how do i find poles of the equation given using laplace transform
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(∂(1+W(s)))/∂s=-1/((m+3) ) E_(1/(m+3)) (s)+1/((m+3) ) ᴦ ((m+2))/((m+3) ) s^(1/((m+3) )) >0
where m=-1/〖log〗_2 (cos(ᴪ_(1/2)))
7 Comments
Walter Roberson
on 26 Aug 2021
I am not clear whether that is
or if it is
where the first one indicates a division by 2, and the second one indicates an unusual variable name that includes "1/2" as part of the name?
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721919/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721924/image.png)
Answers (3)
Walter Roberson
on 26 Aug 2021
Poles of an equation occur at the denominator of the laplace transform become 0.
The first part of your expression ending in
looks to be constants times s -- no denominator. Not unless m == -3
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721834/image.png)
The second part of the equation has two constants times s to a negative power. The constants are not denominator... unless m == -3 .
So you have
which is
and you are looking for a denomininator of 0. Assuming s is non-zero that would require... well, it can't happen for non-zero s. It does exist as a left-limit for m == -3 but only as a limit. If you are relying on a pole at the limit of m = -3 with the multiple m+3 in the expression, you have a lot of close examination to do to figure out the overall limits of the expression. And for the case where m is not -3, then there is no pole.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721839/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721844/image.png)
2 Comments
Walter Roberson
on 26 Aug 2021
I still do not know what
represents. When it appears in the second line, it appears to have a mark over it that I cannot read, maybe a Δ and I do not know what that would represent either.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721934/image.png)
I calculated with your definition for m and it appears to me that m+3 could possibly be 0, if
is 2*arccos(2^(2/3)/2) or oddly-named
is arccos(2^(2/3)/2) . Passing through that point would be a mathematical mess.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721939/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721944/image.png)
Walter Roberson
on 26 Aug 2021
Depending on what
means, there just might be a solution, shown here as sol.s
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/721994/image.png)
syms m s Em4m3s real
denom = 1/(m+3) * Em4m3s + gamma((m+2)/(m+3)) * s^(1/(m+3))
sol = solve(denom == 0, s, 'returnconditions', true)
sol.s
K = sol.parameters;
simplify(sol.conditions)
sk = subs(sol.s, K, 0);
limit(sk, m, -3, 'left')
limit(sk, m, -3, 'right')
ck = simplify(subs(sol.conditions, K, 0))
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