12

First of all the dirac delta is NOT a function, it's a distribution. See for example http://web.mit.edu/8.323/spring08/notes/ft1ln04-08-2up.pdf Treating it as a conventional function can lead to misunderstandings. Example: "informally" the dirac delta is often defined as "infinity at x=0 and zero everywhere else". Now let's look at a ...


11

If you imagine a Dirac delta impulse as the limit of a very narrow very high rectangular impulse with unit area centered at $t=0$, then it's clear that its derivative must be a positive impulse at $0^-$ (because that's where the original impulse goes from zero to a very large value), and a negative impulse at $0^+$ (where the impulse goes from a very large ...


5

Ideal derivative filter Let $f(x)$ be a signal bandlimited to frequencies $(-\pi,\, \pi)$. Given $f(x)$ as input, the same $f(x)$ is given as output by a system that has as its impulse response the sinc function: $$\operatorname{sinc}(x) = \left\{\begin{array}{ll}1&\text{if }x = 0,\\ \frac{\sin(\pi x)}{\pi x}&\text{otherwise.}\end{array}\right.\tag{...


5

I assume you mean the derivative with respect to $t$. In that case, the derivative of $\sin(\omega_0t)$ is not $\cos(\omega_0t)$ but $\omega_0\cos(\omega_0t)$. And luckily, this is also obtained via the Fourier transform relation you mentioned in your question: $$\begin{align}\mathcal{F}\left\{\frac{d}{dt}\sin(\omega_0t)\right\}&=j\omega\cdot \mathcal{F}...


5

The estimation of derivative is straightforward: $$x'(n)~=\frac{x(n+1)-x(n-1)}{2}$$ $$x''(n)~={x(n+1)-2*x({n})+x({n-1})}$$ or if you have a signal sampled at $t_i=i\Delta t$, it is $$x'(t_{i})~=\frac{x(t_{i+1})-x(t_{i-1})}{2*\Delta t}$$ $$x''(t_{i})~=\frac{x(t_{i+1})-2*x(t_{i})+x(t_{i-1})}{(\Delta t)^2}$$ What you are interested in may be how to ...


5

That's a trick which you will also find in a DSP context, that's why I choose to provide an answer here. It is related to the Wirtinger derivative, and you can find more details about it in this answer over at math.SE. In practice this trick is often used to compute the extremum (minimum or maximum) of a real-valued function depending on a complex variable (...


4

2 point discrete differentiation is bound to produce highly noisy results. try the 5-points stencil. you can also generate coefficients (i.e. more points) yourself using derivation of Lagrange polynomials.


4

So, the delta function satisfies: $$\int_{-\infty}^{\infty} f(x) \delta(x - a)\, \mathrm{d}x = f(a)$$ Now, suppose we substitute $f(x) = x$ $$\int_{-\infty}^{\infty} f(x) \delta(x - a)\, \mathrm{d}x = a = \int_{-\infty}^{\infty} a \delta(x - a)\, \mathrm{d}x$$ This means that multiplying by $\omega$ is the same as multiplying by a constant. If you substitute ...


4

You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92). With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B\,\text{Hz}$, and $\text{sup}\,|f(t)| = A$), then $$\left|\frac{\text{d}f(t)}{\text{d}t}\right| \leq 2AB\...


4

Your first suggestion is correct, the derivatives of the local polynomials are being sampled. From the horse's mouth (Savitzky & Golay 1964): Again, if we restrict ourselves to evaluating the function only at the center point of a set of equally spaced observations, then there esist sets of convoluting integers for the first derivative as well. (...


4

It's a matter of perspective if the bilinear transform "fails so miserably" when trying to approximate a derivative. First of all, it's not true that the approximation is quadratic for small $\omega$, it's linear as it should be: $$D(e^{j\omega})=\frac{2}{T}\frac{1-e^{-j\omega}}{1+e^{-j\omega}}=\frac{2j}{T}\tan\left(\frac{\omega}{2}\right)\approx \frac{j\...


4

In general, the time derivative property of the Fourier Transform is given as $$\mathscr{F}[\frac{d}{dt}x(t)] = j\omega X(j\omega) $$ Notice that we can simply multiply by the frequency index in the Fourier Transform result. For the 2D FT result: $$\mathscr{F}[f(x,y)]= F(u,v)$$ Using the same property results in: $$\mathscr{F}[\frac{d}{dx}f(x,y)]= uF(u,...


4

$\delta(t)$ is a distribution, which means it is represented by a limitng set of functions. To find $\delta'(t)$, start with a limiting set of functions for $\delta(t)$ that at least have a first derivative. The triangle function of unit area is the simplest function to chose: $$\delta(t) = \lim_{\epsilon \to 0} \dfrac{\Lambda\left(\frac{t}{\epsilon }\right)...


4

Maybe a picture is worth a thousand words? Here's how a Gaussian pulse of variable width and its derivatives look like: As others have said, Dirac is a distribution, hence the Gaussian pulse, and its width gets narrower and narrower. The derivative of $$\mathrm{e}^{-x^2}=-2x\mathrm{e}^{-x^2}$$ Which says that the derivative is the same as the function, ...


