Second order approximation of a third order system

I have a transfer function with a zero at 259.6 and poles at 0, -1.6, and -33.8. How do I make a second order approximation of this system to implement a step input and analyze the overshoot and settling time?

[Mathieu NOE]

hello

your TF contains a pole at 0 so it contains an integrator - any how a step input will create an output signal like a ramp - no way you can measure overshoot and setling time (like underdamped second order systems)

what kind of second order approximation are you trying to do , and why ? removing the p = 0 term is no more an approximation it's a different system ! and removing other pole(s) will not change the integrator effect either.

all your poles are real so there will be no oscillations, overshoot. For that you need complex conjugated poles.

zz = 259.6;
pp = [ 0 -1.6 -33.8];
[NUM,DEN] = zp2tf(zz,pp,1);
figure,bode(NUM,DEN)
figure,step(NUM,DEN)
figure,impulse(NUM,DEN)

Answer 1

[Sam Chak]

Hi @Keaton Looper

For versions before R2023b release, use the balred() command. Now you can use the newer reducespec() command instead.

Update: The OP has clarified that the value 259.6 in the numerator is actually the 'gain' (k), not a 'zero'.

%% Original 3rd-order LTI model
z    = [];
p    = [0, -1.6, -33.8];
k    = 259.6;
sys  = tf(zpk(z, p, k))

sys =
 
           259.6
  ------------------------
  s^3 + 35.4 s^2 + 54.08 s
 
Continuous-time transfer function.

%% Reduced 2nd-order LTI model
R1   = reducespec(sys, "balanced");
R2   = reducespec(sys, "ncf");      % requires Robust Control Toolbox™ 
rsys = tf(getrom(R2, Order=2))

rsys =
 
      -0.17 s + 7.437
  -----------------------
  s^2 + 1.508 s + 0.03729
 
Continuous-time transfer function.

bode(sys, rsys, 'r--'), grid on
legend("Original", "Order 2")

[Keaton Looper]

the only value for the numerator is 259.6 there is no s in the numerator of my transfer function. therefore there is only 3 poles of the fucntion and no zeros.

[Sam Chak]

Hi @Keaton Looper

Take a look at the updated Answer. If you find the solution helpful, please consider clicking 'Accept' ✔ on the answer and voting 👍 for it. Thanks a bunch!

[Mathieu NOE]

as expected , the second order approximation does not retain the integrator part of the original TF

for closed loop design, that can have some serious impact !