1. sports
2. wrestling

om
y.c
ud
st
al4
ric
ct
Ele
Lecture 2
Dr. Jamal Tariq Mian
[email protected]
om
y.c
ud
st
al4
ric
ct
Ele
Lecture Contents
y Laplace variable s and the s – plane
y Poles and Zeros in Laplace Transform
y Root Location(s) on s‐plane and
Root Location(s) on s plane and corresponding time function.
corresponding time function
y Concept of Bounded Signals
y Stability
y
y Coefficient tests For stability
y Routh – Hurwitz Criterion (Special Cases)
y Zero only in the 1st column
y Entire row is zero
y Stability Design via Routh –
Stability Design via Routh Hurwitz
y Routh – Hurwitz Criterion – Examples
y Assignment Problems
g
ric
ct
Ele
om
y.c
ud
st
al4
Laplace variable s and the s‐plane
y The Laplace variable s is a complex variable defined in the formula of Laplace transform. It is given by:
s = σ + jω
y where is the real part of s, which is also denoted as h σ
h
l
f
h h
l d
d
Re(s).
y σ is related to the boundedness of a signal or the i l t d t th b
d d
f i
l th stability of the system. y variable ω is the imaginary part of s, which is also is the imaginary part of s which is also denoted as Im(s).
is related to the frequency content of the signal y ω is related to the frequency content of the signal. ric
ct
Ele
om
y.c
ud
st
al4
Laplace variable s and the s‐plane
y It is possible to show range of variation of s or the domain of s schematically as a plane. y The horizontal axis of this plane is called as the real axis, The horizontal axis of this plane is called as the real axis since it is defined by:
s = σ + j0
y The vertical axis is known as imaginary axis and is represented by: p
y
s = 0 + jω
y The origin
g is represented by complex zero:
p
y
p
s = 0 + j0
ric
ct
Ele
om
y.c
ud
st
al4
Laplace variable s and the s‐plane
y The region to the left of Th i h l f f the imaginary axis for jω
−∞ < σ < 0 is known as i k
Left Half Plane (LHP). y The region to the right of h
h
h f
jω
the imaginary axis for
0 < σ < ∞ is known as Right Half Plane (RHP).
y The whole complex plane is called as s‐plane.
s − plane s = σ + jω
ric
ct
Ele
om
y.c
ud
st
al4
P l
Poles and Zeros in Laplace Transform
dZ
i L l
T
f
y Very often Laplace transform works out as the ratio of two constant coefficient polynomials, for example:
s+6
G (s) = 2
s + 4s + 3
y This can be written in general form as:
bm sm + bm−1sm−1 + ...... + b1s + b0
X ( s) =
an sn + an−1sn−1 + ...... + a1s + a0
y Or in the format:
num(s)
X ( s) =
den(s)
n≥m
ric
ct
Ele
om
y.c
ud
st
al4
P l
Poles and Zeros in Laplace Transform
dZ
i L l
T
f
y Transform Zeros:
y The Laplace transform zeros are the values of s for which the transform is zero. They are found by setting the numerator polynomial to zero and solving the resulting equation: num(s) = bm s m + bm −1 s m −1 + ...... + b1 s + b0 = 0
y Transform Poles:
y The Laplace transform poles are the values of s for which the transform is infinite. The poles are found by setting the denominator polynomial to zero and solving the resulting equation: den(s) = a n s + an−1s
n
n −1
+ ...... + a1s + a0 = 0
om
y.c
ud
st
al4
ric
ct
Ele
Poles and Zeros
y Example:
s+6
G (s) = 2
s + 4s + 3
We can write this in the form:
num(s)
s+6
= 2
G (s) =
den(s) s + 4 s + 3
Transfer zeros: num(s) = s + 6 = 0 or s = ‐6
Transfer poles: den(s) = s2+4s+3 = 0 or s1 = ‐1 & s2 = ‐3
om
y.c
ud
st
al4
ric
ct
Ele
P l
Poles and Zeros
dZ
y Example:
s +1
G (s) = 2
s +s−6
Poles:
(
)
d(s) = s2 +s−6 =( s−2)( s+3) =0
Poles of the system are at s = 2 and s = ‐3, hence poles are p1 = 2 and
p2 = ‐3
3
Zeroes:
n ( s ) = ( s + 1) = 0
The solution is s = ‐1. Hence the zero of the system is z1 = ‐1
om
y.c
ud
st
al4
ric
ct
Ele
P l
Poles and Zeros
dZ
y Example:
G (s) =
(s − 4 )(
s+3
s 2 + 12 s + 52
)
Draw pole‐zero map of G(s)
Poles:
(
)
d ( s ) = ( s − 4 ) s 2 + 12 s + 52 = 0
Solving: p 1 = 4 ; p 2 , p 3 = − 6 ± 4 j
Zeros: n ( s ) = ( s + 3 ) = 0 ⇒ s = -3
Therefore, z1 = ‐3 is a zero of G(s)
om
y.c
ud
st
al4
ric
ct
Ele
Poles and Zeros
y Multiple Poles
y When the system G(s) has more than one pole or one zero at the same coordinates on the s plane , we say that zero at the same coordinates on the s‐plane
we say that G(s) has multiple poles or multiple zeros. For example, consider the following transfer function:
g
1
G (s) = 2
s
Here denominator polynomial, d (s) = s 2 = 0
This has two roots both at s = 0, hence p
s as t o oots bot at s 0, e ce p1 = 0
0
and p2 = 0, since it is of degree two. Thus the
transform has multiple poles at the origin of
s‐plane This is indicated by double cross in s‐plane. This is indicated by double cross in the figure. ct
Ele
om
y.c
ud
st
al4
ric
Root Location(s) on s‐plane and R
t L ti ( )
l
d
corresponding time function.
