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

(∂(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)))

[Walter Roberson]

Could we get an image of those? Or a latex form of that?

Is E "expected value" ?

[NAGA LAKSHMI KALYANI MOVVA]

Yes sir E is expected value

[NAGA LAKSHMI KALYANI MOVVA]

[Walter Roberson]

I do not know what is in this case, and with that notation I am concerned that it might be indicating a function with parameter s.

Please show the definition of m in better form.

[NAGA LAKSHMI KALYANI MOVVA]

m=-1/〖log〗_2 (cos⁡(ᴪ_(1/2)))

ᴪ_(1/2) denotes the semi-angle

[NAGA LAKSHMI KALYANI MOVVA]

based on ᴪ_(1/2) semi-angle the value of pole 's' varies

[Walter Roberson]

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?

Answer 1

[Walter Roberson]

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.

[Walter Roberson]

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]

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 = 

Answer 2

[NAGA LAKSHMI KALYANI MOVVA]

[NAGA LAKSHMI KALYANI MOVVA]

there is nothing obove the E sir actually it is pen mark. sorry

Answer 3

[NAGA LAKSHMI KALYANI MOVVA]

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