Question

For the given open-loop system, \[ G(s)=\frac{1}{(s+2)(s+4)}, \] the poles of the unity feedback closed-loop system are:

• A. \(-3, 3\)
• B. \(-1, -3\)
• C. \(-3, -3\)
• D. \(1, 3\)

Hint:

For unity feedback configurations with \(G(s) = \frac{1}{s^2 + bs + c}\), the closed-loop characteristic equation simplifies instantly to \(s^2 + bs + (c+1) = 0\). Here, \(s^2 + 6s + 8 \rightarrow s^2 + 6s + 9 = 0\), which yields the real double root \(-3, -3\).

Answer 1

The Correct Option is C

Concept: For a system with an open-loop transfer function \(G(s)\) and a unity negative feedback layout (\(H(s) = 1\)), the closed-loop transfer function is given by: \[ T(s) = \frac{G(s)}{1 + G(s)} \] The positions of the closed-loop poles are found by computing the roots of the characteristic equation: \[ 1 + G(s) = 0 \]

Step 1: Substitute the given open-loop function into the characteristic equation. The open loop transfer function given is: \[ G(s) = \frac{1}{(s+2)(s+4)} \] Set up the equation: \[ 1 + \frac{1}{(s+2)(s+4)} = 0 \]

Step 2: Convert into polynomial form. Multiply through by the denominator term \((s+2)(s+4)\): \[ (s+2)(s+4) + 1 = 0 \] Expanding the product of linear factors: \[ s^2 + 4s + 2s + 8 + 1 = 0 \] \[ s^2 + 6s + 9 = 0 \]

Step 3: Factorize the quadratic equation to find the roots. The resulting quadratic polynomial can be recognized as a perfect square trinomial: \[ s^2 + 2(3)s + 3^2 = 0 \quad \Rightarrow \quad (s+3)^2 = 0 \] Solving for \(s\): \[ (s+3)(s+3) = 0 \quad \Rightarrow \quad s = -3, -3 \] Thus, the system has real, repeated closed-loop poles situated at \(-3\) and \(-3\), corresponding to option (C).