Question

R(s) Q1 (25p). The closed loop control system is shown in the figure. The response of the control system to a unit step reference input is shown in below. C(s) Kp 0.6 S2 + 5s + A 4 a) Find and wn. b) Find Ke and A c) If we use a proportional-integral (PI) controller instead of a proportional (P) controller, how the graph would be expected to change. For two different integral coefficients (bigger Ki_1 and smaller K_2), make an approximate drawing on the graph by hand. Please draw the graphic clearly and show all annotations. 0.9596 0.8 c(t) 0.5 C max CS =e - VICE CS 0 0.5 1.5 2 2.5 3 3.5 t(s)

Answer 1

Solution:

Solutions cus) RU)+ (9) Кр 0.6 s'+537A cis) RLS) GUS) 1 + 608 2.4ke 2.4 KP/es:+594) u Kp/137534) 3?+59 +(A+2.4 kp2/ T(S) => 2.4kp s+55 +LA+ 2-4 kp) Now from Graph. Cgs =0.8 Cmax 0.9596, Cmax - Css -REJTE e: -T/17-52 => 0.1995 = losa Taking => 71.611 = the √1-12 => 2.59(1-3) = NE? =) E = 0.455 Ang CS Scanned with CamScanner

Characteristic en 5 +55 +CA+buke) Genexalised chat eqn.. 3² 228 2²Wnst om? 2 & Wn = 5 => 2 10.455) m = 5 Wn = 5.49 Ane Css 0.8. - EWnit Cgs = Act lim AK (16-ė Sin (Watto) it oo J1-e Css AK where A → Magnitude of Step imput. K-> D.C. Gain. [Tls=o) ] ... Css = K = 0.8 2.410 Given, TLS) = $+55+LA+2.460) with ² +2 & Was tun? Compairing A+2.4Kp = m = 30. 1401 Putting it in eo TO)= 2.4ko s'+ 55+30.1401 Now D. C. Gain = 0.83 TUS=0) 2.4ke O+0+ 30.1407 CS Scanned with CamScanner

Kp= 10.0467 – 10 Ane Hence, A+2.4kg = 30.1401 : A = 6.02 6 Ane S ess w when we replace Kp with 'P. I. controller then, Ge(s) of P.I. = kost KI where Kp → mognitude of proportional KI magnitude of Integrado Now, with P-controller, lim 3R) 30 17665) A A+2.4ko 3+55+A [R) Type is Order is two with P.I - Controller. lim SRUS) 50 2+ 420) In's) (W) [RU)= }]. [403) = 2.4ke ₃ Zero, ess = 5 +5STA! 1 + Type is 1 order is three. Steady State error is zero. CS Scanned with CamScanner

In terms g stability, Let kp=1 and Ky » KP, then for the kq <2&wn w system to be stalle In our case 28 Wn = 5 b OK KI 45 then system will be stable. CS Scanned with CamScanner

Graph:

For simplicity let Kp = 1 and Ki = 1 for the first case. then we get T(s) = (2.4s+2.4)/(s^3+5s^2+8.4s+2.4)

[I've taken value of A = 6, the same value we calculated]

graph of this transfer function will be,

Unit step response plot: 1.0 0.9 0.8 0.7 0.6 0.5 5 10 15 20

Now for the second case, we'll be taking Kp = 1, and Ki = 5, which is the limit of Ki for the system to be stable as we discussed above, then we get T(s) = (2.4s+12)/(s^3+5s^2+8.4s+12), whose graph will be,

Unit step response plot: 1.00010 1.00005 1.00000 у 0.99995 10 20 30 40

We can see it's stability by it's root locus diagram,

Root locus plot: Im 20 10 XX Re -5 4 -2 -1 X -10 -20

As you can see it's on the verge of being unstable, any further increase in value of Ki and the system will become unstable which you can see in the given below root locus diagram in which value of Ki is 6

Root locus plot: Im 20 10 X X Re -6 -5 -4 -3 -2 -1x -10 -20

Comment down below for any doubt