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Laplace transform |
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| Nov27-11, 05:22 PM | #1 |
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Laplace transform
Hello I'm struggling to understand some basics here with the laplace transform..
I'm given the laplace transform of 2/(s + 4)^4 and I need to take the inverse of this to get back to y(t) Looking at my tables the only transform similar to this is 1/(s + a)^2 I understand I can pull out the 2 and write it in the form 2[ 1/(s + 4)^4 ] but this is where i'm stuck can I do this.. 2[ 1/ (s + 4)^2 * 1/(s + 4)^2 ] and then since the inverse of 1/(s + 4)^2 = te^-4t would i just multiply te^-4t * te^-4t ? is that allowed or what other route would should I take? |
| Nov27-11, 05:44 PM | #2 |
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Mentor
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No, you can't do that. In fact, you'll learn later that the product of two Laplace transforms corresponds to the convolution of two functions in the time domain. (Don't worry if you don't know what "convolution" means right now.)
Your table should list general properties of the Laplace transform that will let you relate what you have to known transforms. That's the approach you want to take. |
| Nov27-11, 05:54 PM | #3 |
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ahh thanks.. we actually learned convolution theorem last week...but I was unsure of where to apply it.. I've been trying the convolution method for the last hour and apparently I'm not applying it correctly. and in my table there are general laplace transforms yes but not any for something raised to the power of 4...is there away of achieving a solution to the above example with just basic transforms? I couldnt figure out how to convert it to any other recognizable function...when I played around with partial fractions I just get back the original
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| Nov27-11, 06:05 PM | #4 |
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Mentor
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Laplace transform
Obviously, the table isn't going to list every possible power, so there's usually some property that tells you how to treat the n-th power. You want to use that with n=4.
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