location of the real poles. Hence, the output can be estimated as c(r) = K1 + K2e-

3t+ K3te-3t. This type of response, shown in Figure 5(e), is called critically

damped. Critically damped responses are the fastest possible without the

overshoot that is characteristic of the underdamped response.

Page 36: PART VIII. Transient Response

Summary of ResponsesSummary of Responses

ECE 131 36Michael Calizo Pacis

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Summary of ResponsesSummary of Responses

ECE 131 37Michael Calizo Pacis

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Summary of ResponsesSummary of ResponsesNotice that the critically damped case is the division between the overdamped

cases and the underdamped cases and is the fastest response without overshoot.

ECE 131 38Michael Calizo Pacis

Figure 7. Step responses for second-order system

damping cases

Page 39: PART VIII. Transient Response

The General SecondThe General Second--Order SystemOrder System

In this section, we define two physically meaningful

specifications for second-order systems. These quantities can be

used to describe the characteristics of the second-order

transient response just as time constants describe the first-order

system response. The two quantities are called natural

frequency and damping ratio.

ECE 131 39Michael Calizo Pacis

frequency and damping ratio.

Page 40: PART VIII. Transient Response

Natural Frequency, ����n

The natural frequency of a second-order system is the frequency of

oscillation of the system without damping. For example, the frequency of

oscillation of a series RLC circuit with the resistance shorted would be the

natural frequency.

Damping Ratio, Damping Ratio, ��������

ECE 131 40Michael Calizo Pacis

Damping Ratio, Damping Ratio, ��������

A viable definition for this quantity is one that compares the

exponential decay frequency of the envelope to the natural frequency. This ratio

is constant regardless of the time scale of the response. Also, the reciprocal,

which is proportional to the ratio of the natural period to the exponential time

constant, remains the same regardless of the time base.

Page 41: PART VIII. Transient Response

Damping Ratio, Damping Ratio, ��������

The general second-order system shown in Figure 5(a) can be

transformed to show the quantities �and �n. Consider the general system

6

ECE 131 41Michael Calizo Pacis

transformed to show the quantities �and �n. Consider the general system

7

Page 42: PART VIII. Transient Response

Damping Ratio, Damping Ratio, ��������

Hence,

For a, assuming an underdamped system, the complex

poles have a real part, �, equal to -a/2. The magnitude of this

value is then the exponential decay frequency

8

ECE 131 42Michael Calizo Pacis

from which9

Page 43: PART VIII. Transient Response

Damping Ratio, Damping Ratio, ��������

The general second-order transfer function finally

looks like this:

10

ECE 131 43Michael Calizo Pacis

Page 44: PART VIII. Transient Response

Example 4. Finding Example 4. Finding �������� and and ��������n,n,n,n,n,n,n,n, for a Secondfor a Second--

Order SystemOrder System

PROBLEM: Given the transfer function, find � and �n.

SOLUTION:

ECE 131 44Michael Calizo Pacis

from:

Thus,

Page 45: PART VIII. Transient Response

Example 4. Finding Example 4. Finding �������� and and ��������n,n,n,n,n,n,n,n, for a Secondfor a Second--

Order SystemOrder System

Now that we have defined �, and �n, let us relate these

quantities to the pole location. Solving for the poles of the

transfer function in Eq. (10) yields

ECE 131 45Michael Calizo Pacis

Page 46: PART VIII. Transient Response

SecondSecond--order response as a function of order response as a function of

damping ratiodamping ratio

ECE 131 46Michael Calizo Pacis

Page 47: PART VIII. Transient Response

Example 5. Characterizing Response from Example 5. Characterizing Response from

the value of the value of ��������

PROBLEM: For each of the systems shown in Figure 8, find the

value of � and report the kind of response expected.

ECE 131 47Michael Calizo Pacis

Figure 8. Example 5 Systems

Page 48: PART VIII. Transient Response

Example 5. Characterizing Response from Example 5. Characterizing Response from

the value of the value of ��������

SOLUTION: First match the form of these systems to the forms

shown in Eqs 7 and 9.

Since a = 2��n and �n = √b

ECE 131 48Michael Calizo Pacis

Using the values of a and b from each of the systems

� = 1.155 for system (a), which is thus overdamped, since � > 1

� = 1 for system (b), which is thus critically damped;

� = 0.894 for system (c), which is thus underdamped, since � < 1.

Page 49: PART VIII. Transient Response

UnderdampedUnderdamped SecondSecond--Order SystemsOrder Systems

In the last slides, it is generalized the second-order

transfer function in terms of � and �n, let us analyze the step

response of an underdamped second-order system. Not only will

this response be found in terms of � and �n, but more

specifications indigenous to the underdamped case will be

defined. The underdamped second-order system, a common

ECE 131 49Michael Calizo Pacis

defined. The underdamped second-order system, a common

model for physical problems, displays unique behavior that must

be itemized; a detailed description of the underdamped response

is necessary for both analysis and design.

