Second Order Systems

Aditee Apurvaa

Transfer function modelA standard second order transfer modely(s) =ω02 /(s2 + 2ζωos + ω02)

Where, ζ (zeta) is the relative damping factorand ω0 [rad/s] is the undamped resonance frequency.

Here in the charcteristic equation

b=2ζωoa=1(coefficient of s2 )c= ω02

Here the input is step input

Classification of second order systemFirst, 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.

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.

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

Critically-damped ζ =1In 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).

Over damped ζ ≥1One 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.

General solution: x(t) = 1-(0.5ωn /(ζ

2-1)c1er’t + c2er’’t .

Under damped case (0 < ζ < 1)

The basic real solutions are:-

e−bt/2a cos(ωdt) and e−bt/2a sin(ωdt).

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

Transient Response (Under damped )Rise timeThe 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

Peak timePeak 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

Settling TimeSettling 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

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 )

Thank you