Inverting Shifted Laplace function

In summary: These can be inverted using the general shift of a Laplace transform. The first term will have a solution of e^(X0s) and the second term will have a solution of e^((X0+L)s). In summary, the known conditions at x = 0 are the beam being supported and a block of mass M and length L being placed on the beam. To solve for the beam displacement, the Laplace transform can be used and the general shift of a Laplace transform can be applied to invert the two terms that arise from the multiplication of the first equation.
  • #1
danmel413
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Homework Statement


A beam is supported at one end, as shown in the diagram (PROBLEM 11 page 281 of Lea, 159 of the course pack). A block of mass M and length l is placed on the beam, as shown. Write down the known conditions at x = 0. Use the Laplace transform to solve for the beam displacement.

Homework Equations


I've gotten almost the entire way through the problem using the differential equations for beam deflections (most relevant ones given here: http://www.me.berkeley.edu/~lwlin/me128/BeamDeflection.pdf (8.4 and 8.50), to get to a Laplace transform I now have to invert:

L{y(x)} = Y(x) = [(Mge-sx0)/(EILs5)] * (1-e-sL) - (Mg(x0 + L/2))/(EIs3) - (Mg)/EIs4). (This is correct, I have the solutions).

The general shift of a Laplace transform: L{S(t-t0)f(t-t0)}=e-st0)F(s) for t0 > 0

The Mellin inversion integral: f(t) = 1/(2πi) ∫F(s)estds from γ-i∞ to γ+i∞

The Attempt at a Solution


So up in my first equation, the only part that can't be inverted easily is the first term which has to be shifted. The solution doesn't walk through the inversion, just gives the answer, which is annoying. It's this part:[(Mge-sx0)/(EILs5)] * (1-e-sL). I'm working in x instead of t, of course, and my F(s) would be [(Mg)/(EILs5)] * (1-e-sL), and of course the other part of the integral would be e-sx0. Can someone put me on the first step to getting through this?
 
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  • #2
Multiply it out and you get two terms, one with a shift X0 and one with a shift (X0+L).
 

What is the Inverting Shifted Laplace function?

The Inverting Shifted Laplace function is a mathematical function used in engineering and science to analyze the behavior of linear dynamic systems. It is used to find the time-domain behavior of a system by transforming the complex frequency domain into the time domain.

What is the difference between the Inverting Shifted Laplace function and the regular Laplace function?

The Inverting Shifted Laplace function is similar to the regular Laplace function, but it includes an additional term that shifts the origin of the function. This shift allows for a more accurate analysis of systems with non-zero initial conditions.

How do you calculate the inverse of the Inverting Shifted Laplace function?

To calculate the inverse of the Inverting Shifted Laplace function, you can use a table of Laplace transforms or use the method of partial fraction decomposition. The result will be a function in the time domain that represents the behavior of the system.

What are some applications of the Inverting Shifted Laplace function?

The Inverting Shifted Laplace function is commonly used in engineering and science to analyze the behavior of control systems, electrical circuits, and mechanical systems. It is also used in signal processing to analyze the response of filters and communication systems.

Are there any limitations to using the Inverting Shifted Laplace function?

Like any mathematical model, the Inverting Shifted Laplace function has limitations. It assumes linearity and time-invariance of the system, which may not always hold true in real-world situations. It is also limited in its ability to accurately model systems with high-frequency components.

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