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For a given «visual accuracy», you need to sample the sine at a sufficient number of time-steps per period.
At some point the display pixel density will be to low to render a sine accurately.
For waveform editors (that usually have access to only a finite rate sampled waveform, not a continously defined trigonometric function), I assume that some practical ...
3
OP clarified that the question in the comments as follows:
If we ignore any modulation for now and assume that we are receiving
pure tones plus the band limited noise and we try to improve the SNR
in post processing how much improvement can we expect by oversampling
and is there a limit to it? My original question was about this
aspect.
First consider the ...
2
The OP mentions "demodulating" but I don't actually see that in the process (meaning the phase modulation is not actually removed). Without demodulation the signal can possibly be spread across multiple DFT bins thus reducing the SNR in the bin that was used to measure the phase. If the signal bandwidth is less than one bin, and the noise is ...
1
If you're really generating a pure sine, you know the amplitude and can force the peaks of a half-cycle to draw at the intended peak.
If the waveform is more complex, you can generate it at a sufficiently high sample rate. The issue with drawing is that you can't necessarily plot consecutive samples and still have enough of the waveform in view, horizontally,...
1
Hint: Consider a factor 2 down-sampled sequence : $\tilde{x} = \{x_1, x_3, x_5, ...\}$ and consider another sequence which is circular shifted by $-1$ sample and downsampled by a factor of 2 : $\hat{x} = \{ x_2, x_4, x_6, ...\}$. Observe that the required sequence is difference of $\tilde{x}$ and $\hat{x}$. Each of the DFT can be computed by observation ...
1
In essence filtfilt does the the following
$$y[n] = x[n] \ast h[n] \ast h[-n]$$
It's convolution with the impulse response and then the time reversed impulse response (in any order since the convolution is commutative).
So in the frequency domain it's simply
$$Y(z) = X(z) \cdot H(z) \cdot H'(z) = X(z) \cdot |H(z)|^2 $$
which obviously is zero-phase since $|H(...
1
The trick is to properly compensate for the fact that Frequency Domain multiplication applies a convolution with the circular boundary conditions in the spatial domain.
You may use the following code:
clear('all');
close('all');
gaussianKernelStd = 2;
gaussianKernelRadius = ceil(5 * gaussianKernelStd);
mI = im2double(imread('cameraman.tif'));
mI = ...
1
When referring to the SNR achieved by oversampling, we have to be careful in using the term "SNR". There are essentially two SNR's to consider:
The SNR that is the signal-to-quantization-noise ratio.
The SNR that is your signal-to-noise ratio. Here the noise is produced by your system. For simplicity this can be modeled by the kTB relation that ...
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