Lec 2 stability

Control Systems
LECT. 2 STABILITY
BEHZAD FARZANEGAN
PROVIDED BY: BF(B.FARZANEGAN@AUT.AC.IR) 13/1/2020

Stability of Systems
Three requirements enter into the design of a control system: transient
response, stability, and steady-state errors.
 Stability is the most important factor in a feedback system. If a system
is unstable, transient response and steady-state errors are moot points.
Stability definitions can be classified as
◦ Stability of systems without any input
◦ Stability of systems with external inputs
e
Gp(s)
+ Gc(s)
r y


Definition: A system is said to be BIBO stable if for any bounded
input and any arbitrary initial condition the output is bounded.
BIBO Stable
System
u y

(No Reference Input)
Definition: the origin (or any other equilibrium point) of a system is said
to be stable in the sense of Lyapunov if all solutions of the system that
start near the origin, will stay near the origin
Definition: the origin (or any other equilibrium point) of a system is said
to be asymptotically stable if all solutions that start near the origin will
converge to the origin
Definition: the origin of a system is said to be globally asymptotically
stable, if it is stable and all solutions converge to the origin as t
Starts within  of origin and stays within  of the origin


Could have multiple equilibrium points!Linear System can only have one equilibrium point!


x

Internal Stability
Why do we need to discuss internal stability?
Consider the following system
As can be seen, there is no 𝑎, 𝑏, 𝑐 > 0 such that the closed loop system is BIBO stable (sign of coefficients in
denominator polynomial) unless we have a pole zero cancellation i.e. 𝑐 = 𝑎
 
 
( ) c
c
c
N s
G s
D s

 
 
( ) p
p
p
N s
G s
D s

( )
1
p c
cl
p c
G G
G s
G G


p c
p c p c
N N
D D N N


( )p
s a
G s
s b



1
( )cG s
s c


Let , , 0a b c 
    
( )cl
s a
G s
s b s c s a


       2
1
s a
s b c s a bc


    
e
Gp(s)
+ Gc(s)
r y


Internal Stability
For 𝑐 = 𝑎
So, when there is a pole-zero cancellation, the closed loop system is
BIBO stable for 𝑏 > −1
How about internal stability?
As can be seen  grows without bound, making the system unstable
    
( )cl
s a
G s
s b s c s a


     
1
1s b

 
e
Gp(s)
+ Gc(s)
r y


   c c pG r y G r G     c p
c p c p
N D
r
D D N N
 

( )p
s a
G s
s b



1
( )cG s
s c


  1
s b
r
s a s b


  

Internal Stability
Consider
With =0
With r=0
System is internally stable
e
Gp(s)
+ Gc(s)
r y

10
( )
3
c
s
G s
s



1
( )
2
pG s
s


,
2
10
( )
2 4
y r
cl
s
G s stable
s s


 

 p cy G G y 
1
p
p c
G
y
G G
 

  
2
2 3
2 4
s s
y
s s

 
 
 
  ,
2
2 3
( )
2 4
y
cl
s s
G s stable
s s
  

 
(disturbance, noise, etc.)
c p
c p c p
N D
r
D D N N
 


  
2
10 2
2 4
s s
r stable
s s
 

 

Internal Stability
Investigate the stability of the following system
Note however that in presence of  the system is not stable
e
Gp+ Gc
r

y
+

disturbance, noise, etc.
 
 
 
1
1
p
p
p
N s
G s
D s s
 

 
 
 
1
3
c
c
c
N s s
G s
D s s

 

 
1
p c
cl
p c
G G
G s
G G


  p c
cl
p c p c
N N
G s
D D N N


 
p c p
cl
c pc p c
N N N
G
D ND N N
  

1
4s


The closed loop system
appears to be BIBO stable!
cl py G r G  
  
1 3
( ) ( )
4 4 1
s
r s s
s s s


 
  
since c pN D
 
Unstable pole-zero cancellation causes internal instability and hence overall
instability will occur in presence of noise or disturbance is entered the system!

