time response analysis by a.g.dinesh asso.prof eee dept · 2020. 2. 26. · •by a.g.dinesh...
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Time Response Analysis
• By A.G.DINESH
Asso.Prof EEE dept
Introduction
• In time-domain analysis the response of a dynamicsystem to an input is expressed as a function oftime.
• It is possible to compute the time response of asystem if the nature of input and the mathematicalmodel of the system are known.
• Usually, the input signals to control systems arenot known fully ahead of time.
• It is therefore difficult to express the actual inputsignals mathematically by simple equations.
Standard Test Signals
• The characteristics of actual input signals are asudden shock, a sudden change, a constantvelocity, and constant acceleration.
• The dynamic behavior of a system is thereforejudged and compared under application ofstandard test signals – an impulse, a step, aconstant velocity, and constant acceleration.
• The other standard signal of great importance is asinusoidal signal.
Standard Test Signals
• Impulse signal
– The impulse signal imitate thesudden shock characteristic ofactual input signal.
– If A=1, the impulse signal iscalled unit impulse signal.
0 t
δ(t)
A
00
0
t
tAt
)(
Standard Test Signals
• Step signal
– The step signal imitatethe sudden changecharacteristic of actualinput signal.
– If A=1, the step signal iscalled unit step signal
00
0
t
tAtu
)( 0 t
u(t)
A
Standard Test Signals
• Ramp signal– The ramp signal imitate
the constant velocitycharacteristic of actualinput signal.
– If A=1, the ramp signalis called unit rampsignal
00
0
t
tAttr
)(
0 t
r(t)
r(t)
unit ramp signal
r(t)
ramp signal with slope A
Standard Test Signals
• Parabolic signal
– The parabolic signalimitate the constantacceleration characteristicof actual input signal.
– If A=1, the parabolicsignal is called unitparabolic signal.
00
02
2
t
tAt
tp
)(
0 t
p(t)
parabolic signal with slope A
p(t)
Unit parabolic signal
p(t)
Relation between standard Test Signals
• Impulse
• Step
• Ramp
• Parabolic
00
0
t
tAt
)(
00
0
t
tAtu
)(
00
0
t
tAttr
)(
00
02
2
t
tAt
tp
)(
dt
d
dt
d
dt
d
Laplace Transform of Test Signals
• Impulse
• Step
00
0
t
tAt
)(
AstL )()}({
00
0
t
tAtu
)(
S
AsUtuL )()}({
Laplace Transform of Test Signals
• Ramp
• Parabolic
2s
AsRtrL )()}({
3)()}({
S
AsPtpL
00
0
t
tAttr
)(
00
02
2
t
tAt
tp
)(
Time Response of Control Systems
System
• The time response of any system has two components
• Transient response
• Steady-state response.
• Time response of a dynamic system response to an inputexpressed as a function of time.
Time Response of Control Systems
• When the response of the system is changed from equilibrium ittakes some time to settle down.
• This is called transient response.
0 2 4 6 8 10 12 14 16 18 200
1
2
3
4
5
6x 10
-3
Step Response
Time (sec)
Am
plit
ude Response
Step Input
Transient Response
Stea
dy
Stat
e R
esp
on
se
• The response of thesystem after the transientresponse is called steadystate response.
Time Response of Control Systems
• Transient response depend upon the system poles only and noton the type of input.
• It is therefore sufficient to analyze the transient response using astep input.
• The steady-state response depends on system dynamics and theinput quantity.
• It is then examined using different test signals by final valuetheorem.
Introduction• The first order system has only one pole.
• Where K is the D.C gain and T is the time constantof the system.
• Time constant is a measure of how quickly a 1st
order system responds to a unit step input.
• D.C Gain of the system is ratio between the inputsignal and the steady state value of output.
1
Ts
K
sR
sC
)(
)(
Introduction• The first order system given below.
13
10
ssG )(
5
3
ssG )(
151
53
s/
/
• D.C gain is 10 and time constant is 3 seconds.
• For the following system
• D.C Gain of the system is 3/5 and time constant is 1/5seconds.
