Time Response of Second Order Systems
d 2 y(t )
dy(t )
= f (t ) − B
− Ky(t )
2
dt
dt
M (s 2Y (s) − sy(0) − y' (0)) = F (s) − B(sY (s) − y(0)) − KY (s)
M
Let y' (0) = 0
Ms2Y (s) + BsY (s) + KY (s) = Msy0 + By0 + F (s)
Or
F ( s)
(Ms + B) y0
+ 2
2
Ms + Bs + K Ms + Bs + K
Consider the first term only:
(s + B / M ) y0
(s + 2ςωn ) y0
= 2
Y ( s) = 2
s + ( B / M )s + K / M s + 2ςωn s + ωn 2
Y ( s) =
ς = dimensionless damping ratio
ωn = the natural frequency
Where
ωn =
K
;
M
ς=
B
2 KM
2
The characteristic equation s 2 + 2ςω n s + ω n has two roots :
s1 = −ςω n + jω n 1 − ς 2
s2 = −ςω n − jω n 1 − ς 2
If ς > 1 ⇒ roots are real
If ς < 1 ⇒ roots are complex (under damped)
If ς = 1 ⇒ same roots and real (critically damped)
For unit step response :
1
y (t ) = 1 − e −ςω nt sin( βω nt + θ )
β
β
β = 1 − ς 2 , θ = tan −1 ( ) = cos −1 ς
ς
Step response:
y (t ) = 1 −
1 −ζω nt
e
sin(ω n βt + θ )
β
β = 1 − ζ 2 , θ = cos −1 ζ
where
Showing the step response with different damping
coefficients
Standard Performance measures
„
„
„
„
Performance measures are usually defined in terms of the step response of a
system as below:
Swiftness of the response is measured by rise time Tr , and peak time Tp
For underdamped system, the Rise time Tr (0-100% rise time) is useful,
For overdamped systems, the the Peak time is not defined, and the (10-90 % rise
time) T is normally used
r1
„
„
„
„
Peak time: T p
Steady-state error: ess
Settling time: T s
Percent of
Overshoot:
P.O. =
M
pt
− fv
fv
× 100 %
is the peak value
fv is the final value
of the response
M pt
„
„
„
Percentage overshoot measures the closeness of the
response to the desired response.
The settling time T s is the time required for the
system to settle within a certain percentage δ of the
input amplitude.
For second order system, we seek T s for which the
response remains within 2% of the final value. This
occurs approximately when:
e −ζω nTs < 0.02
or : ζω nTs ≅ 4
Therefore :
Ts ≅ 4τ =
„
4
ζω n
Hence the settling time is defined as 4 time constants.
„
Explicit relations for M pt and T p
„
To find Tp we can either differentiate y(t) directly or
indirectly through Laplace Transform of y’(t):
Tp =
π
π
=
ωn β ωn 1− ς 2
The peak response is :
M pt = 1 + e −ζπ /
1−ς 2
Percentage overshoot :
P.O. = 100e −ζπ /
1−ς 2
:
Impulse response of the second order system:
„
Laplace transform of the unit impulse is R(s)=1
Y (s) =
„
„
„
ω n2
+ 2 ζω n s + ω
2
(s
n )
Impulse response:
Transient response for the impulse function, which is simply is the
derivative of the response to the unit step:
y (t ) =
ω n − ζω
e
β
nt
sin( ω n β t )
Responses and pole locations
2
„
Time Responses and Pole Locations:
2
The characteristic equation s 2 + 2ςω n s + ω n has two roots :
s1 = −ςω n + jω n 1 − ς 2
s2 = −ςω n − jω n 1 − ς 2
„
s1 , s2
are the poles.
Ts ≅ 4τ =
„
4
ζω n
Therefore the settling time is inversely proportional to the real part of the poles.