Determine the gain at the breakaway point for the open-loop transfer function G(s)H(s) = \(\rm \frac{K}{s(s+1)}\)

Explanation:

To determine the gain at the breakaway point for the open-loop transfer function \( G(s)H(s) = \frac{K}{s(s+1)} \), we need to follow the steps involved in root locus analysis. The breakaway point is the point on the real axis where the root locus leaves or enters the real axis.

Step-by-Step Solution:

Step 1: Identify the poles and zeros of the open-loop transfer function.

The open-loop transfer function is given by:

\( G(s)H(s) = \frac{K}{s(s+1)} \)

From this transfer function, we can see that there are two poles:

• Poles: \( s = 0 \) and \( s = -1 \)
• Zeros: None

Step 2: Determine the characteristic equation.

The characteristic equation is obtained by setting the denominator of the closed-loop transfer function to zero:

\( 1 + G(s)H(s) = 0 \)

Substitute \( G(s)H(s) \):

\( 1 + \frac{K}{s(s+1)} = 0 \)

Multiply through by \( s(s+1) \) to clear the fraction:

\( s(s+1) + K = 0 \)

Or:

\( s^2 + s + K = 0 \)

This is the characteristic equation of the system.

Step 3: Find the breakaway point.

The breakaway point occurs where the root locus intersects the real axis between the poles. To find the breakaway point, we need to find the values of \( s \) where the derivative of the characteristic equation with respect to \( s \) is zero:

\( \frac{d(s^2 + s + K)}{ds} = 0 \)

Differentiate the characteristic equation with respect to \( s \):

\( 2s + 1 = 0 \)

Solve for \( s \):

\( 2s + 1 = 0 \)

\( 2s = -1 \)

\( s = -\frac{1}{2} \)

The breakaway point is at \( s = -\frac{1}{2} \).

Step 4: Determine the gain \( K \) at the breakaway point.

Substitute \( s = -\frac{1}{2} \) into the characteristic equation to find the gain \( K \):

\( \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + K = 0 \)

Simplify the equation:

\( \frac{1}{4} - \frac{1}{2} + K = 0 \)

\( \frac{1}{4} - \frac{2}{4} + K = 0 \)

\( -\frac{1}{4} + K = 0 \)

Solve for \( K \):

\( K = \frac{1}{4} \)

Therefore, the gain at the breakaway point is \( K = 0.25 \).

Correct Option:

The correct option is:

Option 1: 0.25

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: -0.25

This option is incorrect because the gain \( K \) at the breakaway point cannot be negative. The calculation shows that the correct value is \( 0.25 \), not \( -0.25 \).

Option 3: -0.5

Similar to option 2, this option is incorrect because the gain \( K \) at the breakaway point cannot be negative. The correct value is \( 0.25 \), not \( -0.5 \).

Option 4: 0.5

This option is incorrect because the gain \( K \) at the breakaway point is \( 0.25 \), not \( 0.5 \). The calculation clearly shows that the value is \( 0.25 \).

Conclusion:

Understanding the root locus method and the process of finding the breakaway point is essential for control system analysis. The breakaway point for the given open-loop transfer function \( G(s)H(s) = \frac{K}{s(s+1)} \) was found to be at \( s = -\frac{1}{2} \), and the corresponding gain \( K \) at this point was determined to be \( 0.25 \). This analysis confirms that the correct option is indeed option 1.