Solved Examples: Root Locus | Control Systems - Electrical Engineering (EE) PDF Download

Before starting with an example, let's first discuss the steps to solve any problem on the root locus.

Procedure to solve a root-locus problem

• Find the number of poles and zeros of the open-loop transfer function; determine the number of branches (P - Z).
• Plot open-loop poles (×) and zeros (o) on the s-plane.
• Determine the sections of the real axis that belong to the root locus using the real-axis rule.
• Compute angles and positions of asymptotes for branches that go to infinity and draw a separate sketch of these asymptotes.
• Find the centroid of the asymptotes and mark it on the real axis.
• Locate breakaway and break-in points on the real axis using the derivative condition or angle condition; complex break points may be verified using the angle condition.
• Find intersections of the root locus with the imaginary axis using the Routh-Hurwitz criterion or by substituting s = jω and solving the magnitude/angle conditions.
• Compute angles of departure from complex poles and angles of arrival at complex zeros if they exist.
• Combine all sketches and rules to draw the final root-locus plot.
• Use the root locus to infer closed-loop stability and basic performance trends as the gain K varies.

Essential root-locus rules and formulae

• Number of branches: equals the number of open-loop poles P. Each branch starts at an open-loop pole and ends at a finite zero or goes to infinity if there are fewer zeros than poles.
• Real-axis rule: A point on the real axis belongs to the root locus if the number of open-loop poles and zeros to its right is odd.
• Angle condition: For s to lie on the root locus, the sum of angles from s to poles minus the sum of angles from s to zeros must equal (2k + 1)×180°.
• Magnitude condition: For a particular K, |G(s)H(s)| = 1 gives the points on the root locus corresponding to that gain.
• Asymptote angles: When P - Z = n asymptotes exist, their angles are given by (2q + 1)×180° / n for q = 0, 1, ..., n-1.
• Centroid of asymptotes: σ = (sum of pole locations - sum of zero locations) / (P - Z).
• Breakaway/break-in points: Obtained from dK/ds = 0 where K is expressed from 1 + G(s)H(s) = 0 as K = -1/(G0(s)) or algebraically by K = -N(s)/D(s) rearrangements; solve dK/ds = 0 for s on the real axis and retain only points that satisfy the angle condition.
• Imaginary-axis crossings: Use the Routh-Hurwitz array on the characteristic polynomial 1 + G(s)H(s) = 0 to find gains at which roots lie on the jω-axis and then solve for ω.
• Angle of departure/arrival: For a complex pole p, angle of departure = 180° - (sum of angles from p to other poles - sum of angles from p to zeros). Use analogous rule for zeros for angle of arrival.

Example 1. Draw the root locus diagram for a closed loop system whose loop transfer function is given by: G(s)H(s) = K/s(s + 5)(s + 10) Also find if the system is stable or not.

The solution follows the standard procedure and the rules listed above.

Find the open-loop poles, zeros and number of branches.

Open-loop denominator gives poles, numerator gives zeros.

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

No finite zeros.

P - Z = 3 - 0 = 3, so there are three branches; three branches must start at the three poles and, since there are no finite zeros, three branches will go to infinity.

Determine real-axis portions that belong to the root locus.

Consider test points on the real axis. Between 0 and -5 there is one pole to the right (odd), so that segment belongs to the root locus. For s < -10 there are three poles to the right (odd), so the region left of -10 also lies on the locus. between -5 and -10 there are two poles to the right (even), so that segment does not belong to the

Asymptote angles.

P - Z = 3, so asymptote angles are given by (2q + 1)×180°/3 for q = 0,1,2.

For q = 0, angle = 180°/3 = 60°.

For q = 1, angle = 3×180°/3 = 180°.

For q = 2, angle = 5×180°/3 = 300° (or -60°).

Centroid of asymptotes.

Centroid σ = (sum of poles - sum of zeros)/(P - Z) = (0 + (-5) + (-10) - 0)/3 = -15/3 = -5.

Breakaway / break-in points on the real axis.

The characteristic equation is 1 + G(s)H(s) = 0.

Substitute G(s)H(s): 1 + K/[s(s + 5)(s + 10)] = 0.

Multiply through: s(s + 5)(s + 10) + K = 0.

Expand polynomial: s(s² + 15s + 50) + K = 0.

Thus the closed-loop characteristic polynomial is s³ + 15s² + 50s + K = 0.

Express K in terms of s: K = -s³ - 15s² - 50s.

Find potential break points by differentiating K with respect to s and setting derivative to zero:

dK/ds = -(3s² + 30s + 50) = 0.

Thus 3s² + 30s + 50 = 0.

Divide by 3: s² + 10s + 16.667 = 0.

Solve for s using quadratic formula: $s = [-10 ± \sqrt{(100 - 66.667)}]/2 = [-10 ± \sqrt{(33.333)}]/2.$

Numerical roots: s ≈ -2.113 and s ≈ -7.88.

