10

You have two problems here. First, as hotpaw2 mentioned, double integration is sensitive to any non-zero bias in your accelerometer, which will certainly be present if you are using a MEMS accelerometer. Suppose you have a bias of $b$; then, your position at time $t$ is $$p(t) = \int_0^t \int_0^{t^\prime} b \ dt^{\prime\prime} \ dt^\prime = \frac{b}{2} t^...


6

Use integral function. q = integral(g,xmin,xmax) For example $g(t) = e^{-t}$ $$q = \int_0^{\infty}e^{-t}\mathrm{d}t = 1$$ g = @(t) exp(-t); q = integral(g,0,Inf) More information Matlab numerical integration


6

You can see this by using integration by parts: $$\begin{align}\int_{-\infty}^{\infty}x(t)\delta'(t-T)dt&=x(t)\delta(t-T){\Big|}_{-\infty}^{\infty}-\int_{-\infty}^{\infty}x'(t)\delta(t-T)dt\\&=x(T)\delta(t-T){\Big|}_{-\infty}^{\infty}-x'(T)\\&=-x'(T)\end{align}$$ where it is assumed that $x(t)$ and $x'(t)$ are continuous at $t=T$.


6

Just to avoid a misunderstanding: the $\mathcal{Z}$-transform is a transform defined for sequences, comparable to the Laplace transform for continuous functions. What you are talking about is not the $\mathcal{Z}$-transform, but methods for converting analog to digital (actually, discrete-time) systems. [And it doesn't help that one of those conversion ...


5

It is the basic property of Fourier Transforms. Given that $g(t)$ and $G(f)$ are Fourier transform pairs i.e. $g(t) \rightleftharpoons G(f)$ then $\int_{-\infty}^tg(\tau)d\tau \rightleftharpoons \frac{1}{j2\pi f}G(f)$ This assumes that $G(0)=0$. If $G(0)$ is not zeros then the integral of $g(t)$ has a Fourier transform that includes a Dirac delta function.


5

There are different methods to approximate integration in discrete time. The most straightforward ones are the forward and backward Euler methods, and the trapezoidal method. A discrete-time system with transfer function $$H(z)=\frac{T}{z-1}\tag{1}$$ implements the forward Euler method. Similarly, the backward Euler method is implemented by $$H(z)=\frac{...


5

The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind: $$J_n(x)=\frac{1}{\pi}\int_0^{\pi}\cos(nt-x\sin t)dt,\qquad n\in\mathbb{Z}\tag{1}$$ Using the cycloid parameterization $$y(t)=1-\cos t,\qquad x(t)=t-\sin t\tag{2}$$ which results in a period of $2\pi$ and a maximum value of $2$, the Fourier series of $y(t)$ ...


4

$\delta(t)$ is an impulse that's infinitely thin and infinitely high. The area under it is 1 though. $\delta(t - nT)$ places the impulse at time $nT$. Now this is being multiplied by some function, in your case $e^{-st}$ and that product is integrated. Integrating is just a glorified sum. Mathematicians try to disguise it as something fancy by using a ...


4

Remember the definition of the delta function: it integrates $1$ all over the t-axis but it is zero for any $t\neq 0$. This means that if we multiply any function by $\delta (t)$, then the function is going to be zero everywhere except for $t=0$. In this case, we have $$ \delta (t-nT)e^{-st}= \left\{ \begin{array}{ll} \delta (t-nT)e^{-snT} & \mbox{...


4

If your square wave has a mean of zero (this is important!), then a simple accumulator can do the job. Its operation is described by $$y[n]=x[n]+y[n-1]$$ where $x[n]$ is the input (square wave) and $y[n]$ is the output (triangular wave). This is a simple Matlab/Octave script showing how it works: sq = [1,-1,1,-1,1,-1,1,-1]'*ones(1,5); sq = sq'(:); ...


4

Note that the antiderivative of a function is only defined up to a constant. Furthermore, note that if you integrate a periodic function, the result is not necessarily periodic. Let $$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}\tag{1}$$ Now we integrate $(1)$ with an arbitrary lower integration limit $t_0$. I'll say more about that later. $$\begin{...


4

I will introduce some terminology and intuition that will be helpful when reading other references. It will be neither complete nor completely rigorous. The measures that we first encounter in real analysis assign sizes (non-negative real numbers) to measurable subsets of $\mathbb{R}$; Lebesgue measure is the measure that agrees with the intuition we build ...


4

It's not necessary to first fit a curve to the data and then compute the integral. You can directly approximate the integral from the data using numerical integration methods. The most straightforward approximation would be a Riemann sum: $$S=\sum_ny_n\Delta x\tag{1}$$ where $y_n$ are your data, and $\Delta x$ is the distance between the individual data ...


4

I haven't managed to completely evaluate the integral, but I've made some progress and perhaps someone can pick up where I leave off. The integral you gave is $$ c_n = \int_0^{2\pi}\left(\frac{1-\cos\theta}{\pi}\right)^2 e^{-jn(\theta-\sin\theta)}d\theta. $$ The first thing I did is perform a Taylor expansion on that exponential term; $$ c_n = \frac{1}{\pi^2}...


