Answer ALL Questions The system shown below has been tested with three different reference inputs 6u(t), 6tu(t), and 6t2u(t). By using steady-state error calculation, identify which could give zero (0) steady state error. The function u(t) is the unit step. R(s) + E(s) 100(s+2)(s+6) s(s+3)(s+4) C(s) Example #1 Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t²u(t) to the system shown below. The function u(t) is the unit step. R(s) + E(s) 120(s+2) (s+3)(s+4) C(s) For step input 5u(t), we must calculate the position error coefficient (Kp): 6 Chapter Eight: Steady-State Error Benha University Faculty of Engineering at Shubra Dr. Ahmed Mustafa Hussein Electrical Engineering Department Dr. Ahmed Mustafa Hussein 120 x 2 3 x 4 = 20 Кр = lim G(S) = 0-S 5 5 = Ess= 1 +20 21 For ramp input 5tu(t), we must calculate the velocity error coefficient (Kv): 0 x 120 x 2 3 x 4 Kv = lim S G(S) = S→0 5 Ess= =∞ 0 = 0 For parabolic input 5t² u(t), we must calculate the acceleration error coefficient (Ka): Kalim S2 G(S) = S-0 Ess = 2 × 5 0 0 x 120 x 2 3 × 4 =00 = 0

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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The system shown below has been tested with three different reference inputs 6u(t),
6tu(t), and 6t2u(t). By using steady-state error calculation, identify which could give
zero (0) steady state error. The function u(t) is the unit step.
R(s) +
E(s)
100(s+2)(s+6)
s(s+3)(s+4)
C(s)
Transcribed Image Text:Answer ALL Questions The system shown below has been tested with three different reference inputs 6u(t), 6tu(t), and 6t2u(t). By using steady-state error calculation, identify which could give zero (0) steady state error. The function u(t) is the unit step. R(s) + E(s) 100(s+2)(s+6) s(s+3)(s+4) C(s)
Example #1
Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t²u(t) to the system shown
below. The function u(t) is the unit step.
R(s) + E(s)
120(s+2)
(s+3)(s+4)
C(s)
For step input 5u(t), we must calculate the position error coefficient (Kp):
6
Chapter Eight: Steady-State Error
Benha University
Faculty of Engineering at Shubra
Dr. Ahmed Mustafa Hussein
Electrical Engineering Department
Dr. Ahmed Mustafa Hussein
120 x 2
3 x 4
= 20
Кр
=
lim G(S)
=
0-S
5
5
=
Ess=
1 +20 21
For ramp input 5tu(t), we must calculate the velocity error coefficient (Kv):
0 x 120 x 2
3 x 4
Kv
=
lim S G(S)
=
S→0
5
Ess=
=∞
0
= 0
For parabolic input 5t² u(t), we must calculate the acceleration error coefficient (Ka):
Kalim S2 G(S) =
S-0
Ess
=
2 × 5
0
0 x 120 x 2
3 × 4
=00
= 0
Transcribed Image Text:Example #1 Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t²u(t) to the system shown below. The function u(t) is the unit step. R(s) + E(s) 120(s+2) (s+3)(s+4) C(s) For step input 5u(t), we must calculate the position error coefficient (Kp): 6 Chapter Eight: Steady-State Error Benha University Faculty of Engineering at Shubra Dr. Ahmed Mustafa Hussein Electrical Engineering Department Dr. Ahmed Mustafa Hussein 120 x 2 3 x 4 = 20 Кр = lim G(S) = 0-S 5 5 = Ess= 1 +20 21 For ramp input 5tu(t), we must calculate the velocity error coefficient (Kv): 0 x 120 x 2 3 x 4 Kv = lim S G(S) = S→0 5 Ess= =∞ 0 = 0 For parabolic input 5t² u(t), we must calculate the acceleration error coefficient (Ka): Kalim S2 G(S) = S-0 Ess = 2 × 5 0 0 x 120 x 2 3 × 4 =00 = 0
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