Module-3:Time Domain Analysis of Linear Systems
Time Domain Specifications
Akshya Swain
Department of Electrical, Computer & Software Engineering, The University of Auckland,
Auckland, New Zealand.
Akshya Swain
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Learning Outcome of this Module
By studying this module the student would learn about the following:
1. Transient and Steady State Response
2. Typical input signals used for time domain analysis.
3. Compute various time domain specifications such as rise time, peak time, peak
overshoot for an under-damped second order control system.
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Transient and Steady State Response
I Note: Systems with energy storage elements ( i.e. dynamic systems) can not respond instantaneously and will exhibit transient responses whenever they are subjected to inputs or disturbances.
I The time response c(t) of a control system is usually divided, into two parts: the transient response and the steady-state response. Thus
c(t) = ctr(t) + css(t)
where ctr(t) is the transient response and css(t) is the steady state response. I The definition of the steady state has not been entirely standardized.
I In circuit analysis, it is sometimes useful to define a steady-state variable as
being a constant with respect to time.
I In control systems, the steady-state response is simply the fixed response when
time reaches infinity. When a response reaches steady state, it can still vary with time. For example, a sine wave is considered as a steady-state response because its behavior is fixed for any time interval, as when time approaches infinity. Similarly, if a response is described by c(t) = t, it may be defined as a steady-state response.
I Transient response is defined as the part of the response that goes to zero as time becomes large. Therefore, ctr(t) has the property of
lim ctr(t) = 0 t→∞
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Test Signals for Time Domain Analysis
Why there is a need for Test Signals ?
1. The input excitation to many practical control systems are not known ahead of time; unlike many electrical circuits and communication systems.
2. In many cases, the actual inputs of a control system may vary in random fashions with respect to time.
– For instance, in a radar tracking system, the position and speed of the target to be tracked may vary in an unpredictable manner, so that they cannot be expressed deterministically by a mathematical expression.
I This is a major problem for the designer, since it is difficult to design the control system so that it will perform satisfactorily to any input signal.
I For the purposes of analysis and design, it is necessary to assume some basic types of input functions so that the performance of a system can be evaluated with respect to these test signals.
I By selecting these basic test signals properly, not only the mathematical treatment of the problem is systematized, but the responses due to these inputs allow the prediction of the system’s performance to other more complex inputs.
I In a design problem, performance criteria may be specified with respect to these test signals so that a system may be designed to meet the criteria.
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Test Signals for Time Domain Analysis
I The general form of the signals used for time domain analysis can be expressed as:
r(t) = Atn; R(s) = n!A sn+1
I When n = 0, this corresponds to a step signal, for n = 1, this represents a ramp
signal and when n = 2, this corresponds to a parabolic signal.
I Note: From the step function to the parabolic function they become
progressively faster with respect to time.
1. The step function is very useful as a test signal since its initial instantaneous jump in amplitude reveals a great deal about a system’s quickness to respond.
2. The ramp function has the ability to test how the system would respond to a signal that changes linearly with time.
3. A parabolic function is one degree faster than a ramp function. In practice, we seldom find it necessary to use a test signal faster than a parabolic function.
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Test Signals for Time Domain Analysis
I The general form of the signals used for time domain analysis can be expressed as:
r(t) = Atn; R(s) = n!A sn+1
I When n = 0, this corresponds to a step input, for n = 1, this represents a ramp input and when n = 2, this corresponds to a parabolic input.
1. Step Input
I Frequently the performance characteristics of a control system are specified in
terms of transient response to a unit-step input. Because
1. They are easy to generate
2. Its initial instantaneous jump in amplitude reveals a great deal about a system’s
quickness to respond.
3. Also, since the step function has, in principle, a wide band of frequencies in its
spectrum, as a result of the jump discontinuity, as a test signal it is equivalent to the application of numerous sinusoidal signals with a wide range of frequencies.
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Test Signals for Time Domain Analysis
2. Ramp Input
I The ramp function is a signal that changes constantly with time.
Mathematically, a ramp function is represented by
r(t) = At
where A is a real constant.
i. The ramp function has the ability to test how the system would respond to a
signal that changes linearly with time.