4

Dirac's $\delta$ is a distribution. Distributions can be interpreted as limits of smooth functions under an integral or as operators acting on functions in ways which are defined by integrals. Both approaches have in common that basic properties of integrals are expected to work, partial integration in particular. Other answers have showed you the ...


3

Differentiation in one axis will amplify high-frequency components. Assuming noise at Fs/2 (i.e maximum frequency), the gain at this frequency will be 6 dB. Intuitively, if you have a sequence alternating between +1 and -1 i.e (1, -1, 1, - 1), Then the difference will be +2, - 2, +2, -2, a gain of 6 dB Another reason is that by differentiating, you remove ...


3

I will answer this with the understanding that $p$ is the Laplace variable (which I will call $s$). As you stated, differentiation in the time domain corresponds to multiplication by $s$ in the Laplace domain. So it seems that we could get what we want by just choosing $F(s)=s$; then the output is the derivative of the input, i.e., $$y(t) = \frac{du}{dt}$...


3

Simply put: take an independent identically distributed Gaussian noise (one observation in blue, left) and its gradient (in green, right). The amplitude is roughly multiplied by the norm of the gradient filter (here $h=[1\,-1]$), which is $\sqrt{2}$, which you can see from the picture. Their average energy $E$ thus differ by a factor of $|h\|^2=\sqrt{2}^...


3

This is not a standard 5-point derivative formula, the corresponding transfer function of which is $$H(z)=\frac{1}{12}\left(-z^2+8z-8z^{-1}+z^{-2}\right)\tag{1}$$ The figure below shows the magnitude responses of an ideal differentiator (red), of the standard 5 point approximation given by (1) (green), and of the approximation in your question (blue): As ...


3

There is a good article in Wikipedia: Numerical Differentiation Since you'd expect differentiation to be (Anti?)symmetric, using IIR filters might not be the wisest of ideas.


3

Well, Since basically the Derivative Operation is Linear Filter you can chose your own optimal trade off between Noise Sensitivity and Bandwidth. If you look at Finite Differences Coefficients page at Wikipedia you can see you can chose higher "Accuracy" filters for 1st Derivative. You can also chose Forward / Central / Backward method.


3

So, first of all: A differentiator is really just a high-pass filter. Digitally (and this is dsp.SE, so I presume this is the case), that typically means a differentiator is really just FIR with taps $[1;-1]$. A butterworth filter being a low-pass IIR, the combination of both does sound like you might want to build a bandpass, or are doing something that ...


3

Let's have a look at: $$\Delta I (x_1,x_2) = I_{actual}(x_2, y) - I_{actual}(x_1, y) = I_{ideal}(x_1, y) + I_{noise}(x_1, y) - I_{ideal}(x_2, y) - I_{noise}(x_2, y)$$ If you assume that for close $x_1$ and $x_2$ the $I_{ideal}$ doesn't change much so that $$ I_{actual}(x_2, y) \approx I_{ideal}(x_2, y)$$ then $$\Delta I (x_1,x_2) = I_{noise}(x_1, y) - ...


3

No, you are not increasing the sampling time. The sample rate stays the same. The Central difference derivative filter is simply longer than the First difference derivative filter. A longer filter just means more delay before you get your result. No, you should not low pass filter without a clear reason for why it is needed to meet your signal processing ...


3

Practically calculating the derivative of a digital signal is straightforward, just convolve the signal samples with the taps of a(n approximate) derivative filter. MatLab has the functions filter() and conv() which can help you do this. The First Difference derivative filter has taps $$h[n] = [1.0, -1.0]$$ The Central Difference derivative filter has ...


3

As shown in the answers to this question, for continuous-time signals, the bound predicted by Bernstein's inequality is achieved with equality by a sinusoidal signal with a frequency equal to the upper frequency limit. In analogy to this, in this answer I'll show a bound on $\big|x[n]-x[n-1]\big|$ for a discrete-time sinusoidal signal $x[n]$ at angular ...


3

Your interpretation is correct: directional derivation operators highlight variation in a given direction. Here, you use the $2$-point discrete derivative in the $x$-direction (along image rows). It may emphasize vertical features. First, such operators indeed extend the initial image range. However, one often uses the absolute value of the derivative to ...


2

You can use differentiator filter that acts as a differentiator in the band of interest, and as an attenuator at all other frequencies, effectively removing high frequency noise. Take a look at the matlab example


2

If there was no noise in the input, then the causal filter: $$y[k] = \frac{\frac{3}{2}x[k] - 2x[k-1] + \frac{1}{2}x[k-2]}{T_s}$$ would give the asymptotically optimal 3-sample linear approximation of the first derivative in the limit of an infinite sampling frequency assuming zero input noise. The asymptotic error as sampling period $T_s\to0$ is a good ...


2

You can represent the convolution as a matrix multiplication with a $N\times N$ Toeplitz matrix, where $N$ is the length of $\mathbf{w}$. Check out this explanation. So you can write $$ E(\mathbf{w})=\frac{1}{2}\|\mathbf{AHw}+\mathbf{b}\|^2_2 $$ where $\mathbf{H}$ is your Teoplitz matrix filter. Now if you define $$ \tilde{\mathbf{A}} = \mathbf{AH} $$ then $$...


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