Ch5 transient and steady state response analyses(control)

CHAPTER 5
TRANSIENT AND STEADY-STATE RESPONSE ANALYSES

5-1 INTRODUCTION
• The first step in analyzing a control system was to derive a mathematical model of the system.
Once such a model is obtained, various methods are available for the analysis of system
performance.
• In practice, the input signal to a control system is not known ahead of time but is random in
nature, and the instantaneous input cannot be expressed analytically. Only in some special cases is
the input signal known in advance and expressible analytically or by curves, such as in the case of
the automatic control of cutting tools.
• In analyzing and designing control systems, we must have a basis of comparison of performance of
various control systems. This basis may be set up by specifying particular test input signals and by
comparing the responses of various systems to these input signals.

5-1 INTRODUCTION (CONT.)
• Many design criteria are based on the response to such test signals or on the response of
systems to changes in initial conditions (without any test signals).
• The use of test signals can be justified because of a correlation existing between the
response characteristics of a system to a typical test input signal and the capability of the
system to cope with actual input signals.

TypicalTest Signals.
• The commonly used test input signals are step functions,ramp functions,acceleration functions,impulse
functions,sinusoidal functions,and white noise.
• With these test signals, mathematical and experimental analyses of control systems can be carried out
easily, since the signals are very simple functions of time.
• Which of these typical input signals to use for analyzing system characteristics may be determined by the
form of the input that the system will be subjected to most frequently under normal operation.
• If the inputs to a control system are gradually changing functions of time, then a ramp function of time
may be a good test signal.

TypicalTest Signals.(Count.)
• Similarly, if a system is subjected to sudden disturbances, a step function of time may be a
good test signal.
• for a system subjected to shock inputs, an impulse function may be best.
• Once a control system is designed on the basis of test signals, the performance of the
system in response to actual inputs is generally satisfactory.The use of such test signals
enables one to compare the performance of many systems on the same basis.

Transient Response and Steady-State Response.
• The time response of a control system consists of two parts: the transient response and the steady-state
response.
• By transient response,we mean that which goes from the initial state to the final state.
• By steady-state response,we mean the manner in which the system output behaves as t approaches infinity.
• Thus the system response c(t) may be written as
where the first term on the right-hand side of the equation is the transient response and the second term is the
steady-state response

 Absolute Stability, Relative Stability, and Steady-State Error.
• In designing a control system, we must be able to predict the dynamic behavior of the system from a knowledge of the
components.
• The most important characteristic of the dynamic behavior of a control system is absolute stability—that is, whether the
system is stable or unstable.
• A control system is in equilibrium if, in the absence of any disturbance or input, the output stays in the same state.
• A linear time-invariant control system is stable if the output eventually comes back to its equilibrium state when the system is
subjected to an initial condition.
• A linear time-invariant control system is critically stable if oscillations of the output continue forever.
• It is unstable if the output diverges without bound from its equilibrium state when the system is subjected to an initial
condition.

Absolute Stability, Relative Stability, and Steady-State Error.(Count.)
• Actually, the output of a physical system may increase to a certain extent but may be limited
by mechanical “stops,” or the system may break down or become nonlinear after the output
exceeds a certain magnitude so that the linear differential equations no longer apply.
• Important system behavior (other than absolute stability) to which we must give careful
consideration includes relative stability and steady-state error.
• Since a physical control system involves energy storage, the output of the system, when
subjected to an input, cannot follow the input immediately but exhibits a transient response
before a steady state can be reached.

Absolute Stability, Relative Stability, and Steady-State
Error.(Count.)
• The transient response of a practical control system often exhibits damped oscillations
before reaching a steady state. If the output of a system at steady state does not exactly
agree with the input, the system is said to have steady state error.
• This error is indicative of the accuracy of the system. In analyzing a control system, we
must examine transient-response behavior and steady-state behavior.

transient response of a practical control system.

5-2 FIRST-ORDER SYSTEM
• Consider the first-order system shown below. Physically, this system may represent an
RC circuit, thermal system, or the like.
• A simplified block diagram is shown below.

5-2 FIRST-ORDER SYSTEM (CONT.)
• The input-output relationship is given by
• In the following, we shall analyze the system responses to such inputs as the unit-step,
unit-ramp, and unit-impulse functions.The initial conditions are assumed to be zero.
• Note that all systems having the same transfer function will exhibit the same output in
response to the same input.
• For any given physical system, the mathematical response can be given a physical
interpretation.

