ee2365 nol part 2
TRANSCRIPT
CONCEPT OF STABILITY
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Bode Plot
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Frequency Response
We will now examine the response of a system to a sinusoidal input
We call this the frequency response of a system
Steady state response of a system to the sinusoidal input
We vary the frequency of the i/p signal over a certain range and study the resulting response
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Frequency Response Consider a LTI system with T.F Y(s) /X(s) = G(s)
x(t) = X Sin wt Y(t)= Y sin (wt + Ø) where, Ø =tan -1 [imm G(jw)/ Real G(jw)] Y(jw) /X(jw) = G(jw)
G(S)X(S) Y(S)
Ø
X
Y
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Bode plot• Plots of the magnitude and phase characteristics are
used to fully describe the frequency response
• A Bode plot is a (semilog) plot of the transfer function
magnitude and phase angle as a function of
frequency
• The gain magnitude is many times expressed in
terms of decibels (dB)
dB = 20 log10 AMay 1, 2023 5BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
There are 4 basic forms in an open-loop transfer function G(jω)H(jω)
Gain factor, K (jω)±p factor: pole and zero at origin (1+jωT)±q factor Quadratic factor
r
nnj
2
2
21
Bode plot procedure
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The magnitude and phase plots for some typical factors
Factor Magnitude: Phase:
Gain, K
jω
dBjG )( )( jG
0dB20logK
0dB
1 20 dB/decade
90º
45º
0º
0dB
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The magnitude and phase plots for some typical factors
Factor Magnitude: Phase:
jω2
1/jω
dBjG )( )( jG
0dB
-20 dB/decade
0dB
1
40logω
40 dB/decade
180º
90º
0º
0º
-45º
-90º
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The magnitude and phase plots for some typical factors
Factor Magnitude: Phase:
1/jω2
1+jωT90º
45º
0º
0dB
-40logω
-40 dB/decade
0.1/T 1/T 10/T
dBjG )( )( jG
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The magnitude and phase plots for some typical factors
Factor Magnitude: Phase:
1/(1+jωT
dBjG )(
0.1/T
)( jG
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Gain margin and Phase margin
Gain margin:The gain margin is the number of dB
that is below 0 dB at the phase crossover frequency (ø=-180º). It can also be increased before the closed-loop system becomes unstable
Phase margin:The phase margin is the number of
degrees the phase of that is above -180º at the gain crossover frequency
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0dB
20dB
-20dB
-40dB
-60dB
-90º
0º
-180º
-270º
Gain margin
Phase marginPhase Crossover
with -180º line
Gain Crossover with 0dB line
Magnitude curve
Phase curve
Magnitude(dB)
Phase(Degree)
Log ω
Gain margin and Phase margin
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Bode Plot - Example For the following T.F draw the Bode plot and obtain Gain cross over
frequency (wgc) , Phase cross over frequency , Gain Margin and Phase Margin.
G(s) = 20 / [s (1+3s) (1+4s)]
Solution: The sinusoidal T.F of G(s) is obtained by replacing s by jw in the given T.F
G(jw) = 20 / [jw (1+j3w) (1+j4w)] Corner frequencies: wc1= 1/4 = 0.25 rad /sec ; wc2 = 1/3 = 0.33 rad /sec Form a table
Term Corner Frequency Slope db/dB
Change in slope
20/jw --- -20
1/(1+j4w) wc1=1/4=0.25 -20 -20-20=-40
1/(1+j3w) wc2=1/3=0.33 -20 -40-20=-60
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Choose a lower corner frequency and a higher Corner frequencywl= 0.025 rad/sec ; wh = 3.3 rad / sec
Calculation of Gain (A) (MAGNITUDE PLOT)
A @ wl ; A= 20 log [ 20 / 0.025 ] = 58 .06 dB
A @ wc1 ; A = [Slope from wl to wc1 x log (wc1 / wl ] + Gain (A)@wl
= - 20 log [ 0.25 / 0.025 ] + 58.06 = 38.06 dB
A @ wc2 ; A = [Slope from wc1 to wc2 x log (wc2 / wc1 ] + Gain (A)@ wc1
= - 40 log [ 0.33 / 0.25 ] + 38 = 33 dB
A @ wh ; A = [Slope from wc2 to wh x log (wh / wc2 ] + Gain (A) @ wc2
= - 60 log [ 3.3 / 0.33 ] + 33 = - 27 dB
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Calculation of Phase angle for different values of frequencies [PHASE PLOT]
Ø = -90O- tan -1 3w – tan -1 4w
when
Frequency in rad / sec
Phase Angle in degrees
w=0 Ø= -90o
w=0.025 Ø= - 99o
w=0.25 Ø= -172o
w=0.33 Ø= -188o
w=3.3 Ø= - 259o
w=∞ Ø= - 270o
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20
40
60
0
-20
-60
-40
Wl =0.025W1 =0.25
W2 =0.33 Wh=3.3
XX X
X
Log w+[G
(jw)]
in d
B
-50o
-100o
-150o
-200o
-250o
x
x x
-300o
x
x
Log w
Ph. A
ngle
- 180o
Magnitude PlotY-Axis 1 unit =20 dB
Phase PlotY-Axis 1 unit = 50o
Wgc =1.1 rad/sec
Wpc =0.3 rad/sec
GM
PMØgc=-240O
32 dB
GM = - [20 log G(jwpc)] dB = - [32] dB = - 32 dB-
=180+gc = 180o+(-240o)=- 60o
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Calculation of Gain cross over frequency The frequency at which the dB magnitude is Zero wgc = 1.1 rad / sec
Calculation of Phase cross over frequencyThe frequency at which the Phase of the system is - 180o
wpc = 0.3 rad / sec
Gain MarginThe gain margin in dB is given by the negative of dB magnitude of G(jw) at phase cross over frequency GM = - { 20 log [G( jwpc )] }
= - { 32 } = -32 dB
Phase Margin Ґ = 180o+ Øgc
= 180o + (- 240o) = -60o
Conclusion For this system GM and PM are negative in Values. Therefore the system is
unstable in nature.
