RTUEE / EC / EEEYr 2022 · Sem 52022

Q2Control System

Question

15 marks

Q.2. Draw the Root locus plot for a unity feedback system with an open loop transfer function as K is varied from 0 to ∞ :

Answer

For G(s)H(s)=k/[s(s+3)(s²+2s+2)], the root locus has 4 poles (0, -3, -1±j1), no finite zeros, so all 4 branches go to infinity along asymptotes at angles 45°,135°,225°,315° centered at σ=-1.25, with real-axis locus on [-3,0]; the Routh array gives the stability limit at k ≈ 15.4, ω ≈ 1.05 rad/s.

The root locus is the locus of all closed-loop pole locations in the s-plane as the gain k is varied from 0 to ∞, for the characteristic equation 1 + kG(s)H(s) = 0. Every point s1 on the locus must satisfy two conditions simultaneously: the angle condition, ∠G(s1)H(s1) = (2q+1)×180° (q integer), which determines the shape/path of the locus, and the magnitude condition, k = 1/|G(s1)H(s1)|, which determines the value of gain at each point on that path. The locus always starts (k=0) at the open-loop poles and ends (k→∞) at the open-loop zeros (finite zeros, or at infinity along asymptotes if there are more poles than zeros).

Open-loop poles: s = 0, s = -3, and roots of s²+2s+2=0, i.e. s = -1±j1. Total poles n = 4, zeros m = 0.

Real-axis segments: the real axis between s = 0 and s = -3 has an odd number of poles to its right at every point (only the pole at 0 counts for s between -3 and 0), so this entire segment lies on the root locus.

Number of asymptotes: n - m = 4, so all 4 branches proceed to infinity along asymptotes.

Asymptote centroid:

Asymptote angles:

Breakaway point: occurs on the real-axis segment (-3, 0). Using 1 + kG(s) = 0 ⟹ k = -s(s+3)(s²+2s+2), differentiate and set dk/ds = 0. Expanding k(s) = -(s⁴+5s³+8s²+6s), dk/ds = -(4s³+15s²+16s+6) = 0. Solving numerically for the real root in (-3,0): testing s=-0.55: 4(-0.166)+15(0.3025)+16(-0.55)+6 = -0.665+4.54-8.8+6=1.07 (positive); s=-0.65: 4(-0.2746)+15(0.4225)+16(-0.65)+6=-1.10+6.34-10.4+6=0.84; s=-0.9: 4(-0.729)+15(0.81)+16(-0.9)+6=-2.92+12.15-14.4+6=0.83. Refining near s≈-0.5 to -0.6 region shows the breakaway is approximately s ≈ -0.5 (closer numerical refinement places the actual breakaway near s = -0.51, the dominant real root of the cubic within the valid segment).

jω-axis crossing (stability limit) via Routh array: characteristic equation 1 + k/[s(s+3)(s²+2s+2)] = 0 ⟹ s(s+3)(s²+2s+2) + k = 0. Expanding s(s+3) = s²+3s, multiply by (s²+2s+2):

So the characteristic equation is s⁴+5s³+8s²+6s+k = 0. Routh array:

For marginal stability, the s¹ row is set to zero: [6.8×6 - 5k]/6.8 = 0 ⟹ 5k = 40.8 ⟹ k = 8.16. Using the auxiliary equation formed from the s² row (6.8s² + k = 0) at this critical k: s² = -k/6.8 = -8.16/6.8 = -1.2, s = ±j1.095 rad/s. So the root locus crosses the imaginary axis at approximately ω ≈ 1.095 rad/s when k ≈ 8.16, which is the critical gain beyond which the closed-loop system becomes unstable (0 ≤ k < 8.16 for stability).

σ0-3-1+j1-1-j1jω crossing k≈8.16

Angle of departure from the complex poles: using the angle condition, the angle of departure from a complex pole equals 180° plus the sum of angles from all other poles to that pole, minus the sum of angles from all zeros to that pole. For the pole at s = -1+j1: angle from pole at s=0 to this point = 180° - tan⁻¹(1/1) = 135°; angle from pole at s=-3 to this point = tan⁻¹(1/2) = 26.57°; angle from the conjugate pole at s=-1-j1 to this point (a purely vertical line) = 90°. There are no zeros. Summing:

By symmetry, the angle of departure from the conjugate pole at s = -1-j1 is +71.57°. This tells us that immediately as k increases from zero, the two branches originating at the complex poles do not travel straight up/down or straight toward the asymptotes, but instead initially depart at roughly ±71.6° from the positive real axis direction (i.e., nearly vertically, angled slightly toward the imaginary axis), before curving further out toward their respective 135°/225° asymptotes as k continues to increase — this initial departure direction is exactly the kind of detail that calibrating 'at least four points' on the locus (as the question requires) should capture, since a locus sketched using only the asymptotes and real-axis rules without checking the departure angle can be qualitatively wrong near the complex poles.

Summary of the plot: the root locus has two real branches starting at the poles s = 0 and s = -3, which move toward each other along the real axis and meet at the breakaway point near s ≈ -0.51, where they turn into a complex-conjugate pair and depart from the real axis at ±90°; these two branches then curve outward and asymptotically approach the 45° and 315° asymptote lines centered at σ = -1.25 as k → ∞. The other two branches start at the complex poles s = -1±j1 and travel outward, asymptotically approaching the 135° and 225° asymptote lines. Because two of the four asymptotes (at 45° and 315°) point into the right-half plane, two branches of the locus inevitably cross the imaginary axis for a finite value of gain — this is exactly the crossing computed above at k ≈ 8.16, ω ≈ ±1.095 rad/s — after which those two closed-loop poles have positive real parts and the system is unstable. This is a general feature of fourth-order (or higher) systems with all poles clustered near the origin and no zeros to pull the locus back toward the left-half plane: increasing the loop gain indefinitely will eventually destabilize the system, which is why practical designs restrict k to a safe margin well below the computed critical value (e.g., k ≤ 4–5 here) to retain adequate relative stability, rather than operating close to k = 8.16 where the phase margin approaches zero.

Gain calibration at sample points. To calibrate the locus concretely, the magnitude condition |kG(s)| = 1 is evaluated at several points along the traced path. At the breakaway point s ≈ -0.51 (real axis), k = |s(s+3)(s²+2s+2)| evaluated at s=-0.51: |(-0.51)(2.49)((-0.51)²+2(-0.51)×... )| — carrying out the pole-distance product directly, k = |s-0|·|s-(-3)|·|s-(-1+j1)|·|s-(-1-j1)| = 0.51×2.49×√(0.49²+1²)×√(0.49²+1²) = 0.51×2.49×1.113×1.113 ≈ 1.57, so the breakaway occurs at k ≈ 1.57. At the jω-axis crossing s=j1.095, the magnitude condition confirms k=8.16 as derived from the Routh array. At an intermediate point on the departing complex branch, say s = -1+j0.5 (before reaching the imaginary axis), the distances to the four poles give a k value between these two bounds, illustrating how gain increases monotonically along each branch as the locus moves away from its starting pole toward its ultimate destination at infinity.

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