 Unit-Step Response of First-Order Systems.
• Since the Laplace transform of the unit-step function is 1/s, substituting R(s)=1/s into
Equation:
• Expanding C(s) into partial fractions gives
• Taking the inverse Laplace transform of Equation, we obtain

 Unit-Step Response of First-Order Systems. (Cont.)
• Equation (inverse Laplace) states that initially the output c(t) is zero and finally it becomes
unity. One important characteristic of such an exponential response curve c(t) is that at t=T
the value of c(t) is 0.632,or the response c(t) has reached 63.2% of its total change.
• This may be easily seen by substituting t=T in c(t).That is,
• Note that the smaller the time constantT, the faster the system response.Another important
characteristic of the exponential response curve is that the slope of the tangent line at t=0 is
1/T, since

 Unit-Step Response of First-Order Systems. (Cont.)
• The output would reach the final value at t=T if it maintained its initial speed of response.
• we see that the slope of the response curve c(t) decreases monotonically from 1/T at t=0 to zero at
t=∞.
• In one time constant, the exponential response curve has gone from 0 to 63.2% of the final value. In two
time constants, the response reaches 86.5% of the final value. At t=3T, 4T, and 5T, the response reaches
95%, 98.2%, and 99.3%, respectively, of the final value. Thus, for t 4T, the response remains within 2% of
the final value.
• In practice, however, a reasonable estimate of the response time is the length of time the response curve
needs to reach and stay within the 2% line of the final value, or four time constants.

 Unit-Ramp Response of First-Order Systems.
• Since the Laplace transform of the unit-ramp function is 1/s2 , we obtain the output of
the system
• Expanding C(s) into partial fractions gives
• Taking the inverse Laplace transform

• The error signal e(t) is then

• As t approaches infinity, e–t/Tapproaches zero, and thus the error signal e(t) approaches
T or
• The error in following the unit-ramp input is equal toT for sufficiently large t.The smaller
the time constantT, the smaller the steady-state error in following the ramp input.

 Unit-Impulse Response of First-Order Systems.
• For the unit-impulse input, R(s)=1 and the output of the system
• The inverse Laplace transform of Equation gives

 Unit-Impulse Response of First-Order Systems.
• For the unit-impulse input, R(s)=1 and the output of the system
• The response curve given by Equation (inverse Laplace) is shown below.

 An Important Property of LinearTime-Invariant Systems.
• In the analysis above, it has been shown that for the unit-ramp input the output c(t) is
• For the unit-step input, which is the derivative of unit-ramp input, the output c(t) is
• Finally, for the unit-impulse input, which is the derivative of unit-step input, the output c(t)
is

 An Important Property of LinearTime-Invariant Systems.
• Comparing the system responses to these three inputs clearly indicates that the
response to the derivative of an input signal can be obtained by differentiating the
response of the system to the original signal. It can also be seen that the response to the
integral of the original signal can be obtained by integrating the response of the system to
the original signal and by determining the integration constant from the zero-output
initial condition. This is a property of linear time-invariant systems. Linear time-varying
systems and nonlinear systems do not possess this property.

• we shall obtain the response of a typical second-order control system to a step input,
ramp input, and impulse input. Here we consider a servo system as an example of
second-order system.
Servo System.
• The servo system shown in Figure (a) consists of a proportional controller and load
elements (inertia and viscous-friction elements). Suppose that we wish to control the
output position c in accordance with the input position r.

Servo System.(Cont.)
• The equation for the load elements is
• where T is the torque produced by the proportional controller whose gain is K. By taking
Laplace transforms of both sides of this last equation, assuming the zero initial conditions,
we obtain
• So the transfer function between C(s) and T(s) is

• By using this transfer function, Figure (a) can be redrawn as in Figure (b), which can be
modified to that shown in Figure(c).

• The closed-loop transfer function is then obtained as
• Such a system where the closed-loop transfer function possesses two poles is called a
second-order system. (Some second-order systems may involve one or two zeros.)

Step Response of Second-Order System.
• The closed-loop transfer function of the system shown in Figure (c) is
• which can be rewritten as

Step Response of Second-Order System.(Cont.)
• The closed-loop poles are complex conjugates if 𝐵2-4JK≥0. In the transient-response
analysis, it is convenient to write
• where s is called the attenuation;𝑤 𝑛, the undamped natural frequency; and , the
damping ratio of the system. The damping ratio z is the ratio of the actual damping B to
the critical damping or

• This form is called the standard form of the second-order system.