3. Parabolic Input
I The parabolic function represents a signal that is one order faster than the ramp
function. Mathematically, it is represented as
r(t) = At2 2
where A is a real constant. The factor 1 , is added for mathematical convenience; 2
because the Laplace transform of r(t) becomes A . s3
Note: From the step function to the parabolic function, the signals become progressively faster with respect to time.
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Time Domain Analysis of First Order System-1
I Determine the step response of the first order system shown in the Figure.
I Since we need to compute the step response of this system, then, from the figure
G(s)= C(s) = a R(s) s+a
,
R(s)= 1 s
a
s(s+a) s s+a
= 1 − 1 c(t) = cf(t)+cn(t) = 1−e−at
Now C(s) = R(s)G(s) =
I Taking the inverse Laplace transform gives
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Time Domain Analysis of First Order System-2
I Determine the step response of the first order system shown in the Figure.
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Time Domain Analysis of First Order System-3
Time Constant T
Computation of Time Constant for a First Order System
1. We know that the step response of first order system c(t) = 1 − e−at
2. Let T = time when the response equals to 0.63 i.e. c(T ) = 0.63. This implies
1−e−aT =0.63 =⇒ aT =ln( 1 )=ln(2.7027)≈1.0 =⇒ T ≈ 1
0.37 a
3. Thus the time constant T equals to
– It is defined as the time required for the step response to reach go from 63% of its final value.
T=1 a
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Time Domain Analysis of First Order System-2
Rise Time Tr
Computation of Rise Time for a First Order System
1. We know that the step response of first order system c(t) = 1 − e−at
2. Let t0= time when the response equals to 0.1 i.e. c(t0) = 0.1. This implies 1−e−at0 = 0.1 =⇒ at0 = ln(10) ≈ 0.11 =⇒ t0 ≈ 0.11
9a
3. Let t1= time when the response equals to 0.9 i.e. c(t0) = 0.9. This implies
1−e−at1 = 0.9 =⇒ at1 = ln(10) ≈ 2.31 =⇒ t1 ≈ 2.31 a
4. Thus the rise time Tr equals to
– It is defined as the time required for the response to go from 10% (0.1) to 90% (0.9) of its final value.
Tr =t1−t0 ≈2.31−0.11=2.2 aaa
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Time Domain Analysis of First Order System-3
Setting Time Ts
Computation of Settling Time for a First Order System
1. We know that the step response of first order system c(t) = 1 − e−at
2. Let Ts= time when the response equals to 0.98 i.e. c(Ts) = 0.98. This implies
1−e−aTs = 0.98 =⇒ aTs = ln(50) ≈ 3.91 =⇒ Ts ≈ 3.91 a
3. Thus the settling time Ts equals to
– It is defined as the time required for the response to reach and stay within 2% of its final value.
Ts = 3.91 ≈ 4 aa
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Time Domain Analysis of a Second Order System-1 I Consider a second order system shown in the Figure.
I The closed loop transfer function of this system is given by C ( s ) = ω n2
R(s) s2 + 2ζωns + ωn2
I The parameter ωn is the natural frequency of the second order system, and the parameter ζ is called the damping ratio.
I The closed loop poles λ1,2 are given by
λ1,2 = −ζωn ± ωnζ2 − 1
I The nature of the response of this system depends on the value of the damping ratio. Let us find the step response of this system for different cases
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Time Domain Analysis of a Second Order System-2
I The nature of the response of this system depends on the value of the damping ratio. Let us find the step response of this system for different cases
1. Case-1 : ζ = 0
– The closed loop poles are located at ±jωn
– The step response under this case will be sinusoidal with frequency ωn and it is expressed as:
c(t) = 1 − cos ωn(t) – This type of response is called undamped.
