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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 is not in $\ell^1(\mathbb Z)$ since
$$\sum_{n\in\mathbb{Z}}|x[n]|=\sum_{n=1}^{\infty}\frac{1}{n}=\infty\tag{3}$$
We know that for sequences in $\ell^1(\...
7
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It's a complex exponential that rotates forever on the complex plane unit circle:
$$e^{-j\omega t} = \cos(\omega t) + j \sin(\omega t).$$
You can think of Fourier transform as calculating correlation between $f(t)$ and a complex exponential of each frequency, comparing how similar they are. Complex exponentials like that have the nice quality that they ...
6
If you don't like thinking about imaginary numbers, complex numbers and functions, you can alternatively think of the complex exponential in the FT as just shorthand for mashing together both a sinewave and a cosine wave (of the same frequency) into a single function that requires less chalk on the chalkboard to write.
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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 formally, take one of the two sums, transform the index by changing its sign ($k=-l$) and see what you get.
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As mentioned in Batman's answer, the condition of the sequence being absolutely summable is only sufficient but not necessary. The Fourier transform can be extended to $\ell_2$ sequences, i.e. sequences for which
$$\sum_{n=-\infty}^{\infty}|f[n]|^2<\infty$$
is satisfied. A further generalization is possible if you allow distributions and their ...
5
$$\begin{align}X(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\\&=\sum_{n=-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\Omega})e^{jn\Omega}d\Omega\right]\;e^{-jn\omega}\\&=\int_{0}^{2\pi}X(e^{j\Omega})\left[\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}e^{jn(\Omega-\omega)}\right]d\Omega\\&=\int_{0}^{2\pi}X(e^{j\Omega})\delta(...
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You defined the signal vector as x = [1 2 3 2 1]. Since the DFT is defined by
$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$
the command fft(x) computes the DFT of the signal
$$x[n]=\delta[n]+2\delta[n-1]+3\delta[n-2]+2\delta[n-3]+\delta[n-4]$$
This signal is not symmetrical with respect to $n=0$, and it is not equal to the signal you computed the DTFT of.
...
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Whether it's the Fourier Transform or the Laplace Transform or the Z Transform, etc. the exponential is the eigenfunction of Linear and Time-invariant (LTI) operators. if an exponential function of "time" goes into an LTI, an exponential just like it (but scaled by the eigenvalue) comes out. what the F.T. does is break down a general function into a sum ...
answered Apr 9 '15 at 18:20
robert bristow-johnson
13.2k3 gold badges21 silver badges55 bronze badges
4
The Fourier Transform:
$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(t)e^{i\omega t} dt\\
F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$$
converts a function to an integral of harmonic functions. You can think of these as sins and cosines because $e^{i\theta} = cos(\theta) + i \sin(\theta)$. The Fourier Transform as a continuous form of the ...
4
According to Oppenheim and Schafer's "Discrete Time Signal Processing", the Goertzel algorithm will be more efficient than the FFT in computing an N point DFT if less than $2 Log_2 N $ DFT coefficients are needed. So if you need to compute 15 frequency points, the Goertzel will be more efficient for total number of samples $N > 65536$. Anything below ...
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Yes your understanding is basically correct.
The 1st paragraph (2 lines) expresses the fundamental relation between the DFS and the DFT of a finite-length sequence $x[n]$ while the 2nd paragraph tries to put down the relation between the DFT $X[k]$ of a sequence and the DTFT $X(e^{j\omega})$ of it (assuming it exists).
However this 2nd paragraph shall ...
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do you see anywhere in your book where this "DTFT" is defined for your w.s.s. process, $x[n]$? the DTFT is normally defined as
$$ X(e^{j \omega}) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] e^{-j \omega n} $$
but for this infinite sum to converge, we normally require that
$$ \sum\limits_{n=-\infty}^{+\infty} \Big|x[n]\Big| < +\infty $$
which is ...
answered Nov 6 '17 at 20:16
robert bristow-johnson
13.2k3 gold badges21 silver badges55 bronze badges
4
4
The DFT is a sampled version of the DTFT only for finite length signals. Otherwise, there is no point in comparing the DTFT with the DFT because you can only compute the DFT for finite length (or periodic) signals. Your example $x[n]=\sin(\omega_0n)$ is an infinitely long sequence. For specific choices of $\omega_0$ it is periodic, but nevertheless, you can'...
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The DTFT relationships
$$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)\Longleftrightarrow j\,\textrm{Im}\left\{X(e^{j\omega})\right\}$$
hold for any sequence $x[n]$ for which the DTFT exists. There is no assumption about $x[n]$ being real-valued ...
