G(S) =K(S(1+S)(1+0.4S)
H(S) =1, unity feedback.
(1) For Ramp input
r (t)=4t,
k=2
Steady state error,
ess=Akv
kv= lim(s→0) S G(S) H(S)
=lim(s→0) Sk(S(1+S)(1+0.4S)
=lim(s→0) S2(S(1+S)(1+0.4S)
=2/(1)(1)
kv = 2
ess=Akv ,
As A=2 for Ramp input
ess=2/2
e_ss=1
(2) ess=0.2, k=?
same input Ramp of magnitude
A=2
kv=lim(s→0) S. G(S) H(S)
=lim(s→0) Sk(S(1+S)(1+0.4S)
kv=k/(1)(1)
kv=k
ess=Akv
=Ak
0.2=2/k
k=2/0.2
k=10
(3) Steady state error for input step of magnitude A_1=2 and Ramp of magnitude A_2=6.
k=10.
For step input, position error coefficient
kp=lim(s→0) G(S) H(S)
=lim(s→0) k(S(1+S)(1+0.4S) . 1
=lim(s→0) 10(S(1+S)(1+0.4S)
=lim(s→0) 10(0(1)(1)
=10/0
kp=∞
For Ramp input velocity error coefficient,
kv=lim(s→0) S G(S) H(S)
=lim(s→0) Sk(S(1+S)(1+0.4S) . 1
=lim(s→0) S10(S(1+S)(1+0.4S) .1
=lim(s→0) 10(1)(1)
=10
kv=10
ess=A1(1+kp)+A2kv
=2∞+610
=0+0.6
ess=0.6