automatic control (1) engineering/871... · 2020. 12. 16. · automatic control (1) by associate...
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Automatic Control (1) By
Associate Prof. / Mohamed Ahmed Ebrahim Mohamed
E-mail: [email protected]
Web site: http://bu.edu.eg/staff/mohamedmohamed033
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Second Order System• We have already discussed the affect of location of poles and zeros on
the 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 can changethe 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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Introduction• A general second-order system is characterized by the
following transfer function.
22
2
2 nn
n
sssR
sC
++=)(
)(
4
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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
5
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Introduction
22
2
2 nn
n
sssR
sC
++=)(
)(
• Two poles of the system are
1
1
2
2
−−−
−+−
nn
nn
6
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
−−−
−+−
nn
nn
1. Overdamped - when the system has two real distinct poles ( >1).
-a-b-cδ
jω
7
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
−−−
−+−
nn
nn
2. Underdamped - when the system has two complex conjugate poles (0 < <1)
-a-b-cδ
jω
8
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
−−−
−+−
nn
nn
3. Undamped - when the system has two imaginary poles ( = 0).
-a-b-cδ
jω
9
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
−−−
−+−
nn
nn
4. Critically damped - when the system has two real but equal poles ( = 1).
-a-b-cδ
jω
10
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Underdamped System
11
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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Delay Time
12
• The delay (td) time is the time required for the response toreach half the final value the very first time.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Rise Time
13
• 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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Peak Time
14
• The peak time is the time required for the response to reach the first peak of the overshoot.
1414
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Maximum Overshoot
15
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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Settling Time
16
• 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%).
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
17
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
++−
++
+−=)(
18
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
19
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Step Response of underdamped System
tetetc dt
dt nn
sincos)(−−
−
−−=21
1
−
+−=−
ttetc ddtn
sincos)(21
1
n
nd
=
−=
21
• When 0=
ttc ncos)( −= 120
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
21
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
22
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
23
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Step Response of underdamped System
−
+−=−
ttetc ddtn
sincos)(21
1
24
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
S-Plane (Underdamped System)
1
1
2
2
−−−
−+−
nn
nn
25
Since 𝜔2𝜁2 − 𝜔2 𝜁2 − 1 = 𝜔2, the distance from the pole to the origin is 𝜔 and 𝜁 = 𝑐𝑜𝑠𝛽
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Analytical Solution
• Rise time: set c(t)=1, we have 𝑡𝑟 =𝜋−𝛽
𝜔𝑑
• Peak time: set 𝑑𝑐(𝑡)
𝑑𝑡= 0, we have 𝑡𝑝 =
𝜋
𝜔𝑑
• Maximum overshoot: M𝑝 = 𝑐 𝑡𝑝 − 1= 𝑒−(𝜁𝜔/𝜔𝑑)𝜋 (for unity output)
• Settling time: the time for the outputs always
within 2% of the final value is approximately 4
𝜁𝜔
−
+−=−
ttetc ddtn
sincos)(21
1
21 −= nd
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Steady State Error• If the output of a control system at steady
state does not exactly match with the input,the system is said to have steady state error.
• Any physical control system inherentlysuffers steady-state error in response tocertain types of inputs.
• A system may have no steady-state error to astep input, but the same system may exhibitnonzero steady-state error to a ramp input.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Classification of Control Systems
• Control systems may be classifiedaccording to their ability to followstep inputs, ramp inputs, parabolicinputs, and so on.
• The magnitudes of the steady-stateerrors due to these individual inputsare indicative of the goodness of thesystem.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Steady State Error of Unity Feedback Systems
• Consider the system shown in following figure.
• The closed-loop transfer function is
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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 → ∞.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
• 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)
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Static Velocity Error Constant (Kv)
• For a Type 0 system
• For Type 1 systems
• For type 2 or higher order systems
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Static Velocity Error Constant (Kv)
• For a ramp input the steady state error ess is
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
• 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)
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Static Acceleration Error Constant (Ka)
• For a parabolic input the steady state error ess is
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Summary
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
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
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Example 2
=pK =vK 410.=aK
0=
0=
090.=
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
System representations
• Continuous-time LTI system
– Ordinary differential equation
– Transfer function (Laplace transform)
– Dynamic equation (Simultaneous first-order ODE)
• Discrete-time LTI system
– Ordinary difference equation
– Transfer function (Z-transform)
– Dynamic equation (Simultaneous first-order ordinary difference equation)
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Continuous-time LTI system
−
+
)(tx
−
+
)(tyF
21 F
31
1 1AV
A
A
AA
AAA
Vdt
tdyty
dt
tdyVty
tytxdt
dVV
tyV
dt
dVtxV
=+
=+−
+=+
=−
++−
)(
3
1)(
0)(
3
1
1
)(
)()(2
12
01
)(
2
1
1
)(
)(6)(6)(7)(
)()()()(
)()()]()([2
1)]()([2
67
61
31
31
txtytyty
txtytyty
tytxtytytyty
=++
=++
+=+++
Ordinary differential equation
Laplace transform
67
6
)(
)()(
)()(6)(7)(
20..
