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The open loop T.F of unity feedback system is G(S)=kS(1+TS), for this system overshoot reduces from 0.6 to 0.2 due to change in k only.

Show that (Tk11)(Tk21)=43.33, where k1 & k2 are values of ‘k’ for 0.6 & 0.2 overshoot respectively

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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=12kT

Now for MP=0.6,

Let ξ=ξ1

Therefore, 0.6=eπξ1(1ξ12)

Therefore,ξ1=0.1602

For MP=0.2,

Let ξ=ξ2

Therefore, 0.2=eπξ2(1ξ22)

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),

(k1T1)(k2T1)=8.74120.2028

=43.33

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