Rules for Sketching Root Locus (with Examples) | Control Systems - Electrical Engineering (EE) PDF Download

The root locus is a graphical representation in the s-domain that shows how the closed-loop poles of a feedback system move in the complex s-plane as a system parameter (usually the loop gain K) varies. The root locus is always symmetric about the real axis because open-loop poles and zeros are either real or occur in complex conjugate pairs. The following sections give the standard rules used to construct (sketch) the root locus and illustrate them with examples.

Root Locus

Rule Number 1 - Symmetry

• A root locus plot is symmetric about the real axis.

Rule Number 2 - Number of Branches (P versus Z)

• Let P = number of open-loop poles and Z = number of open-loop zeros (finite zeros). The number of branches N of the root locus equals the larger of P and Z. Each branch starts at an open-loop pole (when K→0) and terminates at an open-loop zero (when K→∞). If there are fewer finite zeros than poles, the remaining branches go to infinity. If there are more zeros than poles, some branches originate from infinity and terminate at zeros.
• Two common cases:
  • Case 1: P > Z. N = P. All branches start at the P open-loop poles. Z branches terminate at the finite zeros. The remaining P - Z branches go to infinity.
  • Case 2: Z > P. N = Z. Z branches terminate at the finite zeros. P branches start at the open-loop poles. The remaining Z - P branches originate from infinity and end at the extra finite zeros.
• Examples from input:
  • P = 3, Z = 1 → 3 branches start at poles; 1 terminates at finite zero; 2 go to infinity.
  • P = 1, Z = 3 → 3 branches in total; 1 starts at the pole; 2 originate from infinity and terminate at finite zeros.

Rule Number 3 - Root Locus on the Real Axis

• A point on the real axis belongs to the root locus if and only if the total number of real open-loop poles and real open-loop zeros to the right of that point is odd.
• This rule determines which segments of the real axis are part of the locus.
• Worked illustration (retain original values): Poles at -2 and -4; zeros at -1 and -3. Determine whether points -2.2 and -3.4 lie on the locus.

Point -2.2:

The point -2.2 lies between -2 and -3.

The number of poles and zeros to the right of -2.2 is 2 (one pole at -2 and one zero at -1).

2 is even, so -2.2 is not on the root locus.

Point -3.4:

The point -3.4 lies between -3 and -4.

The number of poles and zeros to the right of -3.4 is 3 (one pole at -4? - check positions carefully: actually to the right are poles/zeros at -2, -3, and -1 depending on placement). According to the given example, the count is 3, which is odd.

3 is odd, so -3.4 lies on the root locus.

Short Real-axis Example

Example: Find on which sections of the real axis the root locus exists for a system whose poles are 0, -3, -5 and zeros are -1 and -4.

Poles: 0, -3, -5. Zeros: -1, -4. There are P = 3 poles and Z = 2 zeros.

• Section between 0 and -1 (example point -0.4): points to the right are {0} only → count = 1 (odd). So this section is on the root locus.
• Section between -1 and -3 (example point -2.1): points to the right are {0, -1} → count = 2 (even). So not on the root locus.
• Section between -3 and -4 (example point -3.5): points to the right are {0, -1, -3} → count = 3 (odd). So on the root locus.
• Section between -4 and -5 (example point -4.3): points to the right are {0, -1, -3, -4} → count = 4 (even). So not on the root locus.
• Section left of -5 (example point -8.6): points to the right are {0, -1, -3, -4, -5} → count = 5 (odd). So on the root locus.

Rule Number 4 - Asymptotes (Branches to Infinity)

• If P - Z > 0 there are (P - Z) asymptotes describing directions of branches that go to infinity.
• The angles θq of the asymptotes are given by:

θq = (2q + 1) · 180° / (P - Z)

where q = 0, 1, 2, ..., (P - Z - 1). The asymptotes are symmetric about the real axis.