Impulse Response of 1st Order System
• Consider the following 1st order system
1Ts
K)(sC)(sR
0t
δ(t)
1
1 )()( ssR
1
Ts
KsC )(
Impulse Response of 1st Order System
• Re-arrange following equation as
1
Ts
KsC )(
Ts
TKsC
/
/)(
1
TteT
Ktc /)(
• In order to compute the response of the system in time domainwe need to compute inverse Laplace transform of the aboveequation.
atCeas
CL
1
Impulse Response of 1st Order System
TteT
Ktc /)( • If K=3 and T=2s then
0 2 4 6 8 100
0.5
1
1.5
Time
c(t
)
K/T*exp(-t/T)
Step Response of 1st Order System
• Consider the following 1st order system
1Ts
K)(sC)(sR
ssUsR
1 )()(
1
Tss
KsC )(
1
Ts
KT
s
KsC )(
• In order to find out the inverse Laplace of the above equation, weneed to break it into partial fraction expansion (page 867 in theTextbook)
Step Response of 1st Order System
• Taking Inverse Laplace of above equation
1
1
Ts
T
sKsC )(
TtetuKtc /)()(
• Where u(t)=1 TteKtc /)( 1
KeKtc 63201 1 .)(
• When t=T (time constant)
Step Response of 1st Order System• If K=10 and T=1.5s then TteKtc /)( 1
0 1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
6
7
8
9
10
11
Time
c(t
)
K*(1-exp(-t/T))
Unit Step Input
Step Response
1
10
Input
outputstatesteadyKGainCD
.
%63
Step Response of 1st order System
• System takes five time constants to reach itsfinal value.
Step Response of 1st Order System• If K=10 and T=1, 3, 5, 7 TteKtc /)( 1
0 5 10 150
1
2
3
4
5
6
7
8
9
10
11
Time
c(t
)
K*(1-exp(-t/T))
T=3s
T=5s
T=7s
T=1s
Step Response of 1st Order System• If K=1, 3, 5, 10 and T=1 TteKtc /)( 1
0 5 10 150
1
2
3
4
5
6
7
8
9
10
11
Time
c(t
)
K*(1-exp(-t/T))
K=1
K=3
K=5
K=10
Relation Between Step and impulse response
• The step response of the first order system is
• Differentiating c(t) with respect to t yields
TtTt KeKeKtc //)( 1
TtKeKdt
d
dt
tdc /)(
TteT
K
dt
tdc /)(
Example 1• Impulse response of a 1st order system is given below.
• Find out
– Time constant T
– D.C Gain K
– Transfer Function
– Step Response
tetc 503 .)(
Example 1• The Laplace Transform of Impulse response of a
system is actually the transfer function of the system.
• Therefore taking Laplace Transform of the impulseresponse given by following equation.
tetc 503 .)(
)(..
)( sSS
sC
50
31
50
3
50
3
.)(
)(
)(
)(
SsR
sC
s
sC
12
6
SsR
sC
)(
)(
Example 1• Impulse response of a 1st order system is given below.
• Find out
– Time constant T=2
– D.C Gain K=6
– Transfer Function
– Step Response
tetc 503 .)(
12
6
SsR
sC
)(
)(
Example 1• For step response integrate impulse response
tetc 503 .)(
dtedttc t
503 .)(
Cetc ts 506 .)(
• We can find out C if initial condition is known e.g. cs(0)=0
Ce 05060 .
6C
ts etc 5066 .)(
Example 1• If initial conditions are not known then partial fraction
expansion is a better choice
12
6
SsR
sC
)(
)(
12
6
SssC )(
1212
6
s
B
s
A
Ss
ssRsR
1)(,)( input step a is since
50
66
12
6
.
ssSs
tetc 5066 .)(
Ramp Response of 1st Order System• Consider the following 1st order system
1Ts
K)(sC)(sR
2
1
ssR )(
12
Tss
KsC )(
• The ramp response is given as
TtTeTtKtc /)(
Parabolic Response of 1st Order System• Consider the following 1st order system
1Ts
K)(sC)(sR
3
1
ssR )(
13
Tss
KsC )(Therefore,
Second Order System
• We have already discussed the affect of location of poles and zeros onthe transient response of 1st order systems.
• Compared to the simplicity of a first-order system, a second-order systemexhibits a wide range of responses that must be analyzed and described.
• Varying a first-order system's parameter (T, K) simply changes the speedand offset of the response
• Whereas, changes in the parameters of a second-order system canchange the form of the response.
• A second-order system can display characteristics much like a first-ordersystem or, depending on component values, display damped or pureoscillations for its transient response. 33
Introduction• A general second-order system is characterized by the
following transfer function.
22
2
2 nn
n
sssR
sC
)(
)(
34
un-damped natural frequency of the second order system,which is the frequency of oscillation of the system withoutdamping.
n
damping ratio of the second order system, which is a measureof the degree of resistance to change in the system output.
Example 2
42
42
sssR
sC
)(
)(
• Determine the un-damped natural frequency and damping ratioof the following second order system.
