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 Commenti
Walter Roberson
il 26 Ago 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?
Risposte (3)
Walter Roberson
il 26 Ago 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
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.
2 Commenti
Walter Roberson
il 26 Ago 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.
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
il 26 Ago 2021
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))
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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