It shows the classification

2. Transfer function model
 A standard second order transfer model
y(s) =ω02 /(s2 + 2ζωos + ω02)
Where,
ζ (zeta) is the relative damping factor
and ω0 [rad/s] is the undamped
resonance frequency.

3. Here in the charcteristic
equation
b=2ζωo
a=1(coefficient of s2 )
c= ω02
 Here the input is step input

4. Classification of second order
system
 First, if b = 0, the poles are complex
conjugates on the imaginary axis .This
corresponds to ζ = 0, and is referred to as the
undamped case.
 If b2 − 4ac < 0 then the poles are complex
conjugates lying in the left half of the s-
plane. This corresponds to the range 0 < ζ <
1, and is referred to as the underdamped
case.

5. If b2 − 4ac = 0 then the poles coincide
on the real axis at s1 = s2 = −b/2m. This
corresponds to ζ = 1, and is referred to
as the critically damped case.
Finally, if b2 − 4ac > 0 then the poles
are at distinct locations on the real axis
in the left half of the s-plane. This
corresponds to ζ > 1, and is referred to
as the overdamped case.

6. Undamped case (ζ = 0)
In this case, the poles lie at s1
= jωn and s2 = −jωn. The
homogeneous solution takes
the form
 x(t) = 1-cosωn t

8. Critically-damped ζ =1
In the critically damped case, ζ =1
and the two poles coincide at s =
−ωn. The response is then given
by :-
x(t)=1-e−ωnt –ωn (te−ωnt )(t ≥ 0).

10. Over damped ζ ≥1
One extremely important thing
to notice is that in this case the
roots are both negative.The
term under the square root is
positive by assumption, so the
roots are real.

11. General solution:
x(t) = 1-(0.5ωn /(ζ
2-1)
c1er’t + c2er’’t .

13. Under damped case (0 < ζ < 1)
The basic real solutions are:-
e−bt/2a cos(ωdt) and
e−bt/2a sin(ωdt).

14.  The general real solution is found by
taking linear combinations of the two
basic solutions, that is:
x(t) = 1-c1 ejωntcos(ωdt) - c2ejωntsin(ωdt)
where,
ωd = ωn(1−ζ2)1/2
is the damped natural frequency

17. Transient
Response (Under damped )
 Rise time
 The rise time, tr, is usually specified as the time
required for the output of the system to rise from
either 10% to 90%, 5% to 95%, or 0% to 100% of its
final value.
ωdtr + φ = π
Solving yields,
tr=(π –φ)/ ωd

18. Peak time
 Peak time, tp, is the time required for the output of the
system to reach its first peak. It is found by setting the
derivative of the output equal to zero.
tp =π /ωd

19. Settling Time
 Settling time, ts, is the time required for the output of
the system to reach and stay within a desired
percentage of the final value. Typically 2% or 5% is
used. The settling time is given by equation
ts = −ln(ǫ) /ζωn
where ǫ is the tolerance for 2% ǫ = 0.02, for 5% ǫ = 0.05,
etc

20. Maximum Percent OverShoot
 The maximum percent overshoot, Mp, is the peak
value of the output of the system divided by the final
value of the output. Substituting tp into the output,
equation will produce
Mp = e−ζπ /(√1−ζ2 )

21. Thank
you