Bode form of a transfer function

Suppose I have the transfer function k*(s-z1)/((s - p1)*(s - p2)), i.e., in the form that comes from the zpk() function. This form is convenient in that I can see immediately the location of the poles and zeros, but in order to obtain the gain at s=0, I need to compute k*(-z1)*(-p1)*(-p2). Is there a way to instead have the transfer function displayed in Bode form: k2*(-s/z1 + 1)/((-s/p1 + 1)*(-s/p2 + 1)), so that k2 = (-z1)*(-p1)*(-p2), and this represents the gain at s=0? I find this much nicer because I can see the poles, zeros, and the gain at s=0 without having to do any calculations in my head.

Thanks, Luke

Answer 1

[Arkadiy Turevskiy]

Yes, as descsribed in zpk doc.

>> sys=zpk(1,[2 3],4)

sys =

    4 (s-1)
  -----------
  (s-2) (s-3)

Continuous-time zero/pole/gain model.

>> sys.DisplayFormat='frequency';

>> sys

sys =

-0.66667 (1-s)
---------------
(1-s/2) (1-s/3)

Continuous-time zero/pole/gain model.

[Dale]

Excellent! I knew Matlab must have this capability, thank you for finding it.

Answer 2

[David Sanchez]

s=tf('s');
% dummy values for the parameters
k=1;
z1=1;
p1=1;
p2=1;
H=k*(s-z1)/((s - p1)*(s - p2));
k2= (-z1)*(-p1)*(-p2);
H2=k2*(-s/z1 + 1)/((-s/p1 + 1)*(-s/p2 + 1));
bode(H);
figure;
bode(H2);

[Dale]

My question is not about how to make a Bode plot. It is about how to print the transfer function as it would be written mathematically.

Is there a single function built into Matlab that already implements the code you have written? I can write the function to do this myself for an arbitrary transfer function, but I figured since they already have tf() and zpk(), which put things in the other two common forms of a transfer function, they might have a Bode form of the transfer function as well. If they do have it, I can't find it anywhere in the documentation.