The linearized equation of motion for a rocket launch system

The linearized equation of motion for a rocket launch system is Ml theta - (M + m)g theta = f(t) where f(t) is the control input. M = 5000kg, m = 1500kg, l = 10m, g = 10m/s^2 The block diagram of the closed loop system is shown in Fig.7 When K(s) = 1, find the pole of the closed loop system and discuss the stability of the system. Design a control system K(s) = K_p + K_d s which provides the damping coefficient zeta = 0.3, 5% settling time t_s = 60 sec.

Solution

solution;

1)here for given closed system with unity feedback is given by

G(s)=K/[Mls^2-(M+m)g]

H(s)=1

here transfer function of system is given by

T.f.=C(s)=K/Mls^2-(M+m)g+k

here on putting value we get that

C(s)=K/[50000s^2-65000+K]

this is transfer function for given system

3)for checking stability of sytsem let find roots of numerator and denometer in transfer function equation as follows

for N(s)=

K(s)=0

S=[1-Kp]/Kd

and S=+-1.1401i

hence as two roots of eauation are imaginary with zero real part and due to presence of more root in terms of Kp and Kd system stability would depends on value of Kp and Kd for stability as for stability roots must have negative real part.

hence given system would stable for K=1

if Kp>1

Kd>0

4)here from settling time natural frquency is

wn=4/Ts*zeta

wn=4/.3*60=.222 rad/s

5)here transfer function in standard form as

c(s)=wn^2/[s^%2+2*zeta*wn*s+wn^2]

and our T.f. is

C(s)=Kp/[50000s^2+Kds+(Kp-65000)]+Kds/[50000s^2+Kds+(kp-65000)]

on equating first part of our T.F. with standard function we get that

Kd=2*zeta*wn=.1333

for

T.f.=(K/50000)/[s^2-(1.3-(K/50000))]

on equating with standard we get that

K1=2468.64

K2=67468.64

hence from equation as

K=Kp+Kds

for s to be negative we have value of Kp as

Kp1>2468.64

or Kp2>67468.64

hence we choose

Kp2>67468.64 or

Kp2=67500

hence value of k=Kp2+Kds=67500+.1333s

hence for this value K system would be stable for given damping and natural frequency