How would I determine the gain required for a settling time of 1 second for the system G=(s+4)/((​s+1)(s^2+6​s+13))?

For the following equation:

G=(s+4)/((s+1)(s^2+6s+13))

How would I determin the gain required for the system the have a settling time of 1 second?

I have the code set (see below) to confirm the theory, but the hand calculation is messing me up.

clear

close all

clc

a = input('input a in form [... ... ...]');

b = input('input b in form [... ... ...]');

c = input('input c in form [... ... ...]');

sys = zpk(a,b,c)

step(sys)

Thank-you for your time and help.

Answer 1

[Sam Chak]

I think no gain can make the closed-loop system to settle at 1 second. The fastest is around 1.74 sec with the gain 16.89. You need a high-order controller to achieve that.

s   = tf('s');

% Plant
Gp  = (s + 4)/((s + 1)*(s^2 + 6*s + 13))

Gp =
 
           s + 4
  -----------------------
  s^3 + 7 s^2 + 19 s + 13
 
Continuous-time transfer function.

% Controller
Gc  = (425*s^3 + 2977*s^2 + 8080*s + 5528)/(s^4 + 27*s^3 + 260*s^2 + 681*s)

Gc =
 
  425 s^3 + 2977 s^2 + 8080 s + 5528
  ----------------------------------
    s^4 + 27 s^3 + 260 s^2 + 681 s
 
Continuous-time transfer function.

% Closed-loop system
Gcl = feedback(Gc*Gp, 1)

Gcl =
 
               425 s^4 + 4677 s^3 + 19988 s^2 + 37848 s + 22112
  ---------------------------------------------------------------------------
  s^7 + 34 s^6 + 468 s^5 + 3452 s^4 + 14735 s^3 + 36307 s^2 + 46701 s + 22112
 
Continuous-time transfer function.

% Check Settling Time
S   = stepinfo(Gcl); Ts = S.SettlingTime

Ts = 0.9991

% Plot response
step(Gcl, 3), grid on

[Kez]

Hi sam,

thank-you for the help.

May I ask how you determined the equation for the high order controller?

[Sam Chak]

Hi @Kez

The 7 parameters in the controller are tuned simultaneously. However, I believe the current Control System Toolbox lacks a one-click solution when the 1-second requirement is specified. You can explore the optimization-based graphical approach. Nevertheless, I find the following method more suitable since there is only one design parameter (0 dB gain crossover frequency) to be tuned:

s   = tf('s');

% Plant
Gp  = (s + 4)/((s + 1)*(s^2 + 6*s + 13))

Gp =
 
           s + 4
  -----------------------
  s^3 + 7 s^2 + 19 s + 13
 
Continuous-time transfer function.

% Controller
w   = 6.35;     % <-- tune this 0 dB gain crossover frequency
Gc  = pidtune(Gp, 'pidf', w)

Gc =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = 31.4, Ki = 39.9, Kd = 4.87, Tf = 0.00138
 
Continuous-time PIDF controller in parallel form.

% Closed-loop system
Gcl = feedback(Gc*Gp, 1)

Gcl =
 
           3569 s^3 + 3.711e04 s^2 + 1.203e05 s + 1.158e05
  -----------------------------------------------------------------
  s^5 + 732.7 s^4 + 8668 s^3 + 5.092e04 s^2 + 1.297e05 s + 1.158e05
 
Continuous-time transfer function.

% Check Settling Time
S   = stepinfo(Gcl); Ts = S.SettlingTime

Ts = 0.9519

% Plot response
step(Gcl, 3), grid on