• The dynamic behavior of the second-order system can then be described in terms of two
parameters 𝛿 and 𝑤 𝑛. If 0<𝛿<1, the closed-loop poles are complex conjugates and lie in the
left-half s plane.
• The system is then called underdamped, and the transient response is oscillatory. If𝛿 =0, the
transient response does not die out. If 𝛿 =1, the system is called critically damped.
Overdamped systems correspond to z>1. We shall now solve for the response of the system
shown in Figure in previous page to a unit-step input. We shall consider three different cases:
the underdamped (0<𝛿<1), critically damped (𝛿=1), and overdamped 𝛿>1) cases.

• (1) Underdamped case (0<𝛿<1): In this case, C(s)/R(s) can be written
• where 𝑤 𝑑=𝑤 𝑛 1 − 𝛿2The frequency 𝑤 𝑑 is called the damped natural frequency. For a
unit-step input, C(s) can be written.

• The inverse Laplace transform of Equation can be obtained easily if C(s) is written in the
following form:
• Laplace transform,

• Hence the inverse Laplace transform of Equation is obtained as
• From Equation, it can be seen that the frequency of transient oscillation is the damped natural
frequency 𝑤 𝑑 and thus varies with the damping ratio 𝛿.

• The error signal for this system is the difference between the input and output and is
• This error signal exhibits a damped sinusoidal oscillation. At steady state, or at t=∞, no
error exists between the input and output.

• If the damping ratio 𝛿 is equal to zero, the response becomes undamped and oscillations
continue indefinitely. The response c(t) for the zero damping case may be obtained by
substituting 𝛿=0 in ,yielding

• Thus, from Equation, we see that 𝑤 𝑛represents the undamped natural frequency of the
system. That is, 𝑤 𝑛 is that frequency at which the system output would oscillate if the damping
were decreased to zero. If the linear system has any amount of damping, the undamped
natural frequency cannot be observed experimentally.
• The frequency
• that may be observed is the damped natural frequency 𝑤 𝑑 , which is equal to This frequency is
always lower than the undamped natural frequency. An increase in 𝛿 would reduce the
damped natural frequency 𝑤 𝑑 . If 𝛿 is increased beyond unity, the response becomes
overdamped and will not oscillate.

• (2) Critically damped case (𝛿=1): If the two poles of C(s)/R(s) are equal, the system is
said to be a critically damped one.
• For a unit-step input, R(s)=1/s and C(s) can be written
• The inverse Laplace transform of Equation may be found as

• This result can also be obtained by letting𝛿 approach unity in Equation and by using the
following limit:
• (3) Overdamped case (𝛿>1): In this case, the two poles of C(s)/R(s) are negative real
and unequal. For a unit-step input, R(s)=1/s and C(s) can be written

• The inverse Laplace transform:

• where and Thus, the response c(t) includes
two decaying exponential terms.
• When 𝛿 is appreciably greater than unity, one of the two decaying exponentials decreases
much faster than the other, so the faster-decaying exponential term (which corresponds
to a smaller time constant) may be neglected.

• That is, if –s2 is located very much closer to the jw axis than –s1, then for an
approximate solution we may neglect –s1.This is permissible because the effect of –s1 on
the response is much smaller than that of –s2 , since the term involving s1 in Equation
decays much faster than the term involving s2 .
• Once the faster-decaying exponential term has disappeared, the response is similar to
that of a first-order system, and C(s)/R(s) may be approximated by

• This approximate form is a direct consequence of the fact that the initial values and final
values of both the original C(s)/R(s) and the approximate one agree with each other.
• With the approximate transfer function C(s)/R(s), the unit-step response can be obtained as:
• The time response c(t) is then

• This gives an approximate unit-step response when one of the poles of C(s)/R(s) can be
neglected.

Definitions ofTransient-Response Specifications.
• Frequently, the performance characteristics of a control system are specified in terms of the
transient response to a unit-step input, since it is easy to generate and is sufficiently drastic. (If
the response to a step input is known, it is mathematically possible to compute the response
to any input.)
• The transient response of a system to a unit-step input depends on the initial conditions. For
convenience in comparing transient responses of various systems, it is a common practice to
use the standard initial condition that the system is at rest initially with the output and all
time derivatives thereof zero. Then the response characteristics of many systems can be easily
compared.

Definitions ofTransient-Response Specifications.(Cont.)
• The transient response of a practical control system often exhibits damped oscillations
before reaching steady state. In specifying the transient-response characteristics of a
control system to a unit-step input, it is common to specify the following:
• 1. Delay time, td
• 2. Rise time, tr 3. Peak time, tp
• 4. Maximum overshoot, Mp 5. Settling time, ts

• These specifications are defined in what follows and are shown graphically in Figure.