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The steady-state response of a linear system to sinusoidal excitation may be determined from the system transfer function T(s) by replacing s by j
The resulting frequency response function T(j) has the polar form :
)(je)(A)j(T
)s(X)s(Y
)s(X)s(Yarg
T(s)X(s) Y(s)
For any value of , T(j) reduces to a complex number whose amplitude, A(j) gives , and () gives
POLAR PLOT AND STABILITYPOLAR PLOT AND STABILITY
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Open loop response : Polar plots
Polar plots are more useful in control system design work open-loop polar plot indicates whether the closed-loop system is
stable also gives a measure of how stable the system is
Represent T(j) on an Argand diagram :
Polar plot is traced out by the trajectory of T(j) as increases progressively from zero
ReIm
T(j)
A(j)(j)
T(s)-plane
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BASICS OF POLAR PLOT
The polar plot of a sinusoidal transfer function G(jω) is a plot of the magnitude of G(jω) Vs the phase of G(jω) on polar co-ordinates as ω is varied from 0 to ∞.
(ie) |G(jω)| Vs G(jω) as ω → 0 to ∞. Polar graph sheet has concentric circles and radial lines.
Concentric circles represents the magnitude.
Radial lines represents the phase angles.In polar sheet
+ve phase angle is measured in ACW from 0
-ve phase angle is measured in CW from 0
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PLOTTING PROCEDURE Express the given expression of OLTF in (1+sT) form.
Substitute s = jω in the expression for G(s)H(s) and get G(jω)H(jω).
Get the expressions for | G(jω)H(jω)| & G(jω)H(jω).
Tabulate various values of magnitude and phase angles for different values of ω ranging from 0 to ∞.
Usually the choice of frequencies will be the corner frequency and around corner frequencies.
Choose proper scale for the magnitude circles.
Fix all the points in the polar graph sheet and join the points by a smooth curve.
Write the frequency corresponding to each of the point of the plot.
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MINIMUM PHASE SYSTEMS
Systems with all poles & zeros in the Left half of the s-plane – Minimum Phase Systems.
For Minimum Phase Systems with only poles Type No. determines at what quadrant the polar plot starts. Order determines at what quadrant the polar plot ends.
Type No. → No. of poles lying at the origin Order → Max power of ‘s’ in the denominator polynomial of the transfer function.
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START & END OF POLAR PLOT
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TYPICAL SKETCHES OF POLAR PLOT
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GAIN MARGINGain Margin is defined as “the factor by which the system gain can be increased to drive the system to the verge of instability”.For stable systems,
ωgc ωpc
G(j)H(j) at ω=ωpc 1
GM = in positive dB
More positive the GM, more stable is the system.
For marginally stable systems,
ωgc = ωpc
G(j)H(j) at ω=ωpc = 1 GM = 0 dB
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For Unstable systems,ωgc ωpc
G(j)H(j) at ω=ωpc > 1 GM = in negative dB Gain is to be reduced to make the system stable
Note: If the gain is high, the GM is low and the system’s step response shows high overshoots and long settling time. On the contrary, very low gains give high GM and PM, but also causes higher ess, higher values of rise time and settling time and in general give sluggish response.
Thus we should keep the gain as high as possible to reduce ess and obtain acceptable response speed and yet maintain adequate GM and PM. An adequate GM of 2 ( ie 6 dB ) and a PM of about 30 is generally considered good enough as a thumb rule.
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Contd…
PM
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Contd… At = pc , G(j)H(j) = -180
Let G(j)H(j) at = pc be taken as B
If the gain of the system is increased by a factor 1/B, then the G(j)H(j) at = pc becomes B(1/B) = 1 and hence the G(j)H(j) locus pass through -1+j0 point driving the system to the verge of instability.
GM is defined as the reciprocal of the magnitude of the OLTF evaluated at the phase cross over frequency. GM = Kg = 1 / G(j)H(j) = pc
GM in dB = 20 log (1/B) = - 20 log BMay 1, 2023 29BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
PHASE MARGINPhase Margin is defined as “ the additional phase lag that can be introduced before the system becomes unstable”.
Let ‘A’ be the point of intersection of G(j)H(j) plot and a unit circle centered at the origin.
Draw a line connecting the points ‘O’ & ‘A’ and measure the phase angle between the line OA and +ve real axis.
This angle is the phase angle of the system at the gain cross over frequency.
G(jgc)H(jgc) = gc
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Contd… If an additional phase lag of PM is introduced at this frequency, then the phase angle G(jgc)H(jgc) will become 180 and the point ‘A’ coincides with (-1+j0) driving the system to the verge of instability.
This additional phase lag is known as the Phase Margin.
= 180 + G(jgc)H(jgc)
= 180 + gc
[Since gc is measured in CW direction, it is taken as negative]
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Contd… For a stable system, the phase margin is positive.
A Phase margin close to zero corresponds to highly oscillatory system.
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Polar plots contd.
