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39

You can best look at it in the frequency domain. If $x[n]$ is the input sequence and $h[n]$ is the filter's impulse response, then the result of the first filter pass is $$X(e^{j\omega})H(e^{j\omega})$$ with $X(e^{j\omega})$ and $H(e^{j\omega})$ the Fourier transforms of $x[n]$ and $h[n]$, respectively. Time reversal corresponds to replacing $\omega$ by $-\...


22

No, taking the Fourier transform twice is equivalent to time inversion (or inversion of whatever dimension you're in). You just get $x(-t)$ times a constant which depends on the type of scaling you use for the Fourier transform. The inverse Fourier transform applied to a time domain signal just gives the spectrum with frequency inversion. Have a look at ...


17

I found this video to be very, very helpful (it elaborates on Matt's answer). Here are some key ideas from the video: Zero-phase will result in no phase distortion, but will result in a non-causal filter. This means that if the data is being filtered as it's gathered, this will not be an option (only valid for stored data which we can post-process). When ...


17

Whilst taking the Fourier transform directly twice in a row just gives you a trivial time-inversion that would be much cheaper to implement without FT, there is useful stuff that can be done by taking a Fourier transform, applying some other operation, and then again Fourier transforming the result of that. The best-known example is the autocorrelation, ...


14

"Is there any practical application?" Definitely yes, at least to check code, and bound errors. Especially for huge data or a large number of iterations "In theory, theory and practice match. In practice, they don't." So, mathematically, no, as answered by Matt. Because (as already answered), $\mathcal{F}\left(\mathcal{F}\left(x(t)\right)...


12

2D Fourier transform (2D DFT) is used in image processing since an image can be seen as a 2D signal. E.g. for a grayscale image $I$, $I(x,y)=z$, that means that at the coordinates $x$ and $y$ the image has intensity value z. Look at this for example: https://ch.mathworks.com/help/matlab/ref/fft2.html Try this: x=imread('cameraman.tif'); X=fft2(fft2(x)); ...


10

All eigenfunctions of an LTI system can be described in terms of complex exponentials, and complex exponentials form a complete basis of the signal space. However, if you have a system that is degenerate, meaning you have eigensubspaces of dimension >1, then the eigenvectors to the corresponding eigenvalue are all linear combination of vectors from the ...


10

I'll complete a bit the answer given in a comment above. Intuitively first, to which frequency corresponds a signal constant in time, for exemple $x(t) = 1$ $\forall t$ ? Such a signal shows no variation in time and hence contains only a component with frequency 0 (this is a DC signal). This means that its Fourier transform must be 0 everywhere, except in $...


6

To answer the second question, in digital communications there is a technique in use in cellphones right now that makes good use of applying the IFFT to a time-domain signal. OFDM applies an IFFT to a time-domain sequence of data at the transmitter, then reverses that with an FFT at the receiver. While the literature likes to use IFFT->FFT, it really makes ...


5

Depends on assumptions you are willing to make and what type of signals are you trying to sample, but in theory I think that sampling rate equal to the Planck time would be a gold standard for anything... This translates to sampling frequency of $1.855 \times 10 ^ {43} \mathtt{Hz}$ ($18.55$ tredecillion hertz). Personally I believe that machines will never ...


5

I have found some "proverbs" like: Never use software to compensate for a poor lighting system. It is not cost effective and will result in a poor system design. It is cheaper to add a light-proof shroud to keep sun-light away from the object under inspection than to modify the software. Another universal truth which is often forgotten. ...


4

How to Find a Suitable Lighting? This will be the most important question of an engineer who has to select a right lighting set-up for the Machine Vision application. Probably he remembers some clever Machine Vision proverbs such as "better to light than write (software)", "avoid garbage in (bad lighting) that causes garbage out (bad result)", "...


4

I'd like to point out Heisenberg Uncertainty principle, based on which theoretical achievable precision is limited. It states that one can not measure two complementary qualities (e.t. here time and charge) concurrently and there is a trade off between amount of precision you can get from one or another. In ADCs, for example theoretical limit for resolution ...


4

There is not only one sampling rate that theoretically allows exact reproduction. The sampling theorem states that it is sufficient for perfect reconstruction if the sampling frequency $f_s$ is greater than twice the highest signal frequency: $$f_s>f_{max}\tag{1}$$ Without further restrictions, Eq. $(1)$ is also necessary for baseband signals occupying ...


4

If you're not interested in specific practical aspects of A/D conversion, but if you want to learn basic theory concerning sampling and digital (discrete-time) processing of analog (continuous-time) signals, I'd recommend that you read and study the chapter Sampling of Continuous-Time Signals in Oppenheim en Schafer's book Discrete-Time Signal Processing.


