Split the polynomial and determine the degree of numerator and denominator

Hello All,

I am working with some mathematical GUI , i need to split the polynomial and determine the degree of numerator and denominator . Inputs are fed from user in gui in edit text i have attached a pseudo code where i am calculating the transfer function. I need to split

OLPZData=handles.General_test.PlantInfo.OL.UDPPZForm.PZData;
    p(1)=str2double(OLPZData.p1.String);
    p(2)=str2double(OLPZData.p2.String);
    p(3)=str2double(OLPZData.p3.String);
    p(4)=str2double(OLPZData.p4.String);
    p(5)=str2double(OLPZData.p5.String);
    m=str2double(OLPZData.PolesAtOrigin.String);
    if p(1)==0 || p(2)==0 || p(3)==0 || p(4)==0 || p(5)==0
        hObject.BackgroundColor=[1 0.8 0.8];
        return;
    else
        hObject.BackgroundColor=[1 1 1];
    end
    P= sum(~isnan(p));
    if isnan(m)
        m=0;
        handles.General_test.PlantInfo.OL.UDPPZForm.PZData.PolesAtOrigin.String ='0';
        handles.General_test.PlantInfo.OL.UDPPZForm.PZData.PolesAtOrigin.BackgroundColor=[1 1 1];
    end
    z(1)=str2double(OLPZData.z1.String);
    z(2)=str2double(OLPZData.z2.String);
    z(3)=str2double(OLPZData.z3.String);
    z(4)=str2double(OLPZData.z4.String);
    z(5)=str2double(OLPZData.z5.String);
    Kdc=str2double(OLPZData.Kdc.String);
    Z= sum(~isnan(z));
    
    %Kdc(isnan(Kdc))=1;
    if isnan(Kdc)
        Kdc=1;
        handles.General_test.PlantInfo.OL.UDPPZForm.PZData.Kdc.String ='1';
        handles.General_test.PlantInfo.OL.UDPPZForm.PZData.Kdc.BackgroundColor=[1 1 1];
    end
    if z(1)==0 || z(2)==0 || z(3)==0 || z(4)==0 || z(5)==0 ||Kdc==0
        hObject.BackgroundColor=[1 0.8 0.8];
        return;
    else
        hObject.BackgroundColor=[1 1 1];
    end
	 Gvd=Kdc/((s/(2*pi))^m);
    PmultiFactor=zeros(1,5);
    ZmultiFactor=zeros(1,5);
    DENOMM=zeros(1,5);
    for i=1:5
        PmultiFactor(i)=double(~isnan(p(i)));
        ZmultiFactor(i)=double(~isnan(z(i)));
        Gvd=Gvd*(1+s*ZmultiFactor(i)/(2*pi*z(i)))/(1+s*PmultiFactor(i)/(2*pi*p(i)));
        hip = zpk(Gvd);
        hip.DisplayFormat = 'frequency';
        handles.General_test.PlantInfo.OL.PlantTF=Gvd;
        
        %PlotGeneration(hObject);
    end

extract numerator and denominator of Gvd

Thanks in advance

Answer 1

[Steven Lord]

From the fact that this question is tagged "control systems" and you're calling the zpk function in your code, it sounds like you want to obtain the numerator and denominator of that transfer function. Looking at the documentation for the zpk object it has properties Z and P that contain the zeros and poles of the transfer function. You can use the poly function to create the polynomial with those roots or you could use the zp2tf function to convert your zpk object into a tf object then access that tf object's Numerator and Denominator properties.

[sandeep singh]

Hi Steven,

zp2tf function This is not working as my ZPK are in cell format .Any how i found the way to solve its lengthy but getting needed results.

thanks for all

Answer 2

[Walter Roberson]

By definition, the degree of the denominator of a polynomial is 0 (that is, the denominator is restricted to being a constant.)

Any system in which the independent variable appears in the denominator is not a polynomial -- not even if the expression fully cancels with the numerator.

[sandeep singh]

I have mentioned that inputs are on GUI by the user

[Walter Roberson]

So?

We can guarantee that the answer is either

1. That the degree of the denominator is 0 (a constant); or
2. That the system is not a polynomial.

A ratio of polynomials is never a polynomial (except in the degenerate case where the denominator is the constant polynomial.)

[sandeep singh]

Hello Walter,

First of all i am providing inputs on my GUI to generate transfer function. So definately it has to be polynomial.

Kindly understand the question then comment.

Any how i found the way to solve its lengthy but getting needed results.

thanks for all

[Walter Roberson]

Transfer functions are not polynomials, except in the degenerate case where the denominator is a constant, reflecting a constant gain or diminuation.

Transfer functions are potentially the ratio of polynomials.

[sandeep singh]

Hello Walter,

I agree Transfer functions are not polynomials, but i needed to split or parse however you take . anyhow i have found solution by myself thanks for all support and replies.

[Mohsen]

Dear sandeep singh

I have a challenge same you. Can you send your solution?

Regards