A polar plot may be constructed from experimental data or from a system transfer function
If values of are marked along the contour, a polar plot has the same information as a Bode plot
Usually, the shape of a polar plot is of most interest
-j 0.50
j 0-0.5 0 0.5 1
Re
Im
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ExampleProblem : Construct the polar plot for the (critically damped
system) defined by :
Solution :
Limiting conditions :(i)(ii)
(iii)
0j1)j(T:0
o2
180j
1
e0)j(T.e.i
)j(T:
2j)j(T:1
Re-0.2 0 0.2 0.4 0.6 0.8 1
-0.6
-0.4
-0.2
Im
2)1s(1)s(T
2)j1(1)j(T
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Example (contd.) For accurate work, results are usually calculated for a range of
values of :
0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 5.0
T(j) 1+j0
0.83-j0.44
0.48-j0.64
0.18-j0.61
0-j0.5
-0.12-j0.28
-0.12-j0.16
-0.08-j0.06
-0.04-j0.01
-j 0.50
j 0
-0.5 0 0.5 1
Re
Im
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2. G(s) = [(1+0.2s)(1+0.025s)] / [(s^3(1+0.005s)(1+0.001s)]
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STABILITY OF CONTROL SYSTEM
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Concepts of Stability
Stability - Definition
A system is stable if any bounded input produces a
bounded output for all bounded initial conditions
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Basic concept of stability
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Stability of the system and roots of characteristic equations
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Characteristic EquationCharacteristic Equation
Consider an nth-order system whose the characteristic equation (which is also the denominator of the transfer function) is:
00
11
22
11 sasasasas)s(a n
nnn
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BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING 43
Routh Hurwitz Criterion
May 1, 2023
IntroductionIntroduction Goal: Determining whether the system is stable or
unstable from a characteristic equation in polynomial form without actually solving for the roots
Routh’s stability criterion is useful for determining the ranges of coefficients of polynomials for stability, especially when the coefficients are in symbolic (non numerical) form
To find Kmar & ωn
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A A necessary necessary conditioncondition for Routh’s for Routh’s
StabilityStability A necessary condition for stability of the system is
that all of the roots of its characteristic equation have negative real parts, which in turn requires that all the coefficients be positive
A necessary (but not sufficient) condition for stability is that all the coefficients of the polynomial
characteristic equation be positive & none of the co-efficient vanishes
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A A necessary and sufficient necessary and sufficient conditioncondition
for Stabilityfor Stability Routh’s formulation requires the computation of a
triangular array that is a function of the coefficients of the polynomial characteristic equation
A system is stable if and only if all the elements ofthe first column of the Routh array are positive
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Characteristic EquationCharacteristic Equation Consider an nth-order system whose the
characteristic equation (which is also the denominator of the transfer function) is:
00
11
22
11 sasasasas)s(a n
nnn
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Method for determining the Method for determining the Routh arrayRouth array
Consider the characteristic equation:
5311
42
:1:
aaasaas
n
n
00
11
33
22
111 sasasasasas)s(a n
nnnn
First arrange the coefficients of the
characteristic equation in two rows, beginning
with the first and second coefficients and followed by the even-numbered and odd-
numbered coefficients: May 1, 2023 48BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Routh array: method Routh array: method (cont’d)(cont’d) Then add
subsequent rows to
complete the Routh array:
*:*:
**:
:::
1:
0
1
2
3213
3212
5311
42
sss
cccsbbbsaaasaas
n
n
n
n
?!
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Routh array: method Routh array: method (cont’d)(cont’d)
Compute elements for the 3rd row:
1
1673
1
1452
1
1231
1
,1
,1
aaaab
aaaab
aaaab
*:*:
**:
:::
1:
0
1
2
3213
3212
5311
42
sss
cccsbbbsaaasaas
n
n
n
n
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Routh array: method Routh array: method (cont’d)(cont’d)
Compute elements for the 4th row:
1
17413
1
15312
1
13211
,
,
bbabac
bbabac
bbabac
*:*:
**:
:::
1:
0
1
2
3213
3212
5311
42
sss
cccsbbbsaaasaas
n
n
n
n
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Example 1
Given the characteristic equation,
Is the system described by this characteristic equation stable?
Answer: One coefficient (-2) is negative
Therefore, the system does not satisfy the necessary condition for stability
4s4ss2s3s4s)s(a 23456
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Example 2Given the characteristic equation,
Is the system described by this characteristic equation stable?
Answer:All the coefficients are positive and nonzeroTherefore, the system satisfies the necessary
condition for stabilityWe should determine whether any of the coefficients
of the first column of the Routh array are negative
44234)( 23456 sssssssa
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Example 2 (cont’d): Routh array
?:??:??:
???:???:
0424:4131:
0
1
2
3
4
5
6
sssssss
44234)( 23456 sssssssa
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Example 2 (cont’d): Routh array
?:??:??:
???:4025:
0424:4131:
0
1
2
3
4
5
6
sssssss
25
410
44321
44234)( 23456 sssssssa
040
44141
4416
44401
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Example 2 (cont’d): Routh Example 2 (cont’d): Routh arrayarray
?:??:??:
05122:4025:
0424:4131:
0
1
2
3
4
5
6
sssssss
2)2(25
25204
44234)( 23456 sssssssa
512
52032
2525444
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Example 2 (cont’d): Routh Example 2 (cont’d): Routh arrayarray
4:01576:43:
05122:4025:
0424:4131:
0
1
2
3
4
5
6
sssssss
44234)( 23456 sssssssa
The elements of the 1st column are not all positive:
the system is unstable
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Special cases of Routh’s Special cases of Routh’s criteriacriteria Case 1: All the elements of a row in a RA are
zero Form Auxiliary equation by using the co-efficient of the row
which is just above the row of zeros Find derivative of the A.E. Replace the row of zeros by the co-efficient of dA(s)/ds complete the array in terms of these coefficients analyze for any sign change, if so, unstable no sign change, find the nature of roots of AE non-repeated imaginary roots - marginally stable repeated imaginary roots - unstable
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Special cases of Routh’s Special cases of Routh’s criteriacriteria Case 2: First element of any of the rows
of RA is
Zero and the same remaining row contains atleast one non-zero element
Substitute a small positive no. ‘ε’ in place of zero and complete the array.