Only s ≈ -2.113 lies on the real-axis segment between 0 and -5 where the locus exists, so s ≈ -2.113 is the valid breakaway point.

Verify K at this point by evaluating $ K = -s^3 - 15s^2 - 50s$ with s = -2.113.

Compute K ≈ 48.112 (positive), so this is a valid breakaway for positive K.

Imaginary-axis intersections using the Routh-Hurwitz criterion.

Characteristic polynomial: $s^3 + 15s^2 + 50s + K = 0.$

Construct the Routh array (top rows shown):

From the Routh array condition for a row to become zero (onset of imaginary-axis roots) we find K such that a sign change or zero row appears. For this cubic the condition leads to 750 - K = 0.

Thus K = 750.

With K = 750 substitute into the polynomial row: 15s² + K = 0 gives 15s² + 750 = 0.

Therefore s² = -750/15 = -50.

Hence s = ± j√50 = ± j7.071.

So at K = 750 the root locus crosses the imaginary axis at s = ± j7.071 (marginal stability).

Angles of departure/arrival.

All open-loop poles are on the real axis; there are no complex-conjugate open-loop poles. Therefore explicit angle-of-departure calculations are not required here.

Final root-locus sketch.

Combine real-axis portions, breakaway at s ≈ -2.113, centroid at s = -5, asymptotes at ±60° and 180°, and the imaginary-axis crossings at ± j7.071 (for K = 750) to draw the complete root-locus.

Stability conclusions

• Closed-loop poles lie in the left half-plane for 0 < k < 750, so the system is stable for 0 />< k < />
• At K = 750 the system is marginally stable because a pair of poles is on the imaginary axis (s = ± j7.071).
• For K > 750 one or more roots move into the right half-plane and the closed-loop system becomes unstable.
• Thus stability can be inferred directly from the root-locus locations as K varies: left-half-plane poles indicate stability, right-half-plane poles indicate instability.

Impact of adding a pole or a zero to the open-loop transfer function

Impact of adding a pole

Adding an additional pole to the left half of the s-plane generally increases the number of branches and shifts the asymptote centroid and angles. A pole added to the left half-plane tends to pull nearby branches toward the right half-plane relative to their previous positions, reducing closed-loop stability margins. In practice this often reduces the range of gain K for which the closed-loop system remains stable and decreases gain margin.

Example: Consider G(s)H(s) = K/[s(s + 2)(s + 4)].

Now add a pole at s = -6 so that G(s)H(s) = K/[s(s + 2)(s + 4)(s + 6)].

The added pole increases P, changes P - Z, moves the centroid, and typically shifts parts of the root locus toward the right-half plane compared with the original locus, reducing stability margins.

Impact of adding a zero

• Adding a zero tends to attract nearby branches toward the added zero and, if the zero is in the left half-plane, it generally improves closed-loop stability (moves roots left) and increases the allowable gain range and gain margin.
• A zero near the imaginary axis or in the right half-plane can however have destabilising effects; the sign and location of the zero must be considered.

Example: Start with G(s)H(s) = K(s + 4)/[s(s + 2)].

Now add a zero at s = -6 so the transfer function becomes G(s)H(s) = K(s + 4)(s + 6)/[s(s + 2)].

Comparing the loci shows the branches are pulled toward the finite zeros. If the added zero lies in the left half-plane, the root-locus shifts leftward and closed-loop stability and transient response generally improve.

Additional remarks and practice tips

• When computing breakaway points always check whether the candidate points lie on segments of the real axis that belong to the locus and whether the corresponding K is physically admissible (e.g., positive if gain K is positive).
• Use the Routh-Hurwitz test to determine exact gain values at which imaginary-axis crossings occur; then compute the crossing frequencies by substituting s = jω where required.
• For systems with complex poles or zeros, carefully compute angles of departure and arrival; small numerical errors in angle computations can change the drawn locus significantly.
• Sketch rough loci using pole/zero locations and asymptotes first, then refine using breakaway points, imaginary-axis crossings and angle conditions.
• Root-locus gives qualitative information on stability and dominant pole movement with gain; for precise transient metrics (rise time, overshoot, settling time) compute closed-loop poles for specific K values.

Summary: The root-locus method provides a graphical means to study how closed-loop pole locations change as the open-loop gain varies. Key steps are plotting poles/zeros, applying real-axis and angle rules, determining asymptotes and centroid, locating breakaway points, and finding imaginary-axis crossings. Using these steps one can determine ranges of K for stability and predict qualitative performance trends. The worked example above demonstrates these steps for G(s)H(s) = K/[s(s + 5)(s + 10)], showing stability for 0 < K < 750, marginal stability at K = 750, and instability for K > 750.