Answer:
The values of Kc that render this closed-loop process unstable are in the interval
(Kc < 0)
Explanation:
The transfer function of a PI controller is given as
Gc = Kc {1 + (1/sτI)}
τI = 4
Gc = Kc {1 + (1/4s)}
Gc = Kc {(4s+1)/(4s)}
Divide numerator and denominator by 4
Gc = Kc {(s+0.25)/(s)}
For a second order process, the general transfer function is given by
Gp = Kp {1/(τn²s² + 2ζτns + 1)}
Kp = 2, τn = 5 and ζ = 1.5
Gp = 2/(25s² + 15s + 1)
Divide numerator and denominator by 25
Gp = 0.08/(s² + 0.6s + 0.04)
Ga = 1
Gs = 1
We need to find the value(s) of Kc that makes the closed loop transfer function unstable. Gp*Ga*Gc*Gs + 1 = 0
The closed loop transfer function is unstable when the solution(s) of the characteristic equation obtained is positive.
Gp*Ga*Gc*Gs + 1 = 0
Becomes
[0.08/(s² + 0.6s + 0.04)] × [Kc (s+0.25)/(s)] + 1 = 0
[0.08Kc (s + 0.25)/(s³ + 0.6s² + 0.04s)] = - 1
0.08Kc (s + 0.25) = -s³ - 0.6s² - 0.04s
0.08Kc s + 0.02Kc = -s³ - 0.6s² - 0.04s
s³ + 0.6s² + 0.04s + 0.08Kc s + 0.02Kc = 0
s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0
We will use the direct substitution method to evaluate the values of Kc that matter. The values of Kc at the turning points of the closed loop transfer function.
For the substitution,
We put s = jw into the equation. (frequency analysis)
Note that j = √(-1)
s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0
(jw)³ + 0.6(jw)² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0
-jw³ - 0.6w² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0
we then collect terms with j and terms without.
(0.08Kcw + 0.04w - w³)j + (0.02Kc - 0.6w²) = 0
Meaning,
0.08Kcw + 0.04w - w³ = 0 (eqn 1)
0.02Kc - 0.6w² = 0 (eqn 2)
0.02 Kc = 0.6 w²
Kc = 15w²
Substituting this into eqn 1
0.08Kcw + 0.04w - w³ = 0
Kc = 15w²
0.08(15w²)w + 0.04w - w³ = 0
1.2w³ + 0.04w - w³ = 0
0.2w³ + 0.04w = 0
w = 0 or 0.2w² + 0.04 = 0
0.2w² = -0.04
w² = -0.2
w = ± √(-0.2)
w = ± 0.4472j or w = 0
Recall, Kc = 15w² = 15(-0.2) = -3 or Kc = 0
The turning points for the curve of the closed loop transfer function occur when
Kc = 0 or Kc = -3
To investigate, we pick values around these turning points to investigate the behaviour of the closed loop transfer function at those points.
Kc < -3, Kc = -3, (-3 < Kc < 0), Kc = 0 and Kc > 0
Note that, one positive characteristic root or pole is enough to make the system unstable.
We pick a value for Kc in that interval and evaluate the closed loop transfer function.
s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0
- First of, let Kc = - 4 (Kc < -3)
s³ + 0.6s² - 0.28s - 0.08 = 0
Solving the polynomial
s = (-0.22002), 0.44223, (-0.82221)
One positive pole means the closed loop transfer function is unstable in this region
Let Kc = -3
s³ + 0.6s² - 0.20s - 0.06 = 0
s = 0.37183, (-0.21251) or (-0.75933)
One positive pole still means that the closed loop transfer function is still unstable.
Then the next interval
Let Kc = -1
s³ + 0.6s² - 0.04s - 0.02 = 0
Solving this polynomial,
s = 0.18686, (-0.1749) or (-0.61196)
The function is unstable in the region being investigated.
Let Kc = 0
s³ + 0.6s² + 0.04s = 0
s = 0, -0.0769, -0.5236
One zero, all negative roots, indicate that the closed loop transfer function is marginally stable at this point.
Let Kc = 1, Kc > 0
s³ + 0.6s² + 0.12s + 0.02 = 0
s = (-0.42894), (-0.08553 + 0.1983j) or (-0.08553 - 0.1983j)
All the real negative parts of the poles are all negative, this indicates stability.
Hence, after examining the turning points of the closed loop transfer function, it is evident that, the region's of Kc where the closed loop transfer function is unstable is (Kc < 0)
Hope this Helps!!!