Time Response Analysis - fadhilfadhil.yolasite.com/resources/Chapter Five TRA1.pdfTime Response...

Time Response Analysis

Fourth Academic Year

Electrical Engineering Department

College of Engineering

University of Salahaddin

November 2015

Time Response

Since time is used as an independent variable in most control system, it is usually

of interest to evaluate the output response with respect to time, or simply, the

time response.

In designing a system, the time response behavior may well be the most

important aspect of it’s behavior.

Time response can be obtained if

1. The mathematical model of the system, and

2. The input(s) of the system

are available.

Output response of a system is the sum of forced response (steady-state) and

natural response (homogeneous solution)

The concept of poles and zeros, fundamental to the analysis and design of control

systems, simplifies the evaluation of a system’s response.

Control Engineering Salahaddin University / College of Engineering 2 of 50

Transient and Steady-State Responses

For a stable system, the transient response will decay, usually exponentially, to a

steady state as time increases.

Transient response is a function only of the system dynamics, and it is

independent of the input quantity.

Steady-state is the response of the system after the transient component has

decayed and it is a function of both the system dynamics and input quantity.


Test Input Signals


Poles of a Transfer Function

The poles of a transfer function are

1. The values of the Laplace transform variable s, that cause the transfer

function to become infinite.

2. Any roots of the denominator of the transfer function that are common to

roots of the numerator.

For example, the roots of the characteristic polynomial in the denominator are

values of s that make the transfer function infinite, so they are poles.

If a factor of the denominator can be canceled by the same factor in the

numerator, the root of this factor no longer causes the transfer function to

become infinite.

In control systems, we often refer to the root of the canceled factor in the

denominator as a pole even though the transfer function will not be infinite at

this value.


Zeros of a Transfer Function

The zeros of a transfer function are

1. The values of the Laplace transform variables s, that cause the transfer

function to become zero

2. Any roots of the numerator of the transfer function that are common to roots

of the denominator.

For example, the roots of the numerator are values of s that make the transfer

function zero and are zeros.

However, if a factor of numerator can be canceled by the same factor in the

denominator, the root of this factor no longer causes the transfer function to

become zero.

In control systems, we often refer to the root of the canceled factor in the

numerator as zero even though the transfer function will not be zero at this

value.


System Poles and Zeros

The transfer function provides a basis for determining important system

response characteristics without solving the complete differential equation.

The transfer function is a rational function in the complex variable s = + j

It is often convenient to factor the polynomials in the numerator and

denominator, and to write the transfer function in terms of those factor:

Where the numerator polynomials, N(s) and the denominator polynomials D(s),

have real coefficients defined by the system’s differential equation and K = bm / an

Thus zi’s are the roots of the equation N(s) = 0 which are the system zeros.

And pi’s are the roots of the equation D(s) = 0 which are the system poles.


System Poles and Zeros Cont.

When s = zi the numerator N(s) = 0 and the transfer function vanishes, that is

When s = pi the denominator polynomial D(s) = 0 and the value of the transfer

function becomes unbounded, that is

𝐥𝐢𝐦𝒔 →𝒛𝒊

𝑯 𝒔 = 𝟎

𝐥𝐢𝐦𝒔 →𝒑𝒊

𝑯 𝒔 = ∞

All of the coefficients of polynomials N(s) and D(s) are real, therefore the poles

and zeros must be either purely real, or appear in complex conjugate pairs.

In general for poles, either pi = i , ji ,or i ji

The existence of a single complex pole without a corresponding conjugate pole

would generate complex coefficients in the polynomial D(s).

Similarly, the system zeros are either real or appear complex conjugate pairs.



Example 5.1

Find the system poles and zeros of the linear system which is described by the

following differential equation:

Answer

From the differential equation the transfer function is

which may be written in factorial form as

𝒅𝟐𝒚

𝒅𝒕𝟐+ 𝟓

𝒅𝒚

𝒅𝒕+ 𝟔𝒚 = 𝟐

𝒅𝒖

𝒅𝒕+ 𝒖

2

2Control Engineering Salahaddin University / College of Engineering 9 of 50


The system therefore has a single real zero at s = -1/2 and a pair of real poles at

s = -3 and s = -2.

