Root Locus Breakaway Point Calculator

Find the breakaway point and gain K on the real-axis segment of the root locus for a three-real-pole open-loop system 1/(s(s+a)(s+b)).

📍 Root Locus Breakaway Point Calculator
Pole location (a)2
0.58
Pole location (b)4
112
Breakaway point (s)
Gain K at breakaway
Other root
Valid locus segment
Step-by-step working

📍 What is a Root Locus Breakaway Point?

A root locus breakaway point is the location on the real axis where two branches of a closed-loop root locus meet as gain K increases, then split apart and move off the real axis, typically into the complex plane as a conjugate pair. For an open-loop system with three real poles and no zeros, G(s)H(s) = 1/(s(s+a)(s+b)), the breakaway point marks the specific gain at which the closed-loop system transitions from two real, converging poles to a complex-conjugate pair, the onset of oscillatory behavior in the closed-loop step response.

Control engineers find breakaway points while sketching or verifying a root locus by hand, or while checking a computer-generated plot against the expected qualitative shape. A student working through a classical control systems course locates the breakaway point to confirm where the locus leaves the real axis before it can sketch the rest of the plot correctly. A controls engineer tuning a proportional gain checks how close the operating gain sits to the breakaway value, since gains near or past it introduce oscillatory closed-loop dynamics that were not present at lower gain.

A common misconception is that solving dK/ds = 0 always gives the breakaway point directly. In reality, that derivative is zero at every stationary point of K(s) along the whole real axis, including segments where the root locus does not exist at all, only the root that falls inside a segment satisfying the root locus real-axis existence rule (an odd number of poles and zeros to the right) is a genuine breakaway point; any other root is discarded.

This calculator takes the two non-zero pole locations a and b, solves the breakaway quadratic exactly, identifies which root is the physically valid breakaway point, and returns that point along with the gain K there, the discarded root, and a labeled real-axis diagram of the poles and the valid locus segment.

📐 Formula

3s² + 2(a+b)s + ab  =  0
Open-loop system: G(s)H(s) = 1 / (s(s+a)(s+b)), poles at s = 0, −a, −b
K(s) = −1/(G(s)H(s)) = −s(s+a)(s+b)
dK/ds = −(3s² + 2(a+b)s + ab) = 0 on the real axis
Solve for s using the quadratic formula; the root inside (−a, 0) is the breakaway point
Example: a = 2, b = 4 → 3s² + 12s + 8 = 0 → s ≈ −0.845 (breakaway) and s ≈ −3.155 (not on locus) → K ≈ 3.079 at breakaway.

📖 How to Use This Calculator

Steps

1
Enter the first pole location. Type the pole location a, the pole closer to the origin.
2
Enter the second pole location. Type the pole location b, the pole further from the origin.
3
Read the breakaway point. See the breakaway point on the real axis and the gain K at that point.
4
Check the discarded root. Review the other algebraic root, which falls outside the valid locus segment and is not a physical breakaway point.

💡 Example Calculations

Example 1 — Default Poles at 0, −2, −4

a = 2, b = 4

1
3s² + 2(2+4)s + (2×4) = 0 → 3s² + 12s + 8 = 0
2
Roots: s = −0.845 and s = −3.155
3
−0.845 lies inside (−2, 0), so it is the breakaway point; −3.155 lies outside that segment
Breakaway point s ≈ −0.845, K ≈ 3.079, other root −3.155 (not on locus)
Try this example →

Example 2 — Closely Spaced Poles

a = 1, b = 3

1
3s² + 2(1+3)s + (1×3) = 0 → 3s² + 8s + 3 = 0
2
Roots: s = −0.451 and s = −2.215
3
−0.451 lies inside (−1, 0), so it is the breakaway point; −2.215 lies outside that segment
Breakaway point s ≈ −0.451, K ≈ 0.631, other root −2.215 (not on locus)
Try this example →

Example 3 — Widely Spaced Poles

a = 3, b = 6

1
3s² + 2(3+6)s + (3×6) = 0 → 3s² + 18s + 18 = 0
2
Roots: s = −1.268 and s = −4.732
3
−1.268 lies inside (−3, 0), so it is the breakaway point; −4.732 lies outside that segment
Breakaway point s ≈ −1.268, K ≈ 10.392, other root −4.732 (not on locus)
Try this example →