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Time Domain Analysis of Second Order System-2
Case-1 : ζ =0
– The closed loop poles are located at ±jωn
– The step response under this case will be sinusoidal with frequency ωn and it is expressed as:
c(t)=1−cosωn(t)
– This type of response is called
undamped
Case-2 : 0<ζ <1
- The closed loop poles are located at
−ζωn ± jωn1 − ζ2
- The step response is a damped sinusoid
with an exponential envelop whose time constant is equal to the reciprocal of poles’ real part.It is expressed as:
1 −ζω t
c(t)=1−
2 −1ζ
1−ζ2
- This type of response is called under damped
e n cos(ωdt−φ); where ωd=ωn1−ζ;φ=tan( )
1−ζ2
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Time Domain Analysis of Second Order System-3
Case-3 : ζ =1
- We have two closed loop poles which are real and are located at −ζωn
- The step response under this case will be expressed as:
c(t) = 1 − ζωnte−ζωnt − e−ζωnt = 1 − e−ζωnt (1 + ζωnt)
- This type of response is called critically damped
Case-4 : ζ >1
– The two closed loop poles are real and are located at
−ζωn ± ωnζ2 − 1
– The step response is expressed as: √
c(t) = 1 − e−(ζ− ζ2−1)ωnt
– This type of response is called over damped
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Transient (Time) Response Specifications: Some Notes
I Control systems are generally designed with damping less than one, i.e. oscillatory step response.
I High-order control systems usually have a pair of complex conjugate poles with damping less than one which dominates over other poles ( dominant pole pairs).
I Thus, the time response of second and higher-order control systems to step input, is in general, damped oscillatory nature.
Akshya Swain Module-3:Time Domain Analysis of Linear Systems 17 / 30 Figure: Step Response of a Second Order Underdamped Second Order System
Transient (Time) Response Specifications: Some Notes
I From the plot ( shown last slide), it is observed that the step response has a number of overshoots and undershoots with respect to final steady state value.
I This type of response is expressed mathematically as:
e−ζωnt
c(t)=1−
1−ζ2
cos(ωdt−φ), where
1−ζ2, φ=tan
I The pole plot of the underdamped second order system is shown in the figure
ωd =ωn
−1
ζ 1−ζ2
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Transient (Time) Response Specifications: Some Notes
I This type of response is characterised by following performance indices
1. Rise Time Tr
2. Peak Time Tp
3. Peak Overshoot Mp
4. Settling Time Ts
5. Steady State Error ess
I These indices are qualitatively related to
a. How fast is the system i.e. how fast it moves to follow the input ? b. How oscillatory the system is ? ( indicative of damping)
c. How long does it take to practically reach the final value ?
Note: The various indices are not independent of each other.
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Commonly Used Transient Response Specification
1. Rise Time, Tr:
– It 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 system, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time is commonly used.
– Analytically, it is expressed as:
Tr= 1tan−1ωd =π−θ ωd −σd ωd
2. Peak Time, Tp:
– It is the time required for the response to reach the first peak of the overshoot.
Analytically, it is expressed as:
ππ Tp= 2=ω
ωn 1−ζ d
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Commonly Used Transient Response Specification
3. Maximum (percent) Overshoot %OS, or Mp:
– It is the maximum peak value of the response It is defined by
Mp = c(tp) − c(∞) × 100% c(∞)
– Analytically, it is expressed as:
%OS = e
−√πζ 1−ζ2
4. Settling Time, Ts:
– It is the time required for the response to reach and stay within either 2% or
5% of its final ( steady state) value. Commonly, it is expressed as:
Ts =4T = 4 = 4 (2%criterion) σd ζωn
Ts =3T = 3 = 3 (5%criterion) σd ζωn
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Step Response of Second Order Underdamped System
Figure: Step Response of a Second Order Underdamped Second Order System
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Correlation Between Pole Locations with Time Domain Specifications
1. The radial distance from the origin to the pole is the natural frequency ωn.
2. The damping ratio ζ equal to cosθ i.e cosθ = ζ
3. We know that the peak time and setting time are
ππ Tp= 2=ω
ωn 1−ζ d Ts = 4 = 4
ζωn σd
where ωd is the imaginary part of Figure: Pole Plot for Underdamped Second Order
System
4. Thus Tp is inversely proportional to the imaginary part of the pole .
5. Ts is inversely proportional to the real part of the pole .
the pole and is called the damped frequency of oscillation.
and σd is the magnitude of the real part of the pole and is the exponential damping frequency. .
