Fractional order transfer function

I tried to write a matlab script for the computation of this transfer function but I keep receving an error message saying that the exponent must a scalar integer.

How can I overcome this issue?

Thank you in advance.

[Jakob B. Nielsen]

Post your code, otherwise we have no chance of helping you save for guesswork :)

[Abdelkarim Jabrane]

v=20;
a=5;
C=1;
for k=2:v
    C=C+1/((s/a)^((k-1)/v));
end

Above is the reported the section of the code in question

[Jakob B. Nielsen]

That is odd. It runs just fine for me, and looking at it it shouldnt give an error...

However, you do make a mistake in the evaluation so lets take that; the transfer function is one divided by your sum. Your function is the sum of one divided by each individual component. I think you are looking for this;

v=20;
a=5;
s=2;
clear denom
denom=0; 
for k=1:v
    denom=denom+((s/a)^((k-1)/v));   
end
C=1/denom;

[Abdelkarim Jabrane]

Sorry I missed to paste the definition of s as Laplace transform operator. C(s) is a transfer function in continuos time domain.

Thank you very much for the correction, I appreciate it.

s=tf('s');
v=20;
a=5;
denom=0; 
for k=1:v
    denom=denom+((s/a)^((k-1)/v));   
end
C=1/denom;

In this way it will show an error message.

Answer 1

[Jyothis Gireesh]

A possible workaround to this scenario may to be define the transfer function using symbolic variables. You may use the following code to implement the same.

syms s k;
v = 20;
a = 5;
C = 1/symsum((s/a)^((k-1)/v),k,1,v);

The above symbolic function can be provided as input to “ilaplace ()” to get the time-domain representation of the same. However do note that due to the presence of fractional order terms in the denominator, the final result may be a piecewise approximation of the function.

[Walter Roberson]

The piecewise() approximation is more to avoid a singularity than because of fractional order terms.