Does a food market have any rooms in particular? Also whats units?
2 answers:
You might be interested in
Answer:
The error e(∝) turns 0 when Gc(s) has a double integrator.
(See attachment)
Explanation:
At any point between the disturbance entry point and the error measuring element, there's an integrator which served as an intermediate such that R(s) = 0.
Given that
C(s)/D(s) = G(s)/(1 + Gc(s)G(s))
And
E(s) = D(s) - Gc(s)C(s) --- Divide through by D(s)
E(s)/D(s) = 1 - Gc(s)C(s)/D(s)
E(s)/D(s) = 1 - (C(s)/D(s))Gc(s)
Substitute G(s)/(1 + Gc(s)G(s)) for C(s)/D(s)
E(s)/D(s) = 1 - (G(s)/(1 + Gc(s)G(s)))Gc(s)
E(s)/D(s) = 1 - (G(s)Gc(s))/(1 + Gc(s)G(s))
E(s)/D(s) = (1 + Gc(s)G(s) - G(s)Gc(s))/(1 + Gc(s)G(s))
E(s)/D(s) = 1/((1 + Gc(s)G(s))
E(s) = D(s)/((1 + Gc(s)G(s))
For ramp disturbance at d(t) = at
D(s) = a/s².
From this, the steady state error equals.....
e(∝) = lim s->0 s[E(s)]
Substitute D(s)/((1 + Gc(s)G(s)) for E(s)
e(∝) = lim s->0 s[D(s)/((1 + Gc(s)G(s))]
Substitute a/s² for D(s)
e(∝) = lim s->0 s[(a)/(s²(1 + Gc(s)G(s))]
e(∝) = lim s->0 s[(a)/((s² + s²Gc(s)G(s))]
e(∝) = lim s->0 s[(a)/((s + sGc(s)G(s))]
e(∝) = lim s->0 s[(a)/(sGc(s)G(s))]
The error e(∝) turns 0 when Gc(s) has a double integrator.
I think this is helpful for you
