# Tag Info

6

The total (standard) RMS, continuous time: $$x_{RMS}=\sqrt{\frac 1{T_0} \int_0^{T_0} x^2(\tau)d\tau}$$ And the moving RMS, continuous time: $$x_{RMS}(t)=\sqrt{\frac 1{T} \int_{t-T}^t x^2(\tau)d\tau}$$ Are essentially the same operation, the first one taken over the total number of samples $T_0$ and the second taken over a small number of samples $T$. ...

6

No. It's only LTI (Linear and Time-Invariant) systems that can be modeled with convolution through a unique single impulse response. For example the systems $$y(t) = g(t) x(t)$$ or $$y[n] = \sum_{k=0}^{k < n} x[n-k]$$ are both linear but not time-invariant and their output $y[n]$ cannot be computed with the convolution operation ( $\star$ denoting ...

4

Depending on the exact characteristics of the signal and implementation requirements (if any) I can think of a number of approaches to extract content at a "single" frequency. The most common: Apply an FFT to calculate the spectrum of the signal and extract the frequency of interest. Apply Goertzel's algorithm to effectively extract signal content at a ...

4

This a very complicated question, and I would say a still open topic. The concept of stationarity is manifold, from pure statistics to applied DSP (strict, strong, wide-sense, quasi-stationarity, cyclo-stationarity, to refer to a recently closed question). The lack of access to sound models and faithful realizations renders the quest quite difficult. Non-...

4

The sinc is just fine an example as a signal with infinite support: Support being defined as the smallest interval in which the function has non-zero values, it's trivial to see that $\text{sinc}(x)=\sin(x)/x$ (for all $x$ but a single point) becomes arbitrarily small for large $x$, but never actually constantly $\text{sinc}(x>\xi)=0$ for any finite $\... 4 (Update: I just realized that the first part of this covering CIC structures is basically what Hilmar has already answered-- I'll leave this up since it offers more graphics and details in case that his helpful to anyone but it is indeed the same answer) This may not be optimum and although a feedback structure is involved it is strictly not IIR, but want ... 4 This is not a full answer. It explores some basic approximations. For a moving average as long as 600 samples it is informative to look at impulse responses of continuous-time filters as approximations of those of the desired discrete-time filters. Let's have as the ideal desired impulse response a boxcar function normalized to a width of 1 spanning$x = 0\...

4

I can show you some low order IIR approximations to an FIR moving average filter. In the figure below you see $3$ (infinite) impulse responses that approximate a moving average of length $N=600$. The filter orders are $1$, $2$, and $5$, respectively, and they all approximate the desired response in a least squares sense. I used the equation error method to ...

3

As far as my logic goes, I wont ge able to tell any of musical features just of an amplitude value in the time domain. That's wrong. The amplitude-over-time actually is the way the song is played by your sound card (we call that PCM). So, this is your song. What you really need to understand that your analog audio (i.e. the variations in air pressure over ...

3

Your definitions are not correct. For a Strict Sense Stationary process (signal) the joint distribution of your process' value for all instants of time must be independent of time, in other words if x(t) were your process, P(x(t1),x(t2),x(t3),...) must be independent of time's origin. For a Wide Sense Stationary process the joint probability of process ...

3

You have a finite length sequence $x[n]$, $n=0,1,\ldots, N-1$, with the discrete Fourier transform (DFT) given by $$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{1}$$ and its inverse DFT (IDFT) given by $$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$ From $(1)$ it is clear that $$X[0]=\sum_{n=0}^{N-1}x[n]\tag{3}$$ If you compute another ...

3

If you sample a pure sinewave for an integer number of periods, the samples will sum to zero (it will have equal "ups" and "downs"). If you zero an FFT bin, that's the same as subtracting a sinewave of the same magnitude and phase as was in that bin, or identically, adding a pure sinewave of the same magnitude but opposite phase. So zeroing bins is the ...

3

So, first of all, CSV seems to me the least suitable format imaginable for this amount of data. It needs to be parsed, is memory hungry, and wastes precision, and isn't linearly addressable (ie. to get to the 99999. element, you need to parse the preceeding 99998 elements). So I'd recommend keeping the structure from these files, but converting them to ...

