Classical control 2

Classical
Control
Third Year
Second Semester
Dr. Amr A. Sharawi
T.A. Amira Gaber
T.A. Asmaa Mohamed

STABILITY IN THE FREQUENCY
DOMAIN

Absolute Stability
• The most important characteristic of the dynamic
behavior of a control system is absolute stability.
• A control system is in equilibrium if, in the absence
of any disturbance or input, the output stays in the
same state.
• A linear time-invariant control system is stable if the
output eventually comes back to its equilibrium
state when the system is ONLY subjected to an
initial condition.

Absolute Stability
• A linear time-invariant (LTI) control system is
critically stable if oscillations of the output continue
forever.
• It is unstable if the output diverges without bound
from its equilibrium state when the system is
subjected to an initial condition.
• Thus, a LTI is stable, if it is a bounded-input-
bounded-output (BIBO) system.

Relative Stability
• An important system behavior (other than absolute
stability) to which we must give careful
consideration includes relative stability.
• Since a physical control system involves energy
storage, the output of the system, when subjected
to an input, cannot follow the input immediately
but exhibits a transient response before a steady
state can be reached.
• The duration of that transient response in the
practical sense is a measure of relative stability.

Time &
Frequency
Domain Stability
Analysis
Linear Time Invariant
System
(Frequency domain)
Complex Frequency
Domain
(Routh stability
analysis)
Real Frequency
Domain
(Nyquist stability
analysis)
Time-varying &
Nonlinear System
(Time domain)
Time Domain
Analysis
(Lyapunov stability
analysis)

Routh’s Stability Criterion
• The Routh’s stability criterion tells us whether or
not there are unstable roots in a polynomial
equation without actually solving for them.
• This stability criterion applies to polynomials with
only a finite number of terms.
• When the criterion is applied to a control system,
information about absolute stability can be
obtained directly from the coefficients of the
characteristic equation.

Routh’s Stability Analysis Procedure
• Write the polynomial in s in the following form:
where the coefficients are real quantities.
We assume that an an > 0; that is, any zero
root has been removed.
If any of the coefficients are zero or negative in the
presence of at least one positive coefficient,
a root or roots exist that are imaginary or that have
positive real parts.
Therefore, in such a case, the system is not stable.

• If we are interested in only the absolute stability,
there is no need to follow the procedure further.
Note that all the coefficients must be positive.

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

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

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

This process is continued until the nth row has been
completed.
The complete array of coefficients is quasi-triangular.
In developing the array an entire row may be divided or
multiplied by a positive number in order to simplify the
subsequent numerical calculation without altering the
stability conclusion.

Implication of Routh’s Stability
Criterion
• Routh’s stability criterion states that the number of
roots of the characteristic equation with positive
real parts is equal to the number of changes in sign
of the coefficients of the first column of the array.
• The necessary and sufficient condition that all roots
of the characteristic equation lie in the left-half s-
plane is that all the coefficients of the characteristic
equation be positive and all terms in the first
column of the array have positive signs.

Example 5–11 – p. 214 – Ogata V
• Apply Routh’s stability criterion to the following
third-order polynomial:
• where all the coefficients are positive numbers.

Solution
• Routh’s array:
The condition that all roots have negative real parts is
given by

• Consider the following polynomial:
Solution
The first two rows can be obtained directly from the
given polynomial.
The second row can be divided by 2

Solution
• Having done so we get
𝑠3
1 2 0
We proceed with the remainder of the array as follows
The number of changes in sign of the coefficients in the
first column is 2.
This means that there are two roots with positive real
parts.

Special Cases
• If a first-column term in any row is zero, but the
remaining terms are not zero or there is no
remaining term, then the zero term is replaced by a
very small positive number ε and the rest of the
array is evaluated.

Example
• Consider the following equation:
The array of coefficients is
Now that the sign of the coefficient above the zero (ε) is
the same as that below it, it indicates that there are a pair
of imaginary roots (in that case at s = ±j).

Special Cases (Contd.)
• If, however, the sign of the coefficient above the
zero (ε) is opposite that below it, it indicates that
there are two sign changes of the coefficients in the
first column.
• In that case there are two coincident roots in the
right-half s-plane.

