Hot answers tagged

23

Say you're interested in $$M^{j2\pi f_0 t}. \tag{1}$$ Note that $$M = e^{\log M},$$ so $(1)$ can be written as \begin{align} M^{j2\pi f_0 t} &= \left( e^{\log M} \right) ^ {j2\pi f_0 t} \\ &= e^{j2\pi (f_0\log M) t} \\ &= \cos(2\pi (f_0\log M) t) + j \sin(2\pi (f_0\log M) t), \end{align} which is a complex sinusoid with frequency $f_0 \log M$. ...


19

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 ...


14

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 ...


13

Intuition: The intuition is this: Your noise is some event or events that are rare, and that when compared to other events, look like outliers that shouldn't really be there. For example, if you are measuring the speeds of every car on the highway as they pass by you and plot them, you will see that they are usually in the range of say, $50$ mph to $70$ ...


9

the general polynomial form is: $$\begin{align} f(u) &= \sum\limits_{n=0}^{N} \ a_n \ u^n \\ \\ &= a_{\small{0}} + \Bigg(a_{\small{1}} + \bigg(a_{\small{2}} + \Big(a_{\small{3}} + \,... \big(a_{\small{N-2}} + (a_{\small{N-1}} + a_{\small{N}} \,u \,)u \, \big)u \ ...\Big)u \, \bigg)u \, \Bigg)u\\ \end{align}$$ the latter form is using Horner's ...


9

\begin{align} (1-x)\sum_{k=0}^{n}x^k &= \sum_{k=0}^{n}x^k - x\sum_{k=0}^{n}x^k \\ \\ &= \sum_{k=0}^{n}x^k - \sum_{k=1}^{n+1}x^k \\ \\ &=1 + x+x^2+ \dots+x^n \ \\ & \qquad -x-x^2-\dots-x^n-x^{n+1}\\ \\ &= 1 - x^{n+1} \end{align} Thus if $x \ne 1$, dividing both sides by $(1-x)$ results in $$\sum_{k=0}^{n}x^k = \frac{...


9

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, ...


8

As you have already pointed out in your question, it is not possible (without using optimization methods) to compute an exact L2 solution for the frequency domain design problem of IIR filters due to the non-linear relationship between the filter coefficients and the error function. There is, however, a method which can come close and which transforms the ...


8

If I draw a number uniformly between zero and one, what is the probability that they are equal? Mathematically, it should be zero but I don't recall why? Can somebody please help in explaining why the probability should be zero? Avoiding formal definitions of (Lebesgue) probability measure, an informal way is thinking the probability at a point of a ...


7

Given a set of values $ {\left\{ {s}_{i} \right\}}_{i = 1}^{N} $, we're basically after: $$ \arg \min_{x} \sum_{i = 1}^{N} \left| {s}_{i} - x \right| $$ One should notice that $ \frac{\mathrm{d} \left | x \right | }{\mathrm{d} x} = \operatorname{sign} \left( x \right) $ (Being more rigorous would say it is a Sub Gradient of the non smooth $ {L}_{1} $ Norm ...


6

Here's just a polynomial version: $$ \arcsin(x) = x + \frac{1}{2} \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^5}{5} + \frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{x^7}{7} $$ function y = arcsin_test3(x) y = x.*(1+x.*x.*(1/6+ x.*x.*(3/(2*4*5) + x.*x.*((1*3*5)/(2*4*6*7))))) endfunction which seems to have five multiplies (assuming you can ...


6

Assuming independent random variables with normal distributions, the probability that a value will fall beyond, say, 2 standard deviations will be about 0.01. If you have a median filter of width 3, that triplet must contain two outliers on the same side of the mean in order for an outlier to come through. This event has a probability of $2 \cdot 0.005^2 =$ ...


6

You can obtain a bound on the magnitude of the DFT of $x[n]$ ($|x[n]|\le 1$) as follows: $$\big|X[k]\big|=\left|\sum_{n=0}^{N-1}x[n]e^{-jkn2\pi/N}\right|\le\sum_{n=0}^{N-1}|x[n]|\le N\max_n|x[n]|=N\tag{1}$$ Note that for signals $x[n]=e^{j2\pi nl/N}$, $l\in\mathbb{Z}$, the bound is tight. However, for most other signals with $|x[n]|\le 1$ you will find ...


6

The most straightforward way to see this is to note that for $k=mN$ $$W_N^{kn}=e^{j2\pi mnN/N}=e^{j2\pi mn}=1$$ So the sum for the case $k=mN$ is simply $$\sum_{n=0}^{N-1}1=1+1+\ldots+1=N$$ Note that the solution using L'Hopital's rule is a bit dubious because for $k=mN$ the formula for the geometric series is not valid because the terms are all equal to ...


6

It just means "the transformation that turns $x$ into $y$." You might also see $\mathbf{T}^{-1}$ which means the inverse: turning $y$ into $x$.


6

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 ...