corresponding time function.
y The type of time function depends upon the terms root location in the s‐plane and upon whether or not the root is location in the s
plane and upon whether or not the root is repeated. y Shown in next slides are representative time functions associated with various transform denominator root locations.
ct
Ele
om
y.c
ud
st
al4
ric
Root Location(s) on s‐plane and R
t L ti ( )
l
d
corresponding time function.
corresponding
time function.
y Table
Root location(s) on the
s‐plane
Corresponding time function
ct
Ele
om
y.c
ud
st
al4
ric
Root Location(s) on s‐plane and R
t L ti ( )
l
d
corresponding time function.
corresponding
time function.
y Table
Root location(s) on the
s‐plane
Corresponding time function
ct
Ele
om
y.c
ud
st
al4
ric
Root Location(s) on s‐plane and R
t L ti ( )
l
d
corresponding time function.
corresponding
time function.
y Table
Root location(s) on the
s‐plane
Corresponding time function
ct
Ele
om
y.c
ud
st
al4
ric
Root Location(s) on s‐plane and R
t L ti ( )
l
d
corresponding time function.
corresponding
time function.
y Table
Root location(s) on the
s‐plane
Corresponding time function
ct
Ele
om
y.c
ud
st
al4
ric
Root Location(s) on s‐plane and R
t L ti ( )
l
d
corresponding time function.
corresponding
time function.
y Table
Root location(s) on the
s‐plane
Corresponding time function
ric
ct
Ele
om
y.c
ud
st
al4
What are Bounded Signals ?
y A time domain signal, x(t) is assessed by the behavior of its A time domain signal x(t) is assessed by the behavior of its magnitude over an infinite time interval. As time tends to infinity, the absolute value of the signal magnitude can either:
ih
y continuously decrease and/or increase (or stay constant) but ti
l d
d/ i
( t t t) b t remain within a bounded range.
y continuously increase to very large values without any y
y g
y
bounds.
y Figures on the next slides show examples of bounded Fi
th t lid h l f b
d d exponential signals, bounded sinusoidal signals and unbounded signals.
g
ric
ct
Ele
om
y.c
ud
st
al4
Examples of Bounded Signals
Bounded Exponential Signal
Bounded Sinusoidal Signal
We can see that the magnitude of an exponential function eat, with a < 0, will decrease to zero as time tends to infinity. The magnitude of unit step function is finite since its value is 1 even when the time tends to infinity function is finite since its value is 1, even when the time tends to infinity. These types of signals are called as bounded signals. ric
ct
Ele
Unbounded Exponential
Signal
om
y.c
ud
st
al4
Examples of Unbounded Signals
Unbounded Parabolic and Sinusoidal Signal
d l
l
Above are the examples of unbounded signals. If the exponent in the exponential signal eat is positive, with a > 0, the signal will increase to infinity as time tends to infinity infinity as time tends to infinity. These types of signals are called as unbounded signals.
ct
Ele
om
y.c
ud
st
al4
ric
jω
Poles in LHP, RHP and on axis
y Signals whose transforms have all the poles in the LHP
are bounded.
y Signals
Si
l whose transforms have any poles
h t
f
h l in the RHP i th RHP are unbounded.
