Inverse laplace transform (polynomial division? Complex roots?)

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Homework Statement



Decide the inverse laplace transform of the problem below:

F(s)= [itex]\frac{4s-5}{s^2-4s+8}[/itex]

You're allowed to use s shifting.

Homework Equations





The Attempt at a Solution



By looking at the denominator, I see that it might be factorized easily, so I try that.

I end up struggling and realizing that it's a complex root. Complex roots and inverse laplace transform isn't something we've learned yet, but I'm keen to solve this problem regardless.

So the denominator can be written like this:

##s^2 - 4s +8 = 2+/- 2i##

Looking at my laplace transform table, I can't recognize the pattern to try solving this.

Can I use polynomial division?

Any help here is greatly appreciated.
 

Answers and Replies

  • #2
SteamKing
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Homework Statement



Decide the inverse laplace transform of the problem below:

F(s)= [itex]\frac{4s-5}{s^2-4s+8}[/itex]

You're allowed to use s shifting.

Homework Equations





The Attempt at a Solution



By looking at the denominator, I see that it might be factorized easily, so I try that.

I end up struggling and realizing that it's a complex root. Complex roots and inverse laplace transform isn't something we've learned yet, but I'm keen to solve this problem regardless.

So the denominator can be written like this:

##s^2 - 4s +8 = 2+/- 2i##

Looking at my laplace transform table, I can't recognize the pattern to try solving this.

Can I use polynomial division?

Any help here is greatly appreciated.

I'm not sure what this means:

##s^2 - 4s +8 = 2+/- 2i##

Are you saying that s^2 - 4s + 8 factors into (s - (2 + 2i)) and (s - (2 - 2i)) or what?

Have you tried completing the square of the denominator? This may help to avoid complex factors.
 
  • #3
Ray Vickson
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Homework Statement



Decide the inverse laplace transform of the problem below:

F(s)= [itex]\frac{4s-5}{s^2-4s+8}[/itex]

You're allowed to use s shifting.

Homework Equations





The Attempt at a Solution



By looking at the denominator, I see that it might be factorized easily, so I try that.

I end up struggling and realizing that it's a complex root. Complex roots and inverse laplace transform isn't something we've learned yet, but I'm keen to solve this problem regardless.

So the denominator can be written like this:

##s^2 - 4s +8 = 2+/- 2i##

Looking at my laplace transform table, I can't recognize the pattern to try solving this.

Can I use polynomial division?

Any help here is greatly appreciated.

When you factor the denominator you can write ##F(s) = (4s-5)/(s^2 - 4s + 8)## in partial fraction form. What is that form in your case? From there, you need only know how to find the inverse transform of ##g(s) = 1/(s-a)##, and the same formula applies whether ##a## is real or complex.
 
Last edited:
  • #4
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I'm not sure what this means:

##s^2 - 4s +8 = 2+/- 2i##

Are you saying that s^2 - 4s + 8 factors into (s - (2 + 2i)) and (s - (2 - 2i)) or what?
A friend of mine used a rather advanced calculator and got that as an answer, yes.
Have you tried completing the square of the denominator? This may help to avoid complex factors.
I tried that, but it leaves me with no answer. Using the ABC-rule, the square root turns negative so that makes sense with the answer from the aforementioned calculator.

When you factor the denominator you can write ##F(s) = (4s-5)/(s^2 - 4s + 8)## in partial fraction form. What is that form in your case? From there, you need only know how to find the inverse transform of ##g(s) = 1/(s-a)##, and the same formula applies whether ##a## is real or complex.

Can the form be the following:

[itex]\frac{4s-5}{s^2-4s+8}[/itex] = [itex]\frac{2+\frac{3i}{4}}{(s-(2+2i))}[/itex] - [itex]\frac{2+\frac{3i}{4}}{(s-(2-2i))}[/itex]?

Hey, I'm beginning to see a pattern here :D Thank you.

If my above work make any kind of sense, I'll be attempting to solve this problem even with complex numbers. This is so fun!
 
  • #5
vela
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A friend of mine used a rather advanced calculator and got that as an answer, yes.
The point is that you shouldn't write ##s^2-4s+8 = 2\pm 2i## if what you mean is ##s^2-4s+8 = (s-(2+2i))(s-(2-2i))##.

I tried that, but it leaves me with no answer. Using the ABC-rule, the square root turns negative so that makes sense with the answer from the aforementioned calculator.
SteamKing is suggesting you write the transform in the form
$$\frac{4s-5}{(s-a)^2 + b^2},$$ which you can invert using some of the properties listed in your tables.
 
  • #6
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The point is that you shouldn't write ##s^2-4s+8 = 2\pm 2i## if what you mean is ##s^2-4s+8 = (s-(2+2i))(s-(2-2i))##.


SteamKing is suggesting you write the transform in the form
$$\frac{4s-5}{(s-a)^2 + b^2},$$ which you can invert using some of the properties listed in your tables.

Thanks for replying.

I understand what you mean, although I'm a bit confused as to what a would be in my case. I'm just confused since I've been working on complex numbers today, and my teacher just said that there are two (possibly more) solutions to this problem. One complex and one "normal".

Looking at what you've written, it seems like there should be a sin(wt) + cos(wt) solution. But it doesn't make totally sense.

If I look at the ##e^wt## (both w and t should be "squared") properties, I can see a possible connection, but I'm not sure how I go about to solve it completely?

Thanks for sticking with me on this, I really appreciate it!
 
Last edited:

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