The difference between DFT and DFS

Question

In the literature, I've found that DFS and DFT are one and the same. If they are one and the same why to use two different names for them? If there is really a difference what is it and what is the significance of discrete Fourier series?

Answer 1

We have had this fight many, many, many times at comp.dsp.

The DFT is the same thing as the DFS.

The DFT maps a discrete and periodic sequence of numbers with period length of $N$ to another discrete and periodic sequence of numbers with period length of $N$ and the iDFT (which has the same form as the DFT) maps it back.

Some people don't like anthropomorphizing algorithms or procedures, but I do. The DFT "assumes" that the $N$ samples passed to it are one period of a periodic sequence. The DFT periodically extends the data passed to it.

It is clear in the math, both in the definition of the DFT (and iDFT), and in any theorem applicable to the DFT other than linearity (the periodic nature of the DFT is not evident in the linearity property, but it is evident in anything that causes shifting or convolution in one domain or multiplication by a non-constant in the other domain).

This is why, if periodicity is not assumed (a better word would be "recognized"), then people need to use this clunky modulo notation in the indices, like $x[ ((n))_N ]$ (this is the notation that O&S use) and that, in my opinion, is a pathetic confession from the periodicity deniers that, when it comes down to the bottom line, even they recognize that the DFT is inherently periodic.

To be explicit, the periodic extension of the $N$ samples of $x[n]$ passed to the DFT is:

$$ \tilde{x}[n] = x[ ((n))_N ] = x[n\operatorname{mod}_N] \qquad \forall n \in \mathbb{Z}, \ N \in \mathbb{Z}>0$$

where $ \qquad\qquad\qquad ((n))_N \triangleq n\operatorname{mod}_N = n - N\left\lfloor \frac{n}{N} \right\rfloor $

The notation $\lfloor \cdot \rfloor$ means the floor() function which is the largest integer that does not exceed the argument.

To use any of the shifting or convolution theorems of the DFT, this modulo arithmetic of the indices is absolutely required. for the scaling or superposition theorems, this modulo arithmetic is not required, but does not break those theorems in any case.

Therefore, to be consistent, when using the DFT for any theorems to do any real work with the DFT, one should simply apply the modulo arithmetic all of the time. Doing so explicitly periodically extends the $N$-sample sequence, $x[n]$ passed the DFT.

For me, it's just easier to drop the tilde "$\tilde{\ }$" and simply say that $\tilde{x}[n]$ is the same as $x[n]$ and that $\tilde{X}[k]$ is the same as $X[k]$ and just stop fucking around with this DFT business.

People should read this other answer I wrote a long time ago regarding the inherent periodic nature of the DFT.

Answer 2

I think part of the problem is an awkward and inconsistent naming convention. There are 4 flavors of Fourier Transforms depending on which domain is continuous or discrete (which maps to being aperiodic or perodic in the other domain). So we have

         Name                  Time                  Frequency
Fourier Transform    continous/aperiodic     continous/aperiodic 
Fourier Series       continous/periodic      discrete/aperiodic 
Discrete Time FT     discrete/aperiodic      continous/periodic
DFT or DFS           discrete/periodic       discrete/periodic

A better naming would have been

         Name                  Time                  Frequency
Fourier Transform    continous/aperiodic     continous/aperiodic 
Fourier Series       continous/periodic      discrete/aperiodic 
Discrete FT          discrete/aperiodic      continous/periodic
Discrete FS          discrete/periodic       discrete/periodic

so that discrete refers to "discrete in time and periodic in frequency" and "series" refers "discrete in frequency and periodic in time". In other words "series" means sums and "transform" means integrals. Discrete mean sums and continuous means integrals.

Answer 3

Basically, DFS is used for periodic and infinite sequence. Whereas, DFT is used for non-periodic and finite sequence. Although, They are same Mathematically. But they differ in properties. Practically, we do not have infinite signal. We can say that DFT is extraction of one period from DFS. In other words, DFS is sampling of DFT equally spaced at integer multiple of $\frac{2 \pi}{N}$. DFT is fast and efficient algorithms exits for the computation of the DFT. DFS is adequate for most cases. But FT(Fourier Transform) leads to simpler expression.

Answer 4

DFS coefficient can be computed from DFT coefficients only when certain conditions are met.

1. DFS is a generic term for any discrete harmonic sum

From Wikipedia:

Discrete Fourier series (DFS) is any periodic discrete-time signal comprising harmonically-related discrete real sinusoids or discrete complex exponentials, combined by a weighted summation.

2. IDFT is a case of DFS

IDFT matches with the definition of a DFS. IDFT is a particular case of DFS.

• DFS is the weighted sum of harmonics in the signal.

• IDFT is the weighted sum of harmonics. The weights are the spectral coefficients of DFT. At certain conditions, DFS weights are DFT coefficients divided by the length of the DFT.