Poles on axis:
jω
y Signals whose transforms have poles in the LHP and jω
no multiple poles on the axis are bounded, otherwise they are unbounded.
om
y.c
ud
st
al4
ric
ct
Ele
S bili
Stability
y What is stability?
y There are many definitions of stability, depending upon the kind of the system or point of view Here we limit the kind of the system or point of view. Here we limit ourselves to linear, time‐invariant systems. y The total response of the system to the applied input is a sum of forced and natural response. c(t ) = c forced (t ) + cnatutral (t )
y Using these concepts, presented on next slide is the U i th t t d t lid i th definition of stability, instability, and marginal stability stability. om
y.c
ud
st
al4
ric
ct
Ele
Stability
y A linear, time‐invariant system is stable
A linear time invariant system is stable if the natural response approaches zero as time approaches infinity.
y A linear, time‐invariant system is unstable if the natural response grows without bound as time t l ith t b
d ti approaches infinity. y A linear, time‐invariant system is marginally stable if the natural response neither decays nor grows but h l i h d
b remains constant as time approaches infinity. om
y.c
ud
st
al4
ric
ct
Ele
Stability
y Thus, the definition of the stability implies that only the forced Thus the definition of the stability implies that only the forced response remains as natural response approaches zero.
y These definitions rely on the definition of the natural response. y
p
However, it may be difficult to separate natural response from the forced response. y If the input is bounded
If h i
i b
d d and the total response is not approaching d h l i hi infinity as time approaches infinity, then the natural response is obviously not approaching infinity. y If the input is unbounded, unbounded total response will be there and thus can not arrive at any conclusion about the stability of the system; i.e. can not tell if the total response is unbounded because the forced response is unbounded or because natural response is unbounded because natural response is unbounded. om
y.c
ud
st
al4
ric
ct
Ele
St bilit
Stability y Thus, the alternate definition of stability, one that regards Thus the alternate definition of stability one that regards the total response and implies the first definition based upon natural response, is this:
p
p
,
¾ A system is stable
y
if every
y bounded input yields a bounded p y
output. We call this statement the bounded‐input, bounded‐output (BIBO) definition of stability. om
y.c
ud
st
al4
ric
ct
Ele
St bilit
Stability y Similarly, the alternate definition for instability
Similarly the alternate definition for instability based on the total response rather than the natural response will be as:
¾ A system is unstable
y
if any
y bounded input yields an p y
unbounded output.
These definitions also clarify the definition of marginal stability, which really means that the system is stable for some bounded inputs and unstable for others.
some bounded inputs and unstable for others
om
y.c
ud
st
al4
ric
ct
Ele
S bili
Stability
y How do we determine
How do we determine, if a system is stable?
if a system is stable?
y Poles in the left half plane (LHP) yield pure exponential decay or damped sinusoidal natural responses These decay or damped sinusoidal natural responses. These natural responses decay to zero as time approaches infinity. y
y If the closed‐loop system poles are in the left half of the s‐plane (LHP) and hence have a negative real part, the system is stable. That is, stable systems have closed‐loop transfer functions with poles only in the left half‐plane.
om
y.c
ud
st
al4
ric
ct
Ele
Stability
y How do we determine
How do we determine, if the system is unstable?
if the system is unstable?
y Poles in the right half plane (RHP) yield either pure exponentially increasing or exponentially increasing sinusoidal natural responses. These natural responses approach infinity as time approaches infinity . Thus, if pp
y
pp
y
,
the closed‐loop system poles are in the right half of the s‐plane and hence have a positive real part, the system is unstable. bl Thus, unstable systems have closed‐loop transfer f
functions with at least one pole in the right half‐plane ti
ith t l t l i th i ht h lf l
and/or poles of multiplicity greater than one on the imaginary axis. om
y.c
ud
st
al4
ric
ct
Ele
Stability
y How do we determine, if the system is marginally How do we determine if the system is marginally stable?
y A system that has imaginary axis poles of multiplicity 1 yields pure sinusoidal oscillations as a natural response. ese espo ses e t e c ease o dec ease
These responses neither increase nor decrease in amplitude. Thus, marginally stable systems have closed‐loop transfer functions with only imaginary axis poles of multiplicity 1 and poles in the left half plane. ric
ct
Ele
om
y.c
ud
st
al4
Coefficient Tests For Stability
y For 1st and 2nd order systems, stability is determined by order systems stability is determined by inspection of the characteristic polynomial.
y A 1st or 2nd order polynomial has all roots in the left half plane (LHP) of the complex plane (s‐plane) if and only if all polynomial coefficients have the same
p
y
algebraic sign.
