- #1
ATRIX
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1. Homework Statement [/b]
Hey Guys
Firstly thanks for looking, I really appreciate any help or interest with my problem.
The problem is a closed loop feedback system it's unlike any I have come across before.
I have been given the definition which is as follows:-
G(t)^-1 = 2.5 x 10 ^-3 {(1-400k)}D^2+10D+400
i have to prove it with the following information.
I = Inertia of the turbine blade is = 1kg/m
c.ω= viscous Damped rotation (hysteresis) = 10 rad/s Nm
T = Torque motor= 400[e + k d^2θ/dt^2] Nm
e=(θin-θout) = angular position error in rads between the input and output shaft
k.d^2θout/dt^2= defines the additional feedback signal, accelerometer i am guessing
The definition or the answer that we are trying to work it into is in D notation, or operator D. this is where we can replace the derivatives of angular acceleration or position with D
there fore
D = dθ/dt
D^2 = d^2θ/dt^2
general format of second order system
θout/θin = k/(1/ωn^2)D^2+(2ζ/ωn)D+1
this is a general solution
Zeta is a damping coeff of the system
Newton's 2nd law
∑T=∑I.α
As for the work i have carried out, is that i am trying to reverse engineer the definition of the system to get the numbers, i can see that the answer is quadratic, D^2 and D and 400. but can not seem to get from the original boundary conditions to there. i have also tried to split the loops up into two. this did not work for me.
using basic control algebra i have deduced that the accelerometor should e k. d^2θ/dt^2 not
k.d^2θout/dt^2 this i believe is a mistake on my question paper.
thanks for looking. any advice would be welcomed.
Atrix
Hey Guys
Firstly thanks for looking, I really appreciate any help or interest with my problem.
The problem is a closed loop feedback system it's unlike any I have come across before.
I have been given the definition which is as follows:-
G(t)^-1 = 2.5 x 10 ^-3 {(1-400k)}D^2+10D+400
i have to prove it with the following information.
I = Inertia of the turbine blade is = 1kg/m
c.ω= viscous Damped rotation (hysteresis) = 10 rad/s Nm
T = Torque motor= 400[e + k d^2θ/dt^2] Nm
e=(θin-θout) = angular position error in rads between the input and output shaft
k.d^2θout/dt^2= defines the additional feedback signal, accelerometer i am guessing
The definition or the answer that we are trying to work it into is in D notation, or operator D. this is where we can replace the derivatives of angular acceleration or position with D
there fore
D = dθ/dt
D^2 = d^2θ/dt^2
Homework Equations
general format of second order system
θout/θin = k/(1/ωn^2)D^2+(2ζ/ωn)D+1
this is a general solution
Zeta is a damping coeff of the system
Newton's 2nd law
∑T=∑I.α
The Attempt at a Solution
As for the work i have carried out, is that i am trying to reverse engineer the definition of the system to get the numbers, i can see that the answer is quadratic, D^2 and D and 400. but can not seem to get from the original boundary conditions to there. i have also tried to split the loops up into two. this did not work for me.
using basic control algebra i have deduced that the accelerometor should e k. d^2θ/dt^2 not
k.d^2θout/dt^2 this i believe is a mistake on my question paper.
thanks for looking. any advice would be welcomed.
Atrix