2
++=
=++
=sssX
sYsH
sXsYssYsYs
CI
Transfer function
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)()(
)(6)(6)(7)(
)()(
)()(
)(6)(6)(7)(
2
'
1
12
'
2
2
1
txtx
txtxtxtx
tytx
tytxlet
txtytyty
=
=++
=
=
=++
)(6
0
)(
)(
76
10
)(
)(
2
1
2
1tx
tx
tx
tx
tx
+
−−=
State equation (Simultaneous first-order ODE)
=
)(
)(01)(
2
1
tx
txty
output equation
Dynamic equation
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Stability• Internal behavior–The effect of all characteristic
roots.
• External behavior–The effect by cancellation of
some transfer function poles.
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The Concept of Stability
A stable system is a dynamic system with a
bounded response to a bounded input.
➢ Absolute stability is a stable/not stable
characterization for a closed-loop
feedback system.
➢ Given that a system is stable we can
further characterize the degree of
stability, or the relative stability.
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Definition :
A system is internal (asymptotic)
stable, if the zero-input response
decays to zero, as time approaches
infinity, for all possible initial
conditions.
Asymptotic stable =>All the
characteristic polynomial roots are
located in the LHP (left-half-plane)
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Definition :
A system is external (bounded-input,
bounded-output) stable, if the zero-
state response is bounded, as time
approaches infinity, for all bounded
inputs.
bounded-input, bounded-output stable =>All the poles
of transfer function are located in the LHP (left-half-
plane)
Asymptotic stable => BIBO stable
BIBO stable=> Asymptotic stable
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
System response
(i) First order system response
(ii) Second order system response
(iii)High order system response
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
The Routh-Hurwitz Stability Criterion
➢ It was discovered that all coefficients of the
characteristic polynomial must have the same sign and
non-zero if all the roots are in the left-hand plane.
➢ These requirements are necessary but not sufficient. If
the above requirements are not met, it is known that
the system is unstable. But, if the requirements are
met, we still must investigate the system further to
determine the stability of the system.
➢ The Routh-Hurwitz criterion is a necessary and
sufficient criterion for the stability of linear systems.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
The Routh-Hurwitz Stability Criterion Steps
The method requires two steps:
(1)Generate the data table (Routh table).
(2)Interpret the table to determine the
number of poles in LHP and RHP.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Initial layout for Routh table
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Completed Routh table
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Feedback system and its equivalent closed-loop system
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Completed Routh table
Interpretation of Routh table
The number of roots of the polynomial that are in
the right half-plane is equal to the number of sign
changes in the first column.
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
R-H: Special case. Zero in the first column
Replace the zero with , the value of is then allowed to approach zero
from –ive or +ive side.
Problem Determine the stability of the closed-loop transfer function
5 4 3 2
10( )
2 3 6 5 3T s
s s s s s=
+ + + + +
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
R-H: Special case. Entire row is zero
Problem Determine the number of right-half-plane poles in the closed
transfer function
Solution: Form an auxiliary polynomial, P(s) using the entries of row above
row of zeros as coefficient, then differentiate with respect to s finally use
coefficients to replace the rows of zeros and continue the RH procedure.
5 4 3 2
10( )
7 6 42 8 56T s
s s s s s=
+ + + + +
4 2( ) 6 8P s s s= + + 3( )4 12 0
dP ss s
ds= + +
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Problem Determine the number of poles in the right-half-plane, left-half-
plan and on the axis for the closed transfer functionj
8 7 6 5 4 3 2
20( )
12 22 39 59 48 38 20T s
s s s s s s s s=
+ + + + + + + +
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Feedback Control System
Problem Determine the number of poles in the right-half-plane, left-half-
plan and on the axis for system
Solution: The closed loop transfer function is
j
4 3 2
200( )
6 11 6 200T s
s s s s=
+ + + +
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Routh table
2 poles in RHP, 2 poles in LHP no poles on axis
The system is unstable
j
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Feedback control system
Problem Find the range of gain K for the system that will cuase the
system to be stable, unstabel, and marginally stable. Assume K>0.
Solution: The closed loop transfer function is
3 2( )
18 77
KT s
s s s K=
+ + +
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Routh table for Example 6.9
For K < 1386 the system is stable. For K > 1386 the system is unstable.
For K = 1386 we will have entire row of zeros (s row). We form the even
polynomial and differentiate and continue, no sign changes from the
even polynomial so the 2 roots are on the axis and the system is
marginally stable
j
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Routh table for Example 6.10
Problem Factor the polynomial
Solution: from the Routh table we see that the s1 row is a row of zeros.
So the even polynomial at the s2 row is since this
polynomial is a factor of the original, dividing yields
As the other factor so
4 3 23 30 30 200s s s s+ + + +
2( ) 10P s s= +2( ) 3 20P s s s= + +
4 3 23 30 30 200s s s s+ + + + =2( 10)s + 2( 3 20)s s+ +
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Stability is State Space Example 6.11
Problem Given the system
Find out how many poles in the LHP, RHP and on the axis
Solution: First form (sI-A)
Now find the det (sI-A) =
0 3 1 10
2 8 1 0
10 5 2 0
1 0 0
X X u
y X
= + − − −
=
j
+
−−−
−−
=
−−−
−
=−
2510
182
13
2510
182
130
00
00
00
)(
s
s
s
s
s
s
AsI
5276 23 −−− sss
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Routh table for Example 6.11
One sign change, so 1 pole in the LHP and the
system is unstable
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
With Our Best Wishes
Automatic Control (1)
Course Staff
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Associate Prof. Dr . Mohamed Ahmed Ebrahim
Mohamed Ahmed
Ebrahim