Page 50: PART VIII. Transient Response

UnderdampedUnderdamped SecondSecond--Order SystemsOrder Systems

Begin by finding the step response for the general

second-order system of Eq. (10). The transform of the

response, C(s), is the transform of the input times the

transfer function, or

ECE 131 50Michael Calizo Pacis

where it is assumed that � < 1 (the underdamped case).

Expanding by partial fractions, using the methods described in

Lecture Series 3, Case 3, yields

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UnderdampedUnderdamped SecondSecond--Order SystemsOrder Systems

Taking the inverse Laplace transform

10

ECE 131 51Michael Calizo Pacis

Page 52: PART VIII. Transient Response

UnderdampedUnderdamped SecondSecond--Order SystemsOrder SystemsA plot of this response appears in Figure 9 for various values of

�, plotted along a time axis normalized to the natural frequency.

ECE 131 52Michael Calizo Pacis

Figure 9. Second-order underdamped responses for damping ratio

values

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UnderdampedUnderdamped SecondSecond--Order SystemsOrder Systems

ECE 131 53Michael Calizo Pacis

Figure 10. Second-order underdamped response specifications

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UnderdampedUnderdamped Response SpecificationsResponse Specifications

ECE 131 54Michael Calizo Pacis

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UnderdampedUnderdamped Response SpecificationsResponse Specifications

Notice that the definitions for settling time and rise

time are basically the same as the definitions for the first-order

response. All definitions are also valid for systems of order

higher than 2, although analytical expressions for these

parameters cannot be found unless the response of the higher-

order system can be approximated as a second-order system.

ECE 131 55Michael Calizo Pacis

order system can be approximated as a second-order system.

Rise time, peak time, and settling time yield information

about the speed of the transient response. This information can

help a designer determine if the speed and the nature of the

response do or do not degrade the performance of the system.

Page 56: PART VIII. Transient Response

Evaluation of Evaluation of TpTp

Tp is found by differentiating c(t) in Eq. (11) and finding

the first zero crossing after t=0. This task is simplified by

"differentiating" in the frequency domain by using Item 7 of

Table 2 (Lecture Series 3). Assuming zero initial conditions and

using Eq. (10), thus,

ECE 131 56Michael Calizo Pacis

Completing squares in the denominator,

Page 57: PART VIII. Transient Response

Evaluation of Evaluation of TpTpTherefore,

Setting the derivative equal to zero yields

11

12

ECE 131 57Michael Calizo Pacis

or

12

13

Page 58: PART VIII. Transient Response

Evaluation of Evaluation of TpTp

Each value of n yields the time for local maxima or minima.

Letting n = 0 yields t = 0, the first point on the curve in Figure 10

that has zero slope. The first peak, which occurs at the peak time,

Tp, is found by letting n = 1 in Eq. (13):

14

ECE 131 58Michael Calizo Pacis

14

Page 59: PART VIII. Transient Response

Evaluation of Evaluation of %OS%OS

From Figure 10 the percent overshoot, %OS, is given by

The term cmax is found by evaluating c(t) at the peak time, c(Tp).

Using Eq. (14) for Tp and substituting into Eq. (10) yields

15

ECE 131 59Michael Calizo Pacis

Using Eq. (14) for Tp and substituting into Eq. (10) yields

16

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Evaluation of Evaluation of %OS%OS

For the unit step used for Eq. (10),

17

Substituting Eqs. (16) and (17) into Eq. (15), we finally obtain

ECE 131 60Michael Calizo Pacis

Substituting Eqs. (16) and (17) into Eq. (15), we finally obtain

18

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Evaluation of Evaluation of %OS%OS

allows one to find %OS given �, the inverse of the equation

allows one to solve for � given %OS. The inverse is given by

20

ECE 131 61Michael Calizo Pacis

Figure 11. Percent overshoot versus damping ratio

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Evaluation of Evaluation of TsTs

In order to find the settling time, we must find the time for

which c(t) in Eq. (10) reaches and stays within ±2% of the steady-

state value, Cfjnal. Using our definition, the settling time is the time

it takes for the amplitude of the decaying sinusoid in Eq. (10) to reach

0.02, or

21

ECE 131 62Michael Calizo Pacis

This equation is a conservative estimate, since we are

assuming that cos (�n √l1- �2t - Ө) = 1 at the settling time. Solving

Eq. (21) for t, the settling time is

22

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Evaluation of Evaluation of TsTs

You can verify that the numerator of Eq. (4.41) varies from

3.91 to 4.74 as � varies from 0 to 0.9. Let us agree on an

approximation for the settling time that will be used for all

values of �; let it be

ECE 131 63Michael Calizo Pacis

22

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Evaluation of Evaluation of TrTrA precise analytical relationship between rise time and damping ratio, �,

cannot be found. However, using a computer and Eq. (4.28), the rise time can be found.

