Laplace Transform with Imaginary Roots

  • Thread starter jdawg
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  • #1
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Homework Statement



4(d2x/dt2) +3x = t*e-3tsin(5t)

Homework Equations




The Attempt at a Solution


So I know how to take the Laplace transform and find the function for the Laplace domain:

X(s) = 10(s+3)/(((s+3)2+25)2)(4s2+3) + (10s/(4s2+3)) + (2/(4s2+3))

But trying to convert 10(s+3)/(((s+3)2+25)2)(4s2+3) back into the time domain is giving me a lot of trouble!
Those imaginary roots are really throwing me off. It was suggested to me to use conjugate poles to solve for this part, but I've never learned this method.

Any help is greatly appreciated!

Also not related to the homework question but, is there a fraction button somewhere?? I realize the question is hard to look at when there are so many parenthesis and slashes.
 

Answers and Replies

  • #3
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No. I pretty much just post homework questions, I never really look at anything else on this site. There used to be a really convenient fraction button that you could just click, I was kind of hoping for something like that.
 
  • #4
LCKurtz
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How hard is it to type ##\frac{numerator here}{denominator here}##?
 
  • #5
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Pretty hard.
 
  • #6
Ray Vickson
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Homework Statement



4(d2x/dt2) +3x = t*e-3tsin(5t)

Homework Equations




The Attempt at a Solution


So I know how to take the Laplace transform and find the function for the Laplace domain:

X(s) = 10(s+3)/(((s+3)2+25)2)(4s2+3) + (10s/(4s2+3)) + (2/(4s2+3))

But trying to convert 10(s+3)/(((s+3)2+25)2)(4s2+3) back into the time domain is giving me a lot of trouble!
Those imaginary roots are really throwing me off. It was suggested to me to use conjugate poles to solve for this part, but I've never learned this method.

Any help is greatly appreciated!

Also not related to the homework question but, is there a fraction button somewhere?? I realize the question is hard to look at when there are so many parenthesis and slashes.

You cannot possibly obtain ##X(s)## unless you are told the values of ##x(0)## and ##x'(0)##. You could obtain a formula for ##X(s)## that involves two (symbolic) parameters such as ##a = x(0)## and ##b = x'(0)##. Of course we cannot check your work if you don't tell us what these are.

Anyway, there are no tricks involved; just use
$${\cal L}^{-1} \frac{1}{s+c} = e^{-ct}.$$ It does not matter whether ##c## is real, imaginary or a general complex number.
 
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  • #7
Orodruin
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Indeed, as Ray says, there is nothing particular with using complex numbers in the inverse transformation. Note that your denominator having only real coefficients will lead to its poles being either real or conjugate to each other. Although ##e^{ct}## is generally complex if ##c## is not real, the appearance of the conjugate pole will ensure that your final result is real with the imaginary parts of the poles resulting in trigonometric functions when their contributions are summed.

How hard is it to type ##\frac{numerator here}{denominator here}##?
Pretty hard.
To be honest, I find it much faster than using a button as you can type it without moving your hands from the keyboard. Also, incorporating LaTeX into your posts will make them more readable and thereby increase your chances of getting help faster. The threshold to learning rudimentaty LaTeX is really not that high.
 
  • #8
vela
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You cannot possibly obtain ##X(s)## unless you are told the values of ##x(0)## and ##x'(0)##. You could obtain a formula for ##X(s)## that involves two (symbolic) parameters such as ##a = x(0)## and ##b = x'(0)##. Of course we cannot check your work if you don't tell us what these are.
I assume that's where the last two terms of jdawg's expression for X(s) came from.
 
  • #9
vela
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But trying to convert 10(s+3)/(((s+3)2+25)2)(4s2+3) back into the time domain is giving me a lot of trouble! Those imaginary roots are really throwing me off. It was suggested to me to use conjugate poles to solve for this part, but I've never learned this method.
You don't need to deal with the complex roots, but you will have to do a lot of algebra. Use a partial fractions expansion.

If you're familiar with evaluating the Bromwich integral using residues, that's probably the fastest way to invert the transform.
 
  • #10
Ray Vickson
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I assume that's where the last two terms of jdawg's expression for X(s) came from.

Of course, but how can we tell if he made an error? His problem statement should tell us what are ##x(0), x'(0)##.
 

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