13

You'll need to understand the sampling theorem. In short, each signal has what we call a spectrum¹, which is the Fourier transform of the signal as it comes in time domain (if it is a time signal), or spatial domain (if it is a picture. Since the Fourier transform is bijective, a signal and its transform are equivalent; in fact, one can often interpret the ...


8

The reason is that if it is true for any $m$, it is also true for $m=kn$. I will sketch the proof in another way. Call $f_s = 1/t_s$ sampling frequency where $t_s$ is sampling period, the two signal $x(t) = \sin(2\pi f_0 t)$ and $x_k(t) = \sin(2\pi (f_0 + k f_s) t)$ have the same values at sampling instants (aliasing), i.e. $x[n] = x_k[n]$. Indeed, \...


7

The answer is: yes, sampling in the frequency domain causes aliasing in the time domain, exactly like the dual case: sampling in the time domain causes aliasing in the frequency domain. There are many ways to see this. One standard way is to sample the discrete-time Fourier transform (DTFT) of a discrete-time signal by multiplying it with a Dirac comb and ...


6

Intuitively, if You move the sensor $ N $ steps each at the size of $ \frac{1}{N} $ of its resolution you can get $ \times N $ more resolution. It is like a polyphase representation of the signal. Using estimation methods, any movement which is not an (Event with zero probability) integer multiplication of the resolution of the sensor, namely, fractional ...


6

Regarding example 1: first of all, either the fft or the ifft needs to be normalised by the number of sampling points as you have done (actually you can normalise each by a factor of the square root of the number of points, it is just a meter of definition). However, in your case the ifft is half the length of the fft you performed. Hence your signal is ...


6

I think you mean "images", not "aliases". They become aliases if there is foldover from resampling. It's because you are not adding two signals, $x(t)$ and $\operatorname{III}(t)$, you are multiplying them that these images appear. $$\begin{align} x_\text{s}(t) & \triangleq x(t) \cdot \operatorname{III}(t/T) \\ &= x(t) \cdot \sum\limits_{n=-\...


5

You may not need to explicitly reconstruct. But if you did reconstruct a waveform using the samples that you have, and end up with something different from the actual input, your controller is controlling as if that new different reconstructed waveform was really the input. Depending on what your controller is doing, you may have wanted it to do something ...


5

The concept of reconstruction has nothing to do with the application, rather it has to do with the question: did I get the same signal that is really there. If you cannot recreate the signal back, that means the conversion process is loosing/modifying underlying information, which in most cases you do not want to happen. So the confidence on the Fourier ...


5

In addition to @hotpaw2 explanation, a graphic. There are two analog square waves (red and green), with different lengths. They are depicted with a fine sampling, denoted by crosses. Their actual sampling is denoted by circles. The red one is shorter than the green one, as can be seen in the interval $]0.7\;0.8[$. Yet, the sample points are the same. Thus, ...


5

this paper was done long before i had MATLAB. the drawings are poor, but the math (at least in this revision, which is what you should use) is spot on. send me an email address (to my audioimagination.com, not the wavemechanics.com on the paper) and i will send you a very short C file that shows how to generate and crossfade the wavetables in the synthesis ...


5

No. The aliased component will interfere with the non-aliased components and the interference can constructive or destructive. Trivial example: $$x[n] = \sin\left(\frac\pi2n\right)$$ If you down sample this to $y[n] = x[2n]$, you get all zeros.


4

Think of frequency in this context more in terms of angular velocity. If that velocity is constant, nothing changes from the usual picture. However, if the frequency changes, it is more correct to compute the actual angle that is input into the trig functions as an anti-derivative of the frequency function. Here the interpolated frequency is $$f(t)=(1-\...


4

Edit: I have recently created two Jupyter Notebooks that illustrate this behaviour and let you play around with some actual matrices and actual signals. I find understanding MDCT easiest if we define our transforms as matrix operations. DFT In case of a DFT such a matrix would look like $$ F_M = \left( \sqrt{\frac{1}{M}} e^{-2 \pi i k n / M} \right)_{k,n=...


4

A few approaches to alias-free nonlinear distortion (in increasing order of difficulty): Subband distortion: Use a low pass filter to extract the lower end of the signal. If you choose a cutoff frequency of $\frac{f_s}{2N}$ you can apply any non-linear transfer function $f$ with derivatives starting at $f^{N+1}$ vanishing to avoid aliasing. Add just the ...


4

I think you should plot something like: t = [0:0.05:1]; %20Hz sampling a = sin(2*pi*2*t); %2Hz sine wave b = sin(2*pi*18*t); %18Hz sine wave plot(t, a, 'bo'); hold on; plot(t, b, 'ro'); T = [0:0.001:1]; %1000Hz sampling frequency A = sin(2*pi*2*T); B = sin(2*pi*18*T); % plot for 1000Hz sampling frequency plot(T, A, 'b'); plot(T, B, 'r'); ...