3

Since this is most likely homework, here is a hint. Write the integral you have displayed in the form $\int_{-\infty}^\infty x(\tau)h(t-\tau) d\tau$ where you get to choose what the function $h(\cdot)$ is to make it all work out. Then, $h(t)$ is the impulse response of the LTI system, Do you know the criterion for BIBO stability of an LTI system in terms of ...


3

I can say a couple of things about this problem since I have been working on a similar one for a few weeks. First thing to note is that you should subtract the mean of a section of noise from your data, not the entire section. As below, moving from figure 1 to figure 2: You could also fit a 0 degree polynominal and subtract from the original data: A = ...


3

Double integration amplifies any offsets, non-linearities and noise. These can't be removed without the use of some type of external reference point measurements (e.g. not from the accelerometer) or other information. Calibration might be able to remove most of the ones that do not vary with time (temperature, etc.) For instance, if the size of the room ...


3

Numerical integration may reduce noise under some conditions: the noise is zero-average, ergodic, its properties do not vary too much over time. Since numerical integration seems linear, pre-filtering the acceleration with a linear filter, or post-filtering after integration are likely to yield very similar results (up to border effects), since linear ...


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

Double integration is very sensitive to any offset noise in acceleration measurement. Even a very tiny non-zero acceleration offset will, when double integrated and given enough time, send the calculated position outside the radius of the galaxy. Some sort of position or location measurements, with respect to some fixed reference point, in addition to ...


3

It depends what type of accelerometer you are using and what you want the displacement signal for. I'm not an expert in signal processing, but I have sure used a lot of accelerometers. Most piezoelectric accelerometers do not work down to DC - these accelerometers use charge amplifiers and the Low Frequency response is dependant on the time constant of the ...


3

If you upsample a causal discrete-time signal $x[n]$ by an integer factor $L$ and you use an interpolation filter with a causal impulse response $h[n]$, the upsampled and interpolated signal is given by $$y[n]=\sum_{k=0}^{\infty}x[k]h[n-kL]\tag{1}$$ (Note that the requirement of causality can be easily removed; I just use it to have the lower summation ...


3

Let us try with another hints: could you imagine a bounded input signal which could result in a non bounded output? general suggestion whenever analyzing a system: try a few "simple to compute" input signals, looking a the outputs, this could help you guessing some properties: an impulse, a ramp, a step, etc.


3

The proof comes from the Dirac delta function property: \begin{align} \int\limits_{-\infty}^{\infty} x(t) \delta^{(n)}(t-t_0)dt=(-1)^{n}\frac{d^n}{dt^n}x(t)\bigg\vert_{t=t_0} \end{align} where $x(t)$ is a continuous function of time with a continuous derivative at $t= t_0 $.


3

Indeed you have reached what can be reached, may be the following additional line can be obtained by moving the $t$ function $e^{-2t}$ out of the integral and replacing the $u(t-\tau)$ by $1$ as: $$ y(t) = e^{-2t} \int\limits_{-\infty}^t {x(\tau)e^{2\tau}{}d\tau} $$ In the general case you cannot simplify it any further...


3

As $T\rightarrow \infty$, the integral becomes the convolution integral. You can use the fact that convolution in the time domain is multiplication in the frequency domain and then you have a single carrier being passed through a low pass filter. The term where $\omega=\omega_c$ just has to do with how you define the ideal low pass filter and in this case ...


3

Since, the Taylor series expansion for $e^z$ about $z=0$ is $$e^z = \sum_{n=0}^{\infty}\dfrac{z^n}{n!}$$ then, ignoring the question of convergence, you can say $$\begin{align*} \int_a^b \int_c^d e^{A\sin(x)\cos(y-B)}dx dy &= \int_a^b \int_c^d \sum_{n=0}^{\infty}\dfrac{(A\cos(y-B))^n}{n!}\sin^n x \; dx \, dy\\ \\ &= \int_a^b \sum_{n=0}^{\infty}\...


3

Since $W(t)$ is assumed to be zero-mean, also the RV $Y$ is zero-mean. Hence, the variance of $Y$ is given by $$\begin{align}\sigma_Y^2&=E\left\{Y^2\right\}\\&=E\left\{\int_{-\infty}^{\infty}W(t_1)\phi(t_1)dt_1\int_{-\infty}^{\infty}W(t_2)\phi(t_2)dt_2\right\}\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(t_1)\phi(t_2)E\big\{W(t_1)W(t_2)\...


2

In order to be able to do this, you need to know Cauchy's residue theorem, from which you get $$\frac{1}{2\pi j}\oint_Cf(z)dz=\sum_kR_k\tag{1}$$ where $R_k$ are the residues of the analytic function $f(z)$ at its poles $p_k$, which must lie inside the closed contour $C$. The residue at pole $p_k$ is given by $$R_k=\lim_{z\rightarrow p_k}(z-p_k)f(z)$$ In ...


2

The Fourier series of the square wave tells us that the input signal has harmonics at odd multiples of the fundamental frequency $f_1=349\,\text{Hz}$: $$f_k=kf_1,\,\text{$k$ odd}$$ with amplitudes $$A_k=\frac{4A}{k\pi}.$$ The triangular wave, on the other hand, has harmonics at the same frequencies, but their amplitudes are $$B_k=\frac{4A}{k^2\pi^2}.$$ In ...


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