Hi,
Do a partial fraction expansion of that first equation 1/((s+a)*(s+b)).
Okay, so I have now worked through all the tasks and I got the following results:
Q1)
a) G(s) = 1/((s^2) + s(6+k) -4k + 25))
b) K=0, S = -3±4j
(I asked my lecturer and he said the frequency of oscillation in this question is the imaginary part of the poles, i.e. 4 as shown in the question)
c) When K = 4, S = -1, -9 and to get the
impulse response we do Inverse laplace of the transfer function and get the response shown below:
y1(t) = Inv Laplace G1(t) = g1(t) = (e(^-t) - e^(-9t))/8
see here:
https://www.wolframalpha.com/input/?i=y =(e^(-x) - e^(-9x))/8
When K = 6.25, S = 0, -12.25
y2(t) = Inv Laplace G2(t) = g2(t) = (1/(12.25))*(1-e^(-12.25t))
see here:
https://www.wolframalpha.com/input/?i=y+=+(1/(12.25))*(1-e^(-12.25x))
Steady State error:
Unit step R(s) = 1/s
Since Y(s) = R(s)G(s)
When K = 4:
E(s) = R(s) - Y(s) = R(s)(1-G(s))
E(s) = (1/s)*(1-(1/s^(2)+10s+9))
**Using final value theorem**
e∞ = Lim s->0 sE(s) =(1-(1/9)) = 8/9
When K = 6.25
E(s) = (1/s)*(1-(1/s^(2)+12.25s))
e∞ = (1-(1/0)) = undefined
d)
system is critically damped when ζ = 1.
using the general characteristic equation:
s^(2)+ 2ζwns + wn^(2) = 0
It's critically damped when k = 2.12, -30.12
(if these values are correct then getting the poles are easy, since we only need to substiture them into the characteristic equation and solve for s)
e)
S(T,k) = k(s-4)/((s^2) + s(6+k) -4k + 25))
It's at this point I am unsure what to do... if I did the sensitivity correct, then what happens as K changes? do we look at S(T,k) as k->∞ and k->0?
Thanks for any comments.
Fouad.