Relative Stability Analysis
• Routh’s stability criterion provides the answer to
• the question of absolute stability.
• This, in many practical cases, is not sufficient.
• We usually require information about the relative
stability of the system.
• A useful approach for examining relative stability is to
shift the s-plane axis and apply Routh’s stability
criterion.
• That is, we substitute
𝑠 = 𝑠−σ

Relative Stability Analysis
• into the characteristic equation of the system, write
the polynomial in terms of 𝑠 andapply Routh’s
stability criterion to the new polynomial in 𝑠
• The number of changes of sign in the first column of
the array developed for the polynomial in 𝑠 is equal
to the number of roots that are located to the right
of the vertical line s = –σ.
• Thus, this test reveals the number of roots that lie
to the right of the vertical line s = –σ.

Application of Routh’s Stability Criterion
to Control-System Analysis
• It is possible to determine the effects of changing
one or two parameters of a system by examining
the values that cause instability, particularly by
determining the stability range of one (or more)
parameter value.

Example
• Determine the range of K for stability for the system
shown.
Solution
The closed-loop transfer function is

Example (Contd.)
• The characteristic equation is
Routh’s array becomes
For stability, K must be positive, and all coefficients in
the first column must be positive. Therefore,

Introduction
• The basic characteristic of the transient response of a closed-loop
system is closely related to the location of the closed-loop poles.
• If the system has a variable loop gain, then the location of the
closed-loop poles depends on the value of the loop gain chosen.
• It is important, therefore, that the designer know how the closed-
loop poles move in the s plane as the loop gain is varied.
• From the design viewpoint, in some systems simple gain
adjustment may move the closed-loop poles to desired locations.
• If the gain adjustment alone does not yield a desired result,
addition of a compensator to the system will become necessary.
• The closed-loop poles are the roots of the characteristic equation.

The Concept
• The root-locus method is one in which the roots of the
characteristic equation are plotted for all values of a
system parameter, which will be assumed to be the
gain of the open-loop transfer function, unless
otherwise stated.
• The roots corresponding to a particular value of this
parameter can then be located on the resulting graph.
• It is desired that the designer have a good
understanding of the method for generating the root
loci of the closed-loop system, both by hand and by use
of a computer software program like MATLAB.

Angle and Magnitude Conditions
• For the negative feedback system shown the closed-loop transfer
function is
The characteristic equation for this closed-loop system is
or
Here we assume that G(s)H(s) is a rational polynomial in s.

• Since G(s)H(s) is a complex quantity, the characteristic can be split
into two equations by equating the angles and magnitudes of both
sides, respectively, to obtain

• The values of s that fulfill both the angle and
magnitude conditions are the roots of the
characteristic equation, or the closed-loop poles.
• A locus of the points in the complex plane satisfying
the angle condition alone is the root locus.
• The roots of the characteristic equation (the closed-
loop poles) corresponding to a given value of the
gain can be determined from the magnitude
condition.

The General Implication
• Generally, G(s)H(s) involves a gain parameter K, and the characteristic
equation may be written as
Then the root loci for the system are the loci of the closed-loop poles
as the gain K is varied from 0 to ∞.
The angles of the complex quantities originating from the open-loop
poles and open-loop zeros to a test point s are measured in the
counterclockwise direction.
In order that a test point lie on (i.e., be part of) the root locus, it
must satisfy the angle and magnitude conditions.

• For example, if G(s)H(s) is given by
where –p2 and –p3 are complex-conjugate poles, then the angle
of G(s)H(s) is
where 𝜙1 , θ1 , θ2 , θ3 , and θ4 are measured counterclockwise.
The magnitude of G(s)H(s) for this system is
where A1, A2, A3, A4, and B1 are the magnitudes of the complex
quantities s+p1, s+p2, s+p3, s+p4, and s+z1, respectively.

Diagram showing angle and magnitude measurements from open-loop poles and
open-loop zero to test point s.

Very Important
• Because the open-loop complex-conjugate poles and complex-
conjugate zeros, if any, are always located symmetrically about the
real axis, the root loci are always symmetrical w.r.t. this axis.
• Therefore, we only need to construct the upper half of the root loci
and draw the mirror image of the upper half in the lower-half s plane.
• Because graphical measurements of angles and magnitudes are
involved in the analysis, we find it necessary to use the same
divisions on the abscissa as on the ordinate axis when sketching the
root locus on graph paper.
• The number of individual root loci for a system is the same as the
number of open-loop poles.
• The starting points of the root loci (the points corresponding to K = 0)
are open-loop poles.

