Talk:Discrete-time Fourier transform/Archive 1

Should articles be merged
Should this article be merged with discrete Fourier transform? Michael Hardy 01:34, 2 Jun 2005 (UTC)


 * No, but I do not see anything wrong with merging discrete-time Fourier transform with Fourier series since the forward discrete-time Fourier transform is just the reverse Fourier series.


 * I don't think any of the Fourier transform articles should be merged. I think it is important to make it absolutely clear that there are four separate and distinct types of FT's:
 * Continuous-time Fourier Transform (CTFT)
 * Discrete-time Fourier Transform (DTFT)
 * Discrete Fourier Transform (DFT)
 * Fourier series


 * Obviously, there are relationships between all four of these types of FT, but each one is different from the others, and each serves its own purpose. -- Metacomet 06:02, 24 December 2005 (UTC)
 * I agree with Meta. In a world without hyperlinks, I might feel differently.  But fortunately we no longer live there. --Bob K 14:04, 19 March 2006 (UTC)

Plots and resolution
The plots used to show the increase of resolution actually show something else. They show the effect of the DFT on a frequency that is between the frequency it samples. A clearer example could be produced by transforming a smooth function of frequency instead of the Dirac delta that that example uses.

DTFT vs. DFT (moved from User talk:Stevenj)
At http://en.wikipedia.org/w/index.php?title=Discrete-time_Fourier_transform&diff=32124402&oldid=32123635 I don't think you quite grasped the significance of "DFT". There is one underlying DTFT, which contains an infinite number of non-zero values. Since we cannot compute all of them, we choose a DFT size (N) to sample the DTFT. Different choices produce different DFT's (i.e. different resolution). So "DFT" is really what I meant to say. --Bob K 18:11, 20 December 2005 (UTC)


 * It's not a DFT if you have $$N < L$$, for example. Anyway, please use the Talk page of the article.  —Steven G. Johnson 18:16, 20 December 2005 (UTC)


 * Okay, I read that paragraph through again and I see that you were specifically referring to the zero-padded case of $$N > L$$, where it is a DFT. I've rephrased it to distinguish these two cases more clearly at that point.  (Note, by the way, that there are pruned FFT algorithms that can potentially exploit zeros in the input, although the savings in practice are not generally very great.)  —Steven G. Johnson 18:26, 20 December 2005 (UTC)

Relationship between discrete and continuous
The article does not provide any connection between the use of $$\omega \,$$, in radians per sample, to represent the spectrum of a discrete-time signal, and the use of $$ \omega \, $$, in radians per second, to represent the spectrum of a continuous-time signal. The two are related by the sampling rate, $$ f_s \, $$ in samples per second:


 * $$ \mathrm{radians \ per \ sample} = \mathrm{ radians \ per \ second \over samples \ per \ second }$$

It is difficult to show this relationship explicitly because it is common practice to use a single symbol, $$ \omega \, $$, to represent two very different concepts, angular frequency per unit time and angular frequency per sample.

-- Metacomet 05:38, 24 December 2005 (UTC)


 * The relationship is not at all difficult to show. But I disagree with the original author(s), who [apparently] deemed it obvious (i.e. too easy).  Rather it is the source of much confusion and misunderstanding, so I have added it to the article.  radians per sample is an abstraction called normalized frequency.  It is primarily useful when making a point (as in a textbook) that does not depend on the actual sample-rate.  But the DTFT can be written in terms of angular or ordinary frequency ( as I have done), when the sample-rate is actually important (as when using the DTFT for a frequency measurement).
 * --Bob K 16:56, 20 March 2006 (UTC)

Why introduce both f and &omega; in the introduction?
Is it necessary to include the discussion, in the introduction, relating frequency f in cycles per sample (which is not even correct) to angular frequency &omega; in radians per sample? I do not think it serves any purpose, particularly in the introduciton, to open the whole can of worms about these two different ways of representing frequency. IMO, it would be simpler, more concise, and less confusing simply to define the DTFT using angular frequency (in radians per sample) right at the beginning, and to avoid any mention of frequency at all, which is completely unnecessary. -- Metacomet 05:47, 24 December 2005 (UTC)