3

If the response to $x(t)$ is given by $$y(t)=\int_{-\infty}^{\infty}h(t,\tau)x(\tau)d\tau\tag{1}$$ then the response to $x(t-T)$ is $$\tilde{y}_T(t)=\int_{-\infty}^{\infty}h(t,\tau)x(\tau-T)d\tau=\int_{-\infty}^{\infty}h(t,\tau+T)x(\tau)d\tau\tag{2}$$ If the system is time-invariant we require $$\tilde{y}_T(t)\stackrel{!}{=}y(t-T)=\int_{-\infty}^{\infty}... 3 Suppose that Alice has a vector \mathrm x \in \mathbb R^n. She computes the DFT of \mathrm x$$\mathrm y := \mathrm F \mathrm x \in \mathbb C^n$$where \mathrm F \in \mathbb C^{n \times n} is a Fourier matrix. Alice then tells Bob what \mathrm y is. Since the inverse of the Fourier matrix is \mathrm F^{-1} = \frac 1n \, \mathrm F^*, Bob can ... 3 The two approaches should return the same solution. They are just two different ways to get to the same place. In the ZSR/ZIR method, you are solving two different IVPs - they have the same expression on the left-side of the equality, but different initial conditions and different right-hand sides (one has a 0, the another one has some function of t). ... 3 The coreof the question relies on how to define for the support of a function with x in a domain \mathcal{D}. The most common notion in DSP is the closure of the set where f does not vanish, i.e.$$S=\overline{\{x\in\mathcal{D}:f(x)\ne0\}}\,,$$hence this as a closed set. Here, clearly, for a cardinal sine, the closure of \mathbb{R}/ \mathbb{Z^*} is ... 3 Depends a bit on your application. A moving average filter is a low pass filter and one with many lobes and pretty poor stop band rejection at that. Depending on what specific requirements you have, you may be better off doing a simple butterworth lowpass filter. If you are worried about execution cycles, there is a very efficient way to implement it, if ... 3 i would think that if your system has no complex poles, only real poles, then you could make the impulse response to be monotonic. the impulse response would be the sum of decaying exponential functions. you can always normalize its area to 1 and that would make it a moving weighted average. how do you prefer your moving average to be weighted? 3 The discrepancy between your derivation and matlab computation results because of a mistake you did during the partial fraction expansion: Given that the function to be expanded is H(s)$$ H(s) = \frac{K}{As^2 + Bs + C} $$then you should first convert it into the form$$ H(s) = \frac{K/A}{s^2 + (B/A)s + C/A} $$and then apply the expansion as$$ H(s) ...

3

Really, there's nothing special about time and frequency domain. The math doesn't care whether you're transforming amplitude over time, gravel over mountain height, or smell intensity over fridge temperature. To the math, you map elements from a function space to elements from a function space when you do the Fourier transform. Full stop. Any scalar ...

3

If you have a one dimensional signal $s(\cdot)$, it somehow belongs to a combination of two different domains: the sampling domain: where samples are considered or taken, or how the signal is sampled. When samples are taken in some order, this is often called: the ordinal variable. This is generally related to some "physically-sound" unit, like time (you ...

3

Any LTI system can be completely characterized (among other things) by it's transfer function or it's impulse response. If your filter represents an LTI system, that you can calculate it's output by either convolving the input with the impulse response or multiplying the transfer function with the spectrum of the input signal. In theory these things are ...

3

A plot of the normalized impulse responses, for the n = 2 through 10 Butterworth low pass filters, are given by H.J. Blinchikoff, A.I. Zverev, "Filtering in the Time and Frequency Domains", Wiley-Interscience, John Wiley & Sons, NY, ©1976, p. 113. This is shown below. They do not give the h(t) expressions in the book, at least where I have looked thus ...

3

There's no need to use numerical methods here. The most straightforward way to compute the output is to see that the filter's impulse response is given by $$h(t)=\sum_{k=1}^Nr_ke^{s_kt}u(t)=\sum_{k=1}^Nh_k(t)\tag{1}$$ where $N$ is the filter order, $u(t)$ is the unit step function, and $r_k$ are the coefficients of the partial fraction expansion of $H(s)$: ...

2

It's generally not possible to compute the exact maximum value, but you can compute a bound on the maximum value. Assuming your data are discrete-time, and you're using the discrete Fourier transform (DFT), you have the following relation between time domain and frequency domain: $$x[n]=\frac{1}{N}\sum_{n=0}^{N-1}X[k]e^{j2\pi kn/N}\tag{1}$$ where $N$ is ...

2

I will solve it for T=1. You can solve the others on your own. First, note that my definition of sinc is: $$sinc(t)=\frac{\sin(\pi t)}{\pi t}$$ Definitions of sinc tend do differ, so check what your definition is. Now, for T=1, we have $$f_T: n \mapsto f(nT)=sinc(n)$$ So, \begin{align} [\ldots, -3, -2, -1, 0, 1, 2, 3, \dots] &\mapsto [\ldots, ... 2 Figure 1. Illustration of the problem with Lanczos2 interpolation: The sum of the four time-shifted Lanczos-window multiplied sinc functions (red) do not sum (black) to unity in the middle between horizontal coordinates 1 and 2. A simple fix would be to distribute one fourth of the deviation from unity as function of the fractional position in the middle ... 2 Your proposed condition is NOT true as written. What would be true is\sum_{t=-\infty}^{\infty}x_t = \sum_{t=-\infty}^{\infty}y_t$$or$$\sum_{t=1}^{N}x_t = \sum_{t=-\infty}^{\infty}y_t The sum simply represents the DC component the value of the Fourier Transform at $\omega = 0$. If the value of transfer function of your lowpass filter at \$\omega = ...

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