written 8.2 years ago by | • modified 5.4 years ago |
Show that (Tk1−1)(Tk2−1)=43.33, where k1 & k2 are values of ‘k’ for 0.6 & 0.2 overshoot respectively
written 8.2 years ago by | • modified 5.4 years ago |
Show that (Tk1−1)(Tk2−1)=43.33, where k1 & k2 are values of ‘k’ for 0.6 & 0.2 overshoot respectively
written 8.2 years ago by |
T.F=(C(S))(R(S))
=(G(S))(1+G(S)H(S))
=(k(S(1+TS)(1+k(S(1+TS).1)
=(k/T)(S2+ST+kT)
Comparing with standard form
(w2n)(S2+2ξwnS+w2n)
w2n =kT
Therefore, wn=√kT
2ξwn=1T
Therefore, ξ=12√kT
Now for MP=0.6,
Let ξ=ξ1
Therefore, 0.6=e−πξ1√(1−ξ12)
Therefore,ξ1=0.1602
For MP=0.2,
Let ξ=ξ2
Therefore, 0.2=e−πξ2√(1−ξ22)
Therefore, ξ2=0.4559
ξ1=0.1602, k=k1
&ξ2=0.4559, k=k2
ξ1=12√(k1T)
Therefore, 0.1602=12√(k1T)…………………(1)
ξ2=12√(k2T)
Therefore, 0.4559=12√(k2T))…………………(2)
For equation(1),
√(k2T)=12 x 0.1602 =3.1210
k1T=9.74
k1 T-1=8.7412……………….(3)
From equation (2),
√(k2T)=12 x 0.4559)
=1.09673
k2 T=1.2028
k2 T-1=0.2028…………………..(4)
From equation (3) & (4),
(k1T−1)(k2T−1)=8.74120.2028
=43.33