Fractional order transfer function

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Abdelkarim Jabrane
Abdelkarim Jabrane 2019-12-13
评论： Walter Roberson 2020-1-3
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.
  4 个评论
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Jakob B. Nielsen
Jakob B. Nielsen 2019-12-13
编辑：Jakob B. Nielsen 2019-12-13
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
Abdelkarim Jabrane 2019-12-13
编辑：Abdelkarim Jabrane 2019-12-13
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.

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回答（1 个）

Jyothis Gireesh
Jyothis Gireesh 2020-1-2
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.
  1 个评论
Walter Roberson
Walter Roberson 2020-1-3
The piecewise() approximation is more to avoid a singularity than because of fractional order terms.

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