normal rank calculation with tzero

Consider the following transfer matrix:

s = tf('s')

s =
 
  s
 
Continuous-time transfer function.

G  = [1/(s+2) 0; 0 1/((s+1)*(s+3))]

G =
 
  From input 1 to output...
         1
   1:  -----
       s + 2
 
   2:  0
 
  From input 2 to output...
   1:  0
 
             1
   2:  -------------
       s^2 + 4 s + 3
 
Continuous-time transfer function.

the calcultation of the normal rank of G is done by:

[z, nrank] = tzero(G)

z =

  0x1 empty double column vector

nrank = 1

and the result is

nrank = 1.

I was wondering, because in my understanding G will never lose rank for any and the normal rank should be equal to 2. So far, my observation is, that if the number of poles in the part transfer functions is equal,

i.e.

G  = [1/(s+2) 0; 0 1/(s+1)]

G =
 
  From input 1 to output...
         1
   1:  -----
       s + 2
 
   2:  0
 
  From input 2 to output...
   1:  0
 
         1
   2:  -----
       s + 1
 
Continuous-time transfer function.

[z, nrank] = tzero(G)

z =

  0x1 empty double column vector

nrank = 2

the result meets with my expectations (nrank = 2).

So my question is, am I missing something when using tzero or is there a general problem of understanding regarding the normal rank of a transfer matrix?

Many thanks in advance.

[Paul]

Hi Andreas,

I took a look and I agree that the first case seems odd. I tried lots of variation in the tol input to tzero, but that had no effect. Unfortunately, the heavy lifting for tzero, at least for the first case, is in a .mex file, so I couldn't dig into the code.

If you open a case with Tech Support, would you mind posting back here with a summary of their response?

[Andreas]

Hi Paul,

thank you very much for your answer. sure, I'll do that.

Answer 1

[Christian]

We have identified that this is indeed a bug in "tzero".

We are actively working on a fix, and aim to provide it as soon as possible.