Note:

The poles and zeros are properties of transfer function, and therefore of the

differential equation describing the input-output dynamics. Together with the

gain constant K they completely characterize the differential equation, and

provide a complete description of the system.



Example 5.2

A system has a pair of complex conjugate poles p1, p2 = -1 j2, a single real

zero z1 = -4, and a gain factor K = 3. Find the differential equation

representing the system.

Answer

The transfer function is

and the differential equation is


The Pole-Zero Plot: s-plane

A system is characterized by its poles and zeros in the sense that they allow

reconstruction of the input/output differential equation.

In general, the poles and zeros of a transfer function may be complex, and the

system dynamics may be represented graphically by plotting their locations on

the complex s-plane, whose axes represent the real and imaginary parts of the

complex variables s.

It is usually to mark a zero location by a circle (o) and a pole location a cross ().


The Pole-Zero Plot: s-plane

Example 5.3

Show the locations of poles and zeros on s-plane of the following system

Answer

Re-writing in the factorial form as

Showing pole and zeros on s-plane jw


First-Order Systems

A first-order system without zeros can be described by the following transfer

function

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

C(s), where

Taking the inverse Laplace transform, the step response is given by

where the input pole at the origin generated the forced response (steady-state)

cf(t) = 1, and the system pole at –a generated the natural response (homogeneous

solution) cn(t) = -e-at.


First-Order Systems Cont.

Plotting of c(t) is

Let examine the significant of

parameter a, the only parameter

needed to describe the transient

response. When t = 1/a


First-Order Systems: Time Constant

The time constant is the time for e-at to decay to 37% of its initial value, or the

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

value.

The reciprocal of the time constant has the units (1/seconds), or 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.

The time constant can also be evaluated from the pole on s-plane.

Since the pole of the transfer function is at –a, it can be said that the pole is

located at the reciprocal of the time constant, and the farther the pole from the

imaginary axis, the faster the transient response.


First-Order Systems: Rise Time, Tr

Rise time Tr , is defined as the time for the waveform to go from 0.1 to 0.9 of its

final value.

From previous case, c(t) = 0.9, c(t) = 0.1, and from the curve, hence


First-Order Systems: Settling Time, Ts

Settling time is the time for the response to reach, and stay within, 2% of its final

value.

From previous case, let c(t) = 0.98, and solving c(t) for time, t, the setting time is

Setting time within a specific percentage of the final value could be

𝑻𝒔 =𝟒

𝒂𝒇𝒐𝒓 𝟐%

𝑻𝒔 =𝟑

𝒂𝒇𝒐𝒓 𝟓%

𝑻𝒔 =𝟒. 𝟔

𝒂𝒇𝒐𝒓 𝟏%

𝑻𝒔 =𝟒

𝒂


First-Order Systems

Example 5.4

A system has a transfer function, G(s) = 50/(s+50). Find the time constant, Tc,

settling time, Ts (for 2% of the final value), and rise time, Tr for a unit step input.

Answer

Since a = 50, thus

Tc = 1/a = 1/50 = 0.02 s

Ts = 4/a = 4/50 = 0.08 s

Tr = 2.2/a = 2.2/50 = 0.044 s


Second-Order Systems: General

A general second-order system is characterized by the following transfer function

The transient response of second-order systems can be described by two physical

quantities: undamped natural frequency, n , and damping ratio . Solving the

above transfer function for poles yields

𝒔𝟏,𝟐 = −𝝎𝒏± 𝝎𝒏 𝟐 − 𝟏

Undamped Natural Frequecny, n , of the second-order systems is the frequency of

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

RLC circuit with the resistance shorted would be the natural frequency.

𝒘𝒏𝟐 = 𝒃

or 𝒘𝒏 = 𝒃 𝐢𝐧 𝐫𝐚𝐝/𝐬


Second-Order Systems: General Cont.

Damping Ratio,

Since 𝒂 = 2n

Thus,

=𝐚

𝟐𝛚𝐧


Second-Order Systems: Step Response


Second-Order Systems :Undamped Response = 0

A typical second-order system which has an undamped (oscillatory) response is

Poles: two imaginary poles at jn

Natural Response: undamped sinusoidal with radian frequency equal to the

imaginary part of the poles.


