Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Continuous Fourier transform vs. Discrete Fourier transform. Can anyone tell me what the difference is physics-wise? I know the mathematical way to do both, but when do you use the other instead of the other?

Is the DFS not as accurate, since it relies on discrete values, or has it nothing to do with that? Or is it just because it's easier to do with FFT.

share|improve this question

4 Answers 4

up vote 5 down vote accepted

The DFT is used when all you have available are samples of the function, rather than the function itself. If you are doing an FT on experimental data, it's always (as far as I know) recorded in discrete numbers: an array of floating point numbers, for example. There are a few times when the DFT has some applicability to real systems, for example simple theories of solids in which the ion cores occupy well-defined, regular, periodic locations. Not too many of those.

If you are using a computer to take a FT, and you know the mathematical form of the function, then you might be able to calculate the continuous FT symbolically using a computer algebra system. The output will be the mathematical form of the FT. If you calculate the FT "digitally", either because you have samples of experimental data, or you have evaluated a mathematical function a discrete regular points, then you are using the DFT and your output is an array of values.

share|improve this answer

To be sure, it's the continuous (time) Fourier transform versus the discrete time Fourier transform (DTFT).

The former is a continuous transformation of a continuous signal while the later is a continuous transformation of a discrete signal (a list of numbers).

The discrete Fourier transform (DFT), on the other hand, is a discrete transformation of a discrete signal. It is, in essence, a sampled DTFT.

Since, with a computer, we manipulate finite discrete signals (finite lists of numbers) in either domain, the DFT is the appropriate transform and the FFT is a fast DFT algorithm.

share|improve this answer
1  
+1 for distinguishing the DTFT from the DFT. A computer calculates the DFT, but it's often applied to a non-periodic signal. In such a case the computer doesn't know whether you have a true periodic signal or an infinite signal that you've had to truncate. In the latter case, the signal looks to the computer like a truncated version of your function, repeated periodically to infinity in both directions. This often leads to problems, if, for example, the value (and derivatives) of the function on the left side of the truncation is not the same as its value on the right. DFT needs care. –  garyp Jun 9 at 21:01

Physics-wise: The far field (Fraunhofer region) of a horn or reflector antenna is the (continuous) Fourier transform of its aperture field. The far field of an array of point sources is the Discrete Fourier Transform of their current (or voltage) drives. If instead of point sources we have an array of, say, horns then the resulting far field is the convolution of the two far fields, one DFT and CFT.

share|improve this answer

At the end of your question you mentioned FFT which is Fast Fourier Transform and is actually a method for computing the DFT (which happens to be orders of magnitude faster than computing the DFT using normal methods however it has specific requirements on the number of samples used as input but that is usually fixed using padding in the software implementation).

As for the reason to use the fourier transform or one of the discrete fourier transform implementations in physics will come down to whether you are dealing with symbolic (mathematical function) or sampled data as the two work on different inputs.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.