Understanding the Variables of the FT and DTFT: Intuition and Differences

In summary, the FT and DTFT are both mathematical operations used in signal processing. The FT uses frequency, measured in cycles per second or radians, as its argument, while the DTFT uses digital frequency, ranging from 0 to 2π or -π to π. This is due to the periodicity of the DTFT, which is a periodic extension of the FT. The DTFT is often used in the form of X(e^{j\omega}), which is an algorithm used by computers to approximate the FT. Although they serve similar purposes, the Fourier pairs for the FT and DTFT differ due to their different argument variables.
  • #1
npit
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Could someone explain the intuition behind the variables of the FT and DTFT? Do I understand it correctly ?

For FT being [itex]X(f)[/itex], I understand that [itex]f[/itex] is a possible argument the frequency, as in number of cycles per second.

FT can be alternatively parameterized by [itex]\omega = 2 \pi f [/itex] which specified the number of cycles in radians, which results in also appending a division by [itex]2 \pi [/itex] to the transform.

For DTFT, I am told that we use the "digital frequency" [itex]\Omega[/itex] which ranges from 0 to [itex]2\pi[/itex] (or from[itex]-\pi [/itex] to [itex]\pi[/itex]). I vaguely understand that this is because of the periodicity, since [itex]X(\Omega)[/itex] is a periodic expansion of the FT.

Is that all there is? If so, why do the Fourier pairs differ in the case of the FT and the DTFT?
(For example, see http://www.mechmat.ethz.ch/Lectures/tables.pdf

I have also come across the DTFT in the form of [itex]X(e^{j\omega})[/itex]. What's that about?
 
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In short: FT is a mathematical operation, DTFT is an algorithm for computers to create something that approximates an FT.
 

1. What is the difference between FT and DTFT?

The FT (Fourier Transform) is a continuous function that decomposes a signal into its constituent frequencies, while the DTFT (Discrete Time Fourier Transform) is a discrete function that performs the same decomposition on a sampled version of the signal.

2. Why do we use the FT and DTFT in signal processing?

The FT and DTFT are used to analyze and process signals in the frequency domain, which can provide valuable insights and simplify certain calculations. They are also essential in applications such as image and audio processing, where understanding the frequency components of a signal is crucial.

3. Can the FT and DTFT be applied to any signal?

The FT and DTFT can be applied to any signal that is finite and has a well-defined frequency spectrum. However, certain criteria must be met for the transforms to be accurate, such as the signal being square-integrable for the FT and periodic for the DTFT.

4. What is the inverse transform of the FT and DTFT?

The inverse transform of the FT is the inverse Fourier Transform, which reconstructs the original signal from its frequency components. The inverse transform of the DTFT is the inverse Discrete Time Fourier Transform, which reconstructs the original discrete signal from its frequency components.

5. Are there any limitations to the FT and DTFT?

Yes, there are limitations to the FT and DTFT. For example, the FT cannot be applied to signals that are not square-integrable, and the DTFT cannot be applied to signals that are not periodic. Additionally, both transforms assume that the signal is continuous and infinite, which may not always be the case in practical applications.

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