Unformatted text preview: o ICs)
zero
1 &
Ï„y + y = Ku
â‡’ y ( t ) = K (1 âˆ’ e âˆ’t Normalized Response 0.9 Ï„) N ormalized
(such that as t â†’ âˆž , y n â†’ 1) : â‡’ y n (t ) = y (t )
âˆ’t
= (1 âˆ’ e Ï„ )
K 0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 1Ï„ 0 Ï„ Time t
( 1âˆ’ e âˆ’ t/ Ï„ 2Ï„ 3Ï„ 2Ï„ 3Ï„
Time [ t ] 4Ï„ 4Ï„ 5Ï„ 6Ï„ 5Ï„ )
School of Mechanical Engineering
Purdue University ME575 Session 6 â€“ Dynamic Response Slide 5 Step Response of 1st Order System
Effect of Time Constant Ï„ :
1 &
Ï„y + y = Ku
Slope at t = 0:
0:
d
â‡’
y (t ) =
dt
d
y (0) =
â‡’
dt âˆ’t Ï„) 0.8
Normalized Response â‡’ y (t ) = K (1 âˆ’ e 0.9 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 ME575 Session 6 â€“ Dynamic Response 2 School of Mechanical Engineering
Purdue University 4
6
Time [sec] 8 10 Slide 6 3 ME575 Handouts Example â€“ Vehicle Acceleration
v m
1
&
v + v = F = vmax
b
b F m
b StandingStart Acceleration; Lincoln Aviator SUV StandingStart Acceleration; Dodge Viper SRT10 140 120 120 100 100
Speed (MPH) 160 140 Speed (MPH) 160 80 80 60 60 40 40 20 20 0
0 5 10 15 20
Time (sec) 25 30 35 0
0 40 5 10 15 20
Time (sec) 25 30 35 40 School of Mechanical Engineering
Purdue University ME575 Session 6 â€“ Dynamic Response Slide 7 Example â€“ Automotive Suspension
y
m && &
&
my + by + ky = br + kr g
k b r Response to Initial Conditions
0.02 b
k
&& + y + y = 0
&
y
m
m
&& + 28 y + 400 y = 0
&
y 0 0.02
Amplitude for free response:
&& &
my + by + ky = 0 0.04 0.06 0.08 0.1 ME575 Session 6 â€“ Dynamic Response 0 0.05 0.1 School of Mechanical Engineering
Purdue University 0.15 0.2
Time (sec) 0.25 0.3 0.35 0.4 Slide 8 4 ME575 Handouts Step Response of 2nd Order System
â€¢ Standard Form of Stable 2nd Order System (no zeros)
&& + a1 y + a 0 y = b u
&
y â‡’ && + 2 Î¶ Ï‰ n y + Ï‰ n 2 y = K Ï‰ n 2 u
&
y where Ï‰n > 0 : Natural Frequency [rad/s]
Î¶ > 0 : Damping Ratio
K : Static (Steady State, DC) Gain
Static
Pole locations
s = âˆ’Î¶Ï‰n Â± Ï‰n Img. Ï‰n ( Î¶ 2 âˆ’ 1) Overdamped (Î¶ > 1):
Critically damped (Î¶ = 1): Real Underdamped (Î¶ < 1):
âˆ’Ï‰n
School of Mechanical Engineering
Purdue University ME575 Session 6 â€“ Dynamic Response Slide 9 Step Response of 2nd Order System
Unit Step Response of Underdamped 2nd Order Systems
( u = 1 and zero ICs )
&& + 2 Î¶ Ï‰ n y + Ï‰ n 2 y = K Ï‰ n 2 u
&
y
Y ( s ) = G ( s )U ( s ) = Y (s) = K Ï‰n2
s + 2Î¶ Ï‰ n s + Ï‰ n
2 2 â€¢ K Ï‰n2
1
=
s s âŽ¡ ( s + Î¶ Ï‰ )2 + Ï‰ 2 âŽ¤
n
d
âŽ...
View
Full Document
 Fall '10
 Meckl
 Mechanical Engineering, Impulse response, Purdue University, School of Mechanical Engineering

Click to edit the document details