 * Notice that the definition contains no mention of sample-rate, or equivalently the time-interval between samples. That means $$f\,$$ is a normalized frequency, which is the actual frequency (cycles per second) divided by the sample rate (samples per second).  And therefore its units are indeed cycles per sample.
 * In my opinion, this article (and others) should introduce concepts in terms of real-world units so that beginners can immediately see the relevance. The notational simplifications afforded by normalized units should also be explained.  But many readers will not need that information, so we should not force it on them by putting it at the beginning.
 * --Bob K 20:45, 17 March 2006 (UTC)

Notation for rect and tri functions
What is with this notation for the rect and tri funcitions with the 2-delta subscripts? Is that a common way of representing these functions? Why not simply introduce a scaling factor to the argument, such as:


 * $$\mathrm{rect} \left( { n \over N } \right) $$

or


 * $$\mathrm{rect} \left( { \omega \over W } \right) $$

This approach seems a lot simpler and more intuitve than using the subscripted delta notation.

I would propose changing the definitions to the following, which I believe are far more standard and far simpler to use:


 * Rectangle function
 * $$ \operatorname{rect} (\lambda) =

\begin{cases} 1 & -{1 \over 2} \leq \lambda \leq  { 1 \over 2  } \\ 0 & \mbox{otherwise} \end{cases} $$


 * i think that we will run into problems later (involving the sign function, step function, analytic signal, hilbert, at least in the continuous domain) unless we adjust that definition to be:


 * Rectangle function
 * $$ \operatorname{rect} (\lambda) =

\begin{cases} 1 & |\lambda| <  \frac{1}{2} \\ \frac{1}{2} & |\lambda| =  \frac{1}{2} \\ 0 & \mbox{otherwise} \end{cases} $$

and


 * Triangle function
 * $$\operatorname{tri} ( \lambda ) =

\begin{cases} 1 + 2 \lambda ; & - {1 \over 2 }\leq \lambda \leq 0 \\ 1 - 2 \lambda ; & 0 < \lambda \leq {1 \over 2 } \\ 0 & \mbox{otherwise} \end{cases} $$

where, for example,


 * $$ \lambda = { n \over N } \,$$

or


 * $$ \lambda = { \omega \over W } \,$$

-- Metacomet 07:40, 24 December 2005 (UTC)


 * there are several different notations, and they all have their own good points. The one I used, with the $$2 \Delta$$ as subscript, is the one I have found in several DSP-related books. On the other hand, I found the one you suggest in physics-related books. According to me, the best thing to do would be to start a proper article about the rect and tri functions, choosing only one notation (the one you suggest is more consistent) and link to that, just like we did for the unit step and the sinc. Alessio Damato 09:37, 24 December 2005 (UTC)

I just did a quick search on Wikipedia, and I found an article for the rectangle function but nothing for the triangle function. The notation and definition in the rectangle article are similar if not identical to what I have proposed. As you suggest, it may make sense to create a similar article for the triangle function. -- Metacomet 15:29, 24 December 2005 (UTC)

I am not a big fan of using the symbol &Delta; to represent a number, since &Delta; is, as you know, very often used as an operator to represent, for example, the change in or difference between two numbers:
 * $$ \Delta x = x_2 - x_1 \, $$

Alternatively, it is sometimes used to represent the Laplace operator.

The other drawback to using &Delta; in the context of physics or engineering is that it is not at all clear what dimensions or units of measure are associated with this symbol. If you use a variable such as T to scale the time domain t, or W to scale the frequency domain &omega;, then it is generally clear that the units of measure are seconds in the one case, and radians per second or radians per sample in the other case. -- Metacomet 15:29, 24 December 2005 (UTC)


 * I have created a new article called triangular function as a mathematics-related stub. Please check it out, and expand if you like.  Thanks.   -- Metacomet 18:51, 28 December 2005 (UTC)

Change in DTFT for Unit Step?
I placed a negative in the exponential for the DTFT of the unit step function u[n]. Can anyone verify that this is correct? This is what my textbook had.


 * First off, you're definitely right (I just did it out by hand, and checked in some tables). Also, for some reason 130.63.96.199 decided to flip the frequency domain entry in the table for the unit step with the one below it. I flipped them back. I'm not too sure about the others, though (for example, the DTFT Transform table on Wikibooks has exp(-an) in the fourth row where this has exp(-jan)... pretty sure the other one is at fault here, though) --Dirk Gently 03:54, 6 December 2006 (UTC)

Is the transform for sinc correct?
I am certainly no expert in this area, but it seems that the transform listed for (W/pi)*sinc(W*n) to rect(w/W)*e^(jaw) is incorrect. If rect is 1 from -(1/2) to (1/2) then the transform should have a (1/2) in it. ie. rect(w/(2W), otherwise it won't work correctly.