• 1. Delay time, td: The delay time is the time required for the response to reach half the
final value the very first time.
• 2. 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.

• 3. Peak time,tp: The peak time is the time required for the response to reach the first peak of the
overshoot.
• 4. Maximum (percent) overshoot, Mp: The maximum overshoot is the maximum peak value of the
response curve measured from unity. If the final steady-state value of the response differs from unity,
then it is common to use the maximum percent overshoot. It is defined by
The amount of the maximum (percent) overshoot directly indicates the relative stability of the system.

• 5. Settling time, ts : The settling time is the time required for the response curve to reach
and stay within a range about the final value of size specified by absolute percentage of the
final value (usually 2% or 5%). The settling time is related to the largest time constant of the
control system.Which percentage error criterion to use may be determined from the
objectives of the system design in question.
• The time-domain specifications just given are quite important, since most control systems are
time-domain systems; that is, they must exhibit acceptable time responses. (This means that,
the control system must be modified until the transient response is satisfactory.)

• Note that not all these specifications necessarily apply to any given case.
For example, for an overdamped system, the terms peak time and
maximum overshoot do not apply.

A Few Comments onTransient-Response Specifications.
• Except for certain applications where oscillations cannot be tolerated, it is desirable that the
transient response be sufficiently fast and be sufficiently damped. Thus, for a desirable
transient response of a second-order system, the damping ratio must be between 0.4 and 0.8.
• values of 𝛿 (that is, 𝛿 < 0.4) yield excessive overshoot in the transient response, and a system
with a large value of 𝛿 (that is, 𝛿 >0.8) responds sluggishly.
• We shall see later that the maximum overshoot and the rise time conflict with each other. In
other words, both the maximum overshoot and the rise time cannot be made smaller
simultaneously. If one of them is made smaller, the other necessarily becomes larger.

5-6 ROUTH’S
STABILITY CRITERION

5-6 ROUTH’S STABILITY CRITERION
• The most important problem in linear control systems concerns stability.
• That is, under what conditions will a system become unstable? If it is unstable, how
should we stabilize the system?
• Most linear closed-loop systems have closed-loop transfer functions of the form
where the a’s and b’s are constants and m <= n

5-6 ROUTH’S STABILITY CRITERION (CONT.)
• A simple criterion, known as Routh’s stability criterion, enables us to determine the
number of closed-loop poles that lie in the right-half s plane without having to factor the
denominator polynomial. (The polynomial may include parameters that MATLAB cannot
handle.).

 Routh’s Stability Criterion.
• Routh’s stability criterion tells us whether or not there are unstable roots in a polynomial
equation without actually solving for them.
• This stability criterion applies to polynomials with only a finite number of terms.
• When the criterion is applied to a control system, information about absolute stability can be
obtained directly from the coefficients of the characteristic equation.
• The procedure in Routh’s stability criterion is as follows:
1.The polynomial in s in the following form:
where the coefficients are real quantities. We assume that an Z ; that is, any zero root has been
removed.

2. If any of the coefficients are zero or negative in the presence of at least one positive coefficient, a root
or roots exist that are imaginary or that have positive real parts. Therefore, in such a case, the system is not
stable.
• If we are interested in only the absolute stability,there is no need to follow the procedure further.
• Note that all the coefficients must be positive. A polynomial in s having real coefficients can always be
factored into linear and quadratic factors, such as (s+a) and ( +bs+cB), where a, b, and c are real.
• The linear factors yield real roots and the quadratic factors yield complex-conjugate roots of the
polynomial.
𝑆2

• The factor (𝑠2
+bs+c) yields roots having negative real parts only if b and c are both positive.
• For all roots to have negative real parts, the constants a, b, c, and so on, in all factors must be
positive.
• The product of any number of linear and quadratic factors containing only positive
coefficients always yields a polynomial with positive coefficients.
• It is important to note that the condition that all the coefficients be positive is not sufficient
to assure stability.