3

The integration variable is $\tau$ and the integration limits are determined by the range of $\tau$ over which both step functions under the integral are non-zero. For the first step functions that's $\tau > 1$ and for the second one it's $\tau < t$ . This determines the integration interval. Within this interval the function to integrate is simply 1,...


3

Hint: Your integration by parts is wrong. The second integral looks OK. The first integral should be solved like this (leaving out constant factors): $$\int_0^{\tau}te^{-j\omega t}dt=\frac{te^{-j\omega t}}{-j\omega}\bigg|_0^{\tau}+\frac{1}{j\omega}\int_0^{\tau}e^{-j\omega t}dt$$ I'm sure you can take it from here.


3

No, and the reason is not so much a question of how fast one can sample a continuous-time signal (as the accepted answer and another one says) but rather the impossibility of representing a real number with perfect accuracy via a quantized representation of the real number (as noted in the answer by Marcus Muller). At best, even if we assume an infinite ...


3

Is it theoretically possible to perfectly quantize a continuous signal? No. A quantization has an information content obviously countable as bits. Now, if you have a continuously distributed 1D random variable $X$, then the event that any of these real numbers $x$ occurred is unbounded ("infinite"): $$I(x) = -\log_2\left(P(X=x)\right)$$ So, for ...


3

Both real and complex signals can be passband. Assuming $x(t)$ and $y(t)$ are baseband signals with bandwidth much less than $f_c$, then $f(t)$ is a passband signal, even if it is complex.


3

Well, getting a bit linguistic, according to the Oxford dictionary: stochastic (adj.): Having a random probability distribution or pattern that may be analysed statistically but may not be predicted precisely. So the definition would be the first one (I don't know where you might have found the second one as you didn't put any source about it). ...


3

It really boils down to aliasing. In continuous-time, if you have any two signals $x_1(t) = \sin(2 \pi F_1 t)$ and $x_2(t) = \sin(2 \pi F_2 t)$, then as long as $F_1$ and $F_2$ are distinct, the signals are, too. But consider sampling at some time interval $T_s$, so that the sampled signals are $x_1(k) = \sin(2 \pi F_1 T_s k)$ and $x_2(k) = \sin(2 \pi F_2 ...


3

The autocorrelation function of a random process $x(t)$ is defined as $$R(t,\tau) = {\mathbb E}\{x(t) x(t+\tau)\}.$$ For a stationary random process, this function does not depend on $t$, i.e., we have $R(t,\tau) = R(\tau) \; \forall t$. The correlation only depends on the time difference between two samples, not the "absolute" time $t$ when the ...


3

Complex exponentials are eigenfunctions of LTI systems because they are eigenfunctions of the convolution operator: $$\begin{align}e^{j\omega_0t}\star h(t)&=\int_{-\infty}^{\infty}h(\tau)e^{j\omega_0(t-\tau)}d\tau\\&=e^{j\omega_0t}\int_{-\infty}^{\infty}h(\tau)e^{-j\omega_0\tau}d\tau\\&=e^{j\omega_0t}H(j\omega_0)\end{align}\tag{1}$$ where $h(t)$ ...


2

I'm not sure I can find your quote, but I can mention a few books of the past 30 years that leaned at least somewhat toward the practical advice rather than toward the more purely theoretical/mathematical/snooty. (One of the more "theoretical" textbooks I've read simply regurgitated pages of math from an earlier textbook, complete with the exact same glaring ...


2

ok so here is the answer of my Prof. Hagit Hal-or: if there is such a mask then it must be 5x5. A counter example shows that this can not be. consider the 5x5 region of an image: we fill it with values 0...0,1,2...2 (12 0's and 12 2's) the 5x5 median on this region gives 1, regardless where you place the numbers. Now we build the 5x5 region so that if we ...


2

You can use rectangular pulses, but it is not necessary to do so. Rectangular pulses have advantages and disadvantages. The advantage is that they are very simple and easy to use, both for the modulator and the demodulator. It makes life easier for the demodulator because it makes symbol timing very forgiving- you can be off on where you sample the symbol ...


2

If you are only decimating by four, then a CIC filter is not the way to go. CIC filters have two advantages and one big disadvantage. The advantages are that they don't require any multipliers, just (large) adders, and they can efficiently decimate by really large factors. The no-multipliers advantage was a big deal years ago when multipliers were ...


2

I believe that it will be worth for many people to do it from scratch. Let's call the signal after input and feedback summation $E(s)$: Then question arises what $E(s)$ is equal to? Well that's simple: $$E(s)=5R(s)-(5+K_vs)C(s) $$ So we have our backward path, now the forward one. What $C(s)$ is equal to? $$C(s)=E(s)K_p \dfrac{2}{(Js+B)s} $$ Let's ...


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