Examine the sign change by taking Lt ε 0
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Example 3: Stability versus Parameter Range
Consider a feedback system such as:
The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable
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Example 3 (cont’d)
The characteristic equation for the system is given by:
0)6)(1(
11
ssssK
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Example 3 (cont’d)
Expressing the characteristic equation in polynomial form, we obtain:
0)6(5 23 KsKss
0)6)(1(
11
ssssK
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Example 3 (cont’d)
The corresponding Routh array is:
Therefore, the system is stable if and only if
KsKsssa )6(5)( 23
KsKs
KsKs
:5)304(:
5:61:
0
1
2
3
5.7K
0Kand5.7K
0Kand05
30K4
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Root Locus Technique
jj
XX
XX
––44OO
pp22
pp33
pp11
zz11
XX
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INTRODUCTION TO ROOT LOCUS Introduced by W. R. Evans in 1948
Graphical method, in which movement of poles in the s-plane is sketched when some parameter is varied
The path taken by the roots of the characteristic equation when open loop gain K is varied from 0 to ∞ are called root loci
Direct Root Locus = 0 < k < ∞
Inverse Root Locus = - ∞ < k < 0
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Root Locus Analysis The roots of the closed-loop characteristic equation
define the system characteristic responses
Their location in the complex s-plane lead to prediction of the characteristics of the time domain responses in terms of: damping ratio, natural frequency, wn
damping constant, first-order modes
Consider how these roots change as the loop gain is varied from 0 to
second-order modes
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BASICS OF ROOT LOCUS Symmetrical about real axis
RL branch starts from OL poles and terminates at OL zeroes
No. of RL branches = No. of poles of OLTF
Centroid is common intersection point of all the asymptotes on the real axis
Asymptotes are straight lines which are parallel to RL going to ∞ and meet the RL at ∞
No. of asymptotes = No. of branches going to ∞
At Break Away point , the RL breaks from real axis to enter into the complex plane
At BI point, the RL enters the real axis from the complex plane
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ANGLE AND MAGNITUDE CRITERIA Angle condition
G(s)H(s) = ±(2q+1)180º where q = 0, 1, 2 … = Odd multiple of 180º Use : Any point in s-plane which satisfies the angle
condition has to be on the root locus of the corresponding system
Magnitude condition |G(s)H(s)| = 1 Use: It gives the value of K corresponding to any point
which is on the root locusMay 1, 2023 68BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
PROCEDURE FOR CONSTRUCTING ROOT LOCUS Locate the OL poles & zeros in the plot Find the branches on the real axis Find angle of asymptotes & centroid
Φa= ±180º(2q+1) / (n-m) σa = (Σpoles - Σzeroes) / (n-m) Find BA and BI points Find Angle Of departure (AOD) and Angle Of Arrival (AOA)
AOD = 180º- (sum of angles of vectors to the complex pole from all other poles) + (sum of angles of vectors to the complex pole from all zero)
AOA = 180º- (sum of angles of vectors to the complex zero from all other zeros) + (sum of angles of vectors to the complex zero from poles)
Find the point of intersection of RL with the imaginary axis
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The Root Locus Procedure
0)(1 sF
0)(
)(1
1
1
n
ii
m
jj
ps
zsK
Step 1: Write the characteristic equation as
Step 2: Rewrite preceding equation into the
form of poles and zeros as follows:
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The Root Locus Procedure
Step 3: Locate the poles and zeros with specific symbols, the root locus begins at the open-loop poles and ends at the open-loop zeros as K increases from 0 to infinity
If open-loop system has n-m zeros at infinity, there will be n-m branches of the root locus approaching the n-m zeros at infinity
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The Root Locus Procedure
Step 4: The root locus on the real axis lies in a section of the real axis to the left of an odd number of real poles and zeros
Step 5: The number of separate loci is equal to the number of open-loop poles
Step 6: The root loci must be continuous and symmetrical with respect to the horizontal real axis
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The Root Locus Procedure
Step 7: The loci proceed to zeros at infinity along asymptotes
centered at centroid and with angles
mn
zpn
i
m
jji
a
1 1
)1,2,1,0()12(
mnkmn
ka
May 1, 2023 73BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Step 8: The actual point at which the root locus crosses the imaginary axis is readily evaluated by using Routh’s criterion
Step 9: Determine the breakaway point d (usually on the real axis):
m
j
n
i ij pdzd1 1
11
The Root Locus Procedure
May 1, 2023 74BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
The Root Locus Procedure
Step 11: Plot the root locus that satisfy the phase criterion
Step 12: Determine the parameter value K1 at a specific root using the magnitude criterion
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11
11
)(
)(
ss
m
jj
n
ii
zs
psK
,2,1)12()( kksP
The Root Locus - example
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Root Locus Plot - Example Loop Transfer function:Loop Transfer function:
Roots:Roots:s = 0s = 0, , s = –4s = –4, ,
s = –2 s = –2 4 4jj Real axisReal axis segments: segments:
between 0 and –4 .between 0 and –4 . Asymptotes:Asymptotes:
angles = angles =
))202044)()(44(())(( 22
ssssssssKKssGHGH
2244
))00222244(( aa44
77,,44
55,,44
33,,440044
))1122(( kk
asymptotesasymptotes
jj
XX
––44XX
––22
––22jj
22jj
++11
4545°°
44jj
––44jjXX
XX
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Root Locus Plot - Example Breakaway points:
Three points that breakaway at 90°
solving, jsb 45.22,2
ssss)ss(s
sssKssss
Kdsd
020186or
080368
)8072244(80368
23
234
23
234
jj
XX
––44XX
––22
––22jj
22jj
++11
4545°°
44jj
––44jjXX
XX
2
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Root Locus Plot - Example The imaginary axis crossings:The imaginary axis crossings:
Characteristic equationCharacteristic equation
Routh tableRouth tabless44 1 36 1 36 KKss33 8 80 0 8 80 0ss22 26 26 KK 0 0s 80-8s 80-8KK/26 0 0/26 0 0ss00 KK 0 0 0 0
Condition for Condition for critical critical stabilitystability80-880-8KK/26 > 0 /26 > 0 or or KK<260<260The auxiliary equationThe auxiliary equation26 26 s s 22 + 260 = 0 + 260 = 0solvingsolving
008080363688 223344 KKssssssss
jjjjss 1616..331010
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Root Locus Plot - Example The final root locus plot
is ---------
What is the value of the gain K corresponding to the breakaway point at sb = –2 ± 2.45j ?
XX
––44XX
––22
––22jj
22jj
44jj
––44jjXX
XX3.163.16jj
jj
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May 1, 2023 81BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Used to check the stability of CL system from OL response. OL stable system may become unstable in Closed loop and vice-versa.
Pole – Zero configuration from Nyquist point of view
Concept of Number of encirclements
Mapping theorem or Principle of Argument
Nyquist stability criteria
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Single Valued Function
A single valued function is defined as for each point in the s-plane there is only one corresponding point in the q(s)-plane.
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Let a function q(s) = (s-α1)(s-α2)…….. (s-αm) (s-β1)(s- β2)…….. (s- βn)
Let s = σ + jω, a complex variable in S-Plane, then q(s) is also complex and it may be defined as q(s) = u + jv on the complex plane.
A function q(s) is said to be Analytic in the S-plane if the function and all its derivatives exist.
The points in the S-plane where the function q(s) or its derivatives does not exist are called Singular points.
The poles of the function q(s) are singular points.
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Encircled:If a point or region is found inside a closed path (closed contour) in a complex plane then it is said to be encircled
S1 is encircled by the closed path
S2 is not encircledMay 1, 2023 85BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Enclosed:Any point or region is said to be enclosed by a closed path, if it is found to lie to the right of the path when the path is traversed in the prescribed direction
The shaded region are the region enclosed by the closed path.