Rule Number 5 - Centroid of Asymptotes

• The asymptotes intersect the real axis at a point called the centroid. The centroid σa is given by:

σa = (sum of finite pole locations - sum of finite zero locations) / (P - Z)

The centroid is a real number and may lie on or off the root locus depending on pole/zero placement.

Rule Number 6 - Breakaway and Break-in Points

• A breakaway point is a point on the root locus where two or more branches depart from the real axis toward the complex plane as K varies. A break-in point is where branches on the real axis meet (come in) from complex locations and join the real axis. Both are locations where multiple roots of the characteristic equation occur for a specific value of K.
• The breakaway / break-in points are obtained from the condition that K as a function of s has an extremum with respect to s on the real axis. For a characteristic equation written as 1 + K·N(s)/D(s) = 0, the explicit expression is

K(s) = -D(s) / N(s)

The breakaway/break-in points satisfy

dK(s)/ds = 0

• When N(s) = 1 (or when G(s)H(s) = K/D(s)), the formula simplifies to K(s) = -D(s), and dK/ds = 0 reduces to dD/ds = 0. Solve dK/ds = 0 for s and accept roots that lie on real-axis sections already identified as root-locus segments.
• At a breakaway point where N branches meet or leave the real axis, the branches leave at angle 180°/N (the sign may be positive or negative depending on orientation).
• There is at least one breakaway point between any two adjacent real poles if the real axis segment joining them is part of the root locus (Rule 3).

Key Points - Procedure to Sketch Root Locus

1. Identify number and locations of open-loop poles and zeros from G(s)H(s).
2. Mark poles (×) and zeros (o) on the s-plane.
3. Use Rule 3 to determine which real-axis sections are part of the locus.
4. Compute P - Z. If positive, compute asymptote angles using θq = (2q + 1)·180°/(P - Z).
5. Compute the centroid σa = (sum of poles - sum of zeros)/(P - Z) and draw asymptotes through σa at the computed angles.
6. Find breakaway and break-in points from dK/ds = 0 and verify they lie on real-axis locus segments.
7. Compute imaginary-axis crossings (if any) using Routh-Hurwitz or by substituting s = jω and solving for K and ω.
8. Find angles of arrival/departure at complex zeros/poles when needed by applying the angle condition.
9. Combine all the above to sketch the final root locus and use it to assess closed-loop stability and performance as K varies.

Worked Example - Complete Root Locus for K / [s(s + 5)(s + 10)]

Problem: Draw the root locus diagram for a closed-loop system whose loop transfer function is

Also determine stability ranges of K.

Solution (stepwise reasoning and calculations):

Identify open-loop poles and zeros.

Poles are the roots of s(s + 5)(s + 10) so:

Poles: s = 0, s = -5, s = -10.

There are no finite zeros (Z = 0).

P = 3, Z = 0, so P - Z = 3. There will be three branches, all starting at the poles and all going to infinity.

Determine which real-axis sections belong to the root locus using Rule 3.

Consider a point in each interval and count poles and zeros to the right:

Segment between 0 and -5 (example -3.5): there is one pole to the right → odd → included.

Segment between -5 and -10 (example -7): there are two poles to the right → even → not included.

Segment left of -10 (example -12): there are three poles to the right → odd → included.

Therefore, root locus real segments: between 0 and -5, and left of -10.

Find asymptote angles.

Use θq = (2q + 1)·180°/(P - Z) with P - Z = 3 and q = 0, 1, 2.

For q = 0: θ = 180°/3 = 60°.

For q = 1: θ = 3·180°/3 = 180°.

For q = 2: θ = 5·180°/3 = 300° (or -60° equivalently).

Find centroid σa.

Sum of poles = 0 + (-5) + (-10) = -15.

Sum of zeros = 0 (no finite zeros).

σa = (sum of poles - sum of zeros) / (P - Z) = (-15 - 0) / 3 = -5.