5

The upper bound would be a coherent summation of all samples of $ x \left [ x \right ] $. Specifically: $$\begin{aligned} X \left[ k \right ] & = \sum_{n = 0}^{N - 1} x \left[ n \right] \exp^{-j 2 \pi \frac{nk}{N}} \\ & \leq \left| \sum_{n = 0}^{N - 1} x \left[ n \right] \exp^{-j 2 \pi \frac{nk}{N}} \right| \\ & \leq \sum_{n = 0}^{N - 1} \left| ...


5

Gaussian Kernel is made by using the Normal Distribution for weighing the surrounding pixel in the process of Convolution. Since we're dealing with discrete signals and we are limited to finite length of the Gaussian Kernel usually it is created by discretization of the Normal Distribution and truncation. I created a project in GitHub - Fast Gaussian Blur. ...


5

Defining $ a \left[ k \right] = {2}^{- \left| k \right|} $. Moreover, the Auto Correlation function of $ v $ defined as $ {r}_{vv} \left[ k \right] = \left \langle {v}^{\left( 0 \right)}, {v}^{\left( k \right)} \right \rangle = \sum_{n = -\infty}^{\infty} {v}_{n} {v}_{n - k} $. Pay attention that Auto Correlation is Hermitian Function. Using the definition ...


5

Although the question could belong to SE.math, mastering inequalities for $\ell_p$ norms (for $p\ge 1$) or quasinorms (for $0<p< 1$), and their norm ratios and powers, is quite important in modern signal/image processing: starting from Cauchy-Bunyakovsky-Schwarz, to power means, Rogers-Hölder, Young, Minkowski, Dresher-Gini etc.) There are scalar, ...


5

$\delta(t)$ 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)...


5

First of all, welcome to DSP SE. What you see in the image you have linked is termed (spectral) leakage. When you are dealing with the Fourier series you deal with a periodic continuous function which is "decomposed" into a (possibly) infinite sum (series). Then, when you go to the Fourier transform, you have a non-periodic function (which you ...


4

i have a pretty good implementation of $\arctan()$ here. i think you can use the identity: $$ \arcsin(x) = \arctan\left( \frac{x}{\sqrt{1-x^2}} \right) $$ to get what you want.


4

Q1. As shown by Oppenheim's experiment, the phase spectrum contains most of the structural information about the image. In 2D this are things like lines and edges. In 3D it is things like lines and edges but also movement. Instead of 2D frames and time, imagine the video as a 3D solid where the z-axis is the frame number. If you took a slice along the z-axis ...


4

Since $ M \in \mathbb{S}^{N}_{++} $ (In other convention $ M \succ 0 $) by Cholesky Decomposition there is a Triangular Matrix $ R \in \mathbb{R}^{N \times N} $ such that $ M = {R}^{T} R $. Using this fact one could prove $ 1 \iff 2 $ as following: $$\begin{align*} \arg \min_{\hat{x}} \mathbb{E} \left[ {\left( \hat{x} - x \right)}^{T} M \left( \hat{x} - x \...


4

The key is in the last step of your work: $$ \frac{1 -e^{j2\pi k}}{1-e{\frac{j2\pi k}{N}}} $$ If $k$ is some integer multiple of $N$, then the exponents in the numerator and denominator are both some integer multiple of $j2\pi$. In this case, both exponential functions are equal to 1, meaning that the expression above is equal to $\frac{0}{0}$ for $k$ an ...


4

To complement my part to this question: Here is a somewhat shorted answer based upon a manual expansion of the odd function $f(x)$ \begin{align*} f(x)&=\ln\left(\arctan\left(\alpha e^x\right)\right)-\ln\left(\arctan\left(\alpha e^{-x}\right)\right)\\ &=f_1x+f_3x^3+O\left(x^5\right)\tag{1} \end{align*} into a series up to the third order. Some ...


4

i have no idea what the "reconstruction fidelity term" is or what it's about. Hermitian symmetry is a term usually applied to some form the Fourier Transform of a signal that is purely real. for continuous-time, continuous-frequency Fourier Transform: $$ X(f) = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \ dt $$ if $x(t)$ is purely real (that ...


4

Mathematically, an electrical circuit is as an operator, i.e., a function that takes a function and returns another function. Let this operator be denoted by $\mathcal T$, let $x : \mathbb R \to \mathbb R$ be the input signal and let $y := \mathcal T (x)$ be the output signal. Let $\mathcal D_{t_0}$ be the delay operator that delays its input by $t_0 > 0$....


4

$\DeclareMathOperator{\sgn}{sgn}$ The modulating signal in AM is $$s(t) = C + a(t)\text,$$ where $a(t)$ is the (audio) amplitude, and $C$ is a constant so that $s(t) \ge 0 \;\forall t$. (Otherwise, your audio amplitude would just frequently "switch" the wave's sign, not really modulate the envelope.) That means, $C > - \min_t(s(t))$. Therefore, the ...


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