RTUEE / EC / EEEYr 2024 · Sem 52024

Q4Control System

Question

10 marks

Q.4. Given the unity feedback system, make an accurate plot of the root locus for the following system :

Calibrate the gain for at least four points. Also find the breakaway points, the jω axis crossing and range of gain for stability. Find angle of arrival also.

Answer

For G(s)=K²(s²-2s+2)/[(s+1)(s+2)], the root locus has 2 poles (-1,-2), 2 zeros (1±j1); breakaway/entry points occur on the real axis between the poles, the loci cross the jω-axis at specific points found via the Routh array, and the angle of arrival at the complex zero is computed from the angle condition.

Open-loop poles are at s = -1 and s = -2 (on the real axis). Open-loop zeros are the roots of s² - 2s + 2 = 0, i.e. s = 1 ± j1 (complex, in the right-half plane, since s²-2s+2=0 gives s=(2±√(4-8))/2=1±j1).

Number of branches and asymptotes: number of poles n = 2, zeros m = 2, so n - m = 0; there are no asymptotes going to infinity — both branches terminate on the finite zeros at 1±j1.

Real-axis segments: the segment between s = -1 and s = -2 lies on the root locus (odd number of poles/zeros to its right, namely one pole at -1).

Breakaway/break-in points: found from dK/ds = 0 where the characteristic equation is 1 + K²(s²-2s+2)/[(s+1)(s+2)] = 0, i.e.

Differentiating the right side with respect to s and setting the derivative to zero (standard breakaway condition d/ds[(s+1)(s+2)/(s²-2s+2)] = 0) and solving the resulting polynomial numerically over the real-axis segment (-2,-1) gives a breakaway point at approximately s ≈ -1.45 (the unique real root of the derivative equation lying within the valid real-axis locus segment).

Calibration of gain at four points: using the magnitude condition |K²(s²-2s+2)| = |(s+1)(s+2)| at test points along the locus. At s = -1.45 (breakaway): numerator |(-1.45)²-2(-1.45)+2| = |2.1+2.9+2| = 7.0, denominator |(-0.45)(0.55)| = 0.2475, so K² = 0.2475/7.0 = 0.0354, K ≈ 0.188. At s = -1.2: numerator |1.44+2.4+2|=5.84, denom |(-0.2)(0.8)|=0.16, K²=0.0274, K≈0.166. At s=-1.7: numerator |2.89+3.4+2|=8.29, denom|(-0.7)(0.3)|=0.21,K²=0.0253,K≈0.159. At s=-1.0 (pole, K=0) and s=-2 (pole, K=0) mark the start points of the two loci branches on the real axis, each moving toward the breakaway point before departing into the complex plane toward the zeros at 1±j1.

jω-axis crossing: substitute s = jω into the characteristic equation 1 + K²(s²-2s+2)/[(s+1)(s+2)] = 0, i.e. (s+1)(s+2) + K²(s²-2s+2) = 0:

Separating real and imaginary parts: Real: -ω²+2 + K²(-ω²+2) = 0 ⟹ (2-ω²)(1+K²) = 0. Since (1+K²) ≠ 0, this requires ω² = 2, i.e. ω = ±√2. Imaginary: 3ω - 2K²ω = 0 ⟹ ω(3-2K²)=0. For ω = √2 ≠ 0: K² = 1.5, K = 1.225. So the locus crosses the imaginary axis at ω = ±√2 ≈ ±1.414 rad/s when K ≈ 1.225, and this is also the gain range boundary for stability.

Range of gain for stability: since the loci originate at the stable poles (-1, -2) and move toward the zeros (1±j1, which lie in the right-half plane), the closed-loop poles cross into the RHP once K exceeds the value found above. Hence the system is stable for 0 ≤ K < 1.225 and becomes unstable for K > 1.225 (marginally stable exactly at K = 1.225).

Angle of arrival at the complex zero (1+j1): using the angle condition, angle of arrival at a zero = 180° + Σ(angles from poles to zero) - Σ(angles from other zeros to this zero). Angles from poles -1 and -2 to the zero (1+j1): from -1: angle = tan⁻¹(1/2) = 26.57°; from -2: angle = tan⁻¹(1/3) = 18.43°. Angle from the other zero (1-j1) to (1+j1): this is a vertical line, angle = 90°. Angle of arrival:

σ-1-21+j11-j1breakaway
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