Clueless about the Laplace Operational Method

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wolly

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Hi I'm trying to calculate the order of the poles in the Laplace Transform Operational Method and I have no idea what order they are.


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Can the pole be negative like -1 ? I wanted to solve a Laplace differential equation and I don't know a lot about this!
 

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wolly said:
And 1 and 2 are poles of what order?
Click to expand...
A pole is a value that makes the denominator \(\displaystyle = 0\). A pole can have any order.

For example:

\(\displaystyle F(s) = \frac{1}{(s - 1)} \rightarrow\) here \(\displaystyle s = 1\) is a pole of order \(\displaystyle 1\)

\(\displaystyle F(s) = \frac{1}{(s - 1)^2} \rightarrow\) here \(\displaystyle s = 1\) is a pole of order \(\displaystyle 2\)

\(\displaystyle F(s) = \frac{1}{(s - 1)^3} \rightarrow\) here \(\displaystyle s = 1\) is a pole of order \(\displaystyle 3\)

And

\(\displaystyle F(s) = \frac{1}{(s - 2)} \rightarrow\) here \(\displaystyle s = 2\) is a pole of order \(\displaystyle 1\)

\(\displaystyle F(s) = \frac{1}{(s - 2)^2} \rightarrow\) here \(\displaystyle s = 2\) is a pole of order \(\displaystyle 2\)

\(\displaystyle F(s) = \frac{1}{(s - 2)^{100}} \rightarrow\) here \(\displaystyle s = 2\) is a pole of order \(\displaystyle 100\)


Understood the idea?

🤔
 
logistic_guy said:
A pole is a value that makes the denominator \(\displaystyle = 0\). A pole can have any order.

For example:

\(\displaystyle F(s) = \frac{1}{(s - 1)} \rightarrow\) here \(\displaystyle s = 1\) is a pole of order \(\displaystyle 1\)

\(\displaystyle F(s) = \frac{1}{(s - 1)^2} \rightarrow\) here \(\displaystyle s = 1\) is a pole of order \(\displaystyle 2\)

\(\displaystyle F(s) = \frac{1}{(s - 1)^3} \rightarrow\) here \(\displaystyle s = 1\) is a pole of order \(\displaystyle 3\)

And

\(\displaystyle F(s) = \frac{1}{(s - 2)} \rightarrow\) here \(\displaystyle s = 2\) is a pole of order \(\displaystyle 1\)

\(\displaystyle F(s) = \frac{1}{(s - 2)^2} \rightarrow\) here \(\displaystyle s = 2\) is a pole of order \(\displaystyle 2\)

\(\displaystyle F(s) = \frac{1}{(s - 2)^{100}} \rightarrow\) here \(\displaystyle s = 2\) is a pole of order \(\displaystyle 100\)


Understood the idea?

🤔
Click to expand...
How do you decide it's order?
 
logistic_guy said:
Do you have an example of that in your mind? Or do you mean this:

\(\displaystyle F(s) = \frac{1}{(s + i)^p} \)

And this:

\(\displaystyle F(s) = \frac{1}{(s - i)^p} \)

??
Click to expand...
For example If [math]z=2i[/math] and [math]z=-2i[/math]from the function

[math]f(z)=\frac{1}{z^2+4}[/math]
Which order would be 2i and -2i as a pole?
 
Or

[math]\frac{10}{(z^2+4)(z^3-2z^2-z+2)} + \frac{z^2-z-4}{z^3-2z^2-z+2}[/math]
 
wolly said:
For example If [math]z=2i[/math] and [math]z=-2i[/math]from the function

[math]f(z)=\frac{1}{z^2+4}[/math]
Which order would be 2i and -2i as a pole?
Click to expand...
First factor the denominator.

\(\displaystyle f(z) = \frac{1}{z^2 + 4} = \frac{1}{(z + 2i)(z - 2i)}\)

So, we have two poles, \(\displaystyle z = 2i\) and \(\displaystyle z = -2i\), each one of them is of order one.

wolly said:
Or

[math]\frac{10}{(z^2+4)(z^3-2z^2-z+2)} + \frac{z^2-z-4}{z^3-2z^2-z+2}[/math]
Click to expand...
You always start by factoring first.

\(\displaystyle f(z) = \frac{10}{(z^2+4)(z^3-2z^2-z+2)} + \frac{z^2-z-4}{z^3-2z^2-z+2}\)

\(\displaystyle = \frac{10}{(z + 2i)(z - 2i)(z + 1)(z - 1)(z - 2)} + \frac{z^2-z-4}{(z + 2i)(z - 2i)(z + 1)(z - 1)(z - 2)}\)

Since the numerator \(\displaystyle z^2-z-4\) does not have a nice factor, it will not be cancelled with the denominator, so you leave it alone.

Now we collect the poles. These are the poles:

\(\displaystyle z = 2i\)
\(\displaystyle z = -2i\)
\(\displaystyle z = 1\)
\(\displaystyle z = -1\)
\(\displaystyle z = 2\)

Each one of them is of order one.
 
Last edited:
I have used the residue theorem but this is in a Laplace Operational method with

[math]Θ(p)=Y(p)*e^{pt}[/math]
Can p be irrational?If yes then how can it be solved in this equation?
@logistic_guy ?

It's based on the 2 functions which I wrote today.
 
[math]f(z) = \frac{10}{(z^2+4)(z^3-2z^2-z+2)} + \frac{z^2-z-4}{z^3-2z^2-z+2}[/math]
This one^
 
wolly said:
Can p be irrational?If yes then how can it be solved in this equation?
Click to expand...
Yes it can be.

wolly said:
It's based on the 2 functions which I wrote today.
Click to expand...
The two functions that you wrote had real and imaginary poles, not irrational.

An irrational pole will be something likes this:

\(\displaystyle f(z) = \frac{1}{z - \sqrt{2}}\)

Here \(\displaystyle z = \sqrt{2}\) is an irrational pole.

wolly said:
how can it be solved in this equation?
Click to expand...
You can use the Laplace transform table.

Or use the residue theorem.
 
logistic_guy said:
Yes it can be.


The two functions that you wrote had real and imaginary poles, not irrational.

An irrational pole will be something likes this:

\(\displaystyle f(z) = \frac{1}{z - \sqrt{2}}\)

Here \(\displaystyle z = \sqrt{2}\) is an irrational pole.


You can use the Laplace transform table.

Or use the residue theorem.
Click to expand...
So z tends to [math]-2i[/math] and [math]2i[/math]?

[math]lim_{z \to 2i} f(z)*(z-2i)*e^{pt}[/math]and
[math]lim_{z \to -2i} f(z)*(z+2i)*e^{pt}[/math]
I mean I don't know how to calculate [math]e^{2it}[/math] and [math]e^{-2it}[/math]!
@logistic_guy
 
[math]rezθ(p_k)=\frac{1}{(m_k-1)!}*\bigr[(p-p_k)^{m_k}*Y(p)*e^{pt} \bigr][/math]
I also found this formula in my textbook but I don't know who is [math]m_k[/math]!
 
It says in my book that [math]m_k[/math] is the order of multiplicity!
 
Also can the result of the limit of the residue have values like i,2i and so on?
 
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