# Phasor and Fourier's Transform

Discussion in 'Math' started by htzzz, Feb 6, 2010.

1. ### htzzz Thread Starter New Member

Sep 13, 2009
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I am getting quite confused about these two concepts, is there any relation btwn them?

I read in an EM book says:
general wave could be written: E(r,t)=1/2∏ ∫(E(r) exp(jωt))dω
where E(r) is the time harmonic fields in terms of phasor

Did we do any Fourier tranform when converting to phasor? why doing an inverse-FT could change a phasor term back?

Thanks!

Last edited: Feb 6, 2010
2. ### studiot AAC Fanatic!

Nov 9, 2007
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515
You have actually introduced three concepts here, not two.

The most fundamental is the statement

This is a particular solution to the general wave equation.

http://en.wikipedia.org/wiki/Wave_equation

The general wave equation is a second order diferential equation.

Do you understand differential equations and particular solutions?

Your particular solution is obtained by using the Fourier transform to integrate the wave equation.

http://math.nist.gov/~BAlpert/evolution.pdf

Phasors are another particular solution.

There are actually an infinite number of particular solutions. Picking the one we want depends upon the application and the boundary conditions.

Phasors are particularly suitable for elementary ac circuit analysis.
The Fourier Transform you mention is a basis for certain numerical methods solutions.

3. ### killerfish Member

Feb 27, 2009
17
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how about laplace transform and fourier transform??? they looks similar when work with them...

4. ### studiot AAC Fanatic!

Nov 9, 2007
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515
It is a fairly fundamental principle that if we take an equation and

'do the same thing' to both sides eg add A to both sides, multiply both sides by 52x etc

The equality is preserved. That is the left hand side still equals the right hand side.

Now a differential equation is an equation containing derivatives. We cannot solve such an equation by simple algebra and the solution is not a simple number it is a function.

We solve a differential equation by integrating both sides. This can be difficult.

However it can be made easier by multiplying both sides by an 'integrating factor'.
We gain nothing if we simply multiply by a constant, say A, but if we chose cleverly we can make the integrals easier.

There are a variety of such integrating factors. The ones which help solve the differential equations commonly encountered in engineering and physics are give special prominence. The process of applying the factor and performing the integration is called a transform. The Laplace transform is the most widely encountered but there are many others such as Fourier, Hankel, Mellin, Z and so forth.

These transforms take advantage of the fact that the equations of interest are linear equations. This fact allows us to work on the equations term by term. Because in engineering and physics we encounter the same expressions over and over we have found it worth building up tables of ready integrated expressions we can just 'mix n match' to form our particular case.

5. ### scigeek New Member

Mar 11, 2010
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But, what if we are dealing with non-linear expressions...????

If it's broke
Break it until it's fixed...

6. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Or stochastic processes. Then your options become very constrained. Check Out the Feynman-Kac theorem for an example of solving a PDE, in closed form, by taking a discounted expectation.