We first designate �nt as the normalized time variable and select a value for �. Using the

computer, we solve for the values of �nt that yield c(t) = 0.9 and c(t) = 0.1. Subtracting

the two values of cont yields the normalized rise time, �nt Tr, for that value of �.

Continuing in like fashion with other values of �, we obtain the results plotted in Figure

12

ECE 131 64Michael Calizo Pacis

Figure 12. Normalized rise time versus damping ratio for a second-order

underdamped response

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Example 6. Finding Example 6. Finding TpTp, %OS and , %OS and TrTr from a from a

transfer function transfer function

PROBLEM: Given the transfer function

�n �

ECE 131 65Michael Calizo Pacis

SOLUTION: �n and � are calculated as 10 and 0.75

Substitute � and � n into Eqs. (14), (18), and (22)

Tp = 0.475 second, %OS = 2.838, and Ts = 0.533 second

Using the table in Figure 12, the normalized rise time is approximately

2.3 seconds. Dividing by �n yields Tr = 0.23 second.

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Example 7. Transient Response through Example 7. Transient Response through

component designcomponent design

PROBLEM: Given the system shown in the Figure, find J and D to yield 20%

overshoot and a settling time of 2 seconds for a step input of torque T(t).

ECE 131 66Michael Calizo Pacis

SOLUTION: First, the transfer function for the system is

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Example 7. Transient Response through Example 7. Transient Response through

component designcomponent designFrom the transfer function, 23

24

ECE 131 67Michael Calizo Pacis

25

26

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Example 7. Transient Response through Example 7. Transient Response through

component designcomponent design

Also, from Eqs. (23) and (25),

From Eq. (20), a 20% overshoot implies ����= 0.456. Therefore, from

27

ECE 131 68Michael Calizo Pacis

From Eq. (20), a 20% overshoot implies ����= 0.456. Therefore, from

Eq. (27)

28

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Example 7. Transient Response through Example 7. Transient Response through

component designcomponent design

From the problem statement, K = 5 N-m/rad. Combining

this value with Eqs. (24) and (28), D = 1.04 N-m-s/rad, and J =

0.26 kg-m2.

ECE 131 69Michael Calizo Pacis

Page 70: PART VIII. Transient Response

Analysis and Design of Feedback SystemsAnalysis and Design of Feedback Systems

Consider,

Figure 13. Second-order feedback control system

which can model a control system such as the antenna azimuth position

ECE 131 70Michael Calizo Pacis

which can model a control system such as the antenna azimuth position

control system. For example, the transfer function, K/s(s + a), can model the

amplifiers, motor, load, and gears. From Lecture Series 7, the closed-loop

transfer function, T(s), for this system is

Page 71: PART VIII. Transient Response

Analysis and Design of Feedback SystemsAnalysis and Design of Feedback Systems

where K models the amplifier gain, that is, the ratio of the

output voltage to the input voltage. As K varies, the poles move

through the three ranges of operation of a second-order

system: overdamped, critically damped, and underdamped.

ECE 131 71Michael Calizo Pacis

Page 72: PART VIII. Transient Response

Example 8. Finding Transient ResponseExample 8. Finding Transient Response

PROBLEM: For the system shown in the Figure, find the peak time,

percent overshoot, and settling time.

ECE 131 72Michael Calizo Pacis

SOLUTION: The closed-loop transfer function

Also, and For �

Page 73: PART VIII. Transient Response

Example 8. Finding Transient ResponseExample 8. Finding Transient Response

Using the values for � and �n

ECE 131 73Michael Calizo Pacis

Page 74: PART VIII. Transient Response

Example 9. Gain Design for Transient Example 9. Gain Design for Transient

Response Response

PRO BLEM: Design the value of gain. K, for the feedback

control system so that the system will respond with a 10%

overshoot.

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SOLUTION: The closed-loop transfer function of the

system is

Page 75: PART VIII. Transient Response

Example 9. Gain Design for Transient Example 9. Gain Design for Transient

Response Response 29

30

ECE 131 75Michael Calizo Pacis

Since percent overshoot is a function only of �, Eq 31 shows that

the percent overshoot is a function of K.

31

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Example 9. Gain Design for Transient Example 9. Gain Design for Transient

Response Response

A 10% overshoot implies that � = 0.591. Substituting this

value for the damping ratio into Eq. (31) and solving for K yields

K = 17.9

ECE 131 76Michael Calizo Pacis

Page 77: PART VIII. Transient Response

REFERENCESREFERENCES

Control Systems Engineering, 3rd Edition,

by Norman S. Nise