# LaPlace, Fourier, and Phasors

#### Distort10n

Joined Dec 25, 2006
429
I need some help getting my facts straight regarding LaPlace, Fourier, and phasor transforms since I am getting back into my math to gear up for grad school.

To keep things basic, the LaPlace and Fourier transforms are similar. I believe phasors fall under LaPlace, but I am not sure. Readingthis on pg. 1 the LaPlace transform is "a superset of the phasor representation..."
Both have a real and imaginary part denoted by sigma and j omega, and I think I understand this. I will only ask a few questions because I have read so much that I will probably ask contradicting questions. For now I would like to know the following:

• The LaPlace operator 's' is equal to sigma + j omega and sigma is zero. Meaning that we are only interested in the imaginary part in the complex domain or S-Plane.

• Phasors also have a real and imaginary part. When these are expressed on 2-D cartesian graph: real (x-axis) and imaginary (y-axis). This is not the S-Plane.

• Fourier transforms take the magnitude and phase of a transfer function. A signal is broken into its separate spectral components and phase. Like an FFT.

All true or not true? Apologies of these contradict each other. Too much reading.

#### Dave

Joined Nov 17, 2003
6,969
Hi knightofsolamnus,

Here is my take on your questions:

• The LaPlace operator 's' is equal to sigma + j omega and sigma is zero. Meaning that we are only interested in the imaginary part in the complex domain or S-Plane.
Correct s is a complex number comprising the real and imaginary components, however in the Laplace Transform the real component sigma is not equal to zero - this would be the situation for the Fourier Transform. As an analytical explanation, in the Fourier Transform integral the harmonic components are multiplied to the input signal and the integral produces a resultant energy that corresponds to that frequency, i.e. it is only affected by the imaginary component, omega. The s-plane is similar however takes account of decay which is affected by the real component of s (sigma), as well as the frequency component omega. Think of the fundamental integrals for the Laplace and Fourier Transforms to see this is the case.

• Phasors also have a real and imaginary part. When these are expressed on 2-D cartesian graph: real (x-axis) and imaginary (y-axis). This is not the S-Plane.
I would suggest you look up Argand Diagrams, the idea is that the real component is on the x-axis and the imaginary component is on the y-axis. The fundamental analysis of complex numbers is understood through Argand Diagrams and this is applicable to both Phasor diagrams and the S-plane - how this translates into a practical analysis is another question altogether.

• Fourier transforms take the magnitude and phase of a transfer function. A signal is broken into its separate spectral components and phase. Like an FFT.
Your basic assessment is correct about magnitude, phase and spectral components. Please be aware there are 4 main types of Fourier Transform:

1. Continous Fourier Series
2. Continous Fourier Transform
3. Discrete Fourier Transform
4. Discrete-time Fourier Transform

The Fourier Transform you will most probably be looking at is the Continuous Fourier Transform and this is the one that has the analogies to the Laplace Transform, as per my first answer. This analyses signals in continuous time. The FFT is merely a fast algorithmic implementation of the Discrete Fourier Transform, which looks at signals in discrete-time.

Dave

#### Distort10n

Joined Dec 25, 2006
429
Thanks Dave. That puts things in a clearer perspective. I figured I typed out something wrong after all the reading that I did. Unfortunately, most of the engineers that I work with no longer deal with the mathematics. They 'just know.'

#### Dave

Joined Nov 17, 2003
6,969
No problems. I find an appreciation of the rudiments of Fourier and Laplace theory goes a long way, however we luckily live in an age where we can abstract from the low-level workings of Fourier and Laplace analysis because of the wide availability of softwares that implement them.

If you've any further questions, feel free to post them up.

Dave