Operator '>' is not supported for operands of type 'tf'

%% w1 ,w2 ,X1, X2, X3, X4, X5, u1, u2, u3, y1, y2, y3, y4 are defined variables i did not put them to make the question short

s=tf('s')

Gvd = (Vo*(s*w1 + w2)*((s^4)*x1+(s^3)*x2+(s^2)*x3+s*x4+x5))/(dp*((s^2)*u1+s*u2+u3)*((s^3)*y1+(s^2)*y2+s*y3+y4));

d = Gvd>0;

values = Gvd(d);

objective = -20*log(values);

Operator '>' is not supported for operands of type 'tf'

Answer 1

[Paul]

Hi Salah

Gvd is a transfer function. It looks like you want to evaluate Gvd at a set of frequencies and then operate on that result.

Use freqresp to evaluate Gvd at the frequencies of interest and then compute the objective. Something like this

omega = logspace(-1,3,100);
H = freqresp(Gvd,w);
f = -20*log10(abs(H));  % as defined in Presentation1.pdf, though f is a function of omega, not x

[Salah]

Thanks a lot Paul, it worked!

Much appreciation.

Answer 2

[Image Analyst]

"i did not put them to make the question short" <= that was a mistake. Now you are forcing us to assign variables to them if we want to test it. Most people won't do that. If even you don't want to put in the work, then they won't either.

If Gvd(d) is true or false you don't want to do d = Gvd>0 to just give another true or false variable. And then taking the log of a vector of true or false values is nonsense.

[Salah]

function objective = objectiveFcn(Lc,Cc,Lf,Cf)

%% Fitness Function

Vo = 500;

R = 6;

DutyC=0.45;

s=tf('s');

dp = 1-DutyC;

w1 = -Lc;

w2 = dp*R;

u1 = Lc*Cc*R;

u2 = Lc;

u3 = (dp^2)*R;

x1 = Lf*Cf*(Lc^2)*dp*R*Cc;

x2 = (Lf+Lc*dp*R-(dp^2)*R*Lf*Cf)*Lc*Cc+(Lc^2)*Lf*Cf*dp*R;

x3 = (Lf+Lc*dp*R-(dp^2)*R*Lf*Cf)*Lc+Lf*Cf*Lc*(dp^3)*(R^2);

x4 = (Lf+Lc*dp*R-(dp^2)*R*Lf*Cf)*(dp^2)*R-(dp^2)*R*Lc;

x5 = -(dp^4)*(R^2);

y1 = Lc*Cc*Lf*Cf;

y2 = Lc*Cc + Lc*Lf*Cf;

y3 = Lc+Lf+Lf*Cf*(dp^2)*R;

y4 = R*(dp^2);

Gvd = (Vo*(s*w1 + w2)*((s^4)*x1+(s^3)*x2+(s^2)*x3+s*x4+x5))/(dp*((s^2)*u1+s*u2+u3)*((s^3)*y1+(s^2)*y2+s*y3+y4));

y = Gvd>0;

values = Gvd(y);

objective = -20*log(values);

end

%% 1e-9< Lc,Cc,Cf,Lc < 1e-2 as optimization variables

[Image Analyst]

I don't know that I answered it, but if I did, thanks for accepting it. For people to run your code they need to know what you passed into the function. What are Lc, Cc, Lf, and Cf?

If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:

TUTORIAL: How to ask a question (on Answers) and get a fast answer

[Salah]

Sorry , i am new to the MATLAB community and its my first time trying to ask a qeustion here, i accepted your answer because logically you are correct, but i want to eleminate all the values that causing the problem, firstly i used the function 'abs' because its in the objective function but the same problem appears that 'abs' is not supported for 'tf' , so i changed it to y=Gvd>0, because the domain of the log is from (zero to infinity) but the same problem occured. All the information of this code are attached in the pdf file. My apologies again for any inconvenience.