Find breakaway point(s) on the real axis between 0 and -5.

Characteristic equation: 1 + G(s)H(s) = 0.

Substitute G(s)H(s) = K / [s(s + 5)(s + 10)].

Characteristic: 1 + K / [s(s + 5)(s + 10)] = 0.

Multiply both sides by denominator:

s(s + 5)(s + 10) + K = 0.

Rewriting:

s^3 + 15s^2 + 50s + K = 0.

Express K as a function of s:

K(s) = -s^3 - 15s^2 - 50s.

Differentiate with respect to s and set derivative to zero for breakaway/break-in candidates:

dK/ds = -(3s^2 + 30s + 50) = 0.

Therefore:

3s^2 + 30s + 50 = 0.

Divide by 3:

s^2 + 10s + 16.6666667 = 0.

Use quadratic formula with a = 1, b = 10, c = 16.6666667.

s = [-b ± sqrt(b^2 - 4ac)] / (2a).

Compute discriminant and roots:

Discriminant = 100 - 66.6666668 ≈ 33.3333332.

sqrt(Discriminant) ≈ 5.7735.

Roots: s ≈ (-10 ± 5.7735)/2 → s ≈ -2.1133 and s ≈ -7.8867.

Accept only the root lying on the real-axis segment that belongs to the locus (between 0 and -5): s ≈ -2.1133 is between 0 and -5 and is therefore the breakaway point.

Verify K at the breakaway point by substituting s = -2.1133 into K(s):

K = -(-2.1133)^3 - 15(-2.1133)^2 - 50(-2.1133).

Compute K numerically (as given in the example): K ≈ 48.112 (positive), so this is a valid breakaway point for positive K.

Find imaginary-axis crossings using Routh-Hurwitz criterion applied to the characteristic polynomial s^3 + 15s^2 + 50s + K = 0.

Construct the Routh array (top rows shown):

Row s^3: 1 50

Row s^2: 15 K

Row s^1: (15·50 - 1·K)/15 0

Row s^0: K

Simplify s^1 row first element:

(750 - K)/15.

For a purely imaginary pair to appear (marginal stability), the s^1 row first element must be zero:

(750 - K)/15 = 0.

So K = 750.

Substitute K = 750 into s^2 row to find the imaginary roots.

s^2 row corresponds to the auxiliary equation 15 s^2 + K = 0.

With K = 750:

15 s^2 + 750 = 0.

15 s^2 = -750.

s^2 = -50.

s = ± j√50 = ± j7.071.

Thus, at K = 750 the system has imaginary-axis roots at s = ± j7.071; the system is marginally stable at this value.

Angle of departure/arrival: not required here because there are no finite complex poles or zeros (all poles are real).

Conclusion on stability:

For K in the interval 0 < K < 750 the closed-loop poles remain in the left half of the s-plane and the system is stable.

At K = 750 the system is marginally stable (pure imaginary poles at ± j7.071).

For K > 750 the system becomes unstable because poles cross into the right half of the s-plane.

Additional Remarks and Practical Use

• The root locus gives direct visual insight into how gain K affects closed-loop pole locations and therefore transient performance (damping ratio, natural frequency) and stability.
• Angle condition: a point s0 is on the root locus if the total phase of G(s0)H(s0) is (2k + 1)·180°. Magnitude condition gives the corresponding K at that point.
• For systems with complex poles or zeros, compute angles of departure from complex poles and angles of arrival at complex zeros using the angle condition (sum of angles from zeros minus sum of angles from poles equals (2k + 1)·180°), and use magnitude condition to find K values.
• Use Routh-Hurwitz for determining imaginary-axis crossings and stability margins as shown above.

References and Useful Keywords

• Keywords: root locus, open-loop poles, open-loop zeros, asymptotes, centroid, breakaway point, break-in point, angle of departure, angle of arrival, Routh-Hurwitz, characteristic equation, stability, closed-loop poles, K(s).