In Fig. 1, the point A is enclosed
In Fig. 2, the point B is enclosed
Fig. 1 Fig. 2
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For a contour in s-plane, which does not go through any singular point, there corresponds a contour in the q(s) plane
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A function G(s)H(s) = Open Loop Transfer Function
Poles of G(s)H(s) = Open Loop Poles
Zeros of G(s)H(s) =Open Loop Zeros
A function G(s) / [1+G(s)H(s)] = Closed Loop Transfer Function
Let a function F(s) = 1 + G(s)H(s)
Zeros of {F(s) = 1 + G(s)H(s)} = Closed Loop Poles of a systemPoles of {F(s) = 1 + G(s)H(s)} = Open Loop Poles of a system
For a stable system, the roots of the characteristic equation and hence the zeros of F(s) must lie in the Left Half of the S-Plane.
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If q(s) is a single valued function, that has a finite number of poles & zeros in the s-plane and C is a contour in s-plane that does not go through any of the poles & zeros q(s); then the corresponding locus Cq mapped in the q(s)-plane will encircle the origin as many times as the difference between the number of zeros and poles of q(s) that are encclosed by the s-plane contour C.
N = Z – PIf N is positive, encirclement is CW direction.
If N is negative, encirclement is ACW direction.
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1. N > 0 (Z > P)
If the contour in the s-plane encloses more no. of zeros than poles of q(s) in CW direction then the corresponding locus Cq mapped in the q(s)-plane will encircle the origin N times in the CW direction (same direction).
2. N < 0 (P > Z)
If the contour in the s-plane encloses more no. of poles than poles of q(s) in CW direction then the corresponding locus Cq mapped in the q(s)-plane will encircle the origin N times in the ACW direction (opposite direction).
3. N = 0 (P = Z)
If the contour in the s-plane encloses as many poles as zeros, or no poles and no zeros then the corresponding locus Cq mapped in the q(s)-plane will not encircle the origin.
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j
j
s j R e jθ
R
C1
C2
Along C1, s varies from -j∞ to + j∞
Along C2, s = R e jθ ; where R→∞ and θ varies from + Π/2 to –Π/2
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Let P = no. of poles of q(s)-plane lying on Right Half of s-plane and encircled by s-plane contour.
Let Z = no. of zeros of q(s)-plane lying on Right Half of s-plane and encircled by s-plane contour.
For the CL system to be stable, the no. of zeros of q(s) which are the CL poles, that lie in the right half of s-plane should be zero. That is Z = 0, which gives N = -P.
Therefore, for a stable system the no. of ACW encirclements of the origin in the q(s)-plane by the contour Cq must be equal to P.
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We know that q(s) = 1+G(s)H(s)
Therefore G(s)H(s) = [1+G(s)H(s)] – 1
The contour Cq, which has obtained due to mapping of Nyquist contour from s-plane to q(s)-plane (ie)[1+G(s)H(s)] -plane, will encircle about the origin.
The contour CGH, which has obtained due to mapping of Nyquist contour from s-plane to G(s)H(s) -plane, will encircle about the point (-1+j0).
Therefore encircling the origin in the q(s)-plane is equivalent to encircling the point -1+j0 in the G(s)H(s)-plane.
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If the contour CGH of the open loop transfer function G(s)H(s) in the G(s)H(s)–plane corresponding to Nyquist contour in the s-plane encircles the point -1+j0 in the counterclockwise direction as many times as the right half s-plane poles of G(s)H(s), then the closed loop system is stable
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Nyquist Stability Criteria
Stable
Unstable
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ExampleProblem : Construct the Nyquist stability plot for a feedback system with the following open-loop transfer characteristic :
Solution : For section ab, s = j , : 0
The polar plot may be constructed by calculating G(j )H(j ) for increasing values of :
R
Re
Im
a
b
c
d
e
)1s)(10s2s(20)s(H)s(G 2
)1j)(102j(20)j(H)j(G 2
(rad/s) 0 0.5 1.0 2.0 2.5 3.0 3.5 4.0 5.0
G(j )H(j ) 2.0+j0
1.54-j0.98
0.82-j1.29
-0.15-j1.23
-0.62-j1.02
-0.92-j0.49
-0.75+j0.02
-0.45+j0.19
-0.15+j0.15
0-270o
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Example : contd.
On section bcd, s = RejR ; therefore( ie.), all of section bcd maps onto the origin
0R20)s(H)s(G 3
Re-1 0 1 2
Im
-1
0.5
-0.5
G(s)H(s) - plane
Polar plot of G(s)H(s)
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Example : solution contd.On section de, s = –j , : – 0
Therefore, the map of de is the image of the map of ab : G(–j)H(–j) = [G(j)H(j)]*
Re-1 0 1 2
Im
0.5
1
-0.5 G(s)H(s) - plane
Note : the arrows indicate the direction along the contour on section de ; i.e decreasing frequency from – to zero
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Example : solution contd.The complete Nyquist plot is presented as :
Clearly the point (-1,0) is outside the contour the system is stable
Note : the interior of the Nyquist contour is the region lying to the
right of the path (relative to the direction of the
arrows)G(s)H(s) - plane
Im
-1
1
-0.5
Re-1 1 20
0.5
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Poles lying on imaginary axis
If the open-loop transfer function has poles lying on the imaginary axis, and in particular at the origin, the Nyquist contour must be modified as shown :
» i.e. section efa bypasses the origin rather than passing through it
» On section efa : s = rejr0, : –90o to +90o
RRe
Im
a
b
c
d
ef
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ExampleProblem : Sketch the Nyquist stability plot for a feedback system with the following open-loop transfer function :
Solution : For section ab, s = j, : 0
(i) 0 : G(j )H(j ) –1 – j(ii) = 1 : G(j )H(j ) –1+ j0(iii) : G(j )H(j ) 0–270o
)j1(j1)j(H)j(G 2
)1ss(s1)s(H)s(G 2
RRe
Im
a
b
c
d
ef
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Example : contd.
