• Document: EE480.3 Digital Control Systems. Part 8. Root Locus Method - using the z-transform
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EE480.3 Digital Control Systems Part 8. Root Locus Method - using the z-transform Kunio Takaya Electrical and Computer Engineering University of Saskatchewan March 11, 2008 ** Go to full-screen mode now by hitting CTRL-L 1 Contents 1 CONTROL SYSTEM DESIGN II 3 2 Cruise Control System 18 3 Steady State Error 29 4 Satellite Position Control System 36 2 1 CONTROL SYSTEM DESIGN II Controller Design by Root Locus Method The root locus method is a method to determine gain K such that the dominant poles of unity gain closed loop system come close to a desired location(s). When the adjustable gain K is applied to a open loop transfer function G(z), the closed loop transfer function is given by KG(z) Gc (z) = . 1 + KG(z) The characteristic equation 1 + KG(z) = 0 requires that KG(z) = 16 ± 180◦ (2k + 1), where k is an integer. 3 Example - Drawing the root locus The procedure to plot the root locus of a given system G(z) is the same as you did for a continuous time system G(s). • Find the root locus on the real axis. Make pairs of poles or zeros (mixed) from (1,0) to −∞. • Calculate the angles of asymptotes, ±180/(p − z). dK d 1 • Find break-in or break-away points. = (− ) = 0. dz dz G(z) + Kz - (z-1)(z-0.5) 4 From the angle condition, 0.5 ≤ z ≤ 1 is part of the root locus on the real axis. For a complex z satisfying the above condition, let z = α + jβ. 6 G(z) = 6z − 6 (z − 1) − 6 (z − 0.5) β β β = arctan − arctan − arctan α α−1 α − 0.5 β β β tanfarctan − arctan − arctan g = tan 180◦ = 0 α α−1 α − 0.5 Using the identity, tan x ± tan y tan(x ± y) = , 1 ∓ tan x tan y For this particular example, we obtain α2 + β 2 = 0.5, the equation √ of a circle with radius 0.5. 5 Writing the characteristic equation 1 + KG(z) = 0 for K, 1 (z − 1)(z − 0.5) K=− = G(z) z dK (z − 0.707)(z + 0.707) =− =0 dz z2 yields the break away and break in points as z = ±0.707. The number of poles p = 2 and the number of poles z = 1 yields ±180/(p − z) = ±180/(2 − 1) = ±180◦ . 6 0.5 1 o x x 0.707 Root Locus in z-plane 7 8 % ------------------------------------------- % Root Locus Example of z-plane Root Locus % ------------------------------------------- zeros=[0] poles=[0.5, 1] num=poly(zeros); den=poly(poles); zrltempl; rlocus(num, den); [r,k]=rlocus(num, den) axis([-1.0, 1.0, -1.0, 1.0]); axis square; 9 Root Locus Template for z-plane The familiar grids in parallel with the real and imaginary axis of s-plane represent vertically the frequency, horizontally the inverse of time constant. However, this does not apply to the z-plane. The z-plane is mapped from the s-plane by z = esT where s = σ + jω f z = esT = eσT ejωT = eσT 6 ωT = eωT 6 2π fs 10 z = esT = eσT ejωT A strip zone in the LHP s-plane between ±ωs /2 corresponds to the inside of the unit circle in the z-plane. Note that z-plane is periodic with repect to ωT . 11 The standard guide lines used in s-plane such as constant damping ratio, constant σ, constant ωd are mapped by z = esT . • ζ damping ratio constant From the standard form of the second order system, ωn2 G(s) = 2 s + 2ζωn s + ωn2 p Poles are: s = −ζωn ± jωn 1 − ζ 2 Damping ratio is given by ζ = cospθ. For a constant damping ratio, θ=const, and its slope is − 1 − ζ 2 /ζ. Letting s = σ + jω,

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