Inverse Laplace Transform of 2/(s + 4)^4: Explained

In summary, the conversation is about using the Laplace transform and its properties to find the inverse of a given transform. The speaker is struggling with understanding how to apply the convolution theorem and is looking for a way to solve the problem using basic transforms.
  • #1
rambo5330
84
0
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?
 
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  • #2
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.
 
  • #3
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 couldn't figure out how to convert it to any other recognizable function...when I played around with partial fractions I just get back the original
 
  • #4
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.
 

1. What is the inverse Laplace transform?

The inverse Laplace transform is a mathematical operation that takes a function in the complex frequency domain and converts it back into a function in the time domain. It is the inverse operation of the Laplace transform, which is used to convert a function from the time domain to the frequency domain.

2. What is the significance of the "s + 4" term in the inverse Laplace transform of 2/(s + 4)^4?

The "s + 4" term represents the frequency shift in the function. This means that the function has been shifted 4 units to the left in the frequency domain. In the time domain, this would correspond to a delay of 4 seconds.

3. How do you calculate the inverse Laplace transform of 2/(s + 4)^4?

To calculate the inverse Laplace transform, you can use a table of Laplace transforms or use integration techniques. In this case, the inverse Laplace transform of 2/(s + 4)^4 can be calculated using the formula for the inverse Laplace transform of (1/(s + a)^n), where "a" is the shifting factor and "n" is the power of the denominator.

4. How does the inverse Laplace transform of 2/(s + 4)^4 relate to real-world applications?

The inverse Laplace transform is commonly used in engineering, physics, and other scientific fields to solve differential equations and analyze systems in the time domain. In this specific case, the inverse Laplace transform of 2/(s + 4)^4 can be applied to systems that have a frequency response with a fourth-order polynomial shape.

5. Can the inverse Laplace transform of 2/(s + 4)^4 be simplified further?

Yes, the inverse Laplace transform of 2/(s + 4)^4 can be simplified further using partial fraction decomposition and other algebraic techniques. However, the resulting function may not be as useful or easily interpretable as the original inverse Laplace transform.

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