On section bcd, s = RejR ; therefore
i.e. section bcd maps onto the origin of the G(s)H(s)-plane
Polar plot of G(s)H(s)
0R1)s(H)s(G 3
Re-1.5 -1 -0.5 0 0.5
-6
-4
-2
0
2Im
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Example : solution contd.Section de maps as the complex image of the polar plot as before :
Re-1.5 -1 -0.5 0 0.50
2
-6
-4
-2
Im
4
6
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Example : solution contd.Finally on section efa :
, : –90o to +90o
This defines a semi-circle of infinite radius in the clockwise direction; i.e. from +90o to –90o
» Note : (-1,0) is neither inside nor outside the Nyquist plot The system is marginally stable
s = rejr0
0rjre
1)s(H)s(G
Re-1.5 -1 -0.5 0.5
-6
-4
-2
Im
4
6
00
2
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Polar plot : s = j, : 0
(i) 0 : G(j )H(j ) 0o
(ii) = 1 : G(j )H(j ) = 0.5 – j 0.5(iii) : G(j )H(j ) 0–90o
Example Consider a system with open-loop transfer function :
)1s(s1)s(H)s(G 4
)j1(1))H(jG(j 4
0 0.5 1 1.5
-0.75
-0.5
-0.25
ReIm
-1
RRe
Im
a
b
c
d
ef
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Contd… On section bcd, s = RejR ; therefore Section de maps as the complex image of the polar plot as
before However, on section efa :
s = rejr0, , : –90o to +90o
oo
0r4j4
360to360
er1)s(H)s(G
01)()( 4 R
sHsG
0 0.5 1 1.5
-0.75
-0.5
-0.25
Re
Im
-1
0.25
0.5
Since N=2; CL system is not stable
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1. Nyquist plot for the system G(s)H(s) = K / [s(s+2)(s+10)]
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May 1, 2023 108BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
May 1, 2023 109BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
May 1, 2023 110BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
May 1, 2023 111BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
May 1, 2023 112BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
May 1, 2023 113BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
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SAMPLED DATA SYSTEMS
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What is a compensator?
One that compensates any deficiencies in the system
One that compensates any inadequacies in the system
It is also a sub system placed inside the overall system
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What do a compensator do?
Very simply, a compensator adds a zero or pole to a system
Study of the influence of a pole or zero addition inside a system
Improves STABILITY & PERFORMANCE
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Compensator & Controller
What is the difference between a compensator and controller?
When two elements are doing mathematically the same function, what will be the difference?
PI, PD & PID Controllers
Lag, Lead & Lag Lead Compensator
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Compensator - Definition
An additional sub system added in to the closed loop to improve the stability and performance of the overall system
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Types of Compensators(Based on placement)
Cascade / Series Compensators
Feedback / Parallel Compensators
Input compensators
Output Compensators
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Compensator
Process
Feedback
Compensator
Process
Feedback
Process
Feedback
Compensator
Compensator
Process
FeedbackCompensator
Types of compensators
Feedback Compensator
Cascade Compensator
Output or Load Compensator
Input Compensator
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Types of compensators (Based on working)
Lead Compensators
Lag Compensators
Lag – Lead Compensators
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Lag compensator Have the nature of a lag network
For a sinusoidal input, output will have a phase lag (steady state)
Integrator or PI Controller or Low pass filter
Improves the steady state performance
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What happens when we use a lag compensator in a system? Bandwidth reduced
Sluggish in nature that increases the rise time, settling time
System becomes more sensitive to parameter variations
Improves steady state accuracy
Decreases the stability of the system
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Lead compensator Have the nature of a lead network
For a sinusoidal input, output will have a phase lead (steady state)
Differentiator or PD Controller or High pass filter
Improves the Transient state performance
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What happens when we use a lead compensator in a
system? Addition of Zero and a pole to the system Zero added to the right of the pole Add more damping to the closed loop system Reduction of Rise time and Settling time –
Increase in Bandwidth Improves phase margin Improves stability Improves bandwidth No change in SS Error
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Limitations of Lead Compensators
Single stage Lead Compensator fails if the phase angle required is above 60 degree. In that case use of multi stage compensators
As the bandwidth increases, the possibility of noise to enter in to the system also increases
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Design methods Time Domain – Root Locus Frequency Domain – Bode Plot
Not like a PID tuning where sometimes, system TF not required
Essentially system TF is required for design of Compensators
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Design Procedure1. Analysis of the existing uncompensated
system (Bode Plot / Root Locus)2. Altering the performance parameters
(Ex. Phase margin)3. Finding the compensated system function4. Analysis of the compensated system
(Bode / Root Locus)5. Design Verification
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Lead Compensatoruses phase advance
at the zero-crossingincreases high
frequency gain and shifts wc to the right
increases bandwidthlimited amount of
compensation possible
Lag Compensator uses gain reduction to
shift the zero-crossing to the left
introduces phase lag at frequencies below wc
reduces bandwidth very large amount of
compensation possible
Characteristics of Lead andLag Compensators
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Comparison Lead compensator, achieves
desired results through the merits of its property of phase lead contribution
Lag compensator achieves desired results through the merits of its property of attenuation at high frequency
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Concluding Comment Selection of compensator
depends on the designer’s experience in the process
Another factor for consideration is the availability of components (Ex. Non Electrical systems)
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Gp(s)
H(s)
E(s)R(s) C(s)
+ – cascade compensator
s
sK c
1
1
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Design Criteria and Specifications
Design Criteria:accuracystability (absolute and
relative)speed of response
Time domain: steady-state error% overshoot rise time, Tr
time to 1st peak, Tp
settling time, Ts
Frequency domain:bandwidthphase margingain margin
s-plane:damping ratio, znatural frequency, wn
damping constant, s
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Effects of Cascading a Controller
Increase Kc to reduce steady-state error
Increase Kc to increase the speed of response
What is the effect of increasing Kc on the closed-loop stability?
Examine the Nyquist plot and Bode plot
Gp(s)
H(s)
E(s)R(s) C(s)
+ – cascade compensator
ss
K c
1
1
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Effects of Cascading a Controller Consider the following plant
characteristics.
1)( sH
)255)(2.0(50)( 2
sss
sGp
If Kc = 1, then the position error constant is Kp = 10 and the steady-state error for a step input is
Increasing Kc to 3 will reduce the steady-state error to 3.2%
%1.9%1001
1
p
ss Ke
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Effects of Increasing Kc For Kc = 3, the gain curve is
shifted up by about 10 dB, shifting the zero-crossing to the right and increasing the bandwidth
For Kc = 1, the phase margin is about 55, but with Kc = 3, the phase margin becomes -10, resulting in an unstable closed-loop systemFrequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
M
agni
tude
(dB
)P
hase
(deg
)
Mag
nitu
de (d
B)
Bode Diagrams
-80
-60
-40
-20
0
20
10-2
10-1
100
101
102
-250
-200
-150
-100
-50
0
-180m
Kc=3
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Effects of Increasing Kc
On the Nyquist plot, increasing the gain Kc from 1 to 3 results in encirclements of the –1 point and the system becomes unstable. Note with Kc = 1, the gain margin is about 2.5 or about 8 dB
How can the loop characteristics be changed to allow a gain of Kc = 3 and keep the system stable?
