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Step 2: Calculate the break-in point.

The characteristic equation is:
1 + G(s)H(s) = 0, so 1 + \frac{K(s+2)}{s^2-4s+13} = 0.

This leads to:
s^2 - 4s + 13 + K(s + 2) = 0

K = -\frac{s^2-4s+13}{s+2}

To find the break-in point, we need to find:
\frac{dK}{ds} = 0.

\frac{dK}{ds} = \frac{(2s-4)(s+2)-(s^2-4s+13)(1)}{(s+2)^2} = 0

(2s-4)(s+2) - (s^2 - 4s + 13) = 0

2s^2 + 4s - 4s - 8 - s^2 + 4s - 13 = 0

s^2 + 4s - 21 = 0

(s+7)(s-3) = 0

Thus, s = -7 or s = 3.

Answer :

To solve this problem, we are looking into a control systems question related to determining the break-in points using the characteristic equation derived from the open-loop transfer function. Break-in points occur where portions of the root locus enter the real axis. Here’s a detailed explanation of how to find the break-in point:

  1. Characteristic Equation: We start with the characteristic equation given as:
    [tex]1 + \frac{K(s+2)}{s^2-4s+13} = 0[/tex]
    Rearranging this, we have:
    [tex]s^2 - 4s + 13 + K(s + 2) = 0[/tex]
    From this, we solve for [tex]K[/tex]:
    [tex]K = -\frac{s^2-4s+13}{s+2}[/tex]

  2. Finding the Break-in Point:
    To find where the root locus breaks into the real axis, differentiate [tex]K[/tex] with respect to [tex]s[/tex] and set the derivative to zero:
    [tex]\frac{dK}{ds} = \frac{(2s-4)(s+2)-(s^2-4s+13)(1)}{(s+2)^2} = 0[/tex]

  3. Simplification and Solving:
    Solve the numerator for zero:
    [tex](2s-4)(s+2) - (s^2-4s+13) = 0[/tex]
    Simplifying, we have:
    [tex]2s^2 + 4s - 4s - 8 - s^2 + 4s - 13 = 0[/tex]
    [tex]s^2 + 4s - 21 = 0[/tex]

  4. Finding Break-in Points:
    Solve the quadratic equation above:
    [tex](s+7)(s-3) = 0[/tex]
    Thus, the solutions are:
    [tex]s = -7 \text{ or } s = 3[/tex]

Therefore, the root locus breaks in at [tex]s = -7[/tex] and [tex]s = 3[/tex]. This means the trajectory of the root locus will enter the real axis at these points.

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