# Automatic Control Lab 05-Ait

CONTROL THEORY

Name: Dilesha Herath

ID.No: st20000297

Date 13/03/2013

Exercise

Solve the following problem with the help of Matlab as much as possible. All the Matlab command inputs used in this problem should be listed in the report in accordance with the command results. Only the commands studied in the class are allowed to use. Problem Assume that the relation between input, u, and output, y, of a system shown below is represented by the transfer function, s3 . G( s) 4 3 s 12s 49s 2 78s 40

U(s)

G(s)

Y(s)

(a) By Routh-Hurwitz stability criteria, determine that the system under consideration is stable or not. >> alpha3=b1/c1; >> d1=b2; >> alpha4=c1/d1; >> alpha5=d1; >> alpha1, alpha2, alpha3, alpha4, alpha5 alpha1 = 0.0833 alpha2 = 0.2824 alpha4 = 1.6676 alpha5 = 40

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ASIAN INSTITUTE OF TECHNOLOGY

CONTROL THEORY

Since alpha1, alpha2, alpha3, alpha4, alpha5 are positive the system is stable. (b) Dominant time constant is a number determined when the system is stable. This number informs about speed of system response. It is defined as absolute inverse of real part of the root locating the nearest with the imaginary axis. Find the system’s dominant time constant. >> [x y z] = residue([1 3], [1 12 49 78 40]); >> r = max(y); >> time_constant = 1/(-r) time_constant = 1.0000 (c) For this problem, a choice to improve the system’s dominant time constant is done by equipping an adjustable amplifier, providing proportional gain, with the system. The output signal is then fed back to compare with the original input. The difference of the input and the feedback output is finally used as the amplifier input. This makes the new system become close loop. The compensated system configuration is shown in the below figure. U(s) K G(s) Y(s)

Find roots of the compensated system when the proportional gain is adjusted as K = 0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, and 1024. Then plot all the roots in the complex plane. The closed loop characteristic equation, s4 + 12s3 + 49s2 + (k+78)s + 40 + 3k =0 for k=0, >>k=0; k1=k+78; k2=40+3*k; >>roots([1 12 49 k1 k2]) ans = -5.0000 -4.0000 -2.0000 -1.0000

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ASIAN INSTITUTE OF TECHNOLOGY

CONTROL THEORY

for k=1 to 1024, >>i=[0:10]; k=2.^i; >> x=k+78; y=40+3*k; >> roots([1 12 49 x y]) ans = -4.9166 + 1.1548i -4.9166 - 1.1548i 1.1448 + 0.5410i 1.1448 - 0.5410i 1.0552 + 0.2739i 1.0552 - 0.2739i 0.6889 + 1.1185i 0.6889 - 1.1185i 0.7297 + 0.8189i 0.7297 - 0.8189i 0.0050 + 1.3627i 0.0050 - 1.3627i 0.1670 + 1.0880i 0.1670 - 1.0880i -0.7404 + 1.2369i -0.7404 - 1.2369i -1.6548 -1.3750 + 0.7409i -1.3750 - 0.7409i -0.4527 + 1.0058i -0.4527 - 1.0058i -0.9268 + 0.5986i -0.9268 - 0.5986i -1.1032 Since I got only 24 solutions from the above method I used for loop and the scatter option to compute the roots and to draw the graph. >> for i = 0:10; k= 2.^i; 3

ASIAN INSTITUTE OF TECHNOLOGY

CONTROL THEORY

y = [1 12 49 (78+k) (40+(3*k))]; r =roots(y) scatter(real(r),imag(r)),hold on; grid end r= -5.1439 -3.8584 -1.7817 -1.2160

r= -5.2586 -3.7505 -1.4954 + 0.3099i -1.4954 - 0.3099i

r= -5.4428 -3.5923 -1.4825 + 0.6796i -1.4825 - 0.6796i

r= -5.7203 -3.4022 -1.4387 + 1.1039i -1.4387 - 1.1039i

r= -6.1178 -3.2332

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ASIAN INSTITUTE OF TECHNOLOGY

CONTROL THEORY

-1.3245 + 1.6415i -1.3245 - 1.6415i

r= -6.6651 -3.1227 -1.1061 + 2.3045i -1.1061 - 2.3045i

r= -7.3982 -0.7698 + 3.1061i -0.7698 - 3.1061i -3.0622

r= -8.3608 -0.3040 + 4.0790i -0.3040 - 4.0790i -3.0312

r= -9.6071 0.3114 + 5.2719i 0.3114 - 5.2719i -3.0156

r= -11.2058 1.1068 + 6.7479i 1.1068 - 6.7479i -3.0078

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ASIAN INSTITUTE OF TECHNOLOGY

CONTROL THEORY

r= -13.2434 2.1237 + 8.5858i 2.1237 - 8.5858i -3.0039

(d) The above compensated configuration is the concept used to plot system root locus. Root locus plot needs the following information. - Open loop zeros Zeros=3 Open loop poles >> roots([1 12 49 78 40]) ans = 6...

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