New answers tagged

0

The statement is: eliminating all points where the difference between $x$ and $i$ is at least $r$ so eliminate those points for which $$ |x - i | \gt r \tag{1}$$ Isn't that correct? One could argue whether the words say $$ |x - i | \ge r $$ instead, which might be problematic at the border, but using (1) the statement seems OK to me. The text after the ...


5

If not, what is the most efficient way of implementing a 48 point FFT? Three 16 point FFTs plus one set of 3 point "Butterflies". Matlab example %% Do a 48 point FFT, this is NOT efficient but shows the principle n = 48; nFFT = 16; x0 = randn(n,1); % test vector fx0 = fft(x0); % reference x1 = reshape(x0,3,nFFT)'; % reshape into 3 N-16 vectors fx1 ...


1

No, that's not possible. You can piece together a 48-point FFT from factors-of-48-FFTs, in your case from multiple 16-point FFTs using the radix-N method. You can even use the Split-Radix Algorithm to piece together a 48-point FFT from 32 and 16 point FFTs – but it's not going to be 48-point FFT using a 32-point FFT and 16-point FFT but 48-point FFT using ...


0

Please re-read the answer to Defining a discrete sequence along an arbitrary time axis You don't need to do anything special OTHER then manually using a offset constant for indexing the time shifted array. Do NOT move the actual data. If you must, you can create a class that supports arbitrary indexing but this would be significantly less efficient and more ...


0

Matlab arrays always start at index 1, so technically you can't represent an array from indexed from -12850 : 13975 : 13975 in Matlab directly. Most people work around this by simply keeping manually track of an index offset. Something like yOffset = -12850 +1; % offset tmp = y(yOffset + nn); % access y() starting at -12850 The time reversal is simple ...


2

For what values of the model parameters, does it generate a real signal? That's probably not possible. Roughly speaking, your frequencies are all positive and a real signal must have a conjugate symmetric spectrum, i.e. equal amount of positive and negative frequencies. If you make the sum run from $-K$ to $+K$ and set $a_k = a_{-k}, \alpha_k = \alpha_{-k}, ...


3

You have $$ f\left(\mathbf x, u\right) = \begin{bmatrix}\frac{-1}{T}\tau+\frac{K}{T} u \\ \frac{\tau}{mr} \\ 0 \end{bmatrix} \tag a $$ From which you (eventually) derive $$ \mathbf {A}_d=\begin{bmatrix} 1-\frac{\Delta T}{T} & 0 &0 \\ \frac{\Delta T}{m_{op} r} & 1 & \frac{-\tau_{op} \Delta T}{m_{op}^{2} r}\\ 0& 0 & 1 \end{bmatrix} \tag ...


0

Which means that the system is time-invariant but only when k0 is a multiple of 2. If the system is only time invariant at certain starting times, then it's not time-invariant; it's time-varying.


0

What you have done is absolutely correct. Since the property is not true for every $k$0, the system is not time invariant


1

Frequency response is the eigenvalue of the system when the input is an eigenfunction of the form $e$$j\omega n$. Solve it by yourself to get $\displaystyle \sum_{k=-\infty}^\infty h[k]$e$-j\omega k$, which is the Fourier transform of $h[n]$.


0

The answer is wavelet design. In brief, sampling in frequency domain offers precise control over certain desired filtering properties and is often subject to less discretization error. Discretization error: smoothness Which is easier to misrepresent with uniform sampling? Left is a Morlet of high center frequency, right its Fourier transform. We can imagine,...


2

I think this question comes down to the the concept of sampling, and how discrete samples relate to the continuous-domain signal they were obtained from. This question starts with the assumption that, if one has, for example, 8 samples (|), 1 second apart (----), the samples represent 7 seconds (the space in between the 8 samples): |----|----|----|----|----...


0

Hi, eduardokapp. I think you can understand Fourier transform,dft and fft. so, i am going to explain about the meaning of your time domain integer sequence and its fft result by using following simple image from matlab. here, red line is the spectrum and time-graph when Fs is 500 and blue line is when Fs is 1000. so, your integer sequence is the sample ...


2

(edited): Considering that I only passed amplitudes and not much else, if I interpret those results as the spectrum of $x$, how can that be? You passed it amplitudes in a specific order. That means that you can then do any transform on it that depends on the ordering -- which the DFT does. Without making further assumptions about how the data set was ...


1

Since you don't have any inputs, $x_n$ is only a function of past values of itself, your B an D should indeed be zero. You don't have your output defined either. You could pick the most recent $x_n$ as output, but it could also be something else. A transfer functions describe an input output relation. However, you don't have inputs, so also no input output ...


3

For the world in the whole, with our current technical abilities we can't say for sure whether continuous time processes exist in the real world or not. The problem is degree of difference between time measurements. If your degree of difference is $\Delta t$ then you can't say something happened at the moment $t_0 + \Delta t/2$ or $t_0 + \Delta t$. From your ...


2

It's all about vectorization. N = 8; K = 10; k = 1:K; % row vector f = k * 100; % row vector alpha = k / 10; % row vector a = k / 10; % row vector phi = k * pi; % row vector deltat = 1; n = (0:N-1)'; % column vector b = (1:N)'; % column vector x = sum(a.*exp(1j*phi).*exp((-alpha+1j*2*pi*f)*deltat.*n)+b, 2);...


1

For a camera sensor, light arrive as discrete quanta (photons) according to Poisson statistics. One usually regard that as an «artifact», while the (more) continuous scene reflectance/illuminance is seen as the desired signal that photon quantization corrupts. If the scene allows, one could increase exposure time in order to get more photons in order to get ...


1

I would vote for "non-inherently continuous" for the following reasons. First the notion of discrete time series may apply to a discretization either in time or in the amplitude domain (discrete sampling vs discrete valuation). For the latter, the question is really dependent on the generative or physical process. On the valuation: If you do photon ...


3

The official answer to your question is this : random processes (time-series) can be classified as being either continuous-time or discrete-time. Some of those discrete-time processes happen from sampling of continuous-time processes, but some of the discrete-time processes occur naturally, such as daily stock-market values, or Sun-Spot numbers. This ...


6

Can one argue that discrete time-series coming from stocks or commodities (prices) are derived from a continuous-time process? I would say "no". The price of a stock is determined by a trade and each trade is a discrete event in time. I would argue that between trades the price is undefined since there is no experiment you could do to determine ...


2

Well I think we normally assume that to be the case. And that the continuous-time signal "underlying" is bandlimited to below the Nyquist frequency. If the discrete-time sequence is $x[n]$ (and $n$ can only be an integer), then the "underlying" continuous-time signal is: $$ x(t) = \sum\limits_{n=-\infty}^{\infty} x[n] \operatorname{sinc}(...


Top 50 recent answers are included