Second-Order Systems :Underdamped Response 0 < < 1

A typical second-order system which has an underdamped response is

Equating coefficients

Partial fraction expansion


Second-Order Systems :Underdamped Response 0 < < 1 Cont.

Poles: two complex poles at - n jn 𝟏 − 𝟐

Natural Response: Damped sinusoidal with an exponential envelop whose time

constant is equal to the reciprocal of the pole’s real part. The radian frequency of

the sinusoid, the damped frequency of oscillation, is equal to the imaginary part of

the poles.


Second-Order Systems :Critically Damped Response = 1

A typical second-order system which has a critically damped response is

Poles: two real poles at - n

Natural Response: one term is an exponential whose time constant is equal to the

reciprocal of the pole location. Another term is the product of time, t, and

exponential with time constant equal to the reciprocal of the pole location.


Second-Order Systems :Overdamped Response > 1

A typical second-order system which has an overdamped damped response is

Poles: two real poles at

- n n 𝟐 − 𝟏

Natural Response: two exponentials

with time constants equal to the

reciprocal of the pole locations.


Second-Order Systems :Example

Example 5.5

For each of the following systems do the following: (1) Find the values of damping

ratio and undamped natural frequency; (2) Characterize the nature of the response.


Second-Order Systems :Example

Answers

a. n = 𝟒𝟎𝟎 = 20, 2 n = 12 = 0.3 and system is underdamped.

b. n = 𝟗𝟎𝟎 = 30, 2 n = 90 = 1.5 and system is overdamped.

c. n = 𝟐𝟐𝟓 = 15, 2 n = 30 = 1 and system is critically damped.

d. n = 𝟔𝟐𝟓 = 25, 2 n = 0 = 0 and system is undamped.


Underdamped Second-Order Systems: Specifications of Step Response

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.

The objective is to define transient specifications associated with underdamped

responses, and relating these specification to the pole location, drawing an

association between pole location and the form of the underdamped second-order

response.

Tying the pole location to system parameters, thus closing the loop: desired

response generates required system components.

The underdamped of second-order system occurred when the damping ratio is

between 0 and 1.


Underdamped Second-Order Systems: Specifications of Step Response Cont.

The step response of the general second-order system can be founded as

knowing that for underdamped 0 < < 1, thus using partial fractions yields

Taking the inverse Laplace transform

where

which can be determined from the pole location as in the next graph.Control Engineering Salahaddin University / College of Engineering 31 of 50


The relation between value of and the type response of the second-order system is

shown in the following figure.

As shown, the lower value of , the more oscillatory the response. The natural

frequency is a time-axis scale factor and does not affect the nature of the response

other than to scale it in time,Control Engineering Salahaddin University / College of Engineering 33 of 50


The step response of the second-order system can be characterized by using rise

time, peak time, percent overshoot, and settling time, the following figure shows

these parameters.



Rise Time, Tr

The rise time is the time required for the response to rise from 10% to 90%, 5% to

95%, or 0% to 100% of its final value. For underdamped second-order systems the

0% to 100% rise time is normally used. For overdamped systems, the 10% to 90%

rise time is commonly used.

Peak Time, Tp

The time required to reach the first, or maximum, peak.

Percentage Overshoot, % OS

The amount that the waveform overshoots the steady-state, or final value at the

peak time, expressed as a percentage of the steady-state value.

Settling Time, Ts

The time required for the transient’s damped oscillations to reach and stay within

2% of the steady-state value.



Note:

1. Rise time, peak time, and setting 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.

2. Tp, %OS, and Ts are evaluated as functions of and n .

3. Tr and Tp increase together.


Underdamped Second-Order Systems: Evaluation of Tp

Tp is founded by differentiating C(t), and finding the first zero crossing after t = 0.

