# Laplace transforms and the shifting theorem

1. Dec 9, 2003

### Ed Quanta

Can anyone explain to me the point of Laplace transforms and the shifting theorem in general?

2. Dec 9, 2003

### BoulderHead

Here is my two cents;

The reason is because circuit equations are differential equations and solving some of them can be quite a pain, especially if dealing with the high order stuff. Yes, it is possible to do it but requires more effort. The way I think of it is that circuitry doesn’t 'understand' time the way we do, but it does 'understand' frequency. What makes the Laplace method beautiful is that by transforming from the time-domain to the s-domain you can then treat differential equations like you treat algebra. Sound good so far?

So it goes like this; you plug in your differential circuit equations, find the Laplace Transform by integration, manipulate the problem algebraically (solving for whatever variable you need), then find the inverse transform in order to get back into the time domain and have your answer.

Now the rub; the inverse transform isn’t workable by some simple equation. Instead, it is done through visual recognition, which means you have to memorize some basic transformations in order to be successful (Don’t ask me what they are because I forgot most of them long ago, haha).

Oops, thought this was being asked over in engineering.

Good luck

Last edited by a moderator: Dec 10, 2003
3. Dec 9, 2003

### Integral

Staff Emeritus
Let me make an analogy to your question "Could you explaine to me cars and brakes in general"

A Laplace Transform turns a Differentail Equation into a Algebraic equation which can be solved with "simple" algebraic operations. The inverses Laplace Transform then operates on the answer to retrive the solution to the original problem. Many, other wise, difficult to solve Differentail Equations can be easily solved with these methods.

The shifting property is not granted Theorem status in my copy of Operational Mathematic by Churchill it is given 1 out of 200+ pages devoted to Laplace Transforms.

It is essentially the Laplace transform of the an Impulse function or in the limits the Dirac Delta function. It essentially "selects" the value of a function which is specified by the Impulse.

I need to look at it a bit more to revive some long dead memories, this is a start. Does it help?

4. Dec 9, 2003

### BigRedDot

You might recall that one example of vectors are "arrows on the chalkboard." Now, we can talk about adding or subtracting "arrows" and specifying relationships between them. But it turns out that dealing with the arrows directly (via, say, the parallelogram rule) is very cumbersome. So, we are led to introduce a basis that allows us to represent our arrows by lists of numbers like: (1,5) or (-7, 10). Then, solving for relations between the arrows is reduced to the equivalent easier problem of solving algebraic relations between the basis coeffients.

We can generalize the notion of a basis-vector expansion to N dimensions, where we represent vectors as N-tuples of numbers. And from there even to countably infinite dimensions, where we represent our vectors by infinite series of numbers. Now imagine that instead of a countable number of basis elements indexed by some integer n, we have vectors (functions in this case) that require an uncountable number of basis elements (here, basis functions $$e^{-st}$$) indexed by some real number, say s.

A Laplace transfom (like a Fourier transform) is such a continuous basis vector expansion. We represent function as superpositions of basis functions ("adding" them together by integrating over them). The expansion coefficients are themselves given by the Laplace Transform function f(s) itself -- for each value of s, you get a number, the expansion coefficient for that value of s.

Just as changing to a basis allowed us to manipulate arrows on the board more easily, which is to say, algebraically in terms of the expansion coefficients, the same is true here. The expansion coeffiecients, i.e., the Laplace transform function, can be manipulated algebraically instead of dealing with the messy relationships between the original functions.

Of course, when we are done, we must convert from the basis representation back to "arrows" or "functions" to see the final result. That is what the inverse Laplace transform does.

5. Dec 12, 2003

### HallsofIvy

Of course, the Laplace transform only applies to linear differential equations which can be solved by more elementary methods.

Laplace transforms are mostly used in two distinctly different ways:
1) Engineering applications in which you have the same linear differential operator with a variety of different "right hand sides" and want to be able to "look up" the solution.

2) Very theoretical researchs in which you want to be able to write a formula for a very general problem (even if the formula is generally impossible to evaluate!).

6. Dec 29, 2003

### drachman

Two nice things about the Laplace transform:1)It is convenient for a problem that starts at a certain time and moves forward, since the initial conditions are included. 2) If you have lost your table of inverse transforms (which are defined by a certain contour integral in the complex plane) you may be able to work it out yourself by doing the complex integral, especially if only simple poles are involved. What impresses me is the fact that Heaviside seems to have invented these transforms empirically, without knowing their "mathematical" justification.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook