# Tag Info

9

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}$$where it is assumed that x(t) and x'(t) are continuous at t=T. 5 It is the basic property of Fourier Transforms. Given that g(t) and G(f) are Fourier transform pairs i.e. g(t) \rightleftharpoons G(f) then \int_{-\infty}^tg(\tau)d\tau \rightleftharpoons \frac{1}{j2\pi f}G(f) This assumes that G(0)=0. If G(0) is not zeros then the integral of g(t) has a Fourier transform that includes a Dirac delta function. 4 \delta(t) is an impulse that's infinitely thin and infinitely high. The area under it is 1 though. \delta(t - nT) places the impulse at time nT. Now this is being multiplied by some function, in your case e^{-st} and that product is integrated. Integrating is just a glorified sum. Mathematicians try to disguise it as something fancy by using a ... 4 Remember the definition of the delta function: it integrates 1 all over the t-axis but it is zero for any t\neq 0. This means that if we multiply any function by \delta (t), then the function is going to be zero everywhere except for t=0. In this case, we have$$ \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

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 ...

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

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

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

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

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

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

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 $$H(z)=\frac{... 3 As T\rightarrow \infty, the integral becomes the convolution integral. You can use the fact that convolution in the time domain is multiplication in the frequency domain and then you have a single carrier being passed through a low pass filter. The term where \omega=\omega_c just has to do with how you define the ideal low pass filter and in this case ... 2 It depends what type of accelerometer you are using and what you want the displacement signal for. I'm not an expert in signal processing, but I have sure used a lot of accelerometers. Most piezoelectric accelerometers do not work down to DC - these accelerometers use charge amplifiers and the Low Frequency response is dependant on the time constant of the ... 2 In order to be able to do this, you need to know Cauchy's residue theorem, from which you get$$\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 ... 2 Hint: With z = e^{j\omega}, you have: \begin{align} \displaystyle \frac{dz}{d\omega} &= je^{j\omega}\\ \Rightarrow dz &= je^{j\omega} d\omega \end{align} With -\pi \leq \omega \leq \pi, your contour integral goes from -\pi to \pi. Can somebody tell me intuitively how to simplify this equation to look like the inverse Fourier ... 2 I got an answer here, https://math.stackexchange.com/questions/1554659/mutual-information-for-a-continuous-uniform-distribution The  \text{erf}(x)  function approximates to 1 with an error less than 0.0001 for  x \geq 3 . I plotted  f_{y} (y)  and I saw that the most of probability is within a band denpending on  \sqrt{ 3 \gamma } . Moreover, ... 2 Sincef(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d\omega$$you are looking for the inverse transform of$$G(\omega)=\frac{F(\omega-a)}{i\omega}$$If you look at a table of Fourier transform correspondences you will find$$\frac{F(\omega)}{i\omega}\Longleftrightarrow \int_{-\infty}^x f(y)dy$$and$$F(\omega-a)\Longleftrightarrow ...

2

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 ...

2

As this signal (function) exists inside an integration operator, and also assuming that there are no impulses residing at any discontinuities introduced by $u(t)$ step functions, then we can simply disregard the isolated discontinuity points and rely on the following: [u(\tau)-u(\tau-4)][u(t-\tau)-u(t-4-\tau)] = \begin{cases} \begin{align} &0 &\... 2 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 ... 2 First you'ld describe the quantities in Laplace domain for dealing with algebraic rather than differential equations, which greatly simplifies the labor. Then you have to move those feedback branches into proper nodes and, thereby, simplify the block diagram. Note that all three sub-blocks of your system are integrators whose transfer function is H_k = 1/... 2 Note that you can also use Parseval's theorem. Assuming that f(t) and g(t) are both real-valued we have:\int_{-\infty}^{\infty}f(t)g(t)dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)G^*(\omega)d\omega\tag{1}$$If g(t) is a narrow-band low-pass signal, i.e., if we can model G(\omega) as G(\omega)\approx G(0)\text{rect}(\omega/\Delta\omega), ... 2 Okay... I managed to do it on my own. Here's the solution if you're interested. We know that$$H\{x(t)\} = \sum_{k=0}^{\infty}a_k \ H\{e^{jw_kt}\}$$since H is a linear system. We also know that H\{e^{jw_kt}\} = F(w_k) \ e^{jw_kt}, where F(w_k) is the frequency response of the system. By definition, we have that$$F(w_k) = \int_{-\infty}^{+\...

2

Fourier on exponential functions can indeed make sense. In theory, under specific conditions. And in practical DSP, not likely. There are so many conditions under which Fourier transformations exist that whole books are devoted to them, like D. C. Champeney, A Handbook of Fourier Theorems, Cambridge University Press, 1987. And the domain remains an open ...

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