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We are analyzing a real signal with the DTFT. Since we are using a limited number of samples it's like we are transforming a finite signal.
No. The DTFT takes an infinite discrete time signal as an input but the spectrum is continuous. I think you are confusing the DTFT and the DFT. They are different things.
As I remember, the FT of a finite signal has ...
3
If you have a continuous-time signal $x(t)$, then the two signals you're talking about are
$$\begin{align}
x_c(t) &=x(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT) \\
&=\sum_{n=-\infty}^{\infty}x(t)\delta(t-nT) \\
&=\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) \\ \tag{1}
\end{align}$$
and you define
$$x_d[n] \triangleq x(nT)\tag{2}$$
The first ...
3
Hint: Expand $e^{j \omega}$ using Euler's formula. This will enable you to express $H(\omega) = R(\omega) + j I(\omega)$ where $R$ and $I$ are the real and imaginary parts and then the magnitude of the frequency response is just $\sqrt{ R(\omega)^2 + I(\omega)^2}$.
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The long road computes the modulus of the DTFT. This works in general, but can be tedious.
When the formula possesses some symmetry, like here (the two coefficients are the same), you can more efficiently factor a term, so that you recover a known Euler formula for the sine or the cosine, in the shape of $e^{j\nu}+e^{-j\nu}$ or $e^{j\nu}-e^{-j\nu}$. Here, ...
3
The explanation by @Maximilian Matthé is a standard and formal approach for this question. But I think it is not intuitive and easy to understand the reason. In the following, I will try to explain from another aspect.
First of all, periodicity means infinity. For a signal in the temporal/spatial domain, the periodicity is somewhat easy for realization, ...
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You should consider the DTFT pair:
$$\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\omega n_0}$$
plus the linearity of the FT. That is,
$$ax_1[n]+bx_2[n] \xrightarrow{\text{DTFT}} aX_1(e^{j\omega})+bX_2(e^{j\omega})$$
Now consider the ...
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If the magnitude is $1$ and the phase were $0$, you would get
$$H(e^{j\omega})=1$$
which corresponds to $h[n]=\delta[n]$. So this is obviously wrong.
The fact that the phase of a negative real-valued number equals $\pm \pi$ can be easily seen by drawing the number as a vector in the complex plane and measuring its angle with the positive real axis. ...
3
In the cited text, $x[n]$ is not a signal, but a wide-sense stationary random process.
$P_x(e^{j\theta})$ is not the DTFT of $x[n]$, but the DTFT of the autocorrelation of the process $x$, and it is called power spectral density.
The autocorrelation $R_x[k]$ has units of power; in fact, $R_x[0]$ is the average power of the process.
The PSD, therefore, has ...
3
Cedron Dawg posted an interesting initial point in this answer. It begins with these steps:
$$ \begin{align}
U(\omega) &= \sum\limits_{n=0}^{+\infty} e^{-j \omega n} \\
&= \lim_{ N \to \infty } \sum\limits_{n=0}^{N-1} e^{-j \omega n}\\
&= \lim_{ N \to \infty } \left[ \frac{ 1 - e^{-j \omega N} }{ 1 - e^{-j \omega } } \right] \\
&= \...
3
This is pretty straight forward using the definition of the Discrete Time Fourier Transform (DTFT).
The definition of the DTFT:
$$ X \left( {e}^{j \omega} \right) = \sum_{m = -\infty}^{\infty} x \left[ m \right] {e}^{-j \omega m} $$
Differentiating with respect to $\omega$:
$$\begin{align*}
\frac{d}{d \omega} X \left( {e}^{j \omega} \right) & = \...
3
You're almost there. First, there's a small scaling error in the transform $X_c(\omega)$. Since the Fourier transform of an impulse train is given by
$$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\omega_s\sum_{k=-\infty}^{\infty}\delta(\omega-k\omega_s)\tag{1}$$
with $\omega_s=2\pi/T$, and since multiplication in the time domain ...
3
The integral doesn't exist in the conventional sense. How did you evaluate the limit $\lim_{t\to\infty}e^{-j\omega t}$?
If one allows distributions (generalized functions), such as the Dirac impulse (and its derivatives), as Fourier transforms then the class of functions that can be transformed becomes much wider, and the unit step is one of the functions ...
3
I will assume:
The probe is like a line, infinitely narrow.
We move the probe along the $x$ dimension, parallel to the large-scale surface, and from each position, probe the surface:
The probe is extended to the direction it is pointing to until it meets the surface. (An alternative assumption would be that probing is done to the direction $y$, that of the ...
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