Internal Stability
To investigate the internal stability, the usual feedback
block diagram will not work (where the reference signal is the only
input variable and the controlled signal is the only output variable of interest)
Why? Because the internal behavior of the system cannot
be captured
The internal stability problem can be completely analyzed
by introducing the 2input, 2output system shown below
What if there is an unstable pole zero cancellation?
e1(s)
GP(s)+ GC(s)
r1(s) y1(s)
r2(s) (input, disturbance, noise, etc.)
e2(s) y2(s)
+
+
+Fig. ()
Lag compensator

Internal Stability
e1 and e2 are considered as the output signals that are used to obtain the transfer
function, however, signals y1 and y2 could be used instead
1 1 2 1( ) ( ) ( ) ( ) ( ) ( ) ( )P P Ce s r s G s r s G s G s e s  
  1 1 2( ) ( ) ( ) ( ) ( ) ( )P C PI G s G s e s r s G s r s  
   
1
1 1 2( ) ( ) ( ) ( ) ( ) ( )P C Pe s I G s G s r s G s r s

  
2 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( )C C Pe s r s G s r s G s G s e s  
  2 2 1( ) ( ) ( ) ( ) ( ) ( )C P CI G s G s e s r s G s r s  
   
1
2 2 1( ) ( ) ( ) ( ) ( ) ( )C P Ce s I G s G s r s G s r s

  
1 1
2 2
( ) ( )
( )
( ) ( )
e s r s
H s
e s r s
   
   
   
If we take y1 and y2 as the output
signals, we will have
1 1
2 2
( ) ( )
( )
( ) ( )
y s r s
H s
y s r s
   
   
   
   
   
1 1
1 1
C P C C P C P
P C P C P C P
I G G G I G G G G
H
I G G G G I G G G
 
 
  
 
   
Define
   
   
1 1
1 1
P C P C P
C P C C P
I G G I G G G
H
I G G G I G G
 
 
  
 
   

Internal Stability
Definition (exponential stability): a rational transfer
matrix H(s) is exponentially stable if it is proper and has
no poles in the closed RHP
◦ For a matrix to be proper (bounded at infinity) each element of
the matrix must be proper
Each element of H(s) must be stable and proper for H(s)
to be exponentially stable why?
◦ Since each root of the pole polynomial appears as a pole in at
least one of the elements of H(s)
This leads to the definition of internal stability!
Example: G(s)=s, u(t)=sint2, y(t)=2tcost2
Properness is necessary since an improper transfer function with no poles in the
closed RHP could still lead to an unstable system!
A bounded input but an unbounded output!
For LTI systems,
Exponential Stability  Asymptotic Stability
http://en.wikipedia.org/wiki/Exponential_stability#Real-world_example
In control theory, a continuous linear time-invariant system is exponentially stable if
and only if the system has eigenvalues (i.e., the poles of input-to-output systems)
with strictly negative real parts. (i.e., in the left half of the complex plane).
Exponential stability is a form of asymptotic stability. Systems that are not LTI are
exponentially stable if their convergence is bounded by exponential decay.
Of course non-causal

Internal Stability
Definition: the 2input 2output system shown in the picture is internally stable
iff the transfer matrix H(s) is exponentially stable
Internal stability is an important concept since unstable pole zero cancellations
that might occur in the open loop transfer matrix, GC(s)GP(s) or GP(s)GC(s), are not
captured in stability analysis such as Nyquist
Using internal stability, any unstable pole zero cancellations in the open loop
transfer matrix, ifany, will appear in H12 or H21
To show this, assume we are dealing with a SISO system (the SISO assumption
makes the derivation simpler!)
◦ In this case, H(s) becomes
1
1 1
1
1 1
P
P C P C
C
C P C P
G
G G G G
H
G
G G G G
 
  
 
 
   
   
   
1 1
1 1
P C P C P
C P C C P
I G G I G G G
H
I G G G I G G
 
 
  
 
   
Having an unstable pole in GC(s) is pretty much unrealistic!
e1(s)
GP(s)+ GC(s)
r1(s) y1(s)
e2(s) y2(s)
+
+
+

Internal Stability
Suppose the plant has an unstable pole. Therefore
As can be seen, the unstable pole appears in H12(s)
 1
( ) P P
P
P P
N N
G s
D s p D
 

1
1 1
1
1 1
P
P C P C
C
C P C P
G
G G G G
H
G
G G G G
 
  
 
 
   
12
1
P
P C
G
H
G G


 
 
 
1
1
1
1
P
P
CP
P C
N
s p D
s z NN
s p D D





   1 1
P C
P C P C
N D
s p D D N s z N

  
  1
12
P C
P C P Cs p
N D
H
D D N N


 1
( ) CC
C
C C
s z NN
G s
D D

 
1 1 0zas ume ps  
Claim: any unstable pole zero cancellations that might occur in the
open loop transfer matrix will be captured by either H12 or H21!
( ) ( ) P C
P C
P C
N N
G s G s
D D

Doesn’t reveal the unstable pole!