• Consider the negative feedback system given by:
1. Sketch the root-locus plot
2. Determine the value of K such that the damping ratio of the pair
of dominant complex-conjugate closed-loop poles is 0.5.
Solution
For the given system, the angle condition becomes

• The magnitude condition is
Computational Steps
1. Determine the root loci on the real axis.
To determine the root loci on the real axis, we select a test point, s.
a) If s > 0, then:
(angle condition not satisfied)
b) If 0 > s > -1, then:
Thus
(angle condition satisfied)

c) If -1 > s > -2, then:
Thus
d) If -2 > s > -∞, then:
= 180°
Thus
-540° (angle condition satisfied)
Thus, the portion of the real axis that lies on the root locus is given by:
 ]0,1[]2,[ S

2. Determine the asymptotes of the root loci.
The asymptotes of the root loci as s approaches infinity must lie on the real axis.
This is because the root loci are symmetrical about the real axis.
They can be determined as follows:
a) Number of asymptotes = n – m,
where n = number of poles & m = number of zeros.
In our case n = 3 & m = 0, so there are three asymptotes.
b) The angles of the asymptotes are given by:
 ,
12180
mn
k
k


 k = 0, 1, 2, . . ., n - m
In our case the angles are given by θo = 60°, θ1 = 180°, θ2 = 300°,
And the asymptotes with the three negative angles coincide with the other
three.

c) The point where they intersect the real axis is given by:
mn
zp
n
i
m
j
ji
A



 
 1 1

In our case
  1
3
0210


A
The three asymptotes

3. Determine the breakaway point.
To plot root loci accurately, we must find the breakaway point, where the root-locus
branches originating from the poles break away from the real axis and move into the
complex plane, as K is increased.
The breakaway point corresponds to a point in the s plane where multiple roots of
the characteristic equation occur.
To find the breakaway point let us write the characteristic equation as
where A(s) and B(s) do not contain K.
f(s) = 0 has multiple roots at points where
Proof
Suppose that f(s) has multiple roots of order r, where r ≥ 2.
Then f(s) may be written as

     
     
     121
31
2
1
1
...
......
...
)(





n
r
n
r
n
r
ssssss
ssssss
ssssssr
ds
sdf
0
)(
1

ssds
sdf
Now
where

The particular value of K that will yield multiple roots of the characteristic equation
is obtained from
Thus
or
0)(')()(')()(')()(')(  sKAsAsAsBsBsAsAsB
0)()(  sKAsB
Thus

Next we have
= 0
This confirms that the breakaway points can be simply determined from the roots of
If a point at which dK/ds = 0 is on a root locus, it is an actual breakaway or break-
in point. Otherwise, it is not.
In our case the characteristic equation G(s) + 1 = 0 is given by
or
so

The two roots are given by
Since the breakaway point must lie on a root locus between 0 and –1, it is clear
that s = –0.4226 corresponds to the actual breakaway point. Point s = –1.5774 is not
on the root locus. Hence, this point is not an actual breakaway or break-in point.
In fact
4. Determine the points where the root loci cross the imaginary axis.
let s = jω in the characteristic equation, equate both the real part and the imaginary
part to zero, and then solve for ω and K. For the present system, the characteristic
equation, with s = jω, is
or

Equating both the real and imaginary parts of this last equation to zero, respectively,
we obtain
from which
Thus, root loci cross the imaginary axis at 𝜔 = ± 2 and the value of K at the
crossing points is 6.
Also, a root-locus branch on the real axis touches the imaginary axis at ω = 0.
The value of K is zero at this point.

5. Sketch the root loci.
Asymptote
Asymptote

6. Determine the point on the root locus where the damping ration of the
dominant poles is 0.5.
Closed-loop poles with ζ = 0.5 lie on lines passing through the origin and making
angles with the negative real axis given by ±cos-1 ζ = ±cos-1 0.5 = ±60° with the
negative real axis.
From the previous slide such closed loop poles are approximately given by
s = -0.33 ± j 0.58 complex rad/s
The value of K that yields such poles is found from the magnitude condition as
follows:
𝐾 = 𝑠 𝑠 + 1 𝑠 + 2 𝑠=−0.33+𝑗0.58 ≈ 1.05
The basic construction can be found on the next slide.