Real AxisReal Axis
Imag
inar
y A
xis
Imag
inar
y A
xis
Nyquist DiagramsNyquist Diagrams
-1.5-1.5 -1-1 -0.5-0.5
-1.5-1.5
-1-1
-0.5-0.5
00
0.50.5
11
1.51.5
––11
Kc = 3
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Reshaping the Loop Frequency Characteristics
To achieve a stable system while keeping Kc = 3, phase advance may be added to the system in the frequency range where |GH(jw)| = 1
This may be done through cascade phase-lead compensation
Real AxisReal Axis
Imag
inar
y A
xis
Imag
inar
y A
xis
Nyquist DiagramsNyquist Diagrams
-1.5 -1 -0.5
-1.5
-1
-0.5
0
0.5
1
1.5
–1
Kc = 3
m
|GH|=1
May 1, 2023 139BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lead Compensation
Add phase-lead compensation to the controller.
This introduces phase advance of 70 and a gain increase of 15 dB at the new zero-crossing frequency, wc = 9.9 rad/s
sssGc 017.01
6.013)(
Frequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
M
agni
tude
(dB
)P
hase
(deg
)
Mag
nitu
de (d
B)
-100
-80
-60
-40
-20
0
20
Gm=8.1303 dB (at 15.324 rad/sec)Gm=8.1303 dB (at 15.324 rad/sec) Pm=15.772 deg. (at 9.9897 rad/sec)Pm=15.772 deg. (at 9.9897 rad/sec)
10-2 10-1 100 101 102 103
-250
-200
-150
-100
-50
0
50
uncompensated
compensator
compensated
m = 16 at c = 9.9 rad/s
May 1, 2023 140BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Compensation
Add phase-lag compensation to the controller
This introduces gain reduction at and a small phase lag at the new zero-crossing frequency, wc = 3.8 rad/s
Frequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
Mag
nitu
de (d
B)
Pha
se (d
eg)
Mag
nitu
de (d
B)
-80
-60
-40
-20
0
20
Gm=3.5624 dB (at 4.9746 rad/sec)Gm=3.5624 dB (at 4.9746 rad/sec) Pm=28.6 deg. (at 3.7882 rad/sec)Pm=28.6 deg. (at 3.7882 rad/sec)
10-2
10-1
100
101
102
-250
-200
-150
-100
-50
0
compensator
uncompensated
compensated
sssGc 3.31
8.113)(
m = 29 at c = 3.8 rad/s
May 1, 2023 141BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lead Compensator Characteristics
Time constant form
where a > 1
Pole-zero form
Note: | z | < | p |
ssKsG cc
1
11)(
passivepassivenetworknetwork
amp.amp.gaingain
wherewhere1 , 1 pz
)(
pszsKsG cc
May 1, 2023 142BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lead Compensators
Three parameters to be chosen in the design
Kc : the compensator amplifier gain, chosen to meet the required low-frequency gain, or bandwidth requirements
t : the time constant of the compensator, chosen to place the phase advance at the desired frequency
a : the spread of the pole and zero of the compensator, defining the maximum phase advance
ss
KsG cc
1
11)(
May 1, 2023 143BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lead Compensators: Design Steps
1. Loop gain requirement: Determine the loop gain required to meet the
steady-state error specifications
2. Bode plot: Plot the loop transfer function including the
required gain increase from step 1
3.Required phase advance: (calculate a) Determine the required qm to meet the phase
margin specification at the predicted zero-crossing frequency
May 1, 2023 144BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lead Compensators: Design Steps
Calculate from,
Check the new zero-crossing frequency based on the gain increase of ½| |dB to see if the phase margin will be met with the phase advance, m . Some iteration may be required
11sin
m
c , m m ½| |dB
specs.11sin
m
h.f. gainincreaseBode
plot
May 1, 2023 145BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lead Compensators: Design Steps4. Required t :
Select t such that, is at the new predicted zero-crossing frequency
5. Required Kc : Select Kc to give the required gain increase in the loop transfer
function. Remember, the compensator low-frequency gain is Kc/a
6. Implement:
1m
ss
KsG cc
1
11)(
May 1, 2023 146BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lead Design - Example 1
The plant
Specifications:
steady-state error <5% for a ramp input
Phase margin >45°
Steady-state error:
Draw the Bode plot of
21)(
ss
sGp
40 K
05.02/
1)(lim0
K
sEsesss
2
1)()(1
1)(2
sKss
sRsKG
sEp
240)(
ss
sGp
May 1, 2023 147BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lead Design - Example 1
Initial m = 18° , so choose m = 45° – 18° + 3° = 30° .
Then
The gain increase is½| |dB =4.8 dB .
The new predicted c = 8.5 rad/s and m(uncomp) = 13°
0.3
50.0)30sin(
Frequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
Mag
nitu
de (d
B)
Pha
se (d
eg)
Mag
nitu
de (d
B)
Phase-Lead Design ExamplePhase-Lead Design Example
-40
-20
0
20
10 0 10 1 10 2-180
-160
-140
-120
18°
27°
-135
13°
4.8dB
8.5
May 1, 2023 148BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lead Design - Example 1
Second iteration: choose qm = 45° – 13° + 3° = 35°
Then
The gain increase is½| a |dB =5.7 dB
The new predicted wc = 8.7 rad/s and fm(uncomp) = 12°
Calculate t
Calculate the gain Kc
The final compensator
OK
11
57.0)35sin(
7.3
0606..0077..3377..88
11
1481484040404011 cccc KKKK
amplifierssssssGGcc 0606..0011
2222..001177..33
11148148))((
May 1, 2023 149BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lead Design - Example 1Final Bode Plot
Frequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
Mag
nitu
de (d
B)
Pha
se (d
eg)
Mag
nitu
de (d
B)
Phase-Lead Compensation DesignPhase-Lead Compensation Design
-40
-20
0
20
40
10 --1 10 0 10 1 10 2-180
-160
-140
-120
-100
47°
8.7rad/s
compensateduncompensated
May 1, 2023 150BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Time (sec.)Time (sec.)