Internal Stability
Similarly, if the compensator has an unstable pole then
As can be seen, the unstable pole appears in H21(s)
 1
( ) C C
C
C C
N N
G s
D s p D
 

1
1 1
1
1 1
P
P C P C
C
C P C P
G
G G G G
H
G
G G G G
 
  
 
 
   
21
1
C
C P
G
H
G G


 
 
 
1
1
1
1
C
C
PC
C P
N
s p D
s z NN
s p D D





   1 1
C P
C P C P
N D
D Ds zNp s N 


  1
21
C P
C P C Ps p
N D
H
D D N N


 1
( ) PP
P
P P
s z NN
G s
D D

 
1 1 0zas ume ps  
( ) ( ) P C
P C
P C
N N
G s G s
D D

Doesn’t reveal the unstable pole!

Internal Stability
Fact: for H(s) to be exponentially stable, all elements of H(s) must be
exponentially stable
Does one need to check all 4 elements of H(s) for internal stability?
◦ If no information about GC(s) and GP(s) is known, then all 4 elements of H(s)
must be checked for internal stability (as soon as one is unstable, we are
done)
◦ If it is known that GP(s) is stable, then a necessary and sufficient condition for
internal stability is stability of H21 (Theorem)
◦ If it is known that GC(s) is stable, then a necessary and sufficient condition for
internal stability is stability of H12 (Theorem)
◦ If it is known that both GC(s) and GP(s) are stable, then any of the 4 elements
can be checked for internal stability
Next theorem will help one to determine internal stability in less time
and effort
1
1 1
1
1 1
P
P C P C
C
C P C P
G
G G G G
H
G
G G G G
 
  
 
 
   
The answer is “it depends”!
By the way, think of MIMO!

Internal Stability
Theorem: for an exponentially stable GP(s), the feedback system shown in
the picture is internally stable iff H21(s) is exponentially stable
Proof: (the if part)
Also,
Need to show that if H21(s) is exponentially stable, so are all
other elements of H(s) and vice versa!
 
1
21 C P CH I G G G

 
   
   
1 1
1 1
P C P C P
C P C C P
I G G I G G G
H
I G G G I G G
 
 
  
 
   
 
1
21( )P P C P CI G H s I G I G G G

     
1
P C P CI G G I G G

  
 
1
P CI G G

     
1 1
P C P C P C P C
I
I G G I G G G G I G G
 
    
1 4 4 4 4 2 4 4 4 43
HW
12 11 PH H G
It only remains to show that H22(s) is exponentially stable as well!
 
1
22 ( ) C PH s I G G

 
 
1
C P C PI I G G G G

  
    1
C P C P
I
P CI G G I G G G G

   
1 4 4 4 2 4 4 4 3
 H11(s) is exponentially stable!
21 PI H G 
The only if part: if the feedback system is exponentially stable, then H(s) is
exponentially stable  H21(s) is exponentially stable as well. 

Suppose H21(s) is exponentially stable
e1(s)
GP(s)+ GC(s)
r1(s) y1(s)
e2(s) y2(s)
+
+
+

Internal Stability
Theorem: If GP(s) is exponentially stable, then H21(s) is
exponentially stable iff
1. det(IGC(s)GP(s)) has no zeros (including ) in the Closed RHP
(CRHP)
2. H21(s)=(IGC(s)GP(s))
1
GC(s) is analytic (i.e. has no poles) at every
CRHP pole of GC(s) (including )
◦ This condition is needed to avoid any unstable pole-zero cancelation (example on next slide)
‫:اثبات‬ ‫یک‬MFD ‫از‬ ‫اول‬ ‫هم‬ ‫به‬ ‫نسبت‬ GCGP ‫صورت‬ ‫به‬ N(s)D
1
(s) ‫درنظر‬ ‫را‬
‫آنگاه‬ ،‫بگیرید‬ 
1
21 C P CH I G G G

   
1
1
CI ND G


   
1
CD D N G

 
H21(s) is exp. stable  D(s)N(s) cannot have any zero in the RHP
See the proof in the Book!
   