42 n
22
2
2 nn
n
sssR
sC
)(
)(
• Compare the numerator and denominator of the given transferfunction with the general 2nd order transfer function.
2 n ssn 22
422 222 ssss nn 50.
1 n
35
Introduction
22
2
2 nn
n
sssR
sC
)(
)(
• Two poles of the system are
1
1
2
2
nn
nn
36
Introduction
• According the value of , a second-order system can be set intoone of the four categories (page 169 in the textbook):
1
1
2
2
nn
nn
1. Overdamped - when the system has two real distinct poles ( >1).
-a-b-cδ
jω
37
Introduction
• According the value of , a second-order system can be set intoone of the four categories (page 169 in the textbook):
1
1
2
2
nn
nn
2. Underdamped - when the system has two complex conjugate poles (0 < <1)
-a-b-cδ
jω
38
Introduction
• According the value of , a second-order system can be set intoone of the four categories (page 169 in the textbook):
1
1
2
2
nn
nn
3. Undamped - when the system has two imaginary poles ( = 0).
-a-b-cδ
jω
39
Introduction
• According the value of , a second-order system can be set intoone of the four categories (page 169 in the textbook):
1
1
2
2
nn
nn
4. Critically damped - when the system has two real but equal poles ( = 1).
-a-b-cδ
jω
40
Underdamped System
41
For 0< <1 and ωn > 0, the 2nd order system’s response due to a unit step input is as follows.Important timing characteristics: delay time, rise time, peak time, maximum overshoot, and settling time.
Delay Time
42
• The delay (td) time is the time required for the response toreach half the final value the very first time.
Rise Time
43
• 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 isnormally used. For overdamped systems, the 10% to 90% rise time iscommonly used.
43
Peak Time
44
• The peak time is the time required for the response to reach the first peak of the overshoot.
4444
Maximum Overshoot
45
The maximum overshoot is the maximum peak value of theresponse curve measured from unity. If the final steady-statevalue of the response differs from unity, then it is common touse the maximum percent overshoot. It is defined by
The amount of the maximum (percent) overshoot directlyindicates the relative stability of the system.
Settling Time
46
• The settling time is the time required for the response curveto reach and stay within a range about the final value of sizespecified by absolute percentage of the final value (usually 2%or 5%).
Step Response of underdamped System
222222 2
21
nnnn
n
ss
s
ssC
)(
• The partial fraction expansion of above equation is given as
22 2
21
nn
n
ss
s
ssC
)(
22 ns
22 1 n
2221
21
nn
n
s
s
ssC )(
22
2
2 nn
n
sssR
sC
)(
)(
22
2
2 nn
n
ssssC
)(
Step Response
47
Step Response of underdamped System
• Above equation can be written as
2221
21
nn
n
s
s
ssC )(
22
21
dn
n
s
s
ssC
)(
21 nd• Where , is the frequency of transient oscillationsand is called damped natural frequency.
• The inverse Laplace transform of above equation can be obtainedeasily if C(s) is written in the following form:
2222
1
dn
n
dn
n
ss
s
ssC
)(
48
Step Response of underdamped System
2222
1
dn
n
dn
n
ss
s
ssC
)(
22
2
2
22
111
dn
n
dn
n
ss
s
ssC
)(
222221
1
dn
d
dn
n
ss
s
ssC
)(
tetetc dt
dt nn
sincos)(
21
1
49
Step Response of underdamped System
tetetc dt
dt nn
sincos)(
21
1
ttetc ddtn
sincos)(21
1
n
nd
21
• When 0
ttc ncos)( 150
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
3 and 1.0 if n
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
51
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
3 and 5.0 if n
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
52
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
3 and 9.0 if n
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
53
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
54
S-Plane (Underdamped System)
1
1
2
2
nn
nn
55
Since 𝜔2𝜁2 − 𝜔2 𝜁2 − 1 = 𝜔2, the distance from the pole to the origin is 𝜔 and 𝜁 = 𝑐𝑜𝑠𝛽
Steady State Error
• If the output of a control system at steady state does notexactly match with the input, the system is said to havesteady state error
• Any physical control system inherently suffers steady-state error in response to certain types of inputs.
• A system may have no steady-state error to a step input,but the same system may exhibit nonzero steady-stateerror to a ramp input.
Classification of Control Systems
• Control systems may be classified according totheir ability to follow step inputs, ramp inputs,parabolic inputs, and so on.
• The magnitudes of the steady-state errors dueto these individual inputs are indicative of thegoodness of the system.
