Question

B2. Find the steady-state errors for inputs of 5 u(t), 5t-u(t), and 5t-u(t) to the system shown in the following figure. The function u(t) is the unit step. R(S) + E(S) C($) 120(s + 2) (s + 3)(s +4)

Answer 1

For the given G(s) and H(s) values in the diagram, with the given inputs mentioned, the value of steady state error is calculated as below:

As G(s) is having type as Type 0, since no pole at zero.

E(S) Given closed loop system is Ris) C(s) 120(8+2) (S +35+4) Where G(s) = and 120 (5+2) (5+3)(3+4) H(s) = 1. state error, éss= Lim sto 1 if if input input = R() = R4) = sult) SR(S) Steady 1 +G(s) H) R (S) – Laplace of [5 ult)] R (s) = 38 = f, by comparing [A=5 Lim Gi(s) H(s) Static position error constant Kp = [12015+2) Kp sto [(5+3) (s+4)- Kp = 12012) (3) (4) Steady state error 1 tkp sto Lim 1 = 20 A 5 ess 0.238 ' s 1t20 21

€ if input; r(t) = 5t ult) R(S) = 5 S2 - A -) By comparing (A=5 S2 ess A Kv Steady state error Ky = Static , velocity Lim S G(s)H(s) error constant sto 120(8+2) S 1 Lim sto C+2)(534) ] Kv=0 ess Kv 2 6 if input, rct) = struct) = 10 & ult) 10 R(s) = A S3 By comparing A=10 S3 state error = A Ka Ess, Steady ka = static acceleration error constant. Toolste) ]:(2) Lim s Lim Ž Ge(s) H(s) so sto (8+3)(3+4) ko ess A ka For inputs Stult) and stult), ess is Zero as it is Type - Zero (o) system.

So for type zero system, steady state error is infinity for ramp and parabolic inputs.