inverse laplace transform, example 3 | Summary and Q&A

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April 20, 2017
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inverse laplace transform, example 3

TL;DR

This content explains how to calculate the Laas transform for expressions involving S - 3 and S + 10.

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Questions & Answers

Q: How can the Laas transform be calculated for expressions involving S - 3?

To calculate the Laas transform for expressions with S - 3, multiply the Laas transform of cosine BT by e^(3 t) and replace S with S - 3. This will give us (S - 3) / ((S - 3)^2 + B S^2).

Q: What if the expression involves S + 10 instead of S - 3?

If the expression involves S + 10, we can still use the same method by replacing S with S - a. In this case, add 3 to both sides of the expression to make it S - 3. The resulting Laas transform will be (S - 3) / ((S - 3)^2 + B S^2).

Q: How can the Laas transform be calculated for expressions involving a constant on the top?

If there is a constant on the top of the expression, use the formula for the Laas transform of s BT, which is equal to B / (S^2 + B S). Multiply the top by the constant, and divide the equation by the constant to maintain the same values. This will give us the appropriate Laas transform.

Q: How do we handle the situation when the expressions involve a constant and S - 3 or S + 10?

To handle the situation involving a constant on the top and S - 3 or S + 10, use the formula for the Laas transform of s BT and multiply the top by the constant. Be sure to divide by the constant as well to cancel it out. This will give us the correct Laas transform.

Summary & Key Takeaways

  • The Laas transform of cosine BT is equal to S / (S^2 + B S), but if it involves S - 3, it needs to be translated by multiplying with e^(a t) and replacing S with S - a.

  • The explanation shows how to calculate the Laas transform for expressions involving S - 3 and S + 10 by using the appropriate formula and shifting the variables.

  • If the expression involves S + 10 instead of S - 3, the same method can be applied by replacing S with S - a and shifting it accordingly.

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