Completing the squares in the denominator, yields to

Referring to tables in Chapter 2;

Setting the derivative equal to zero yields or

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 previous figure) that has zero slope. The first peak,

which occurs at the peak time, Tp, is found by letting n = 1 as


Underdamped Second-Order Systems: Evaluation of %OS

From the response of the underdamped second-order

system, the percentage overshoot, %OS, is giving by

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

For unit step, cfinal = 1, substituting cmax and cfinal into %OS yields


Underdamped Second-Order Systems: Evaluation of %OS Cont.

Since the percentage overshoot is a function only of the damping ratio, . Thus,

can be founded for the given %OS (another way for finding );

The relation between %OS and is depicted in the following curve.


Underdamped Second-Order Systems: Evaluation of Ts

In order to find the settling time, we must find the time for which c(t) reaches and

stays within 2% of the steady-state value, cfinal. Thus

This equation is conservative estimate, since we assuming that 𝒔𝒊𝒏 𝒘𝒏 𝟏− 𝟐𝒕 − 𝜽

= 𝟏 at the settling time. Solving the above expression for t, the settling time is

Since the numerator varies from 3.91 to 4.74 as varies from 0 to 0.9. Thus,

simplifying it for all values of yields

For 5%, 𝑻𝒔 =𝟑

𝒏

and for 1%, 𝑻𝒔 =𝟓

𝒏


Underdamped Second-Order Systems: Evaluation of Tr

A precise analytical relationship between rise time and damping ratio, , cannot be

found. However, the approximation value of Tr can be founded using the following

equation:


Underdamped Second-Order Systems: Example

Consider the following closed-loop control system:

For unity step input function, find rise time, delay time, peak time, percentage

overshoot, and settling time when damping ratio and undamped natural frequency are

0.6 and 5 rad/sec, respectively.

𝑤𝑛2

𝑠 + 2𝜁𝑤𝑛

1

𝑠

R(s) C(s)+

_


Underdamped Second-Order Systems: Assignment

A servo system block diagram is shown in the following figure:

Determine the values of gain, K, and velocity feedback constant, Kh, so that the

percentage overshoot in the unit-step response is 20% and the peak time is 1 sec. With

these values of K and Kh, obtain the rise time and settling time. Assume J = 1 kg.m2

and B = 1 N.m/rad/sec.

𝐾

𝐽𝑠 + 𝐵

1

𝑠

R(s) C(s)+

_

𝐾ℎ

_


Higher-Order Systems: General Form

The general form of the higher-order control system can be expressed by the following

transfer function:

𝑪(𝒔)

𝑹(𝒔)=

𝒃𝟎𝒔𝒎+𝒃𝟏𝒔

𝒎−𝟏+⋯+𝒃𝒎−𝟏𝒔+𝒃𝒎

𝒂𝟎𝒔𝒏+𝒂𝟏𝒔

𝒏−𝟏+⋯+𝒂𝒏−𝟏𝒔+𝒂𝒏where m ≤ n

Once the numerator and denominator have been factored, C(s)/R(s) can be written as

the following:

𝑪(𝒔)

𝑹(𝒔)= 𝑲

(𝒔+𝒛𝟏)(𝒔+𝒛𝟐)⋯(𝒔+𝒛𝒎)

(𝒔+𝒑𝟏)(𝒔+𝒑𝟐)⋯(𝒔+𝒑𝒏)(K = b0/a0)


Higher-Order Systems: Case 1

Case 1: Poles are real:

For a unit-step input function, the transfer function becomes as

𝑪 𝒔 =𝒂

𝒔+

𝒊=𝟏

𝒏𝒂𝒊

𝒔 + 𝒑𝒊

Where ai is the residue of the pole at s = – pi , ai can be found by using partial-fraction

expansion.