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Example-1: Time Domain Specifications
Consider the system shown in the Figure where ζ = 0.6 and ωn = 5 rad/sec. Compute the rise time Tr, peak time Tp, maximum overshoot Mp, an d settling time Ts when the system is subjected to step input.
Figure
Solution:
From the given values of ζ and ωn, we get ωd = ωn1−ζ2 = 4 and σ = ζωn = 3.
Rise Time Tr
Tr = π−β = 3.14−β, β=tan−1ωd =tan−14 =0.93rad ωd 4 σ 3
Tr = 3.14 − 0.93 = 0.55sec 4
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Example-1: Solution (contd) Peak Time Tp
Tp = π = 3.14 = 0.785sec ωd 4
Maximum Overshoot Mp −πζ
√
Mp =e 1−ζ2 =e ωd , =⇒ =0.095=9.5%
−πσ
Settling Time Ts
For the 2% criterion, the setting time is
Ts = 4 = 4 = 1.33sec σ3
For the 5% criterion, the setting time is
Ts = 3 = 3 = 1sec σ3
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Example-2: Time Domain Specifications
When the system shown in Figure-a is subjected to a unit-step input, the system output responds as shown in Figure-b.Determine the values of K and T from the response curve
1. From the response, the maximum overshoot Mp of 25.4%. Hence
−ln(%Mp /100)
ζ = π2 + ln2(%Mp/100)
−ln(0.254)
= 2 2 = 0.4
π +ln(0.254)
2. From the response Tp = 3sec.Now
ππ
Tp = = ω √1 − 0.42 = 3
ωn 1−ζ2 n
This gives ωn = 1.14 rad/sec
Figure: a. Closed loop system, b. Unit-step response
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Example-2: Solution(contd) I From the block diagram
C(s) = K = R(s) Ts2 +s+K
Comparing we get
(K/T) = s2 +(1/T)+(K/T)
ωn2
s2 +2ζωns+ωn2
I The values of T and K are determined as:
K 1 ωn= T,2ζωn=T
T= 1 = 1 =1.09 2ζωn 2 × 0.4 × 1.14
K = ωn2T = 1.142 ×1.09 = 1.42
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Example-3: Time Domain Specifications
Determine the values of K and k of the closed-loop system shown, so that the maximum overshoot in unit-step response is 25% and the peak time is 2sec.Assume that J = 1kg − m2.
Solution:
Figure
K
Js2 +Kks+K
The closed-loop transfer function is
C(s) = R(s)
By substituting J = 1kg − m2 into this gives
C(s) = K
R(s) s2 +Kks+K
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Example-3: Solution(contd)
By comparing this with standard closed loop transfer function of second order
system gives ωn =
The maximum overshoot Mp = 25%
√
K, 2ζωn = Kk −πzeta
√
1−ζ2
−ln(%Mp/100)
=0.25, =⇒ ζ=2 2 =0.404
Mp=e
I ThepeaktimeTp = π =0.2. Thisgivesωd =1.57
π + ln (%Mp/100)
ωd
I The undamped natural frequency
ωn =
ωd 1−ζ2
= 1.72
Therefore
K=ωn2 =1.722 =2.95N−m
k = 2ζωn = 2×0.404×1.72 = 0.471sec K 2.95
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Effects of Changing Exponential Damping Frequency ωd and Damped Frequency σd on Time Response
– The poles are located at −σd ± jωd. The peak time Tp and setting time Ts are
Tp = π , & Ts = 4 ωd σd
1. Vary ωd and fix σd (see Fig-a.) The frequency of oscillation and the peak time change ; but the settling time remains unchanged.
2. Vary σd and fix ωd (see Fig b).The frequency of oscillation and peak time remains unchanged; but the settling time changes.
3. Vary the natural frequency ωn and fix Figure: Step Response of second order
time changes.
ζ (see Fig-c). The percent overshoot
underdamped system. a. with constant real remains unchanged; but the settling part; b. with constant imaginary part; c. with
constant damping ratio.
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4. Thus Tp is inversely proportional to