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It is actually not distorted, it is sampled at high enough rate. What fools you is the straight lines drawn between sample points, it gives you a false impression of the waveform. It shows you a linear interpolation of the signal. It does not represent how the signal would actually look like. A sampled signal exists only at the sample points, and to convert ...
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The actual requirement is to sample at GREATER then twice the bandwidth, not at a rate equal to it...
So only your 80Hz same set actually meets the requirement, because the 60Hz case is ambiguous in general, consider if you were sampling sin (2PiFt) instead then you would get a flat line at zero amplitude.... And changing the angle between sin and cos would ...
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There is no aliasing as 𝑓 = 30 Hz is less than or equal to the folding frequency, 30 Hz and 40 Hz, respectively.
Yes and no. There isn't significant aliasing when you're sampling at 80Hz, because the resulting signal has frequency components at 30Hz and 50Hz. The result is thus unambiguous as long as you take that 50Hz signal into account.
There is ...
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So the point is that "classically", communications theory tends to be done in complex baseband, i.e. signals centered around 0 Hz, but not necessarily symmetric in spectrum.
When you want to represent such signals, you need complex values. When you want to transmit such signals over the air, you need to:
Convert them from digital to analog,
mix the I(...
4
SOLUTION
Bottom Line
$$(\theta_2-\theta_1) = 2\pi f(T_2-T_1)n -(\phi_2[n]-\phi_1[n]) \tag{1}$$
$f$: frequency in Hz of two tones of the same frequency and fixed phase offset
$(\theta_2-\theta_1)$: phase difference in radians of tones being sampled
$T_1$: period of sampling clock 1 with sampling rate $f_{s1}$ in seconds
$T_2$: period of sampling clock 2 ...
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From an ADC perspective, it is just taking a sample of the voltage in time. I fail to see how a "misinterpretation" could be made since there is no "turning car wheel" to take pictures of at the wrong time. Do the harmonics alias in such a way that the wave shape is preserved?
You can reason this out yourself, in the time domain. Consider a square wave ...
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Sampling is the process of making the x-axis (time) discrete and quantization is the process of making the y-axis (magnitude) discrete. You can sample without quantization (such as done with an analog sample and hold circuit). Quantization is introduced through rounding or truncation when the sampled analog signal is mapped to a digital representation.
...
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Remembering from my 1970 Signal Processing lectures we have ...
The crucial thing is the filter used to reconstruct the signal. Let's do the theory first for ideal sampling a perfect sine wave at 2x its frequency and filtering with an ideal low pass filter.
The samples are infinitely thin - they are delta functions separated by time t.
The filter is an ...
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Since this is a pure sinusoid, it has a bandwidth of 0 Hz. You can multiply it by a carrier signal of the same frequency, pass it through a low pass filter then take only a few samples. What matters is NOT the frequency of the signal, rather the bandwidth. Consider for example a voice signal modulating a 1 GHz carrier. It will be very costly, to sample this ...
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sampling guarantees periodicity
Not exactly. Sampling in one domain guarantees periodicity in the other domain. So sampling in time creates a periodic spectrum and sampling in frequency creates a periodic time signal. Or the other one around: a periodic time signal has a discrete spectrum and a periodic spectrum has a discrete time signal.
That's why ...
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View your frequency response after your low pass filter on a dB scale to better show the limitations of your filter.
Use a multiband filter with the least squares algorithm for an optimized rejection filter for zero-fill interpolation. This will concentrate the rejection to be specifically where the images are that need to be removed.
Given your original ...
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As long the mean sampling frequency fulfills the Nyquist criteria, you're good to go. So, for well behaved distributions with known mean, you won't need any extra math, aside from the required special handling of the non-uniform samples themselves.
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For the first part of your question, perhaps this will shed some light:
Phase difference measurement of a signal sampled with two different sampling frequencies
The answer to your second part of your question is yes for a single pure tone. It will appear as a lower frequency alias in the DFT, but if you know the actual frequency range, you can calculate ...
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Square waves in a practical (and analog) video signal should always be bandlimited. May be they seem infinetely sharp at first, but if you zoom in you would see that their edges are actually rounded, indicating bandlimitedness.
So if you use high enough sampling rate then you will avoid aliasing without an anti-aliasing filter. However, for a bandlimited ...
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The nyquist criteria says that sampling rate should be at least 2 times the highest frequency present in the baseband signal. If you are lucky the samples of the given sine wave signal could be aligned at quarter of T i.e. T/4 and 3T/4 (with T/2 second interval between samples) and you get accurate representatIon of the given continuous time signal. For any ...
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