I've found another table which seems to support this. However I'm not sure so I don't want to make a change.


 * I agree, assuming "sinc" means unnormalized sinc (see below).


 * The Fourier transform of the sinc function is:


 * $$ \mathrm{sinc}(W t) \quad \Longleftrightarrow \quad \frac{1}{a}\cdot \mathrm{rect}\left(\frac{1}{a} f \right),$$


 * where a = W, if the sinc is normalized, or  a = W/π, if the sinc is unnormalized.


 * And:


 * $$\sum_{n=-\infty}^{\infty} \delta(t - n) \quad \Longleftrightarrow \quad \sum_{k=-\infty}^{\infty} \delta \left( f - k \right),$$


 * from which this Fourier transform (and DTFT) follows (convolution theorem):



\mathrm{sinc}(W t) \cdot \sum_{n=-\infty}^{\infty} \delta(t-n) \quad \Longleftrightarrow \quad \sum_{k=-\infty}^{\infty} \frac{1}{a}\cdot \mathrm{rect}\left(\frac{1}{a} (f-k) \right) = \sum_{k=-\infty}^{\infty} \frac{1}{a}\cdot \mathrm{rect}\left(\frac{\omega -2\pi k}{2\pi a} \right). $$


 * The width of the $$\mathrm{rect}\,$$ is $$2\pi a\,$$. And the spacing of $$\mathrm{rect}'s\,$$ is $$2\pi\,$$.  So when $$2\pi a \le 2\pi\,$$, the $$-\pi < \omega < \pi \,$$ region of the DTFT comprises just the $$k=0\,$$ term of the summation:


 * $$\frac{1}{a}\cdot \mathrm{rect}\left(\frac{\omega}{2\pi a} \right),$$


 * which matches your result for the case a = W/π (unnormalized sinc function).


 * --Bob K 23:56, 16 March 2006 (UTC)
 * --Bob K (talk) 14:22, 24 February 2018 (UTC) (revision)

Is there a source for the sinc(n + a) transform? Where does this come from? 2A00:5BA0:10:3E5:25D8:1514:DCDB:2269 (talk) 15:13, 23 February 2018 (UTC)


 * Slightly generalizing the derivation above, for the normalized sinc function:


 * $$ \mathcal{F}\left \{\mathrm{sinc}(W (t+a))\right \} = \frac{1}{W}\cdot \mathrm{rect}\left(\frac{1}{W} f \right) e^{i2\pi fa}$$



\begin{align} X_{2\pi}(\omega) &= \mathcal{F}\left \{\mathrm{sinc}(W (t+a))\cdot \sum_{n=-\infty}^{\infty} \delta(t-n)\right \} \\ &= \sum_{k=-\infty}^{\infty} \frac{1}{W}\cdot \mathrm{rect}\left(\frac{1}{W} (f-k) \right) e^{i2\pi (f-k)a}, \quad \text{where}\ f = \omega/(2\pi)\\ &= \sum_{k=-\infty}^{\infty} \frac{1}{W}\cdot \mathrm{rect}\left(\frac{\omega -2\pi k}{2\pi W} \right) e^{i(\omega - 2\pi k) a}. \end{align} $$


 * For the W=1 case you asked about, the width and spacing of the $$\mathrm{rect}'s\,$$ is $$2\pi\,$$.


 * So the $$-\pi < \omega < \pi \,$$ region of the DTFT comprises just the $$k=0\,$$ term of the summation:


 * $$X_o(\omega) = \underbrace{\mathrm{rect}\left(\frac{\omega}{2\pi} \right)}_{=\ 1,\ \text{for}\ -\pi < \omega < \pi} e^{i\omega a} = e^{i\omega a}$$      QED
 * --Bob K (talk) 15:00, 26 February 2018 (UTC)

Notation for $$X$$
The notation: $$X(e^{i \omega})\,$$ does not make sense for 9 of the 13 entries in Table 1. $$X(\omega)\,$$ works for all 13 entries, and it is also the notation used in Table 2. --Bob K 18:47, 17 March 2006 (UTC)