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1

That depends. But typically, yes, you'll flush out your transmit signal; that basically "costs" nearly nothing, but makes sure your spectral mask stays clean. At the receiver, this part still contributes to the amplitude of the matched filter output for the last couple of symbols, so, albeit not contributing very much, omitting it would just introduce an ...


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So the original signal has a spectrum looking like this: So the 40Hz to 50Hz components are also shown at 50Hz to 60Hz. Assuming that the DAC just outputs the raw data without any interpolation (so it's just a zero-order hold between data points), then the spectrum of that signal will look like this: Note that all I've really done is sampled this at 10 $\...


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I know this is very old, and @Matt L. long since gave an excellent and informative answer. I had no idea that total variation denoising existed, so I learned something quite useful. Accordingly, I upvoted both the question and answer and want to give a little something, such as it is, back to the site. The basic idea is to use a simple digital version of the ...


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I just came across one article related to the compand command that might be useful (just in case someone looking for help sees this article). From the article: To test this, and my understanding of compand, I added a very simple filter to remove the quiet section of the audio by decreasing their volume dramatically: ffmpeg -i in.mp3 -filter_complex \ ...


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has some compensation in the signal chain to compensate for the DC spike which would appear at 370MHz the center frequency. Yes, a typical problem of IQ demodulators is the LO leakage/DC spike. You really can't do much about that. You don't mention the SDR you're using, but many can be used with offset tuning, completely moving the LO out of your band of ...


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Very similar to what Marcus Müller suggested, this is what I did: The raw noisy waveform is compared (via comparators) with upper and lower thresholds. The upper threshold is simply a constant (0.8 here) plus the long term average (LTA). Likewise, the lower threshold is the LTA minus 0.8. The value 0.8 was chosen by inspection of the raw waveform. If a ...


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First, the circular option relates to the treatment of the borders of the image. Then, standard image kernels can be any $[r,c]$ matrix. If either $r$ or $c$ is equal to $1$, then this is a very flat $2D$ filter, that acts only across one direction: across lines if horizontal, across columns if vertical (with the transpose). Filtering is a linear operation: ...


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If I'm not mistaken, a column vector will filter the image across its columns, treating each row independently of the others. Likewise, a row vector will filter across rows, treating all columns the same. edit: Regarding an example - consider the simple image [1,1,1;0,0,0;-1,-1,-1]. It's constant along its rows (i.e., all the columns are the same) and a ...


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Approximation by the real part of a weighted sum of separable complex Gaussian component kernels Figure 1. The proposed scheme illustrated as 1-d real convolutions ($*$) and additions ($+$), for cut-off frequency $\omega_c = \pi/4$ and kernel width $N=41$. Each the upper and the lower half of the diagram is equivalent to taking the real part of a 1-d ...


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To preserve the talker's identity for privacy reasons, I would make available for listening a distorted version of the original audio file, while using the original one for real recognition to improve the system. This won't work. The point is that the whole job of your speech recognition algorithm is to figure out which "components" in your audio signal ...


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A classical result states that a (non-trivial) time-limited signal does not have a limited bandwidth (in terms of support). The converse happens as well: a bandwidth-limited signal cannot be of finite length. And there is a similar rule-of-thumb principle: when you try to reduce the support of a signal in one domain (time or frequency), you generally ...


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As MattL already mentioned, they are the same block diagrams as Direct Form II (canonical) implementation of a third order (on the FIR side) LCCDE. However, the adders in the first block diagram are two input adders, while the the ones in the second block diagram are multiple input adders. Theoretically there is no difference. In a popular DSP text, ...


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Sounds like a job for dedicatedly identifying these surges and then actually subtracting them. I'd start with a low-pass filter to find the slightly time-varying mean of the signal. Use that to define lower and upper thresholds above or below you count something as surge. Identify the samples lying outside the thresholds. Find a signal model for surges, e....


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The "alias" frequencies you see appearing above the Nyquist limit $f_s/2$ do not exist in a proper interpretation of the digitised signal; they are appearing due to incorrect interpolation of the samples during reconstruction. If you correctly interpolate the signal from the samples, those frequences will not be there. Remember, a key condition of the ...


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The problem is the definition of the phase. The command angle() computes the phase $\phi(\omega)$ according to $$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$ where $H(e^{j\omega})$ is the complex frequency response. At frequencies $\omega$ where the frequency response has zeros, the phase jumps by $\pi$. This is shown in the two left figures ...


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How comfortable are you with the concept of negative frequencies? Your 5 kHz cosine consists of two lines in the spectrum, one at +5kHz one at -5kHz. If you made your first diagram such that it goes from $-fs/2$ to $+fs/2$, you'd see exactly these two. I find this way of displaying the spectrum much more intuitive. Outside this band, you will see periodic ...


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You can solve your problem by an fftshift of the impulse response, as this line shows: h = fftshift( real( ifft(H)) ); now it should work. NOTE that there will be a transient of length about $60$ samples (half the linear phase FIR filter length) at the beginning of the convolution output, after which the steady state result emerges...


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If you can generate a grey noise signal to put through the filter, the amplitude can be plotted as the resulting graph. Also if you pitche a sine input to the function at every frequency that you wish to measure for, the sine's amplitude after the filter will be the result for every frequency. I understand that i in this graph is the audio sample instance......


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Since you said you wanted the frequency range to be from 1 $\mu$Hz to 10 Hz, I ran my simulations with 20 Hz sampling frequency and did 100 32k FFT PSDs. My results looked what you have. So I lowered the sampling frequency to 0.2 Hz. The sims started at 0 s, ended at 163835 s, with $\Delta$t = 5 s. The WGN was simply unit normal. My results are shown below: ...


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