Clueless about the Laplace Operational Method

[wolly]

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.




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

 

[logistic_guy]

Can the pole be negative like -1 ?

Yes.
Example of a negative pole:

$\displaystyle F(s) = \frac{1}{(s + 1)^2}$

$\displaystyle s = -1$ is a negative pole of order $\displaystyle 2$

 

[wolly]

Yes.
Example of a negative pole:

$\displaystyle F(s) = \frac{1}{(s + 1)^2}$

$\displaystyle s = -1$ is a negative pole of order $\displaystyle 2$

And 1 and 2 are poles of what order?

 

[logistic_guy]

And 1 and 2 are poles of what order?

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?

 

[wolly]

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?

How do you decide it's order?

 

[logistic_guy]

How do you decide it's order?

The power of the bracket.

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

Here $\displaystyle p$ is the order of the pole.

 

[wolly]

The power of the bracket.

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

Here $\displaystyle p$ is the order of the pole.

And i and -i?

 

[logistic_guy]

And i and -i?

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:

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

??

 

[wolly]

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:

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

??

For example If $$z=2i$$ and $$z=-2i$$from the function

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

 

[wolly]

Or

$$\frac{10}{(z^2+4)(z^3-2z^2-z+2)} + \frac{z^2-z-4}{z^3-2z^2-z+2}$$

 

[logistic_guy]

For example If $$z=2i$$ and $$z=-2i$$from the function

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

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.

Or

$$\frac{10}{(z^2+4)(z^3-2z^2-z+2)} + \frac{z^2-z-4}{z^3-2z^2-z+2}$$

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.

 

[wolly]

I have used the residue theorem but this is in a Laplace Operational method with

$$Θ(p)=Y(p)*e^{pt}$$
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.

 

[wolly]

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

 

[logistic_guy]

Can p be irrational?If yes then how can it be solved in this equation?

Yes it can be.

It's based on the 2 functions which I wrote today.

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.

how can it be solved in this equation?

You can use the Laplace transform table.

Or use the residue theorem.

 

[wolly]

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.

So z tends to $$-2i$$ and $$2i$$?

$$lim_{z \to 2i} f(z)*(z-2i)*e^{pt}$$and
$$lim_{z \to -2i} f(z)*(z+2i)*e^{pt}$$
I mean I don't know how to calculate $$e^{2it}$$ and $$e^{-2it}$$!
@logistic_guy

 

[wolly]

$$rezθ(p_k)=\frac{1}{(m_k-1)!}*\bigr[(p-p_k)^{m_k}*Y(p)*e^{pt} \bigr]$$
I also found this formula in my textbook but I don't know who is $$m_k$$!

 

[wolly]

It says in my book that $$m_k$$ is the order of multiplicity!

 

[wolly]

 

[wolly]

How does the denominator have these values?

$$0!$$$$1!$$$$2!$$@logistic_guy
Can you please explain?

 

[wolly]

Also can the result of the limit of the residue have values like i,2i and so on?