Why does the System Identification Toolbox return a different transfer function than one produced analytically?

Using the system identification toolbox I inserted 2 small data series that were calculated from the following input and output:

Input: u(t) = 6*cos(2*t) Output: y(t) = -exp(-t)-3*exp(-3*t)+4*cos(2*t)+4*sin(2*t)

Analytically I calculated the following transfer function:

H(s) = (9*s+14)/(3*(s+1)*(s+3))

However, when prompted to determine a transfer function with 2 poles and 1 zero, the System Identification toolbox produced a much different looking transfer function than the one that I had calculated.

Does anyone know the reason behind this?

Thanks -Pat

Answer 1

[Arkadiy Turevskiy]

I get very good match:

t=0:0.01:100;
u=6*cos(2*t)';
y=(-exp(-t)-3*exp(-3*t)+4*cos(2*t)+4*sin(2*t))';
data=iddata(y,u,t(2));
sys=(9*s+14)/(3*(s+1)*(s+3));
sys_id=tfest(data,2,1);

Checking:

>>sys

sys =

      9 s + 14
  ----------------
  3 s^2 + 12 s + 9

>> sys_id

sys_id =

From input "u1" to output "y1":
3.023 s + 4.606
---------------
 s^2 + 4 s + 3

Continuous-time identified transfer function.

As you can see, sys_id gets the poles exactly, the zero and gain are slightly off.

If we do the bode and step, the differences are really small:

bode(sys);
hold on;
bode(sys_id,'r');

figure;
step(sys,sys_id,'r');