g
g
y For example:
3s2 + s + 10
is a characteristic equation of the stable system, while
3s2 + s − 10
represents an unstable system.
ric
ct
Ele
om
y.c
ud
st
al4
Coefficient Tests For Stability
y For higher order polynomials, representing higher‐order For higher order polynomials representing higher order systems, the algebraic signs of polynomial coefficients may or may not yield information as to stability.
y
y
y
y A polynomial with all roots in the left half plane (LHP) has f
factors of the form:
f
f
(s + a)
a>0
( s 2 + bs
b + c)
b > 0 and
d c>0
real axis root in the LHP
and
two LHP roots, perhaps
l j
complex conjugate
ric
ct
Ele
om
y.c
ud
st
al4
Coefficient Tests For Stability
y When multiplied out, such a higher order polynomial must When multiplied out such a higher order polynomial must have all the coefficients of the same algebraic sign, all p
positive or all negative. g
y No coefficient may be zero (missing) in a system with LHP roots because there are no minus signs involved and g
no way for a coefficient to be cancelled.
y If imaginary axis roots exist in the polynomial, factors of the following forms maybe present, in addition to the others: (s)
(s 2 + a)
roots at the origin
a>0
complex conjugate roots on the imaginary axis.
ric
ct
Ele
om
y.c
ud
st
al4
Coefficient Tests For Stability
y With such roots present, all polynomial With such roots present all pol nomial coefficients must be of the same algebraic sign , but some coefficients maybe zero.
but some coefficients maybe zero
ric
ct
Ele
om
y.c
ud
st
al4
Coefficient Tests For Stability
y Right half plane (RHP) root involve factors of the form:
(s − a)
a>0
real axis root in the RHP
ea a s oo
e
a > 0 and b > 0
two RHP roots, perhaps
complex conjugate
and
( s 2 − as + b )
y The presence of such factors may or may not cause differing algebraic signs of the coefficient and (by cancellation) zero
ffi i
coefficients.
ric
ct
Ele
om
y.c
ud
st
al4
Coefficient Tests For Stability
S
Summary
Properties of the Polynomial Coefficients
Conclusion about Roots from the Coefficient Test
All of the same algebraic sign.
none zero.
No direct information about the roots; f h
further testing (i.e. Routh
(
h Hurwitz))
necessary.
Differing
g algebraic signs, perhaps some g
g p
p
At least one RHP root; possible p
zero coefficients also
imaginary axis roots also.
All of the same algebraic sign and one or more coefficients zero.
Imaginary axis or RHP roots or both.
ric
ct
Ele
om
y.c
ud
st
al4
C ffi i
Coefficient Tests For Stability
T
F S bili
y Example: Consider the polynomial
7 s6 + 5s 4 − 3s3 − 2 s 2 + s + 10
It definitely has one or more RHP roots, indicated by the differing algebraic signs of the coefficients differing algebraic signs of the coefficients. y Example: Consider the polynomial
8s5 + 6 s 4 + 3s3 + 2 s 2 + 7 s + 10
Examination of coefficients signs yields no information. Further testing is necessary Further testing is necessary. ric
ct
Ele
om
y.c
ud
st
al4
C ffi i
Coefficient Tests For Stability
T
F S bili
y Example: Consider the polynomial
s6 + 3s5 + 2 s 4 + 8s 2 + 3s + 17
It has imaginary axis roots or RHP roots or both, indicated by missing s3. ric
ct
Ele
om
y.c
ud
st
al4
S bili
Stability ‐
E
Example
l
y Example (Stable System):
ric
ct
Ele
om
y.c
ud
st
al4
S bili
Stability ‐
E
Example
l
y Example (Unstable System):
ric
ct
Ele
om
y.c
ud
y Stability of a cone
st
al4
Stability ‐ Example
ric
ct
Ele
om
y Stability in s‐plane
Stability in s plane
y.c
ud
st
al4
Stability in s ‐ plane
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y It is a method that yields stability information without the need to solve for the closed‐loop system poles. y Using this method, we can tell how many closed‐loop U i thi th d t ll h l dl
system poles are in the left half plane (LHP), in the jω
right half plane (RHP) and on the axis
right half plane (RHP), and on the axis.
y This method is called as Routh‐Hurwitz criterion for stability The method requires two steps:
stability. The method requires two steps:
Generate data table called a Routh table.
2. Interpret Routh table to determine how many closed‐
table to determine how many closed
loop system poles are in the left half plane (LHP), the jω
right half plane (RHP), and on axis.
