Question

Please code on MATLAB and explain D) only. Thank you

The block diagram of a linear control system is shown in the Fig., where r(t) is the reference input and n(t) is the disturbance. (a) Find the steady-state value of e(t) when n(t) = 0 and r(t) tuz(t). Find the conditions on the values of a and K so that the solution is valid. N(s) R(S) E(S) S + a K(s + 3) Y(s) S (5² - 1) Controller Process Using MATLAB (b) Construct the root loci for K> 0 with a = 2. Analyze the results. (c) Construct a Bode plot and only the phase margin and the gain margin (d) Construct the Nyquist Diagram and analyze the system stability (e) Show ine system response to a unit step command for n(t) = 0 and a unit step disturbance input for r(t) = 0.

Answer 1

$1596098226997_IMG_20200730_130439_2.png$
R15) Sta Yis) K(st) (S²-1) S Y(S) RIS) E(S) KS+) (5+3) S (5²-1) E(S) s RIS) 1+ K ( S+w) (5+3) S (5²1) R (t) = tult) R(S) = 1² Els) (1/2) (s (s2-1)] S(S-1) + KLS+x) (5+3) → E(S): 1 (53-5) 52 53-5 + [s?+ (x+3) 5 +3«] » E1S) = I (5²1) S 53-5+k [s2+(x+3)S +3«]} Steady state value of €13) (ess)> lim S E(S) S > 0 ess= lim s (3²-1) sys!-s+k[sº++3]5+3]} SO

ess = lim (S²-1) 530 53 5+ k[s"+(x+3)S +3«] ess w) Cow 12 (K) (34) 32 ess = -1 3kX Characteristic equation of System is nio moboto soltolaiturin orodo Birofos 1 + K (5+) CS +3) S (52-1) 7 s (s2-1) +k[s+ (x+3)s + 3] 20 Ź 53- 5+ *S*+(x +3)ks + 3 «K= 0 > s}+ks +[(« +2)k-1]s +3«K = 0 Applying Routh Hurwitz Stability criteria

७२ o 2x K. buvo 531 (x+3)K-1 K 34K six K²+36²-k -30k (1-22 30K For system to be stable element of first column should be positive (JE kyo 3xk 70 2 (out - (AA Xx² + 3k²-K-30k yo - 29 12] » Xk? +3k?-K-3x470 +(-3) = (x +3) k>> XL 3x+1) = (x+3) k > ( 31) 2-2 KY 3x+1 21+ So, Ky 30+1 x+3 K 2x+3 KYOxxo,

For x=2 RIS) S+2 S Y(S) + K(5+3) (S²-1) RIS) YIS) k (5+2)(5+3) S (5²-1)

b.) Matlab code to plot root locus for alpha =2

clc;
clear all;
close all;
G=tf([1 5 6],[1 0 -1 0])
rlocus(G)

OUTPUT

Root Locus 4 3 2 Imaginary Axis -1 -2 -3 -4 -5 -16 -14 -12 -10 0 Real Axis

C.) Matlab code to plot bode plot for alpha=2 and k=2

clc;
clear all;
close all;
G=tf([2 10 12],[1 0 -1 0])
bode(G)
[Gm,Pm]=margin(G)

OUTPUT

Transfer function: 2 s^2 + 10 + 12 S^3 - 3 Gm = 0.7000 Pm = 11.9829

Bode Diagram 80 60 40 Magnitude (dB) 20 -20 -40 -90 -135 Phase (deg) -180 -225 -270 102 10- 101 102 10° Frequency (rad/sec)

D.) Matlab Code for nyquist plot for alpha =2 and k=2

clc;
clear all;
close all;
G=tf([2 10 12],[1 0 -1 0])
nyquist(G)

OUTPUT

Nyquist Diagram 200 150 100 50 Imaginary Axis -50 -100 -150 Ft -200 -10 -8 -6 0 -4 Real Axis

System has no open loop zeroes in right half of s plane.Also nyquist plot do not encircle the (-1,0) point .So system has no closed loop poles in right half of s plane.System is stable

e.) Matlab code to plot output for step input for alpha=2 and k=2

clc;
clear all;
close all;
G=tf([2 10 12],[1 0 -1 0])
T=feedback(G,1) %closed loop transfer function having unity negative feedback
step(T)

OUTPUT

Step Response 1.8 1.6 1.4 1.2 Amplitude mm 0.8 0.6 0.4 0.2 16 6 8 12 14 18 20 10 Time (sec)

System is oscillatory but if we increase the value of K then system will less oscillate.

$1596099447319_IMG_20200730_133742_4.png$

YIS) = [X (5+3)/65-] *5184) NIS) S(S-1) + K (5 +3)(5+2) » YIS) 11 NIS) K (5+3)s. 53-5+k(52+55+6) Y(S) k (52+3) ) N(S) 53-5+k18°+55+6) putk=2 Yrs) – 2 (5²+36) 53-5 +25° +105 +12 A tbs x x 534545+8 NOS) Nis Y(S) NIS) 25²+6s 53+252+ 95+12

Matlab Code

clc;
clear all;
close all;
G=tf([2 6 0],[1 2 9 12])

OUTPUT

Step Response 0.8 06 041 Amplitude 02 0 -0.2. 04 -0.6 6 8 12 14 18 20 10 Time (sec)