• Consider the negative feedback system shown
Solution
For this system
K ≥ 0
G(s) has a pair of complex-conjugate poles at

Computational Steps
1. Determine the root loci on the real axis.
For any test point s on the real axis, the sum of the angular contributions of the
complex-conjugate poles is 360°, as shown
Thus the net effect of the complex-conjugate poles is zero on the real axis.

The location of the root locus on the real axis is determined from the open-loop
zero on the negative real axis.
a) If s > -2, then:
b) If s < -2, then:
= 180° (angle condition satisfied)
Thus, the portion of the real axis that lies on the root locus is given by:
𝑆 = −∞, −2
2. Determine the asymptotes of the root loci.
Since there are two open-loop poles and one zero, there is one asymptote, which
coincides with the negative real axis.

3. Determine the break-in point.
A break-in point exists on the real axis where a pair of root-locus branches
coalesces as K is increased. This is true in this example, because as K increases, the
root loci tend to approach rather than move away from the real axis (which is true
in case of a break-away point).
The break-in point can be found as follows:
which gives

Notice that point s =–3.7320 is on the root locus. Hence this point is an actual break-
in point.
Also at point s = –3.7320 the corresponding gain value is K = 5.4641.
Since point s = –0.2680 is not on the root locus, it cannot be a break-in point.
4. Determine the points where the root loci cross the imaginary axis.
It can be immediately seen that K = 0 at the complex conjugate poles.
Since both poles lie in the negative half plane, then the root locus cannot cross the
Imaginary axis.
5. Determine the angle of departure from the complex-conjugate open-loop poles.
It is clear that the root locus begins its trip here at the complex-conjugate, where
K = 0.
This requires the determination of the angle of departure from these poles, whose
Knowledge is important.

Referring to the figure below, we choose a test point and move it in the very vicinity
of the complex open-loop pole at s = –p1 , we find the following:
Angle of departure = θ1.
2tan 1   7.541
 902

Angle condition:
 12180211  k
 180907.54 1
 7.144907.541801
or
 3.215180907.541
Which is the same angle.
The value of the gain K at any point on root locus can be found by applying the
magnitude condition or by use of MATLAB.

Analytic Approach
We shall prove that in this system the root locus in the complex plane is a part of a
circle. Such a circular root locus will not occur in most systems.
Circular root loci may occur in systems that involve two poles and one zero, two
poles and two zeros, or one pole and two zeros.
Even in such systems, whether circular root loci occur depends on the locations of
the poles and zeros involved.
First we need to derive the equation for the root locus.
For the present system, the angle condition is

which can be written as
or
Taking tangents of both sides of this last equation using the relationship
we obtain

Thus
which can be simplified to
or
This last equation is equivalent to
Notice that the first equation, ω = 0, is the equation for the real axis. The real axis
from s = –2 to s = –∞ corresponds to a root locus for K ≥ 0.

The second equation for the root locus is an equation of a circle with a center at
σ = –2, ω = 0 and a radius equal to
That part of the circle to the left of the complex-conjugate poles corresponds to a
root locus for K ≥ 0. The remaining part of the circle corresponds to a root locus when
K is negative.
3

Very Important
• Where applicable, the analytic method provides a true
mathematical model for the root loci. MATLAB does not
offer this facility.
• Easily interpretable equations for the root locus can be
derived for simple systems only.
• For complicated systems having many poles and zeros,
any attempt to derive equations for the root loci is
discouraged.
• Such derived equations are very complicated and their
configuration in the complex plane is difficult to
visualize.

Introduction
• In preceding chapters, the response and
performance of a system have been described in
terms of the complex frequency variable s and the
location of the poles and zeros on the s-plane.
• We examined the use of test signals such as an
impulse and a step signal to analyze the
performance and consider the design of the system.

Introduction
• A very practical and important alternative approach to
the analysis and design of a system is the so-called
frequency response method.
• In the context of the remaining parts of the Classical
Control course, we define the frequency response of a
system as the steady-state response of the system to a
sinusoidal input signal.
• We will see that the response of a linear constant
coefficient system to a sinusoidal input signal is an
output sinusoidal signal at the same frequency as the
input.
• However, the magnitude and phase of the output signal
differ from those of the input sinusoidal signal, and the
amount of difference is a function of the input
frequency.

