AE 230 - Modeling and Simulation Laboratory
Second order systems
dq o +a o +a q =b i +b i +b q a 2 dt 2 1 dt o o 2 dt 2 1 dt 0 i d 2q dq d 2q
Most general form of second order system
dq d 2q o +a q =b q o +a a o o 2 1 dt 0 i dt 2
b a d 2q a dq o + 1 o +q = 0 q 2 o a i a dt 2 a dt 0 0 0
Special case and most practical in nature
where
2 D 2D 2 + + 1qo = Kq i n n
0 undamped natural frequency rad n a time 2 a 1 damping ratio dimensionl ess a a 2 0 b K 0 system steady state gain (sensitivity) a 0
Cascaded Second order systems
Cascaded Second order systems
( 2 D + 1)qo 2 = K 2 qi 2 = K 2
K1 qi1 ( 1D + 1)
Output of one system going to input of other;
Cascaded Second order systems
( 2 D + 1)qo 2 = K 2 qi 2 = K 2
K1 qi1 ( 1D + 1)
Output of one system going to input of other; second order system but not as a single system
( D + 1)( D + 1)q = [ D 2 + ( + ) D + 1]q = K K q 1 2 12 2 1 1 2 i1 o2 o2
Second order systems
f i1 - B1x 01 = M1 01 x B2 ( x 01 - x 02 ) = K s 2 x 02
Bottom damper and spring can be assumed to be a velocity sensor. Interested in finding X02
Second order systems
x02 ( D) = TF1 x TF2 f i1
1 / B1 TF1 = M1 D +1 B1
B2 / K s 2 TF2 = B2 D +1 Ks2
x02 B2 /( B1 K s 2 ) ( D) = M 1 B2 2 M 1 K s 2 + B1 B2 f i1 D + D +1 K s 2 B1 K s 2 B1
Second order systems
f i1 B1 x01 B2 ( x01 x02 ) = M 1 01 x B2 ( x01 x02 ) = K s 2 x02
B2 D + K s 2 B1 f i1 ( B2 D + K s 2 ) x02 B2 B D 1 x02 B2 2 B2 D + K s 2 = M1 B D x02 2
x02 ( D) = f i1
B2 /( K s 2 ( B1 + B2 )) M 1 K s 2 + B1 B2 M 1 B2 2 D + D +1 K s 2 ( B1 + B2 ) K s 2 ( B1 + B2 )
Second order systems
x02 B2 /( B1 K s 2 ) ( D) = M 1 B2 2 M 1 K s 2 + B1 B2 f i1 D + D +1 K s 2 B1 K s 2 B1
Approximate (no interaction/no loading effect)
x02 ( D) = f i1
B2 /( K s 2 ( B1 + B2 )) M 1 K s 2 + B1 B2 M 1 B2 2 D + D +1 K s 2 ( B1 + B2 ) K s 2 ( B1 + B2 )
Exact (with interaction/loading effect)
Second order systems
x02 ( D) = f i1 B2 /( K s 2 ( B1 + B2 )) M 1 K s 2 + B1 B2 M 1 B2 2 D + D +1 K s 2 ( B1 + B2 ) K s 2 ( B1 + B2 )
If B2 is small compared to B1 then individual transfer function will be a good approximation. System 2 is not significantly loading system1. Assuming all parameter are equal to 1, except B2 = 0.05
x02 0.05 ( D) = f i1 0.05 D 2 + 1.05 D + 1 x02 0.0476 ( D) = f i1 0.0476 D 2 + 1.0 D + 1
No Loading
Exact
Second order systems
Two isolated first order system when joined together draws power from one of the system, if it is significant, system equation cannot be developed using just two first order system
Second order systems
f + W ( W + K x ) Bx = M x i s o 0 o
x Force balance Mo + Bx0 + K s xo = fi
K 1 meter B s rad , ,K M time 2 K M K Newton s s
2 D 2 D + + 1q = Kf 2 i o n n
Second order systems
2 D 2 D + 1q = Kf 2+ i o n n
+ n 2 1 and n 2 1 n n
Two roots of the characteristic equation
Undamped : B = 0, = 0, roots = i x = CSin( t + ) oc n n
Underdamped : 0 < B < 2 K M , 0 < < 1.0, roots = i n 1 2 s n t n Sin ( 1 2 t + ) x = Ce oc n
Second order systems
Mass M = 1 kg;
n K
K = 1 N/m and B = 0.2 N/(m/sec)
s rad = 1 rad / sec M time
B = 0.1 2 K M s
The system is subjected to i) Initial displacement (1m)
ii) Initial velocity (1 m/sec) iii) Step force (1 N)
Second order systems - underdamped
Response for initial displacement, initial velocity, step force input
Second order systems Step input
Step response of undamped second order system =0
x = Kf (1 cos t ) o is n
Second order systems Step input
Step response of critically second order system =1
x = Kf (1 (1 + t )e o is n
nt
Second order systems - Overdamped
Mass M = 1 kg;
n K
K = 1 N/m and B = 20 N/(m/sec)
s rad = 1 rad / sec M time
B = 10.0 2 K M s
1 1 Over damped : B > 2 K M , > 1.0, roots = 2 1 = , s n 1 2 x =C e oc 1 t / 1 +C e 2 t / 2
+ 2 1 t / + 2 1 t / 1 2 x = Kf is 1 e e + o 2 2 1 2 2 1
Second order systems
Mass M = 1 kg;
n K
K = 1 N/m and B = 20 N/(m/sec)
s rad = 1 rad / sec M time
B = 10.0 2 K M s
+ 2 1 t / 2 1 t / 1 2 x = Kf is 1 e e + o 2 2 1 2 2 1
1 1.002e 0.05nt + 0.002e 19.95nt x = Kf is o
Second order systems
Mass M = 1 kg;
n K
K = 1 N/m and B = 20 N/(m/sec)
s rad = 1 rad / sec M time
B = 10.0 2 K M s
+ 2 1 t / 2 1 t / 1 2 x = Kf is 1 e e + o 2 2 1 2 2 1
1 1.002e 0.05nt + 0.002e 19.95nt x = Kf is o 1 1.002e 0.05nt Kf is
Second order systems Step input
Non-dimensional step response of second order system
Second order systems Step input
Effect of damping on overshoot
Significance of K, , n
Steady state gain is only dependent of K n largely governs the speed of response due to product (nt). Doubling the natural frequency will half the response time. To speed up by a factor n, natural frequency has to be increased by a factor of n (when is constant). For step response, when < 1.0 overshooting. To control the overshoot should be adjusted. when = 1.0, least time to reach steady state without overshoot. when > 1.0, no overshoot, time to reach steady state is more than when = 1.0
Lab testing of second order systems Step inputs
Lab testing of second order systems Step inputs
Peaks occur at
0,
2 2
n ,d n ,d n ,d
etc
Time period
n,d
Amplitude ratio of two successive peak is constant for an under damped system, this can be used for finding the damping ratio.
x p ,n +1 x p ,n
=e
1 2
Assignment
Write system equation and transfer function for any two mechanical and two electrical systems given in subsequent slides
Second order systems
Second order systems
Second order Electrical systems
Second order Electrical systems