3. If all coefficients are positive, arrange the coefficients of the polynomial in rows and
columns according to the following pattern:

• The process of forming rows continues until we run out of elements. (The total number
of rows is n+1.)The coefficients 𝑏1, 𝑏2,𝑏3 and so on, are evaluated as follows:

• The evaluation of the b’s is continued until the remaining ones are all zero. The same
pattern of cross-multiplying the coefficients of the two previous rows is followed in
evaluating the c’s, d’s, e’s, and so on.That is,

EXAMPLE 5–11

 EXAMPLE 5–12
• Consider the following polynomial:
• Let us follow the procedure just presented and construct the array of coefficients. (The first two rows can be obtained directly from the
given polynomial.The remaining terms are obtained from these. If any coefficients are missing, they may be replaced by zeros in the
array.).
• In this example, the number of changes in sign of the coefficients in the first column is 2.This means that there are two roots with
positive real parts. Note that the result is unchanged when the coefficients of any row are multiplied or divided by a positive number in
order to simplify the computation.

 Special Cases.
1. If a first-column term in any row is zero, but the remaining terms are not zero or there is
no remaining term, then the zero term is replaced by a very small positive number € and
the rest of the array is evaluated. For example, consider the following equation:
The array of coefficients is

 Special Cases (Cont).
• If the sign of the coefficient above the zero (€) is the same as that below it, it indicates
that there are a pair of imaginary roots.
• If, however, the sign of the coefficient above the zero (€) is opposite that below it, it
indicates that there is one sign change. For example, for the equation.
• the array of coefficients is

 Special Cases (Cont.).
There are two sign changes of the coefficients in the first column. So there are two roots in
the right-half s plane.This agrees with the correct result indicated by the factored form of
the polynomial equation.

2. If all the coefficients in any derived row are zero, it indicates that there are roots of equal
magnitude lying radially opposite in the s plane—that is, two real roots with equal magnitudes
and opposite signs and/or two conjugate imaginary roots. In such a case, the evaluation of the
rest of the array can be continued by forming an auxiliary polynomial with the coefficients of the
last row and by using the coefficients of the derivative of this polynomial in the next row. Such
roots with equal magnitudes and lying radially opposite in the s plane can be found by solving
the auxiliary polynomial, which is always even. For a 2n-degree auxiliary polynomial, there are n
pairs of equal and opposite roots. For example, consider the following equation:

• The array of coefficients is
• The terms in the s3 row are all zero. (Note that such a case occurs only in an odd numbered
row.) The auxiliary polynomial is then formed from the coefficients of the s4 row. The
auxiliary polynomial P(s) is

The terms in the 𝑠3 row are replaced by the coefficients of the last equation—that is, 8 and
96.The array of coefficients then becomes

 Application of Routh’s Stability Criterion to Control-
System Analysis.
In the following, we shall consider the problem of determining the stability range of a
parameter value. Let us determine the range of K for stability.

 Application of Routh’s Stability Criterion to Control-
System Analysis.
The closed-loop transfer function is
The characteristic equation is

 Application of Routh’s Stability Criterion to Control-SystemAnalysis.
The array of coefficients becomes
For stability,K must be positive, and all coefficients in the first column must be positive.Therefore,
When the system becomes oscillatory and, mathematically,the oscillation is sustained at constant amplitude
Note that the ranges of design parameters that lead to stability may be determined by use of Routh’s stability
criterion.

5-8 STEADY-STATE ERRORS IN UNITY-FEEDBACK
CONTROL SYSTEMS
• Errors in a control system can be attributed to many factors.
• Changes in the reference input will cause unavoidable errors during transient periods and may also cause
steady state errors.
• Imperfections in the system components, such as static friction, backlash, and amplifier drift, as well as
aging or deterioration,will cause errors at steady state.
• In this section, however, we shall not discuss errors due to imperfections in the system components.
Rather, we shall investigate a type of steady-state error that is caused by the incapability of a system to
follow particular types of inputs.
• (The only way we may be able to eliminate this error is to modify the system structure.) Whether a
given system will exhibit steady-state error for a given type of input depends on the type of open-loop
transfer function of the system.

CONTROL SYSTEMS. (CONT.)
 Classification of Control Systems.
• Control systems may be classified according to their ability to follow step inputs, ramp
inputs, parabolic inputs, and so on. This is a reasonable classification scheme, because
actual inputs may frequently be considered combinations of such inputs.
• The magnitudes of the steady-state errors due to these individual inputs are indicative of
the goodness of the system.
• Consider the unity-feedback control system with the following open-loop transfer
function G(s):

 Classification of Control Systems. (Count.)
• It involves the term 𝑆 𝑁in the denominator, representing a pole of multiplicity N at the origin.
• The present classification scheme is based on the number of integrations indicated by the
open-loop transfer function.
• A system is called type 0, type 1, type 2, … , if N=0, N=1, N=2, ... , respectively. Note that this
classification is different from that of the order of a system.
• As the type number is increased, accuracy is improved; however, increasing the type number
aggravates the stability problem.
• A compromise between steady-state accuracy and relative stability is always necessary.