Classification of Control Systems
• Consider the unity-feedback control systemwith the following open-loop transfer function
• It involves the term sN in the denominator,representing N poles at the origin.
• A system is called type 0, type 1, type 2, ... , ifN=0, N=1, N=2, ... , respectively.
Classification of Control Systems
• As the type number is increased, accuracy isimproved.
• However, increasing the type numberaggravates the stability problem.
• A compromise between steady-state accuracyand relative stability is always necessary.
Steady State Error of Unity Feedback Systems
• Consider the system shown in following figure.
• The closed-loop transfer function is
Steady State Error of Unity Feedback Systems
• The transfer function between the error signal E(s) and theinput signal R(s) is
)()(
)(
sGsR
sE
1
1
• The final-value theorem provides a convenient way to findthe steady-state performance of a stable system.
• Since E(s) is
• The steady state error is
• Steady state error is defined as the error between theinput signal and the output signal when 𝑡 → ∞.
Static Error Constants
• The static error constants are figures of merit ofcontrol systems. The higher the constants, thesmaller the steady-state error.
• In a given system, the output may be the position,velocity, pressure, temperature, or the like.
• 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 theoutput temperature, and so on.
Static Position Error Constant (Kp)
• The steady-state error of the system for a unit-step input is
• The static position error constant Kp is defined by
• Thus, the steady-state error in terms of the static position error constant Kp is given by
Static Position Error Constant (Kp)
• For a Type 0 system
• For Type 1 or higher order systems
• For a unit step input the steady state error ess is
• The steady-state error of the system for a unit-ramp input is
• The static velocity error constant Kv is defined by
• Thus, the steady-state error in terms of the static velocity error constant Kv is given by
Static Velocity Error Constant (Kv)
Static Velocity Error Constant (Kv)
• For a Type 0 system
• For Type 1 systems
• For type 2 or higher order systems
Static Velocity Error Constant (Kv)
• For a ramp input the steady state error ess is
• The steady-state error of the system for parabolic input is
• The static acceleration error constant Ka is defined by
• Thus, the steady-state error in terms of the static accelerationerror constant Ka is given by
Static Acceleration Error Constant (Ka)
Static Acceleration Error Constant (Ka)
• For a Type 0 system
• For Type 1 systems
• For type 2 systems
• For type 3 or higher order systems
Static Acceleration Error Constant (Ka)
• For a parabolic input the steady state error ess is
Summary
Example 2
• For the system shown in figure below evaluate the staticerror constants and find the expected steady state errorsfor the standard step, ramp and parabolic inputs.
C(S)R(S)-
))((
))((
128
521002
sss
ss
Example 2
))((
))(()(
128
521002
sss
sssG
)(lim sGKs
p0
))((
))((lim
128
521002
0 sss
ssK
sp
pK
)(lim ssGKs
v0
))((
))((lim
128
521002
0 sss
sssK
sv
vK
)(lim sGsKs
a2
0
))((
))((lim
128
521002
2
0 sss
sssK
sa
41012080
5020100.
))((
))((
aK
Example 2
pK vK 410.aK
0
0
090.
Controllers• A controller is a device used to modify the
error signal and to produce a control signal.
• Depending on the control action provided, the controllers can be classified as follows:
1)Proportional controllers
2)Integral controllers
3)Proportional plus integral controllers
4)Proportional plus derivative controllers
5)Proportional plus integral plus derivative controllers
Effect of P-D controller• The proportional plus derivative controller produces an
output signal consisting of two terms: One proportional to error signal and the other proportional to derivative of error signal.
• In PD controller,
where Kp=Proportional gain,
Td=Derivative time.
On taking Laplace transform of the above equation with zero initial conditions we get
Therefore Transfer function of PD controller
• The block diagram of PD controller is as shown in figure
• The derivative control action is effective only during transient periods and so it does not produce corrective measures for any constant error.
• The derivative controller does not affect the steady state error directly but anticipates the error, initiates the early corrective action and tends to increase the stability of the system.
• While derivative control action has advantage of being anticipatory it has disadvantage that it amplifies noise signals and may cause saturation effect in the actuator.
Effect of PI controller• The PI controller produces an output signal consisting of two
terms: one proportional to error signal and the other proportional to integral of error signal.
• In PI controller
where Kp=Proportional gain
Ti=Integral time
On taking Laplace transform of the above equations with zero initial conditions we get
• The block diagram of PI controller is as shown in figure
• The advantages of both P controller and I controller are combined in PI controller.
• The proportional action increases the loop gain and makes the system less sensitive to variations of system parameters.
• The integral action eliminates or reduces the steady state error.
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