10
votes
Accepted
7
votes
Link between DFS, DFT, DTFT
Yes your understanding is basically correct.
The 1st paragraph (2 lines) expresses the fundamental relation between the DFS and the DFT of a finite-length sequence $x[n]$ while the 2nd paragraph tries ...
- 28k
7
votes
Does the DTFT of $\frac{u[n-1]}{n}$ exist?
Note that the sequence
$$x[n]=\frac{u[n-1]}{n}\tag{1}$$
is in $\ell^2(\mathbb Z)$ because
$$\sum_{n\in\mathbb{Z}}|x[n]|^2=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}<\infty\tag{2}$$
but it ...
- 88.9k
7
votes
Accepted
Difference between CTFT and DTFT?
The difference is pretty quickly explained: the CTFT is for continuous-time signals, i.e., for functions $x(t)$ with a continuous variable $t\in\mathbb{R}$, whereas the DTFT is for discrete-time ...
- 88.9k
7
votes
Difference between CTFT and DTFT?
I'll add a couple more facts, to sorta complete the definitions of things. As Matt represented the CTFT and DTFT, they are both shown as special cases of the Laplace Transform and Z Transform (...
- 20.1k
6
votes
Accepted
Derive DTFT of $x[2n]$
HINT:
If you have a sequence
$$\hat{x}[n]=\begin{cases}x[n],&n\text{ even}\\0,&n\text{ odd}\end{cases}$$
then the DTFT of $x[2n]$ can be written as
$$\sum_{n=-\infty}^{\infty}x[2n]e^{-jn\...
- 88.9k
6
votes
$|X(e^{jω})|^2$ - Power or Energy Density?
do you see anywhere in your book where this "DTFT" is defined for your w.s.s. process, $x[n]$? the DTFT is normally defined as
$$ X(e^{j \omega}) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] e^{...
- 20.1k
6
votes
6
votes
Accepted
Proving that the IDTFT is the inverse of the DTFT?
$$\begin{align}X(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\\&=\sum_{n=-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\Omega})e^{jn\Omega}d\Omega\right]\;e^{-jn\omega}\...
- 88.9k
5
votes
Accepted
Relation between the DTFT and the spectrum of a sampled signal
If you have a continuous-time signal $x(t)$, then the two signals you're talking about are
$$\begin{align}
x_c(t) &=x(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT) \\
&=\sum_{n=-\infty}^{\...
- 88.9k
5
votes
Accepted
why $-$ sign in DTFT pair for constant
It can also have a $+$ sign, there's no difference. Write down a part of the sum (around index $l=0$) and try to see that in both cases you're summing the same terms, just in a different order.
More ...
- 88.9k
5
votes
Accepted
Is possible reach the DFT if I have the DTFT?
The DFT is a sampled version of the DTFT only for finite length signals. Otherwise, there is no point in comparing the DTFT with the DFT because you can only compute the DFT for finite length (or ...
- 88.9k
5
votes
Accepted
Why DFT is used for approximating CTFT when you can approximate CTFT-integral itself?
Assuming that future vistors won't take the time to read all the comments, I'd like to give a very simple and straightforward interpretation of the discrete Fourier transform (DFT) as an approximation ...
- 88.9k
5
votes
Accepted
Difference in having even number and odd number of samples in DFT?
We know that DFT is just a sampled version of the DTFT.
Only if there is no time-domain aliasing (see below)
My thoughts are that if we use an odd number of samples of the DTFT in our DFT, the ...
- 42.6k
5
votes
Accepted
Gaussian filter: Plotting DTFT and DFT (by hand) from the continuous-time impulsive response
Before anything, let me just rewrite your TF from linear frequency ($f$, in Hertz) to angular frequency ($\Omega = 2\pi f$, in rad/sec). The notation $\Omega$ is usually adopted in the field of DSP to ...
- 337
4
votes
What does the exponential term in the Fourier transform mean?
Consider the case $\ f(t) = 2 \cos(\omega_0 t) = e^{+i \omega_0 t} + e^{-i \omega_0 t}.\ $ Then
$$
F(\omega) = \int\limits_{-\infty}^{+\infty} e^{i (-\omega + \omega_0) t} \ dt + \int\limits_{-\infty}...
