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Where is the flaw in this derivation of the DTFT of the unit step sequence $u[n]$?

Tendero's user avatar
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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 ...
Fat32's user avatar
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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 ...
Matt L.'s user avatar
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7 votes
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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 ...
Matt L.'s user avatar
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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 (...
robert bristow-johnson's user avatar
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^{...
robert bristow-johnson's user avatar
6 votes

Where is the flaw in this derivation of the DTFT of the unit step sequence $u[n]$?

Matt L.'s user avatar
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6 votes
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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}\...
Matt L.'s user avatar
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5 votes
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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 \...
Tendero's user avatar
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5 votes
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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 ...
Matt L.'s user avatar
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5 votes
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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 ...
Matt L.'s user avatar
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5 votes
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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 ...
Hilmar's user avatar
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5 votes
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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 ...
Rubem Pacelli's user avatar
4 votes
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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 ...
Matt L.'s user avatar
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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}...
Bob K's user avatar
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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 ...
lxg's user avatar
  • 465
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 ...
Dan Boschen's user avatar
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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 ...
Matt L.'s user avatar
  • 90.3k
4 votes

Where is the flaw in this derivation of the DTFT of the unit step sequence $u[n]$?

robert bristow-johnson's user avatar
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)\...
Matt L.'s user avatar
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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 ...
Hilmar's user avatar
  • 45.3k
4 votes
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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 ...
DSP Rookie's user avatar
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4 votes
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Z-transform of the Unit Step and DTFT

Assuming that the $\mathcal{Z}$-transform $X(z)$ of a sequence $x[n]$ exists, there are three cases we need to distinguish when considering the relation between $X(z)$ and the corresponding DTFT $X_F(...
Matt L.'s user avatar
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3 votes
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Magnitude and phase of $-\delta[n]$?

If the magnitude is $1$ and the phase were $0$, you would get $$H(e^{j\omega})=1$$ which corresponds to $h[n]=\delta[n]$. So this is obviously wrong. The fact that the phase of a negative real-...
Matt L.'s user avatar
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3 votes
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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 ...
Laurent Duval's user avatar
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 ...
Atul Ingle's user avatar
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3 votes
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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 ...
Maximilian Matthé's user avatar
3 votes
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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\...
msm's user avatar
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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 $\mbox{asinc}$ ...
Peter K.'s user avatar
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