why is the zpk function not producing the transfer function in MATLAB?

Hello guys,

I am trying to use the zpk function, where I input the zeros, poles and gain and it should return the factored form of the transfer function but it is not giving me that output. The code I used is below with the output I am getting from zpk.

a = [2,5,9,5,3];
b = [5,45,2,1,1];
[zer,pol,t] = zplane(a,b);
z = zeros(1,4);
p = zeros(1,4);
%disp('Zeros of the Transfer function are: ');
for i = 1:4
    X = ['(',num2str(zer.XData(i)),',',num2str(zer.YData(i)),')'];
    z(i) = zer.XData(i) + 1i*zer.YData(i);
    %disp(X)
end
%disp('Poles of the transfer function are: ');
for i = 1:(length(b)-1)
    X = ['(',num2str(pol.XData(i)),',',num2str(pol.YData(i)),')'];
    p(i) = pol.XData(i) + 1i*pol.YData(i);
    %disp(X);
end

%disp(z);
G = zpk([z],[p],1);
disp(G);

Output:

zpk with properties:

Z: {[4×1 double]}

P: {[4×1 double]}

K: 1

DisplayFormat: 'roots'

Variable: 's'

IODelay: 0

InputDelay: 0

OutputDelay: 0

Ts: 0

TimeUnit: 'seconds'

InputName: {''}

InputUnit: {''}

InputGroup: [1×1 struct]

OutputName: {''}

OutputUnit: {''}

OutputGroup: [1×1 struct]

Notes: [0×1 string]

UserData: []

Name: ''

SamplingGrid: [1×1 struct]

Answer 1

[Star Strider]

Remove the closing parentheses in the ‘G’ assignment to see the result as a sort of ‘prettyprint’ output:

a = [2,5,9,5,3];
b = [5,45,2,1,1];
[zer,pol,t] = zplane(a,b);

z = zeros(1,4);
p = zeros(1,4);
%disp('Zeros of the Transfer function are: ');
for i = 1:4
    X = ['(',num2str(zer.XData(i)),',',num2str(zer.YData(i)),')'];
    z(i) = zer.XData(i) + 1i*zer.YData(i);
    %disp(X)
end
%disp('Poles of the transfer function are: ');
for i = 1:(length(b)-1)
    X = ['(',num2str(pol.XData(i)),',',num2str(pol.YData(i)),')'];
    p(i) = pol.XData(i) + 1i*pol.YData(i);
    %disp(X);
end

%disp(z);
G = zpk([z],[p],1)

G =
 
        (s^2 + 0.5s + 0.5) (s^2 + 2s + 3)
  ----------------------------------------------
  (s+8.958) (s+0.2718) (s^2 - 0.2293s + 0.08216)
 
Continuous-time zero/pole/gain model.

Zv = G.Z{:}

Zv = 
  -1.0000 + 1.4142i
  -1.0000 - 1.4142i
  -0.2500 + 0.6614i
  -0.2500 - 0.6614i

Pv = G.P{:}

Pv = 
  -8.9576 + 0.0000i
  -0.2718 + 0.0000i
   0.1147 + 0.2627i
   0.1147 - 0.2627i

Kv = G.K

Kv = 1

disp(G)

  zpk with properties:

                Z: {[4×1 double]}
                P: {[4×1 double]}
                K: 1
    DisplayFormat: 'roots'
         Variable: 's'
          IODelay: 0
       InputDelay: 0
      OutputDelay: 0
               Ts: 0
         TimeUnit: 'seconds'
        InputName: {''}
        InputUnit: {''}
       InputGroup: [1×1 struct]
       OutputName: {''}
       OutputUnit: {''}
      OutputGroup: [1×1 struct]
            Notes: [0×1 string]
         UserData: []
             Name: ''
     SamplingGrid: [1×1 struct]

.

[Laxman Chinnannavar]

Thanks this works

[Star Strider]

As always, my pleasure!

[Laxman Chinnannavar]

Hey, One more small doubt bro why doesn't the transfer function shown in proper factorized form. Like there is still (s^2 + 0.5s +0.5) term right? Cant that get factorized into imaginary roots?