From Wikipedia:

When the coefficients are derived from an $N$-length DFT, and a factor of $1 / N$ is inserted, this becomes an inverse DFT.

This is the crucial point, for DFS and DFT coefficients to be exactly related by $1/N$, $N$ must be equal to the periodicity of the signal, and $1/N$ must correspond to the normalized fundamental frequency.

Practical case

To illustrate, consider a signal which fundamental frequency is 100Hz, and contains harmonics 2, 4, 5, 16, 17 with the following amplitude and phases:

f = [200, 400, 500, 1600, 1700]
a = [2, 4, 6, 8, 10]
p = [1, 0, 0.5, -0.5, 0]

So for this signal the DFS coefficients are 1, 2, 3, 4 and 5 and the same for the conjugates.

In order to have the DFT containing these harmonics, N must be at least 2x17=34 and the sampling frequency must be N*100Hz. These constraints are satisfied by selecting N=40 and fs=4kHz. The corresponding samples and the DFT:

Now let's take either the sampling frequency or the number of samples in a way fs=100*N is not satisfied:

DFS coefficients are not a scaled version of DFT coefficients.

Let's satisfy again the relationship with N=41 and fs=4100Hz:

The correct scaling is back.

Answer 5

This answer documents an important exchange in the periodicity and DFT-DFS debate, for "meta reasons". Comments are from under the accepted answer, with my votes removed:

Answer 6

So @OverLordDragon found a way to see one of the old comp.dsp discussions about this and reminded me of something.

So the Discrete Fourier Transform (DFT) is being defined as:

$$ X[k] = \begin{cases} \sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N}, & 0 \le k \le N-1 \\ 0, & \text{otherwise} \end{cases} $$

$$ x[n] = \begin{cases} \frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N}, & 0 \le n \le N-1 \\ 0, & \text{otherwise} \end{cases} $$

And the Discrete Fourier Series (DFS) is defined as:

$$\tilde{X}[k]= \sum\limits^{N-1}_{n=0} \tilde{x}[n] \ e^{-j2\pi n k/N} \qquad \forall k \in \mathbb{Z}$$

$$\tilde{x}[n] = \tfrac{1}{N} \sum\limits^{N-1}_{k=0} \tilde{X}[k] \ e^{+j2\pi n k/N} \qquad \forall n \in \mathbb{Z} $$

---

Now consider this relationship:

$$ Y[k] = e^{-j 2 \pi k/N} X[k] $$

What is $y[n]$ in terms of $x[n]$? Specifically what is $y[0]$?

Is it?

A: $\qquad \qquad y[n] = x[n-1] \qquad \qquad y[0] = x[-1] $

or

B: $\qquad \qquad y[n] = x[n-1] \qquad \qquad y[0] = 0 $

or

C: $\qquad \qquad y[n] = x[(n-1)\operatorname{mod}_N] \qquad \qquad y[0] = x[N-1] $

$\qquad \qquad$ where $ \qquad n\operatorname{mod}_N = n - N\left\lfloor \frac{n}{N} \right\rfloor $

or

D: $\qquad \qquad \text{none of the above}$

Which is it? A, B, C or D?

---

(If you answer B or D, you're just wrong.)

Now if you say if it's the DFS, it's A and if it's the DFT it's C, then what about the fact that C works for both the DFS and DFT?

So, would it kill the DFS if the time-shifting and convolution properties where changed to:

$$ Y[k] = e^{-j 2 \pi \frac{mk}{N}} \, X[k] \quad \Rightarrow \quad y[n] = x[(n-m)\operatorname{mod}_N] \qquad m \in \mathbb{Z} $$

$$ Y[k] = H[k]\, X[k] \quad \Rightarrow \quad y[n] = \sum\limits_{m=0}^{N-1} x[(n-m)\operatorname{mod}_N] \ h[m] $$ ?

Would that break the Discrete Fourier Series? If the answer is "no", then that means every theorem that the DFS and DFT have are identical to the other corresponding theorem. The DFS and DFT have these different initial definitions above but exactly the same theorems. So, tilde or not, they behave exactly the same.

So, is that "$0$" in the DFT definition ever getting used in the DFT or any theorems associated with the DFT? If that "$0$" changed to:

$$ X[k] = \begin{cases} \sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N}, & 0 \le k \le N-1 \\ \text{ten guzzillion ...} & \text{otherwise} \end{cases} $$

$$ x[n] = \begin{cases} \frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N}, & 0 \le n \le N-1 \\ \text{... something else}, & \text{otherwise} \end{cases} $$

Would it change anything w.r.t. the DFT? Where would that "otherwise" argument find use in any of the DFT mathematics?

Now, if you concede that changing 0 to ten gazzillion in the DFT and iDFT definitions make no difference with any DFT theorem, then what else is different, semantically, between the DFT and the DFS? What other semantic is different between the DFT and the DFS?