   
1 1
1 1
P C P C P
C P C C P
I G G I G G G
H
I G G G I G G
 
 
  
 
   
21
1
;
2 1
3
;
4
C
P
C P
C
Gs
G H
s G G
s
G
s

 
 



 
  
3 1 2 5
det 1
4 2 4 2
C P
s s s
I G G
s s s s
  
   
   
 
  1
21
2 3
( )
2 5
C P C
s s
H s I G G G
s
  
  

One root at   Unstable
Improper! Not Exp. Stable!
Not to have a zero at  implies the determinat
Numerator Degree MUST EQUAL To its Denominator Degree
Validation
If the designer is careful not to have an unstable pole-zero cancelation
when designing a compensator, then only 1. must be checked.
‫کردن‬h21‫که‬ ‫چرا‬ ‫بود‬ ‫خواهد‬
‫بود‬ ‫خواهد‬

Internal Stability Example
Investigate the internal stability of the following system
GP(s) is stable
GC(s) has a pole at +1 (unstable)
◦ Expect that H21 will not be stable
Since unstable pole-zero cancellation exists, the second condition MUST be
checked!
1 1
( ) , ( )
2 1
P C
s
G s G s
s s

 
 
1
1 ( ) ( )
2
C P
s
G s G s
s

 

No root in the CRHP, (s+1) 
First condition is satisfied 
 
  
1
1
1
2
1 ( ) ( ) ( )
1 1
C P C
s
s
s
G s G s G s
s s




   
 
(1GC(s)GP(s))
1
GC(s) is not
analytic at +1 (has a pole)!
Second condition is not satisfied
and hence the system is not
internally stable!
Unstable pole-zero cancellation!

Internal Stability
Note 1: each of the previous theorems can be stated for GC(s)
Note 2: we have used positive feedback in our discussion here, however,
if one uses a negative feedback, all results are valid by changing the sign
of GP(s) (or GC(s))
Note 3: The conditions given for internal stability is applicable to any
two systems that are connected by feedback
◦ i.e. the compensator can be in front of the plant or in the feedback path
Note 4: If a designer can guarantee the absence of an unstable pole zero
cancellation, then the second condition of the previous theorem
doesn’t need to be checked

Internal Stability
What happens when both GP(s) and GC(s) are exponentially stable?
◦ any element of H(s) can be checked
◦ If the element that is checked is exponentially stable, then the closed-loop
system will be internally stable
Do we really need to check the stability of H(s) when both GP(s) & GC(s)
are stable? Yes
3 3
1 1 2( 1)
closed
s G s
G G
s G s
 
   
  
Example of a stable plant but an
unstable closed loop system for
a unity feedback system!
zeros of GP(s) or GC(s) might cause the system to be unstable!
1 7 7
;
1 2 1 ( 1)( 5)
P C
P C closed
P C
G Gs s
G G G
s s G G s s
 
    
    
Another example
Both plant & compensator are stable but the closed loop is not!

Principle of Argument (Cauchy)
Suppose F(s) is a single valued rational function that is analytic everywhere in the s-plane
except possibly at few points
Corresponding to every point on the s-plane where F(s) is analytic, there exists a point in
the F(s) plane
Now consider an arbitrary closed contour S (usually CW) in the s-plane that does not go
through the zeros and poles of F(s)
Corresponding to this closed contour, there exists a closed contour in the F(s) plane that
encircles the origin N=ZP times
Z=zeros of F(s) inside S
P=poles of F(s) inside S
σ
s-planej
s
S1
S3
F
F(s1)
F(s2)
F(s3)
F(s)-plane
Re F
Im F
Negative N implies direction
of F is opposite of that of S
S2
Z
P N

Principle of Argument
A simple way to obtain N is as shown in the following pictures
◦ Depending on F(s), direction of F may be in the same or opposite direction of s
Assuming this corresponds
to clockwise s!
N=2
F
F(s)plane
N=0
F
F(s)plane

Principle of Argument
What if F(s) is of the form 1+k(s)?
In this case,
So we can use the plot of (s) instead of 1+k(s) and count the number
of encirclements around (1/k,0)
In other words
◦ N: number of (1/k,0) encirclements made by (s)
◦ Z: number of zeros of 1+k(s) inside s contour
◦ P: number of poles of 1+k(s) inside s contour
Note: poles of 1+k(s) are identical to the poles of k(s)
◦ Just write k(s) as kp(s)/q(s) then
( ) ( )
1 ( )
( )
q s kp s
k s
q s

  
Poles of 1+(s) 
1+k(s) encircling (0,0)  (s) encircling (1/k,0)