- 127
4
votes
Why DTFT coefficients are periodic and why continuous Fourier transform coefficients are not periodic?
The explanation by @Maximilian Matthé is a standard and formal approach for this question. But I think it is not intuitive and easy to understand the reason. In the following, I will try to explain ...
- 465
4
votes
Accepted
Why is this DFT of a real symmetric signal resulting in complex valued coefficients?
You defined the signal vector as x = [1 2 3 2 1]. Since the DFT is defined by
$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$
the command ...
- 88.9k
4
votes
FFT-like algorithm for fast DTFT computation?
According to Oppenheim and Schafer's "Discrete Time Signal Processing", the Goertzel algorithm will be more efficient than the FFT in computing an N point DFT if less than $2 Log_2 N $ DFT ...
- 48.9k
4
votes
Accepted
Discrete-time Fourier Transform of the unit step sequence $u[n]$
Cedron Dawg posted an interesting initial point in this answer. It begins with these steps:
$$ \begin{align}
U(\omega) &= \sum\limits_{n=0}^{+\infty} e^{-j \omega n} \\
&= \lim_{ N \to \...
- 4,980
4
votes
Discrete-time Fourier Transform of the unit step sequence $u[n]$
I'll provide two relatively simple proofs that do not require any knowledge of distribution theory. For a proof that computes the DTFT by a limit process using results from distribution theory, see ...
- 88.9k
4
votes
4
votes
system function $H(\omega)$ relationship to odd and even components of h[n]
The DTFT relationships
$$x_{even}[n]=\frac12\left(x[n]+x^*[-n]\right)\Longleftrightarrow\textrm{Re}\left\{X(e^{j\omega})\right\}$$
and
$$x_{odd}[n]=\frac12\left(x[n]-x^*[-n]\right)\...
- 88.9k
4
votes
Aliasing and DTFT of a real signal
We are analyzing a real signal with the DTFT. Since we are using a limited number of samples it's like we are transforming a finite signal.
No. The DTFT takes an infinite discrete time signal as an ...
- 42.6k
4
votes
Accepted
Getting the DTFT from the DFT samples
Myth: DTFT is Sinc-interpolated DFT.
Problem with the above statement: Sinc is not $2\pi$-Periodic function, but all DTFTs are.
Correct Answer:
Theoretical, Continuous-$\omega$ $2\pi$-Periodic DTFT ...
- 2,591
3
votes
Accepted
Frequency Response with Delta Function?
The long road computes the modulus of the DTFT. This works in general, but can be tedious.
When the formula possesses some symmetry, like here (the two coefficients are the same), you can more ...
- 31.7k
3
votes
Frequency Response with Delta Function?
Hint: Expand $e^{j \omega}$ using Euler's formula. This will enable you to express $H(\omega) = R(\omega) + j I(\omega)$ where $R$ and $I$ are the real and imaginary parts and then the magnitude of ...
- 4,124
3
votes
Accepted
Why DTFT coefficients are periodic and why continuous Fourier transform coefficients are not periodic?
The result of a DTFT is periodic, because any discrete-time signal has a continuous spectrum. This can be e.g. explained by the following:
Let $x(t)$ be a time-continuous signal. Now, making it ...
- 6,208
3
votes
Difference between Fourier Transform and DFT? - Example
The difference is between $\mbox{sinc}$ and the periodic version obtained using the DFT.
See this answer for a comparison.
It strikes me that asinc = ratio of two sincs.
The $\mbox{asinc}$ ...
- 25.2k
3
votes
Accepted
The DTFT of $\{1,1\}$ is $1+e^{-j\omega}$ but what is the DTFT of $\{1,-1\}$?
You should consider the DTFT pair:
$$\delta[n] \xrightarrow{\text{DTFT}} 1$$
and the time-shifting property of the Fourier transform
$$x[n-n_0] \xrightarrow{\text{DTFT}} X(e^{j\omega})e^{-j\...
- 4,225
Only top scored, non community-wiki answers of a minimum length are eligible