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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 ...


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If your code is working correctly, than there should be no aliasing. Most likely this is a problem with your playback system. Most sound cards in computers are terrible and will create a lot of noise and other artifacts. Since the 19 kHz tone is inaudible for most people, you just hear the artifacts. It's also possible that you turn up the volume too high....


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I think you are confused by negative frequencies and what they mean so let me add this explanation. When you see a spectrum that contains "positive" and "negative" frequencies, each of the frequencies are of the form: $$e^{j\omega t}$$ Where $\omega$ is the frequency (in this case angular frequency as $2\pi f$ with f being the frequency in Hz. The ...


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The digital frequency represented with $\pi$ is the Normalized Angular Frequency, and has units of frequency given as radians/sample. This signal can be represented as a rotating phasor on the complex IQ plane: $Ae^{j\omega n}$, where $A$ is the amplitude and $\omega$ is the normalized angular frequency. With this representation the normalized frequency is ...


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Check this figure. Consider the signal $x[n]=A$ which is constant but can be written as $x[n]=A\cos(0n)$, so its frequency is zero. There is no oscillation. Let's increase the frequency. For the signal $x[n]=A\cos(\pi n/4)$, it oscillates at frequency $\omega_0=\pi/4$. The oscillation frequency increases. For the frequency $\omega_0=\pi$, the signal is $x[...


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(1) is it possible that the point-wise multiplication of two discrete, band-limited functions is aliased itself? Yes. Or at least it's possible that the result won't fit in a tidy manner into your sampling rate. Because multiplying two signals together creates energy at different frequencies than the frequencies of the original (this is easiest to do by a ...


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Yes that is correct. The width of the transition band is inversely proportional to the length (or memory) of the filter which means to say that you would need an infinite amount of time to achieve your perfect filter. Therefore for practical reasons you decide how much aliasing would be tolerable (similar in many ways to deciding how many decimal places you ...


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Your plot is correct. You are sampling three waves $f_1 = 10\textrm{Hz}, f_2 = 30\textrm{Hz}$, and $f_3 = 70\textrm{Hz}$, with a sample frequency of $F_s = 1.5\times70\textrm{Hz} = 105\textrm{Hz}$. This means that your Nyquist frequency is $F_N = F_s/2 = 52.5\textrm{Hz}$, and corresponds to the maximum value of your frequency axis. As such, the signals $...


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Sound absorption is an example of filtering. You can build an enclosure (room) to lessen sound coming from your clothes washer. Adding insulation inside the wall helps further, as does offsetting studs on each side to minimize vibration transfer. This comes at a cost, but at some point you’ve made the problem unnoticeable—you don’t need 100% eradication. ...


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I don’t think that you can do multi-frame super resolution in the traditional sense if there is no camera rel scene movement, nor if the camera applies «proper» Nyquistian spatial prefiltering? I think of good old interlacing as an example of potential super resolution. You get a 2-frame cadence consisting of separate time and spatial samples, usually ...


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Your premise is due to the physical nature of the device under test it is possible that occurrence of higher frequencies vibrations can be predicted by lower frequencies and this is important as it will allow you to use a lower bandwidth sensor that can only respond to lower frequencies. Interesting problem and I would suggest this approach to find evidence ...


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does aliasing occur always if i sample a vibration in real world applications? Yes. The aliasing always occurs. The sampling theorem assumes band limited signals, but these strictly band limited signal do not exist in reality (as they would be infinitely long). Of course any signal can be low pass filtered to be reduce the aliasing to an acceptable level ...


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If I have a vibrations sensor that has a max sample rate of 8kHz -> It can reconstruct signals till 4kHZ perfectly right? Theoretically, yes. Though I would like to add that all the noise signals beyond $+/-4kHz$ will alias back into your sampled signal. But what about frequencies which occur also in the measurements with much higher frequencies? If ...


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That is why, before sampling, a (steep) lowpass filter with cutoff frequency $f_c \leq \frac{f_s}{2}$ shall be applied. Thus, the amount of aliasing will be insignificant.


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I think you're confusing IQ sampling with oversampling. Your first scenario is oversampling, or taking twice the number of samples on a real signal. Other than stating the sampling takes place 90 degrees apart (it would actually be 180 degrees), your analysis is sound. In the second section, you describe IQ sampling. This is usually done by mixing a ...


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Given strictly real data as input, the "upper half" of an FFT is simply a redundant mirrored complex conjugate of the lower half. And in order for an IFFT to produce a strictly real result, you have to maintain this symmetry or you will probably end up with lots of complex numbers in your downsampled result, which is likely not what you want. So you can'...


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Note that the result of a DFT is inherently periodic, so the second half of the resulting output vector does not correspond to the highest frequencies, but to the negative frequencies. The highest frequency (Nyquist) is in the middle of the vector, after that you get the most negative frequency and the last element is the smallest negative frequency just ...


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Yes you are right. Aliasing only happens when you sample (either analog to digital conversion or during digital downsampling) a signal and refers to the overlap of signal spectrum due to an inadequate sampling rate; sampling rate below the Nyquist rate. Aliasing cannot happen at the output of a DAC (digital to analog converter). What happens is, if you do ...


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Suppose if your sampling rate is fs, then the corner frequency of the Anti Aliasing filter would be fs/2


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Peter said it all in his comment, but I'll try to explain it here in a bit more detail. Your thinking is correct, but incomplete. Of course, the component at $f=300$ (unit depends on normalization of $t$) will not cause aliasing, so there will be no additional component below Nyquist, but what you forget is that sampling makes the spectrum periodic with ...


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If you do spectral analysis is typically better to use WAY more points then 10 in the time domain Look at your y axis and make sure your graphs are properly scaled. $f=10$ aliases indeed back to $f=0$. All you see is numerical noise and your y-scale is $10^{-15}$. This would be different if you choose a a different phase, cosine instead of sine.


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There are few problems in the code and question. In your question 'given a sine wave at a given frequency for example, how would you proceed to downsample it to a lower frequency' this statement is not right. We are not reducing the frequency of sinewave, we are reducing the sampling frequency with which we sampled it. Second, the code just inserts zeros ...


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For the Hanning window, the zero crossings occur at a much wider bandwidth (trading for lower side-lobe height) compared to a rectangular window (sharper main-lobe width but higher side-lobe). You need to be careful with the requirements regarding the choice of window given the trade off between side-lobe height and main-lobe width.


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