Nyquist Theorem
The Nyquist Theorem is based on the principle of argument
Using Nyquist, the stability of the closed loop system, for different
values of k, can be investigated from the open loop information
How?
◦ The closed loop TF is given by
◦ For asymptotic stability, poles of Gcl(s) must lie in LHP
◦ Now note the following
◦ Nyquist uses the fact that
( )
( )
1 ( )
cl
kG s
G s
kG s


poles of Gcl(s) = zeros of (1+kG(s))
poles of kG(s) = poles of (1+kG(s))
N: number of encirclements
of the origin by 1+kG(s)
N can also be interpreted as follows:
N: number of encirclements of -1 by kG(s)
N: number of encirclements of -1/k by G(s)
y
G(s)
r
+
_
k
For closed loop stability, we must have
Z=0 OR N=P
if S is selected as Nyquist path!
N=ZP
Z=?

Nyquist Theorem
To use Nyquist for stability analysis, s must be selected
as shown in the picture (common direction is CW)
If ∆(s) has a pole or a zero on the jω axis, then we
modify the contour as shown
jω
σ
Γ(s)
-∞←jωjω→∞
R
→
∞
ε→
0
s: Nyquist path
(CW)

So for the closed loop system to be stable, the Nyquist
plot of GH must not encircle the point (-1,0)
Nyquist Theorem
Example:
There are no open loop poles in the RHP  P=0
For stability we must have Z=0
Nyquist plot of GH does NOT encircle (1,0)
500
( )
( 10)( 3)( 1)
GH s
s s s

  
The closed loop
system is stable!
1K 
 N=Z-P=0
=

Finding closed-loop poles
 Stable systems have closed-loop transfer functions with poles in the
left half-plane.
Unstable systems have closed-loop transfer functions with at least one
pole in the right half-plane, and/or poles of multiplicity greater than
one on the imaginary axis.
 Marginally stable systems have closed-loop transfer functions with
only imaginary axis poles of multiplicity one and poles in the left half-
plane.

Routh-Hurwitz Criterion
Using this method we can tell how many closed-loop poles are in the
left half-plane, in the right half-plane and on the imaginary axis.
 It does not find the exact locations of the roots.
Other methods find the exact locations of the roots. For first and
second order systems, analytical method can be used. For higher order
systems, computer programs or simulation are required.
The method requires two steps:
 Generate the data table (Routh table)
 Interpret the table to determine the number of poles in LHP and RHP.

Some basic results
Second order system:
For third order system:
We see that the coefficients of the polynomial are given by:
o 𝑎 𝑛−1 = negative of the sum of all roots.
o 𝑎 𝑛−2 = sum of the products of all possible combinations of roots taken 2 at a
time.
o 𝑎 𝑛−3 = negative of the sum of the products of all possible combinations of
roots taken 3 at a time
Suppose that all the roots are real and on the left half plane, then all coefficients of
the polynomial are positive.
If all the roots are real and in the left half plane then no coefficient can be zero.
The only case for which a coefficient can be negative is when there is at least one
root in the right half plane.
𝑃2(𝑠 = 𝑠2
+ 𝑎1 𝑠 + 𝑎0 = (𝑠 − 𝑝1 (𝑠 − 𝑝2
= 𝑠2
− (𝑝1 + 𝑝2 𝑠 + 𝑝1 𝑝2
𝑃2(𝑠 = 𝑠3
+ 𝑎2 𝑠2
+ 𝑎1 𝑠 + 𝑎0 = (𝑠 − 𝑝1 (𝑠 − 𝑝2 (𝑠 − 𝑝3
= 𝑠3
− (𝑝1 + 𝑝2 + 𝑝 𝑠2
+ (𝑝1 𝑝2 + 𝑝1 𝑝3 + 𝑝2 𝑝3 𝑠 − 𝑝1 𝑝2 𝑝3

Some basic results
The below is also true for complex roots.
o If any coefficient is equal to zero, then not all roots are in the left half plane.
o If any coefficient is negative, then at least one root is in the right half plane.
o The converse of above rule is not always true.
Example:
 all coefficients are positive. But two roots(complex) are in the right half plane.
𝑷(𝒔 = 𝒔 𝟑 + 𝒔 𝟐 + 𝟐𝒔 + 𝟖 = (𝒔 + 𝟐 (𝒔 𝟐 − 𝒔 + 𝟒

Routh-Hurwitz Stability Criterion
All the coefficients must be positive if all the roots are in the left half
plane.
 It is necessary that all the coefficients for a stable system be nonzero.
These requirements are necessary but not sufficient.
Example :
The Routh-Hurwitz is a necessary and sufficient criterion for the
stability of linear systems.
𝑞(𝑠 = 𝑠3 + 𝑠2 + 2𝑠 + 8 = (𝑠 + 2 (𝑠2 − 𝑠 + 4
the system is unstable yet all coefficients are positive.