If the only semantic difference between the DFT and DFS makes absolutely no difference in any of the theorems that the DFT and DFS share (and they are all shared because we slipped that little modulo into the DFS theorems because it doesn't break them), then this is precisely a distinction without a difference. It makes no operational difference. It doesn't change any mathematics.

It makes no difference.

Then, since we normally express the DFT exactly like the DFS above (but without the tildes), and their theorems are exactly the same, and the only semantic difference makes no operational difference in any manner contained within the context of the DFT, then to claim there is any meaningful difference has no basis.

Answer 7

Conceptually, DFS $\neq$ DFT.

DFS is, by definition, the Fourier transform of discrete-time, infinite-length, and periodic signals, which are denoted as $\tilde{x}[n]$. Therefore, DFS is given by $$ \tilde{X}[k] = \frac{1}{N} \sum_{n= 0}^{N-1} \tilde{x}[n] e^{-\frac{2\pi n k}{N}} $$

Although $\tilde{x}[n]$ is a infinite-length signal, we can store it in a computer by using only the samples within a period, and then computing the DFS. Likewise, since $\tilde{X}[k]$ is also periodic, we can store only the samples within its period as well. Although the DFS allow us to compute the Fourier transform of discrete signals, it is restricted to periodic and infinite-length signals.

The DFT, on the other hand, extends this computation beyond infinite-length and periodic signals: It computes the Fourier transform for a non-periodic and discrete-time sequence by assuming that this sequence is one period of a periodic signal. Consequently, all signal shifting is made on the basis of the so-called "circular shifting" (denoted by Oppenheim as $((n))_N$).

The fact that DFT constructs a circularly periodic signal from a finite-length and non-periodic sequence leads some people to claim that the DFT is also defined for periodic signals, and therefore would have no difference. But it is wrong.

Stick with the fundamentals

Sadly, there is a lot of misinformation on the internet. The best you can find is quotes of authoritative source that might enlighten this issue. From the fact that Oppenheim is an authoritative author in the field of DSP,. He clearly defines that $x[n]$ is 0 outside $\{0,1, \dots, N-1\}$:

In recasting Eqs. (8.11) and (8.12) [DFS equations] in the form of Eqs. (8.67) and (8.68) [DFT equations] for finite-duration sequences, we have not eliminated the inherent periodicity. As with the DFS, the DFT $X [k]$ is equal to samples of the periodic Fourier transform $X(e^{j\omega})$, and if Eq. (8.68) is evaluated for values of $n$ outside the interval $0 ≤ n ≤ N − 1$, the result will not be zero, but rather a periodic extension of $x[n]$. The inherent periodicity is always present. Sometimes, it causes us difficulty, and sometimes we can exploit it, but to totally ignore it is to invite trouble. In defining the DFT representation, we are simply recognizing that we are interested in values of $x[n]$ only in the interval $0 ≤ n ≤ N − 1$, because $x[n]$ is really zero outside that interval, and we are interested in values of $X [k]$ only in the interval $0 ≤ k ≤ N −1$ because these are the only values needed in Eq. (8.68) to reconstruct $x[n]$

Therefore, $x[n]$ is not $5$,$5000$, "man on the moon", or any other flat-earth-like crazy "theory" you came across in this community. Stick with the fundamentals.

FAQ

"Why should $x[n]$ be $0$ outside $\{0,1, \dots, N-1\}$?

That is an axiom which is given in the context of expanding the DFS to a broader class of signals which may not be periodic: the rationale for setting it to zero makes $x[n]$ finite-length and, therefore, nonperiodic.

For instance, in probability theory, It is not because $0\geq P[X] \geq 1$ is an axiom that we can state that $0\geq P[X] \geq \text{his hot girlfriend}$. That would be as aberrant as misleading. The same thing applies to DFS and DFT. So stop messing well-established theories up.

"This is a meaningless difference."

By assuming that there is no difference between DFT and DFS, you are presuming that Oppenheim, Proakis, Simon Haykin, and many other authoritative source have been doing tautology over the decades by sheer will. And yes, authoritative source is important, it is not a fallacy. Otherwise we create nonsense conjectures about the fundamentals. And the reason to have such a distinct difference is obvious: $x[n]$ in nonperiodic and finite-length, which enable us to use DFT for a much broader class of signals beyond the periodic and infinite-length ones.

"If you apply the DFS to $\tilde{x}[n]$, you would get the same result if would applied DFT to $x[n]$.

Of course you will. That is due to the inherent periodicity of the DFT, which extends a finite-length to a periodic in infinite-length signal.

What is true about the relationship of DFT and DFS?

• The mathematical computation of DFT and DFS and absolutely identical. There is no different in the mathematical computation between the them. However, this does not make them equal.

• The DFT inherently extends the periodicity of the finite length signal. By applying the DFT, we extends the periodicity of the finite-length signal.

• The symbol $((n))_N$ is ugly and I really dislike it. So do I.