how do i find poles of the equation given using laplace transform

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NAGA LAKSHMI KALYANI MOVVA
评论： Walter Roberson 2021-8-26
(∂(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)))
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Walter Roberson
Walter Roberson 2021-8-26
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?

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回答（3 个）

Walter Roberson
Walter Roberson 2021-8-26
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
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.
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Walter Roberson
Walter Roberson 2021-8-26
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.
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.
Walter Roberson
Walter Roberson 2021-8-26
Ran in:
Depending on what means, there just might be a solution, shown here as sol.s
syms m s Em4m3s real
denom = 1/(m+3) * Em4m3s + gamma((m+2)/(m+3)) * s^(1/(m+3))
denom = 
sol = solve(denom == 0, s, 'returnconditions', true)
sol = struct with fields:
s: [1×1 sym] parameters: [1×1 sym] conditions: [1×1 sym]
sol.s
ans = 
K = sol.parameters;
simplify(sol.conditions)
ans = 
sk = subs(sol.s, K, 0);
limit(sk, m, -3, 'left')
ans = 
∞
limit(sk, m, -3, 'right')
ans = 
ck = simplify(subs(sol.conditions, K, 0))
ck = 

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NAGA LAKSHMI KALYANI MOVVA
NAGA LAKSHMI KALYANI MOVVA
sir can you help me to solve this to find poles using laplace transform. actually denominator i have menctioned previously.

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