[scilab-Users] transfer function, z^-1 form

[David PICARD]

Ooops, sorry, you're right. I mistyped the denominator (0.86z^-1 instead 
of -0.86z^-1) . Of course, I was confused by the weird step response...

Is it possible to specify the transfer function with polynoms ? In this 
case, it would be something like :
num = [0 0.38 0.31]
den = [1 -0.86 0.55]

Regards.


Collette yann a écrit :
>
> For me it corresponds to the good transfert function:
>
>     0.31 + 0.38z                    z^2* (0.31*z^-2 + 0.38*z^-1)
>   ----------------         =       ------------------------------------
>   0.55 + 0.86z + z ^2            z^2*(0.55*z^-2 + 0.86z^-1 + 1)
>
> Scilab seems to have simplified in a way your transfert function
>
> YC
>
>
> David PICARD a écrit :
>> I need some help to define the following transfer function :
>>
>>  0.381417 z^-1 + 0.310306 z^-2
>> ---------------------------------
>> 1 - 0.857089 z^-1 + 0.548812 z^-2
>>
>> This is a 2nd order system : w0=1, Damp=1.3, Gain=1.0, Te=1s
>>
>> I tried :
>> -->z=%z;
>> -->sys=syslin(1.0,(0.38*z^-1+0.31*z^-2)/(1+0.86*z^-1+0.55*z^-2))
>> sys  =
>>
>>      0.31 + 0.38z
>>    ----------------
>>                    2
>>    0.55 + 0.86z + z
>>
>>
>> But it NOT what I asked !!! Help !
>>
>> It is so straightforward with GNU Octave, that I am really frustrated !
>>> sys=tf([0.000000 0.381417 0.310306],[1.000000 -0.857089 0.548812],1.0);
>>> step(sys)   // shows the oscillating, stabilizing step response
>>
>>
>>
>
>
>