Delay Time
The time required for the response
to reach 50% of the final value in the first time is called the delay time.
Rise Time
The
time required for response to rising from 10% to 90% of final value, for an
overdamped system and 0 to 100% for an underdamped system is called the rise
time of the system.
Peak Time
The
time required for the response to reach the 1st peak of the time response or
1st peak overshoot is called the Peak time.
Maximum overshoot
The
difference between the peak of 1st time and steady output is called the maximum
overshoot. It is defined by
Settling Time (ts)
The
time that is required for the response to reach and stay within the specified
range (2% to 5%) of its final value is called the settling time.
Steady State Error (ess)
The
difference between actual output and desired output as time't' tends to
infinity is called the steady state error of the system.
Example - 1
When
a second-order system is subjected to a unit step input, the values of ξ = 0.5
and ωn = 6 rad/sec. Determine the rise time, peak time, settling time and peak
overshoot.
Solution: Given-
1.
ξ = 0.5 ω n = 6 rad/sec
Rise Time:
Peak Time:
Settling Time:
Maximum overshoot:
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 left 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 right 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.
4. Step 4: We
shall continue this procedure of forming a new rows:
5.
Example
6.
Check the stability of the system whose characteristic equation
is given by
7. s4 + 2s3+6s2+4s+1 = 0
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.
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