[Star Strider]

Apparently not, at least in the ‘prettyprint’ display. It only factors real roots, however the ‘P’ and ‘Z’ outputs display the complex roots in complex-conjugate pairs.

See the documentation section on Examples for an illustration of how it displays the various factors.

[Laxman Chinnannavar]

I had already checked the examples, just wanted to know if there was any way to get the proper factorized form. Really appreciate the help.

[Star Strider]

As always, my pleasure!

It might be possible to add that as a format option for the ‘prettyprint’ output. The best way to approach that is to Contact Support and request that as an enhancement.

Answer 2

[Paul]

What output is expected? It looks like zpk() is returning a zpk object with poles and zeros at locations shown on the plot.

a = [2,5,9,5,3];
b = [5,45,2,1,1];
[zer,pol,t] = zplane(a,b);

z = zeros(1,4);
p = zeros(1,4);
%disp('Zeros of the Transfer function are: ');
for i = 1:4
    X = ['(',num2str(zer.XData(i)),',',num2str(zer.YData(i)),')'];
    z(i) = zer.XData(i) + 1i*zer.YData(i);
    %disp(X)
end
%disp('Poles of the transfer function are: ');
for i = 1:(length(b)-1)
    X = ['(',num2str(pol.XData(i)),',',num2str(pol.YData(i)),')'];
    p(i) = pol.XData(i) + 1i*pol.YData(i);
    %disp(X);
end
%disp(z);
G = zpk([z],[p],1);
G

G =
 
        (s^2 + 0.5s + 0.5) (s^2 + 2s + 3)
  ----------------------------------------------
  (s+8.958) (s+0.2718) (s^2 - 0.2293s + 0.08216)
 
Continuous-time zero/pole/gain model.

G.Z{:}

ans = 
  -1.0000 + 1.4142i
  -1.0000 - 1.4142i
  -0.2500 + 0.6614i
  -0.2500 - 0.6614i

G.P{:}

ans = 
  -8.9576 + 0.0000i
  -0.2718 + 0.0000i
   0.1147 + 0.2627i
   0.1147 - 0.2627i

The order of the inputs to zplane are zplane(numerator,denominator). Is a really the numerator and b really the denominator? Only asking because the typical naming convention in Matlab is that b is the numerator and a is the denominator. Will continue assuing a is the num and b is the den, but I suggest rechecking that.

If the goal is just to get the zpk representation with poles and zeros governed by a and b with a gain of k=1, that code seems like a really complicated way to go about it. Could just do

G1 = zpk(roots(a),roots(b),1)

G1 =
 
        (s^2 + 0.5s + 0.5) (s^2 + 2s + 3)
  ----------------------------------------------
  (s+8.958) (s+0.2718) (s^2 - 0.2293s + 0.08216)
 
Continuous-time zero/pole/gain model.

However, if the goal is get the zpk representation of the transfer function a(s)/b(s), then there is an error because the gain is not unity

G2 = zpk(tf(a,b))

G2 =
 
      0.4 (s^2 + 0.5s + 0.5) (s^2 + 2s + 3)
  ----------------------------------------------
  (s+8.958) (s+0.2718) (s^2 - 0.2293s + 0.08216)
 
Continuous-time zero/pole/gain model.

Note the gain k = 0.4.

Finally, the code is uzing zplane which suggests these are poles and zeros of a discrete time transfer function. As can be seen, zpk returns a continuous time model unless the sample time argument is provided.

G3 = zpk(tf(a,b,-1))

G3 =
 
      0.4 (z^2 + 0.5z + 0.5) (z^2 + 2z + 3)
  ----------------------------------------------
  (z+8.958) (z+0.2718) (z^2 - 0.2293z + 0.08216)
 
Sample time: unspecified
Discrete-time zero/pole/gain model.

[Laxman Chinnannavar]

Thanks @Paul, for providing all this useful facts.

Actually I dont know the general convention for naming denominators and numerators, so switched those up bymistakely but yeah whatever I did in zpk is correct.

Also yes I wanted the transfer function to be z transform rather than in s domain. That helped a lot