Theorem
The output of a linear time-invariant system whose input is a
steady-state sinusoidal signal differs from the input waveform only
in amplitude and phase angle.
Proof:
Consider the system with
Thus
and
where -pi are assumed to be distinct poles. Then, in partial fraction
form, we have
22
1
)(






 
 s
s
ps
k
sY
n
i i
i

Theorem
Taking the inverse Laplace transform yields








 

 22
1
1
)(


s
s
ekty tp
n
i
i
i
L
Where 𝜶 and β are constants which are problem dependent.
If the system is stable, then all pi have positive real parts and
But at steady state
























 
 22
1
22
1
22
1
)(lim








ss
s
s
s
tyy
t
ss LLL
     


  tKtt sinsincos
 


 TAK 






2
2
 


 tan
 


 tan
where

Advantages of Frequency Response
Methods
• The ready availability of sinusoid test signals for various
ranges of frequencies and amplitudes.
• The unknown transfer function of a system can be
deduced from the experimentally determined
frequency of a system.
• The transfer function describing the sinusoidal steady-
state behavior of a system can be obtained by replacing
s with jω in the system transfer function T(s).
• The magnitude and phase angle of T(jω) are readily
represented by graphical plots that provide significant
insight into the analysis and design of control systems.

Disadvantages of Frequency Response
Methods
• The basic disadvantage of the frequency response
method for analysis and design is the indirect link
between the frequency and the time domain.

Laplace Transform Pair
where the complex variable s = σ + jω.

Fourier Transform Pair
The Fourier transform exists for f(t) when

Relationship between Fourier and Laplace
Transforms
• The Fourier and Laplace transforms are closely related,
as can be readily seen.
• When the function f(t) is defined only for t > 0, as is
often the case, the lower limits on the integrals for
both transforms are the same.
• Then we note that the two equations differ only in the
complex variable.
• Thus, if the Laplace transform of a function f1(t) is
known to be F1(s), we can obtain the Fourier transform
of the same time function by setting s = jω in F1(s).

Preference of Using the Fourier Transform
• Since the Fourier and Laplace transforms are so
closely related, why can't we always use the Laplace
transform? Why use the Fourier transform at all?
– The Laplace transform permits us to investigate the s-
plane location of the poles and zeros of a transfer
function T(s).
– The Fourier transform allows us to consider the transfer
function T(ω) and to concern ourselves with the
amplitude and phase characteristics of the system.
– This ability to investigate and represent the character of
a system by amplitude, phase equations, and curves is
an advantage for the analysis and design of control
systems.

FREQUENCY RESPONSE PLOTS
• The transfer function of a system G(s) can be described in
the frequency domain by the relation
This is called the polar plot representation of the
frequency response.

FREQUENCY RESPONSE PLOTS
• Alternatively, the transfer function can be represented by a
magnitude |G(ω)| and a phase 𝜙(ω) as
This representation produces a set of logarithmic plots,
often called Bode plots, used to simplify the determination
of the graphical illustration of the frequency response.

Example 8–1 – p. 497 – Dorf XI
Polar Plot of the Frequency Response of an RC Filter
The transfer function of the RC filter shown is
and the sinusoidal steady-state transfer function is

• The polar plot is obtained from the relation
 
 
 
 2
1
1
2
1 /1
/
,
/1
1








 XR
Let
1

 
    22
1
,
1
1








 XR
R
X
R
X
- 

22
2
2
2
1
1
XR
R
R
X
R




22
1
XR
R


  02222
 XRRRXR
0
4
1
2
1 2
2






 XR
4
1
2
1 2
2






 XR
Thus, the polar plot is a circle centered at (0.5, 0) and of
radius = 0.5.

• At ω = 0, we have R(ω) = 1 and X(ω) = 0.
• At ω = ∞, we have R(ω) = 0 and X(ω) = 0.
• When ω = ω1 the real and imaginary parts are equal in magnitude,
and the angle 𝜙(ω) = -45°.
• We conclude that the polar plot is more specifically a semi-circle
centered at (0.5, 0) and of radius = 0.5, but occupying the lower half
of the complex plane.

Classical control 2

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Classical control 2