NAME: AAMIR IRSHAD SECTION:B
SAB NO:70065601
REG NO:BSME-016-047
ASSIGNMENT NO:02
QNO 1:
Ruth-hurwitz stability criteria and its applications
ANSWER:
next →← prev
Routh- Hurwitz Criterion
Before discussing the Routh-Hurwitz Criterion, firstly we will study the stable, unstable and marginally stable system.
1. Stable System: If all the roots of the characteristic equation lie on the right half of the 'S' plane then the system is said
to be a stable system.
2. Marginally Stable System: If all the roots of the system lie on the imaginary axis of the 'S' plane then the system is
said to be marginally stable.
3. Unstable System: If all the roots of the system lie on the left half of the 'S' plane then the system is said to be an
unstable system.
Statement of Routh-Hurwitz Criterion
Routh Hurwitz criterion states that any system can be stable if and only if all the roots of the first column have the same sign and
if it does not has the same sign or there is a sign change then the number of sign changes in the first column is equal to the
number of roots of the characteristic equation in the right half of the s-plane i.e. equals to the number of roots with positive real
parts.
Necessary but not sufficient conditions for Stability
We have to follow some conditions to make any system stable, or we can say that there are some necessary conditions to make
the system stable.
Consider a system with characteristic equation:
1. All the coefficients of the equation should have the same sign.
2. There should be no missing term.
If all the coefficients have the same sign and there are no missing terms, we have no guarantee that the system will be stable.
For this, we use Routh Hurwitz Criterion to check the stability of the system. If the above-given conditions are not satisfied,
then the system is said to be unstable. This criterion is given by A. Hurwitz and E.J. Routh.
Advantages of Routh- Hurwitz Criterion
1. We can find the stability of the system without solving the equation.
2. We can easily determine the relative stability of the system.
3. By this method, we can determine the range of K for stability.
4. By this method, we can also determine the point of intersection for root locus with an imaginary axis.
Limitations of Routh- Hurwitz Criterion
1. This criterion is applicable only for a linear system.
2. It does not provide the exact location of poles on the right and left half of the S plane.
3. In case of the characteristic equation, it is valid only for real coefficients.
The Routh- Hurwitz Criterion
Consider the following characteristic Polynomial
When the coefficients a0, a1, ......................an are all of the same sign, and none is zero.
Step 1: Arrange all the coefficients of the above equation in two rows:
Step 2: From these two rows we will form the third row:
Step 3: Now, we shall form fourth row by using second and third row:
Step 4: We shall continue this procedure of forming a new rows:
Example
Check the stability of the system whose characteristic equation is given by
s4 + 2s3+6s2+4s+1 = 0
Solution
Obtain the arrow of coefficients as follows
Since all the coefficients in the first column are of the same sign, i.e., positive, the given equation has no roots with positive real
parts; therefore, the system is said to be stable.
The application of the Routh Hurwitz Criterion is discussed in the following by means of an example.
Example
The open-loop transfer function of a unity feedback control system is given
Use Routh Hurwitz Criterion to establish a relation between K,T 1,T2 so that the system is stable.
Solution:
The characteristic equation of the system is
The Routh Hurwitz Criterion array is
For the system to be stable
QNO 2:
NYQUIST STABILITY CRITERION
AMSWER:
Nyquist stability criterion (or Nyquist criteria) is a graphical technique used in control
engineering for determining the stability of a dynamical system. As Nyquist stability criteria only considers
the Nyquist plot of open-loop control systems, it can be applied without explicitly computing the poles and
zeros of either the closed-loop or open-loop system.
As a result, Nyquist criteria can be applied to systems defined by non-rational functions (such as systems
with delays). Unlike Bode plots, it can handle transfer functions with singularities in the right half-plane.