Am
plitu
deA
mpl
itude
Closed-Loop Step ResponseClosed-Loop Step Response
0 1 2 3 4 50
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
compensateduncompensated, K=40
uncompensated, K=1
Phase-Lead Design - Example 1Closed-Loop Step Response
May 1, 2023 151BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Compensator Characteristics
Time constant form
where ’ > 1
Pole-zero form
Note: | p | < | z |
passivenetwork
amp.gain where
ss
KsG cc '11
)(
))((
ppsszzssKKssGG cc
cc ''
11 ,,zz 11pp ''
May 1, 2023 152BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lag Compensators
Three parameters to be chosen in the designKc : the compensator amplifier gain, chosen to
meet the required low-frequency gain requirements
t : the time constant of the compensator, chosen to place the zero of the compensator, one decade below the new zero-crossing frequency
a : the required high frequency gain reduction
ssss
KKssGG cccc ''1111
))((
May 1, 2023 153BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lag Compensators: Design Steps
1. Loop gain requirement:Determine the loop gain required to meet the steady-
state error specifications
2. Bode plot:Plot the loop transfer function including the required
gain increase from step 1
3. Required zero-crossing frequency: (calculate a’)Find the frequency, wc where the specified phase
margin + 5 would be met, if it were the zero-crossing frequency
May 1, 2023 154BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lag Compensators: Design Steps
Determine the gain reduction required to make wc the zero-crossing frequency
Calculate a from the required gain reduction = | a |dB
4. Required t :Select t such that the zero, 1/t is one decade
below the new zero-crossing frequency1010
''11 cc
May 1, 2023 155BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Phase-Lag Compensators: Design Steps
5. Required Kc :Select Kc to give the required low-frequency
gain increase
6. Implement:
ssss
KKssGG cccc ''1111
))((
May 1, 2023 156BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Design - Example 1
The plant
Specifications: steady-state error <5% for a
ramp input. Phase margin >45°
The first part of the design is the same as the phase-lead design
Steady-state error:
Draw the Bode plot of
21)(
ss
sGp
4040 KK
0505..0022//
11))((limlim00
KK
ssEEsseessssss
22
11))(())((11
11))((22
ssKKssss
ssRRssKGKG
ssEEpp
224040
))((
ssss
ssGGpp
May 1, 2023 157BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Design - Example 1
Initial fm = 18° at wc = 6.5 rad/s
Frequency at which fm = 45° + 5° = 50° (where the phase plot = –130°)wc = 1.7 rad/s
The required high frequency gain reduction is | a |dB = 20dB, a = 10
Frequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
M
agni
tude
(dB
)P
hase
(deg
)
Mag
nitu
de (d
B)
Phase-Lag Design ExamplePhase-Lag Design Example
0
20
40
60
10--2 10--1 10 0 10 1
-160
-140
-120
-100
1.7 6.5
20dB
May 1, 2023 158BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Design - Example 1
Calculate t
Set the compensator zero at one decade below wc
The final compensator
17.010
7.1110
c
sssGc 8.581
88.5140)(
May 1, 2023 159BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Design - Example 1Final Bode Plot
Frequency (rad/sec)Frequency (rad/sec)
Pha
se (d
eg)
M
agni
tude
(dB
)P
hase
(deg
)
Mag
nitu
de (d
B)
Phase-Lag Design ExamplePhase-Lag Design Example
-20
0
20
40
60
10--2 10--1 10 0 10 1
-160
-140
-120
-100
-135
compensated
uncompensated
20dB
May 1, 2023 160BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Phase-Lag Design - Example 1Closed-Loop Step Response
Time (sec.)Time (sec.)
Am
plitu
deA
mpl
itude
Lead and Lag Design Step ResponseLead and Lag Design Step Response
0 1 2 3 4 50
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
uncompensated, K=40
uncompensated, K=1
lag compensatedlead compensated
May 1, 2023 161BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Bode Plot Design of Lead-Lag Compensators
Five parameters to be chosen in the design are,Kc : amplifier gain, low-frequency gain requirements
a : the desired phase advance, qm
t : lead time constant, to place qm at the design zero- crossing frequency, wc´
a : the required high frequency gain reduction
t : lag time constant to place the lag zero, one decade below wc´
May 1, 2023 162BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
1. Loop gain requirementDetermine the loop gain required to meet the steady-
state error specifications
2. Bode plotPlot the loop transfer function including the required
gain increase from step 1
3. Desired Phase Advance (calculate a)Select a desirable amount of phase advance qm and
calculate the corresponding a
May 1, 2023 163BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
4. Required Gain Reduction (calculate a )Determine the frequency, wc where the design
phase margin would be achieved, accounting for the phase advance, qm and the 5° of the tail of the lag
fm(uncomp) = fm(spec) – qm + 5°
Determine the gain reduction required to make wc the zero-crossing frequency and calculate a
| gain reduction |dB = |1/ a |dB + ½| a |dB
May 1, 2023 164BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
5. Required t and t :Select t to place the qm at wc ; wc =
Select t such that the zero, 1/t is one decade below wc ;
6. Required Kc :Select Kc to give the required low-frequency gain increase.
7. Implement:
1010''11 ''
cc
KKssGG cccc ))((ss
ss
1111
''''
''ssss
11
11
May 1, 2023 165BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
BIBILIOGRAPHYTEXT BOOKS1. I.J. Nagrath and M. Gopal, ‘Control Systems Engineering’,
New Age International Publishers, 2003.2. Benjamin C. Kuo, Automatic Control systems, Pearson
Education, New Delhi, 2003.
REFERENCE BOOKS1. K. Ogata, ‘Modern Control Engineering’, 4th edition, PHI,
New Delhi, 2002.2. Norman S. Nise, Control Systems Engineering, 4th Edition,
John Wiley, New Delhi, 2007.3. Samarajit Ghosh, Control systems, Pearson Education, New
Delhi, 20044. M. Gopal, ‘Control Systems, Principles and Design’, Tata
McGraw Hill, New Delhi, 2002.
May 1, 2023 166BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING
Thank You
May 1, 2023 167BE (AERO) - V SEM - EE2365 CONTROL ENGINEERING