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Unformatted text preview: . The closed–loop characteristic equation is s 3 + ( K +4) s 2 + s + ( K –6) = 0. Substituting s = jω ( 29 ( 29 3 2 4 6 j K j K ϖ+ + += (1) Root Locus The real and imaginary parts, respectively, must be simultaneously zero ( 29 2 4 6 K K ϖ+ + = (2) 2 1 + = (3) Solving equation (3), 0, 1 = ± . For ω = 0, K = 6; therefore, for K = 6, a root locus branch crosses the imaginary axis at the origin. Substituting ω 2 = 1 into equation (2), –2 = 0. While this equation does not make sense, it is known that the root locus branches will touch the imaginary axis at ±1 when K = ∞. The Matlab code for producing the Root Locus Diagram is sys = zpk([sqrt(1) sqrt(1)],[0 0],1); figure; rlocus(sys), title('G_C(s)G(s) = [s^2+1]/s^2'); The Root Locus Diagram is shown below.642 21.510.5 0.5 1 1.5 G C (s)G(s) = (s 2 +1)/[(s1)(s+2)(s+3)] Real Axis Imaginary Axis 2...
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 Spring '11
 LANDERS
 Root Locus, Complex number

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