Routh-Hurwitz Stability Criterion
The Routh-Hurwitz criterion applies to a polynomial of the form:
The Routh-Hurwitz array:
The number of polynomial roots in the right half plane is equal to the
number of sign changes in the first column of the array.
Note: The Routh-Hurwitz criterion shows only the stability of the system, it
does not give the locations of the roots, therefore no information about the
transient response of a stable system is derived from the R-H criterion. Also
it gives no information about the steady state response. Obviously other
analysis techniques in addition to the R-H criterion are needed.
𝑷(𝒔 = 𝒂 𝒏 𝒔 𝒏 + 𝒂 𝒏−𝟏 𝒔 𝒏−𝟏+. . . . . . . +𝒂 𝟏 𝒔 + 𝒂 𝟎 ; 𝒂 𝟎 ≠ 𝟎
2 4
1 2
1 3 1 51 1
1 51 3
1 2
1 31 21 1
1 1
, .......
1 1
, ......
n n n n
n n n nn n
n nn n
a a a a
b b
a a a aa a
a aa a
c c
b bb bb b
 
    
  
   
   
2 4 6
1
1 3 5 7
2
1 2 3 4
3
1
....
....
....
n
n n n n
n
n n n n
n
n
s a a a a
s a a a a
s b b b b
s c
  

   


2 3 4
2
1 2
1
1
0
1
....
. . .
. . .
c c c
s k k
s l
s m

Initial layout for Routh table

Example
3 2 2
3
2
1
0
( ) 2 8 ( 2)( 4)
The Routh array is:
s 1 2
s 1 8
s -6
s 8
P s s s s s s s       
Since there are two sign changes on the first column, there are two roots of the
polynomial in the right half plane: system is unstable.

From the equations, the array cannot be completed if the first element
in a row is zero. Because the calculations require divisions by zero. We
have 3 cases:
Case 1: none of the elements in the first column of the array is zero.
This is the simplest case. Follow the algorithm as shown in the previous
slides.
Case 2: The first element in a row is zero, with at least one nonzero
element in the same row. In this case, replace the first element which is
zero by a small number . All the elements that follow will be
functions of . After all the elements are calculated, the signs of the
elements in the first column are determined by letting approach
zero.

Example
3 2 2
3
2
1
0
( ) 2 8 ( 2)( 4)
The Routh array is:
s 1 2
s 1 8
s -6
s 8
P s s s s s s s       

Example
𝑃(𝑠 = 𝑠5
+ 2𝑠4
+ 2𝑠3
+ 4𝑠2
+ 11𝑠 + 10
5
4
3
2
1
0
s 1 2 11
s 2 4 10
s 6
12
s - 10
s 6
s 10



Case 3: All elements in a row are zero.
Example:
Here the array cannot be completed because of the zero element in
the first column.
Case 3 polynomial may be analyzed as follows:
 Suppose that the row of zeros is the 𝑠 𝑖 row, then the auxiliary
polynomial is differentiated with respect to s, and the coefficients of the
resulting polynomial used to replace the zeros in the 𝑠 𝑖row. The
calculation of the array then continues as in the case 1.

Example
4 3 2
( ) 3 2 2P s s s s s    
4
3
2
1
0
1 3 2
1 2
1 2
0
s
s
s
s
s
4
3
2
1 2
0
1 3 2
1 2
1 2
2
s
s
s
s
s

Hence there are no roots in the right half plane.

5 4 3 2
10
( )
2 3 6 5 3
T s
s s s s s

    

5 4 3 2
10
( )
2 3 6 5 3
T s
s s s s s

    
5 4 3 2
( ) 3 5 6 3 2 1D s s s s s s     

5 4 3 2
10
( )
7 6 42 8 56
T s
s s s s s

    
4 2
( ) 6 8P s s s   3( )
4 12 0
dP s
s s
ds
  

8 7 6 5 4 3 2
20
( )
12 22 39 59 48 38 20
T s
s s s s s s s s

       

5 4 3 2
1
( )
2 3 2 3 2 1
T s
s s s s s

    

5 4 3 2
( ) 2 3 2 3 2P s s s s s s     

8 7 6 5 4 3 2
128
( )
3 10 24 48 96 128 192 128
T s
s s s s s s s s

       

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