❓ Frequently Asked Questions

What is a breakaway point on a root locus?+
A breakaway point is where two branches of a root locus meet on the real axis and then split apart, typically diverging into the complex plane as gain K continues to increase. It marks a local maximum of K(s) along a real-axis segment where the locus exists, found by solving dK/ds = 0.
How do you find the breakaway point for G(s)H(s) = 1/(s(s+a)(s+b))?+
Set K(s) = -1/(G(s)H(s)) = -s(s+a)(s+b), then solve dK/ds = 0, which expands to the quadratic 3s^2 + 2(a+b)s + ab = 0. This quadratic has two real roots; the physically valid breakaway point is whichever root falls strictly inside a real-axis segment where the root locus actually exists.
Why does the quadratic give two roots but only one breakaway point?+
Setting dK/ds = 0 finds every stationary point of K(s) along the entire real axis, including segments where the root locus does not actually exist (an even number of poles and zeros to the right). Only the root landing inside a valid locus segment corresponds to a real breakaway point; the other root is discarded as not physically meaningful for this configuration.
Which real-axis segments does the root locus exist on for three real poles?+
For poles at 0, -a, and -b (with a under b, no zeros), the root locus exists on the segment between 0 and -a (one pole to the right, an odd count) and on the segment from -b to negative infinity (three poles to the right, also odd). The segment between -a and -b has two poles to the right, an even count, and is not part of the locus.
What does the gain K at the breakaway point represent?+
K at the breakaway point is the specific loop gain value where two closed-loop root locus branches, both still on the real axis at slightly lower gain, meet and then move off the real axis into the complex plane as complex-conjugate roots. Below this K, the closed-loop response has two real poles converging; above it, the response starts to show oscillatory (complex-pole) behavior from these two roots.
Why must a be less than b in this calculator?+
The calculator assumes a labels the pole closer to the origin and b labels the pole further away, so the valid root locus segment for the breakaway search is unambiguously (-a, 0). If a were allowed to be greater than b, the same physical system would just have its pole labels swapped, entering a less than b keeps the segment identification consistent.
Can a system like this have more than one breakaway point?+
Not for exactly three real poles and no zeros; the derivative dK/ds is a quadratic (from cubing the characteristic polynomial factors down by one degree), so it has at most two roots total, and at most one of those two roots lies inside a valid locus segment for this specific pole arrangement.
How does moving the poles further apart affect the breakaway point?+
Increasing the gap between a and b generally shifts the breakaway point closer to the pole at the origin (s = 0) and increases the gain K at which it occurs, since the segment between 0 and -a effectively has more room and the K(s) = -s(s+a)(s+b) product grows larger before its peak.
What happens to the closed-loop poles right after the breakaway point?+
Immediately after gain K exceeds the breakaway value, the two closed-loop poles that met at the breakaway point become a complex-conjugate pair and move away from the real axis, typically curving toward the imaginary axis as K increases further, which is when the closed-loop step response starts to show overshoot and oscillation.
Does this calculator work for open-loop systems with zeros?+
No. This calculator is scoped specifically to a three-real-pole, zero-free open-loop transfer function 1/(s(s+a)(s+b)). Adding zeros changes both the shape of K(s) along the real axis and the degree of the derivative equation, which can produce different numbers of breakaway and break-in points, a more general case not covered here.

What is a breakaway point on a root locus?

A breakaway point is where two branches of a root locus meet on the real axis and then split apart, typically diverging into the complex plane as gain K continues to increase. It marks a local maximum of K(s) along a real-axis segment where the locus exists, found by solving dK/ds = 0.

How do you find the breakaway point for G(s)H(s) = 1/(s(s+a)(s+b))?

Set K(s) = -1/(G(s)H(s)) = -s(s+a)(s+b), then solve dK/ds = 0, which expands to the quadratic 3s^2 + 2(a+b)s + ab = 0. This quadratic has two real roots; the physically valid breakaway point is whichever root falls strictly inside a real-axis segment where the root locus actually exists.

Why does the quadratic give two roots but only one breakaway point?

Setting dK/ds = 0 finds every stationary point of K(s) along the entire real axis, including segments where the root locus does not actually exist (an even number of poles and zeros to the right). Only the root landing inside a valid locus segment corresponds to a real breakaway point; the other root is discarded as not physically meaningful for this configuration.

Which real-axis segments does the root locus exist on for three real poles?

For poles at 0, -a, and -b (with a under b, no zeros), the root locus exists on the segment between 0 and -a (one pole to the right, an odd count) and on the segment from -b to negative infinity (three poles to the right, also odd). The segment between -a and -b has two poles to the right, an even count, and is not part of the locus.

What does the gain K at the breakaway point represent?

K at the breakaway point is the specific loop gain value where two closed-loop root locus branches, both still on the real axis at slightly lower gain, meet and then move off the real axis into the complex plane as complex-conjugate roots. Below this K, the closed-loop response has two real poles converging; above it, the response starts to show oscillatory (complex-pole) behavior from these two roots.

Why must a be less than b in this calculator?

The calculator assumes a labels the pole closer to the origin and b labels the pole further away, so the valid root locus segment for the breakaway search is unambiguously (-a, 0). If a were allowed to be greater than b, the same physical system would just have its pole labels swapped, entering a less than b keeps the segment identification consistent.

Can a system like this have more than one breakaway point?

Not for exactly three real poles and no zeros; the derivative dK/ds is a quadratic (from cubing the characteristic polynomial factors down by one degree), so it has at most two roots total, and at most one of those two roots lies inside a valid locus segment for this specific pole arrangement.

How does moving the poles further apart affect the breakaway point?

Holding a fixed and moving b further away shifts the breakaway point away from the origin, settling toward roughly the midpoint of the (-a, 0) segment rather than toward s = 0, and increases the gain K at which it occurs, since the K(s) = -s(s+a)(s+b) product grows larger before its peak.

What happens to the closed-loop poles right after the breakaway point?

Immediately after gain K exceeds the breakaway value, the two closed-loop poles that met at the breakaway point become a complex-conjugate pair and move away from the real axis, typically curving toward the imaginary axis as K increases further, which is when the closed-loop step response starts to show overshoot and oscillation.

Does this calculator work for open-loop systems with zeros?

No. This calculator is scoped specifically to a three-real-pole, zero-free open-loop transfer function 1/(s(s+a)(s+b)). Adding zeros changes both the shape of K(s) along the real axis and the degree of the derivative equation, which can produce different numbers of breakaway and break-in points, a more general case not covered here.