# HW4_sol - 1 Ziegler-Nichols Tuning G s = Gc s)Gm s)G p s 80 285 = Gc s 2 0.02 s 1 s 8s 400 22,800 = Gc s 3 2 0.02 s 1.16 s 16 s 400 The root locus

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1. Ziegler-Nichols Tuning 2 32 () () () () 80 285 8 400 0.02 1 22,800 0.02 1.16 16 400 cmp c c Gs G sG sG s ss s s = ⎡⎤ = ⎢⎥ ++ + ⎣⎦ = + The root locus of G O.L. is shown in Figure below. From the root locus, we can obtain K=0.02316. With the gain K=0.02316, we can simulate the response of the closed loop system, from which P u =0.0222 0.6 0.6 0.2316 0.0129 0.5 0.5 0.222 0.111 0.125 0.125 0.222 0.02775 1 ( ) 0.0139 1 0.02775 0.111 pu iu du c KK TP s s =×=× = = =×= ⎛⎞ =+ + ⎜⎟ ⎝⎠ System response

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Root locus of the open loop system System response at the stability margin
2. (a) Bollinger 2.4. Obtain the first five responses of the system with y(0)=0 and x(0)=x(1)=x(2)=1 and x(k)=0, k=3,4, Assume τ =2 and K=1. (b) Repeat part (a) with the second order derivative expression covered in class. 2 2 / 12 0 00 1 2 20 () (a) follow the procedure in example 2.2 the complete solution of the above differential eq. is ( ) ( ) ( ) 0 1 0 hp t dy d y Kx t dt dt yt y t CC e K x T ty y y C C y y C K x += =+ + == = + = + ±± ± 200 10 0 0 / 0 0 0 0 / 000 0 21 2 2 2 2 t t Cy K x Cy y K x y t y y Kx y Kx e Kx t For K y y x y x e xt ττ

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HW4_sol - 1 Ziegler-Nichols Tuning G s = Gc s)Gm s)G p s 80 285 = Gc s 2 0.02 s 1 s 8s 400 22,800 = Gc s 3 2 0.02 s 1.16 s 16 s 400 The root locus

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