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Unformatted text preview: EE302 HW4 Solution Q1. Sketch root loci (by obtaining all relevant features) as K : 0 → ∞ for the unity feedback closedloop systems with openloop transfer functions given below. (a) G ( s ) = K s + 3 s ( s 2 + 2 s + 2) (b) G ( s ) = K s + 2 s 4 1 (c) G ( s ) = K ( s + 1) 2 s 3 + 8 Sol’n. (a) We can write the characteristic equation (of the closed loop) as 1 + K ( s + 3) s ( s + 1 + j )( s + 1 j ) = 0 which has three roots. Therefore the root locus diagram (RLD) will have three branches. Open loop has m = 1 zero and n = 3 poles. Three branches will start from openloop poles and two of them will sail to infinity along the asymptotes and the remaining one will approach the zero at s = 3 as K → ∞ . Asymptotes will have angles ϕ = 180 ◦ (2 ℓ + 1) n m = ± 90 ◦ . Asymptotes will meet at the centroid σ ◦ which we calculate as σ ◦ = ∑ n i =1 p i ∑ m i =1 z i n m = (0 1 j 1 + j ) ( 3) 3 1 = 1 2 where p i denotes openloop pole and z i openloop zero. Angle condition reveals that RLD will reside on the real axis only at the interval s ∈ [ 3 , 0]. To find where (if) the RLD crosses jω axis we exploit the characteristic equation. 0 = s 3 + 2 s 2 + ( K + 2) s + 3 K s = jω = jω 3 2 ω 2 + j ( K + 2) ω + 3 K = ( 2 ω 2 + 3 K ) j ( ω 3 ( K + 2) ω ) which implies 2 ω 2 + 3 K = 0 ω 3 ( K + 2) ω = 0 whence either ω = ± √ 6 with K = 4 or ω = 0 with K = 0. Finally, we compute the departure angles. From the open loop pole p = 1 + j RLD will depart by ψ which we obtain from ± 180 ◦ (2 ℓ + 1) = ∠ ( p + 3) ( ∠ ( p + 0) + ∠ ( p + 1 + j ) + ψ ) = arctan(1 / 2) (135 ◦ + 90 ◦ + ψ ) .3.532.521.510.5 0.515105 5 10 15 Root Locus Real Axis Imaginary Axis Figure 1: Root locus diagram for Q1(a). We compute ψ ≈  18 ◦ . Due to symmetry with respect to real axis, the departure angle from the conjugate pole p = 1 j will be ψ . This is all we have to have figured out to plot RLD for this example. The range of K for stability is 0 < K < 4....
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 Spring '10
 erkmen
 Asymptotes, Root Locus, Root Locus Diagram, KS2

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