Question

In the feedback control system shown in the figure below, \[ G(s) = \frac{6}{s(s+1)(s+2)}. \]\(R(s)\), \(Y(s)\), and \(E(s)\) are the Laplace transforms of \(r(t)\), \(y(t)\), and \(e(t)\), respectively. If the input \(r(t)\) is a unit step function, then:

 

• A. \(\lim_{t \to \infty} e(t) = 0\)
• B. \(\lim_{t \to \infty} e(t) = \frac{1}{3}\)
• C. \(\lim_{t \to \infty} e(t) = \frac{1}{4}\)
• D. \(\lim_{t \to \infty} e(t) { does not exist, } e(t) { is oscillatory.}\)

Hint:

To analyze the steady-state behavior in control systems, examine the poles of the error signal's transfer function. Complex poles indicate persistent oscillations in \(e(t)\).

Answer 1

The Correct Option is D

Step 1: Represent the transfer function and input.
The open-loop transfer function is: \[ G(s) = \frac{6}{s(s+1)(s+2)}. \] The input \(r(t)\) is a unit step function, so its Laplace transform is: \[ R(s) = \frac{1}{s}. \] Step 2: Derive the error signal \(E(s)\).
In a feedback system: \[ E(s) = R(s) - Y(s), \] where \(Y(s)\) is the output, given by: \[ Y(s) = G(s)E(s) \quad \Rightarrow \quad E(s) = \frac{R(s)}{1 + G(s)}. \] Substituting \(G(s)\) and \(R(s)\): \[ E(s) = \frac{\frac{1}{s}}{1 + \frac{6}{s(s+1)(s+2)}} = \frac{\frac{1}{s}}{\frac{s(s+1)(s+2) + 6}{s(s+1)(s+2)}} = \frac{s^2 + 3s + 2}{s(s^2 + 3s + 8)}. \] Step 3: Analyze the poles of \(E(s)\).
The denominator of \(E(s)\) is: \[ s(s^2 + 3s + 8). \] The quadratic term \(s^2 + 3s + 8\) has complex conjugate roots because the discriminant is negative: \[ \Delta = 3^2 - 4(1)(8) = 9 - 32 = -23. \] Hence, the roots are: \[ s = -\frac{3}{2} \pm j\frac{\sqrt{23}}{2}. \] Step 4: Determine the behavior of \(e(t)\).
Since \(E(s)\) has complex conjugate poles, the error signal \(e(t)\) contains oscillatory components. These oscillations persist as \(t \to \infty\), so the limit of \(e(t)\) does not exist. Final Answer: \[\boxed{{(4) } \lim_{t \to \infty} e(t) { does not exist, } e(t) { is oscillatory.}}\]