a problem with pade approximation

Hi

I have te code below

And i want to approxiamte delay with pade of first order :

clc 
clear all 
close all 

s = tf('s');
G = [(-21.6*exp(-1*s))/(8.5*s+1) (1.26*exp(-0.3*s))/(7.05*s+1);...
   (-2.75*exp(-1.8*s))/(8.2*s+1) (-4.28*exp(-0.35*s))/(9.0*s+1)]
 
G_new = pade (G,1)

but the answers is like this :

My tf was first order and with first order pade it should be second order

why some of it is third order ?

Thanks

Answer 1

[Paul]

tf objects have three delay properties: InputDelay, OutputDelay, and IODelay. Look at all three of these properties for each element of G:

s = tf('s');
G = [(-21.6*exp(-1*s))/(8.5*s+1) (1.26*exp(-0.3*s))/(7.05*s+1);...
   (-2.75*exp(-1.8*s))/(8.2*s+1) (-4.28*exp(-0.35*s))/(9.0*s+1)]

G =
 
  From input 1 to output...
                     -21.6
   1:  exp(-1*s) * ---------
                   8.5 s + 1
 
                       -2.75
   2:  exp(-1.8*s) * ---------
                     8.2 s + 1
 
  From input 2 to output...
                        1.26
   1:  exp(-0.3*s) * ----------
                     7.05 s + 1
 
                       -4.28
   2:  exp(-0.35*s) * -------
                      9 s + 1
 
Continuous-time transfer function.

[G(1,1).InputDelay G(1,1).OutputDelay G(1,1).IODelay G(1,2).InputDelay G(1,2).OutputDelay G(1,2).IODelay;  G(2,1).InputDelay G(2,1).OutputDelay G(2,1).IODelay G(2,2).InputDelay G(2,2).OutputDelay G(2,2).IODelay]

ans = 2×6

0 0.3000 0.7000 0 0.3000 0 0 0.3500 1.4500 0 0.3500 0

As shown, the exp(-Ts) terms in G11 and G21 were divided into two different types of delays. It looks like the software is trying capture a common delay at the two outputs, and then makes up for the additional delay as needed with the IODelay. Then pade() comes along and substitutes a first order approximant for both of those delays in G11 and G21, resulting in third order models for Gnew11 and Gnew21

Gnew = pade(G,1)

Gnew =
 
  From input 1 to output...
            -21.6 s^2 + 205.7 s - 411.4
   1:  -------------------------------------
       8.5 s^3 + 81.95 s^2 + 171.4 s + 19.05
 
            -2.75 s^2 + 19.51 s - 21.67
   2:  -------------------------------------
       8.2 s^3 + 59.17 s^2 + 71.72 s + 7.882
 
  From input 2 to output...
            -1.26 s + 8.4
   1:  -----------------------
       7.05 s^2 + 48 s + 6.667
 
           4.28 s - 24.46
   2:  -----------------------
       9 s^2 + 52.43 s + 5.714
 
Continuous-time transfer function.

Verify that Gnew11 is based on two first order approximants:

[num1,den1]=pade(.7,1);[num2,den2]=pade(.3,1);
Gnew(1,1), tf(-21.6,[8.5 1])*tf(num1,den1)*tf(num2,den2)

ans =
 
       -21.6 s^2 + 205.7 s - 411.4
  -------------------------------------
  8.5 s^3 + 81.95 s^2 + 171.4 s + 19.05
 
Continuous-time transfer function.


ans =
 
       -21.6 s^2 + 205.7 s - 411.4
  -------------------------------------
  8.5 s^3 + 81.95 s^2 + 171.4 s + 19.05
 
Continuous-time transfer function.

Explicitly using IODelay on the individual elements looks like it yields what you might have expected.

G2 = [-21.6/(8.5*s+1) (1.26)/(7.05*s+1);...
   (-2.75)/(8.2*s+1) (-4.28)/(9.0*s+1)];
G2(1,1).IODelay = 1; G2(1,2).IODelay = 0.3; G2(2,1).IODelay = 1.8; G2(2,2).IODelay = 0.35;
G2

G2 =
 
  From input 1 to output...
                     -21.6
   1:  exp(-1*s) * ---------
                   8.5 s + 1
 
                       -2.75
   2:  exp(-1.8*s) * ---------
                     8.2 s + 1
 
  From input 2 to output...
                        1.26
   1:  exp(-0.3*s) * ----------
                     7.05 s + 1
 
                       -4.28
   2:  exp(-0.35*s) * -------
                      9 s + 1
 
Continuous-time transfer function.

pade(G2,1)

ans =
 
  From input 1 to output...
         21.6 s - 43.2
   1:  ------------------
       8.5 s^2 + 18 s + 2
 
            2.75 s - 3.056
   2:  -------------------------
       8.2 s^2 + 10.11 s + 1.111
 
  From input 2 to output...
            -1.26 s + 8.4
   1:  -----------------------
       7.05 s^2 + 48 s + 6.667
 
           4.28 s - 24.46
   2:  -----------------------
       9 s^2 + 52.43 s + 5.714
 
Continuous-time transfer function.

Make sure that G and G2 have the same IO characteristics:

bode(G,G2)

[arash rad]

Thank you very much

your explanation is complete