4

There are two things to consider here: the noise power after sampling, and the fact whether the sampled noise is uncorrelated or not. Note that the autocorrelation function of the sampled noise equals the sampled autocorrelation function of the continuous-time noise. For a wide-sense stationary (WSS) band-limited white noise signal $X(t)$ with bandwidth $B$ ...


4

Looking at the spectrogram, the prominent artifacts go up or down in frequency synchronously to the 138 MHz signal but have larger bandwidths. That is an indication that they are its harmonics, due to a nonlinearity somewhere in the system. With the sampling frequency of 500 MHz some of the positive and negative harmonics alias to the following (bold) ...


4

This question, and a number of similar others, from DTSP book can be a little tricky to recognise the fact that it's not actually asking alias free operation (which would require 8 khz cutoff signal as you expected), rather it is actually asking how much aliasing (due to initial sampling block) is tolerable: you can allow aliasing in those regions of the DT ...


4

I read Footnote 8, "The complication is that because of the sampling, the total system is not time-translation invariant and so does not have a unique ‘impulse response’ – the response is slightly different according to the position of an original impulse relative to the sampling points." as a cumbersome way of saying that aliasing corrputs an LTI ...


4

This may not fully answer your question, but for getting a feeling of aliasing, maybe a simple demonstration can be helpful... Some initial setup: import numpy as np from matplotlib import pyplot as plt sr = 44100 sig_len = 4096 So, it is well known, that time domain undersampling causes frequency domain aliasing. # sine wav and its spectrum t = np....


4

Hints: Can an LTI system generate components in some frequency $\omega_0$ if the input signal $x(n)$ was such that $X(e^{j\omega_0})=0$? Does aliasing do such thing? The answers to these questions are straightforward and, combined, they answer the original question.


4

The name of this effect is Spectral Leakage. Remember the relation $$\frac{k}{N} = \frac{F}{F_S}$$ where $k$ is the bin number, $N$ is the FFT size, $F$ is the continuous frequency in Hz and $F_S$ is the sample rate in Hz. It can be seen that $k$ varies from $-N/2$ to $N/2-1$ (or from $0$ to $N-1$). So there are only $N$ continuous frequencies $F$ for which ...


4

When discussing DFT, you have to remember two things: you're windowing your true signal $x[n]$ (which is periodic, and then infinitely long) with a window $w[n]$ (here a rectangular window) to obtain a truncated version $x'[n]$ of it $$x'[n] = x[n]w[n].$$ you're sampling the Fourier Transform $X'(e^{j\Omega})$ of your signal at normalized pulsation $\...


4

Decimating a signal (selecting every Dth sample and discarding the rest) does not distort the signal within the passband in any way other than to cause aliases from higher frequencies to fold into the signal bandwidth. Depending on how we model the system the phase may be effected since $z^{-n}$ is replaced with $z^{-n/D}$, but the phase will still be ...


4

This visual phenomenon appears because the maximum frequency is close to the Nyquist frequency, or half the sampling frequency. Sampling begins to approach the limit of $2$ samples per period, and thus the linear interpolation performed by Matlab becomes highly inaccurate. However, samples are correctly located, as you can see from the code where an higher ...


4

it seems like the bigger problem is that subsampling the signal would result in aliasing frequencies No, subsampling in frequency domain corresponds to aliasing in time domain. So the idea here is to purposefully alias in time domain so that you get a sub-sampled FFT. That is exactly why 'Claim 3.7' of paper mentions $y_i=\sum_{j=0}^{n/B-1}x_{i+Bj}$. These ...


4

In contrast to common misunderstanding, aliasing does not affect the process of sampling, rather it affects the process of reconstruction. i.e., the samples themselves do not contain error, but their interpretation at the reconstruction (or anything related with it) will be in error. So, if you sample a sine wave $$x(t) = A \sin( 2 \pi f_0 t + \theta) $$ at ...


4

There really is no practical difference between the two. The important thing to understand about aliasing is not the exact definition of the word, but rather the concept: when two frequency bands alias and there is a signal in one of these bands, after sampling you can't tell which band the signal came from. In a sense, by the time you're asking what is or ...


3

You could use another approach: allow negative frequencies, do a complex IFFT, discard the imaginary parts of the time domain samples, and multiply the result by 2. Let's try it in Octave (MATLAB clone) with a motif $[1 + 2i, 3 + 4i, 5 + 6i, 7 + 8i]$ shifted so that its leftmost bin ($1 + 2i$) lands on a negative frequency. FFT length is 8. (I rewrote the ...


Only top voted, non community-wiki answers of a minimum length are eligible