Example: Find the time response for a unit-step input function of the following closed-

loop control system:

Answer: The inverse Laplace transform is

𝒄 𝒕 = 𝟏 − 𝟐. 𝟓𝒆−𝟐𝒕 − 𝟐. 𝟓𝒆−𝟔𝒕 + 𝟒𝒆−𝟓𝒕

𝑪(𝒔)

𝑹(𝒔)=

𝟔𝟎

𝒔𝟑+𝟏𝟑𝒔𝟐+𝟓𝟐𝒔+𝟔𝟎



Case 2: Real and complex conjugate poles

For a unit-step input function, the transfer function becomes as

𝑪 𝒔 =𝒂

𝒔+

𝒋=𝟏

𝒒𝒂𝒋

𝒔 + 𝒑𝒋

+

𝒌=𝟏

𝒓𝒃𝒌 𝒔 + 𝜻𝒌𝒘𝒌 + 𝒄𝒌𝒘𝒌 𝟏 − 𝜻𝒌

𝟐

𝒔𝟐 + 𝟐𝜻𝒌𝒘𝒌𝒔 + 𝒘𝒌𝟐

where n= q+2r, and the time response is given by:

𝒄 𝒕 = 𝒂 +

𝒋=𝒌

𝒒

𝒂𝒋𝒆−𝒑𝒋𝒕 +

𝒌=𝟏

𝒓

𝒃𝒌𝒆−𝜻𝒌𝒘𝒌𝒕𝒄𝒐𝒔 𝒘𝒌 𝟏 − 𝜻𝒌

𝟐 𝒕 +

𝒌=𝟏

𝒓

𝒄𝒌𝒆−𝜻𝒌𝒘𝒌𝒕𝒔𝒊𝒏 𝒘𝒌 𝟏 − 𝜻𝒌

𝟐 𝒕



Example: Find the time response for a unit-step input function of the following closed-

loop control system:

Answer: factoring the denominator:

𝑪(𝒔)

𝑹(𝒔)=

𝟖

(𝒔+𝟏)(𝒔𝟐+𝟒𝒔+𝟖)=

𝟖

(𝒔+𝟏)(𝒔+𝟐+𝒋𝟐)(𝒔+𝟐−𝟐𝒋)

The inverse Laplace transform for a unit-step input is

𝒄 𝒕 = 𝟏 −𝟖

𝟓𝒆−𝒕 +

𝟑

𝟓𝒆−𝟐𝒕𝒄𝒐𝒔 𝟐𝒕 −

𝟏

𝟓𝒆−𝟐𝒕𝒔𝒊𝒏 𝟐𝒕

𝑪(𝒔)

𝑹(𝒔)=

𝟖

𝒔𝟑+𝟓𝒔𝟐+𝟏𝟐𝒔+𝟖


Laplace Transform Solution of State Equations

Consider the general form of the state-space equations:

ሶ𝒙 = 𝑨𝒙 + 𝑩𝒖

and the output equation

𝒚 = 𝑪𝒙 + 𝑫𝒖

Taking the Laplace transform of both sides of the state equation yields

𝒔𝑿 𝒔 − 𝒙 𝟎 = 𝑨𝑿 𝒔 + 𝑩𝑼(𝒔)

In order to separate X(s), replace sX(s) with sIX(s). Where I is an n n identity

matrix, and n is the order of the system. Combining all of the X(s) terms, we get

𝒔𝑰 − 𝑨 𝑿 𝒔 = 𝒙 𝟎 + 𝑩𝑼(𝒔)

Solving for X(s) by pre-multiplying both sides of the above equation by 𝒔𝑰 − 𝑨 −𝟏, the

final solution for X(s) is

𝑿 𝒔 = 𝒔𝑰 − 𝑨 −𝟏𝒙 𝟎 + 𝒔𝑰 − 𝑨 −𝟏𝑩𝑼 𝒔

=𝒂𝒅𝒋(𝒔𝑰 − 𝑨)

𝒅𝒆𝒕(𝒔𝑰 − 𝑨)[𝒙 𝟎 + 𝑩𝑼 𝒔 ]

Taking the Laplace transform of the output equation yields

𝒀 𝒔 = 𝑪𝑿 𝒔 + 𝑫𝑼(𝒔)


Laplace Transform Solution of State Equations: Eigenvalues and Transfer Function Poles

Knowing that the transfer function of the state-space representation is given by

The roots of the denominator are the poles of the system and the system poles equal

the eigenvalues. The eigenvalues (poles) from the state-space model can be founded

from

𝒅𝒆𝒕 𝒔𝝀 − 𝑨 = 𝟎

Note, here I substituted by which denotes an eigenvalue.


End of Chapter Five!


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