Nyquist Stability Criterion can be expressed as:
Z=N+P
Where:
Z = number of roots of 1+G(s)H(s) in right-hand side (RHS) of s-plane (It is also called zeros of
characteristics equation)
N = number of encirclement of critical point 1+j0 in the clockwise direction
P = number of poles of open loop transfer function (OLTF) [i.e. G(s)H(s)] in RHS of s-plane.
The above condition (i.e. Z=N+P) is valid for all the systems whether stable or unstable.
Now we will explain this criterion with examples of Nyquist stability criterion.
Nyquist Stability Criterion Examples
Example 1
Consider an open loop transfer function (OLTF) as Is it a stable system or
unstable. Perhaps most of you will say it is an unstable system because one pole is at +2. However, note that
stability depends on the denominator of closed-loop transfer function If any root of the denominator of the
closed-loop transfer function (also called characteristics equation) is at RHS of s-plane then the system is
unstable. So in the case above, a pole at +2 will try to bring the system towards instability, but the system
may be stable. Here Nyquist plot is useful to find the stability.
According to Nyquist theory Z=N+P (for any system, whether it is stable or unstable).
For the stable system, Z=0, i.e. No roots of characteristics equation should be at RHS.
So for the stable system N = –P.
The Nyquist plot of the above system is as shown below
Nyquist Plot Matlab Code
s = tf('s')
G1 = 120 / ((s-2)*(s+6)*(s+8))
nyquist(G1, 'red')
As per the diagram, Nyquist plot encircle the point –1+j0 (also called critical point) once in a counter clock
wise direction. Therefore N= –1, In OLTF, one pole (at +2) is at RHS, hence P =1. You can see N= –P, hence
system is stable.
If you will find roots of characteristics equation, it will be –10.3, –0.86±j1.24. (i.e. system is stable), and Z=0.
One question can be asked, if roots of characteristics equation can be found, so we can comment on the
stability on that basis, then what is the need of Nyquist plot. The answer is, when software’s were not
available, in those days Nyquist plot was very useful.
Example 2
Now take another example:
Nyquist plot is as follows:
Nyquist Plot Matlab Code
s = tf('s')
G2 = 100 / ((s-2)*(s+6)*(s+8))
nyquist(G2, 'red')
From the figure, it can be found that N= –1. (Encirclement of Nyquist plot of critical point is one in counter
clock wise direction)
In this example also P=1. (one pole of OLTF at RHS)
So, N=–P. Hence system is stable.
(Roots of characteristics equation are –10.04, –1.72, – 0.23)
Next Topic
next →
QNO:3
Nyquist criterion for system with minimum phase transfer
Functions
Answer
Nyquist criterion minimum phase transfer function
Systems with Minimum phase transfer function: Minimum phase
transfer function is one in which the all the poles and zeros of closed loop
transfer function will be on left half of s-plane including s=0.
For minimum phase transfer function the nyquist criterion for stability
simplifies to N =0 I.e. since none of the zeros and poles of L(s) lie on right
half of s-plane or on j ω axis Z=0 and P=0 hence N=0. Hence For a closed
loop system with loop transfer function L(s) is of the minimum phase type,
the system is closed loop stable if the plot of L(s)that corresponds to the
nyquist path does not encircle the critical point (-1,j0) in the L(s) plane.
Even if the system is unstable Z>0 hence N is a positive integer for such a
system to be stable the L(s) plot that corresponds to nyquist path does not
enclose the point (-1, j0) because if N>0 the point (-1, j0) is encircled in
same direction as that of nyquist path (In nyquist path the poles are
enclosed and hence in L(s) plot the point (-1, j0) is enclosed). If it is
enclosed the system is unstable.
STEPS TO DRAW NYQUIST PLOT:
Substitute s=j ω in L(s)
Substitute ω=0 to get the zero frequency property of L(j* ω)
Substitute ω=infinite to get the property of nyquist plot at infinite
frequency
To find the intersects of the nyquist plot with the real axis, rationalize
the denominator and make the imaginary part of rationalized L(j* ω) to
zero