1.
ric
ct
Ele
om
y.c
ud
st
al4
Routh‐Hurwitz Criterion
y Generating a Basic Routh Table
y Consider the following transfer function:
R(s)
( )
N(s)
a4s4 +a3s3 +a2s2 +as
1 +a0
C(s)
( )
y Focus attention on denominator. Create Routh table as shown in next slide. h
i t lid ct
Ele
om
y.c
ud
st
al4
ric
R th H
Routh‐Hurwitz Criterion
it C it i
y Initial layout for Routh table:
ric
ct
Ele
om
y.c
ud
st
al4
Routh‐Hurwitz Criterion
y Steps to create Routh Table:
y Begin by labeling the rows with powers of s from the highest power of denominator of the closed‐loop transfer function to s0. y Next start with the coefficient of the highest power of s
N i h h ffi i f h hi h f in the i h denominator and list, horizontally in the first row, every other coefficient.
y In the 2
h ndd row list horizontally, starting with the next highest power l h
ll
h h
h h
of s, every coefficient that was skipped in the first row.
y The remaining entries are filled in as follows. Each entry is a negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row. The left hand column of the determinant is always the first column of the previous two rows, and the right hand column is the elements of h i d h i h h d l
i h l
f the column above and to the right.
y The table is completed when all of the rows are completed down to s0. ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Completed Routh table:
ric
ct
Ele
om
y.c
ud
st
al4
Routh‐Hurwitz Criterion
y Interpretation of basic Routh table
y Simply stated, the Routh‐Hurwitz criterion declares that Si l d h R h H
i i i d l
h the number of roots of the polynomial that are in the right half‐plane is equal to the number of sign changes in i ht h lf l
i l t th b f i h
i the first column. y If the closed‐loop transfer function has all poles in the left half of the s plane, the system is stable. left half of the s‐plane
the system is stable Thus a system is stable, if there is no sign change in the first column of Routh table.
f
f
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Routh Table:
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Stability Criterion:
y The roots of the equation are all in the left half plane of Th f h i ll i h l f h lf l
f the s – plane if all the elements of the first column of the Routh s tabulation are of same sign.
Routh’s tabulation are of same sign
y The number of changes of signs in the elements of the first column equals the number of roots with positive real parts or in the right half s –
p
g
f
plane. p
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Example
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Example
Solving for the roots of given
equation, we have four roots at:
s = −1.005 ± j0.933
s = 0.755 ± j1.444
Two last roots are in RHP, which
Two last roots are in RHP
which
cause system to be unstable.
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Example
d ( s ) = s 4 + 2 s 3 + 3s 2 + 4 s + 1
No sign change
System is stable
System is stable.
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Example: Make the Routh table for the system shown in Make the Routh table for the s stem sho n in the figure below:
Solution:
The first step is to find the equivalent closed‐loop system because we want
to test the denominator of this function for pole location Using the feedback
to test the denominator of this function, for pole location. Using the feedback
formula, obtain equivalent system as shown above. ric
ct
Ele
om
y.c
ud
st
al4
Routh‐Hurwitz Criterion
y Build the Routh table for the denominator:
d ( s ) = s 3 + 1 0 s 2 + 3 1s + 1 0 3 0
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Conclusion from Table:
y
y
y
y
Table has two sign changes in the first column. Table has two sign changes in the first column
The first sign change occurs from 1 in the s2 row to ‐72 in the s1
row.
The second sign change occurs from ‐72 in the s1 row to 103 in the s0 row. Th f
Therefore, the system is unstable
h i bl because two poles exist in b
l i i the right half plane. om
y.c
ud
st
al4
ric
ct
Ele
Remember this
y 2ndd order system
d d (s) = as + bs + c
2
System is stable, if all the coefficients have
the same sign
om
y.c
ud
System is stable
2
st
d (s ) = as + bs + cs + d
3
al4
y 3rdd order system
d ric
ct
Ele
Remember this
ric
ct
Ele
om
y.c
ud
st
al4
R hH
Routh‐Hurwitz Criterion
i Ci i
y Example: Make a Routh table and tell how many roots of the following polynomial are in the right half‐plane and in the left half‐plane. p
P ( s ) = 3s + 9 s + 6 s + 4 s + 7 s + 8 s + 2 s + 6
7
6
5
y Solution:
Try yourself …..
Try yourself 4
3
2
ric
ct
Ele
om
y.c
ud
st
al4
Routh‐Hurwitz Criterion: Special Cases
y Two special cases can occur:
y The Routh table sometimes will have a zero
Th R h bl i
ill h only in the l i h first column of a row. y The Routh table will sometimes have an entire row that consists of zero.
consists of zero