 Steady-State Errors.
• The transfer function between the error signal e(t) and the input signal r(t) is

 Steady-State Errors. (Count.)
• where the error e(t) is the difference between the input signal and the output signal. The
final-value theorem provides a convenient way to find the steady-state performance of a
stable system. Since E(s) is
• the steady-state error is

 Steady-State Errors. (Count.)
• The static error constants defined in the following are figures of merit of control systems. The
higher the constants, the smaller the steady-state error.
• In a given system, the output may be the position, velocity, pressure, temperature, or the like.
• The physical form of the output, however, is immaterial to the present analysis.
• Therefore, in what follows, we shall call the output “position,” the rate of change of the
output “velocity,” and so on. This means that in a temperature control system “position”
represents the output temperature, "velocity” represents the rate of change of the output
temperature, and so on.

1. Static Position Error Constant 𝑲 𝒑
• The steady-state error of the system for a unit-step input is
• The static position error constant Kp is defined by

1. Static Position Error Constant 𝑲 𝒑. (Cont.)
• Thus, the steady-state error in terms of the static position error constant Kp is given by
• For a type 0 system,

• For a type 1 or higher system,
• Hence, for a type 0 system, the static position error constant Kp is finite, while for a type
1 or higher system, Kp is infinite.
• For a unit-step input, the steady-state error ess may be summarized as follows:

• From the foregoing analysis, it is seen that the response of a feedback control system to a
step input involves a steady-state error if there is no integration in the feedforward path.
• If zero steady-state error for a step input is desired, the type of the system must be one
or higher.

2. StaticVelocity Position Error Constant 𝑲 𝒗. (Cont.)
• The steady-state error of the system with a unit-ramp input is given by
• The static velocity error constant 𝐾𝑣is defined by

• Thus, the steady-state error in terms of the static velocity error constant 𝐾𝑣is given by
• The term velocity error is used here to express the steady-state error for a ramp input.
The dimension of the velocity error is the same as the system error. That is, velocity
error is not an error in velocity, but it is an error in position due to a ramp input. For a
type 0 system,

• The steady-state error ess for the unit-ramp input can be summarized as follows:

• The foregoing analysis indicates that a type 0 system is incapable of following a ramp
input in the steady state.
• The type 1 system with unity feedback can follow the ramp input with a finite error.
• In steady-state operation, the output velocity is exactly the same as the input velocity, but
there is a positional error.
• This error is proportional to the velocity of the input and is inversely proportional to the
gain K.

2. Static Acceleration Error Constant Ka. (Cont.)
• The steady-state error of the system with a unit-parabolic input (acceleration input),
which is defined by
• is given by

• The static acceleration error constant Ka is defined by the equation
• The steady-state error is then
• Note that the acceleration error, the steady-state error due to a parabolic input, is an
error in position.

• The values of Ka are obtained as follows: For a type 0 system,

• Thus, the steady-state error for the unit parabolic input is

• Note that both type 0 and type 1 systems are incapable of following a parabolic input in
the steady state. The type 2 system with unity feedback can follow a parabolic input with
a finite error signal.

 Summary.
Remember that the terms position error, velocity error, and acceleration error mean steady-state
deviations in the output position.

 Summary. (Cont.)
• A finite velocity error implies that after transients have died out, the input and output move at the same
velocity but have a finite position difference.
• The error constants Kp, Kv , and Ka describe the ability of a unity-feedback system to reduce or
eliminate steady-state error.Therefore, they are indicative of the steady-state performance.
• . It is generally desirable to increase the error constants, while maintaining the transient response within
an acceptable range. It is noted that to improve the steady state performance we can increase the type
of the system by adding an integrator or integrators to the feedforward path. This, however, introduces
an additional stability problem. The design of a satisfactory system with more than two integrators in
series in the feedforward path is generally not easy.

Eng. Elaf Ahmed Saeed
Email: elafe1888@gmail.com
LinkedIn: https://www.linkedin.com/in/elaf-a-saeed
Facebook: https://www.facebook.com/profile.php?id=100004305557442
GitHub: https://github.com/ElafAhmedSaeed

Ch5 transient and steady state response analyses(control)

More Related Content

What's hot (20)

Similar to Ch5 transient and steady state response analyses(control) (20)

More from Elaf A.Saeed (20)

Recently uploaded (20)

Ch5 transient and steady state response analyses(control)