Question:medium

The value of K for which the unity feedback system \( G(s) = \frac{K{S(S+2)(S+4)} \) crosses the imaginary axis is

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For any standard third-order characteristic equation \(as^3 + bs^2 + cs + d = 0\), the marginal stability boundary condition for crossing the imaginary axis can be found directly using the shorthand product formula: \[ \text{Inner Coefficients Product} = \text{Outer Coefficients Product} \implies bc = ad \] Applying it here: \(6 \times 8 = 1 \times K \implies K = 48\). You can solve this in under 5 seconds!
Updated On: Jul 4, 2026
  • 2
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  • 6
  • 48
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The Correct Option is D

Solution and Explanation

Understanding the Concept: A continuous control system crosses the imaginary axis when its closed-loop poles move from the left half of the s-plane onto the \(j\omega\) axis, signifying the boundary condition for marginal stability. This critical gain value can be found by evaluating the system's characteristic equation using the Routh-Hurwitz stability criterion. For a unity feedback configuration, the characteristic equation is: \[ 1 + G(s)H(s) = 0 \implies 1 + G(s) = 0 \]

Step 1: Construct the characteristic polynomial equation.

Given the open-loop transfer function: \[ G(s) = \frac{K}{s(s+2)(s+4)} \] Setting up the closed-loop characteristic relation: \[ 1 + \frac{K}{s(s+2)(s+4)} = 0 \] \[ s(s+2)(s+4) + K = 0 \] Expanding the polynomial factor terms: \[ s(s^2 + 6s + 8) + K = 0 \] \[ s^3 + 6s^2 + 8s + K = 0 \]

Step 2: Construct the Routh Hurwitz array.

Using the coefficients of our third-order characteristic equation, we fill out the standard Routh rows: {c|cc} \(s^3\) & 1 & 8
\(s^2\) & 6 & \(K\)
\(s^1\) & \(\frac{(6 \times 8) - (1 \times K)}{6}\) & 0
\(s^0\) & \(K\) & Simplifying the term in the \(s^1\) row: \[ \frac{48 - K}{6} \]

Step 3: Solve for marginal stability boundary condition.

For the system to cross the imaginary axis and sustain marginal oscillations, a complete row of zeros must appear in the array, or a coefficient in the primary column must equal zero. Setting the \(s^1\) row element to zero: \[ \frac{48 - K}{6} = 0 \] \[ 48 - K = 0 \implies K = 48 \] Thus, when the amplifier loop gain \(K\) reaches exactly 48, the system crosses the imaginary axis. This matches option (D).
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