3

I can't simply just add Asn1 and Asn2 to calculate the total noise, because some signal samples will have opposite phases. On the contrary! Uncorrelated noise is the only thing where adding the noise simply leads to noise with twice the power. So, yes, you simply add the noises, and if you're doing it right, then the resulting noise will have variance ...


3

Given that RMS means Root Mean Square the answer is rather obvious: it's a root power quantity. The RMS of a signal has the same units that the signal itself. Saying that "RMS is proportional to power" is just sloppy phrasing. Correct would be "the square of the RMS is proportional to power". If you by "power" you mean "...


3

The z-transform is the discrete version of the Laplace transform and in both cases z and s are the set of all complex numbers, and as such we map with the transform the time domain function into the domain of complex frequencies; signals that change in rotation only which is the Fourier Transform and in addition to that such signals that can grow and decay ...


3

Why does repeating periodically a signal means removing some frequency intervals from the single pulse spectrum? Because your signal is periodic. A periodic signal only contains frequencies that are integer multiples of the signal's repetition frequency. In other words, if the signal is periodic all of it's constituent parts must be periodic as well. In ...


2

The slope of the lowpass is 20dB/decade or 6dB/octave, that means it's a simple first order lowpass filter. At the corner frequency the gain is -3dB. Same of the phase. It goes from 0 to -90, so it's a first order filter. At the corner frequency, it's 45 degrees. Looking at both phase and level we conclude that it's a first order lowpass with $f_c = 1000 rad/...


1

The figure means to illustrate a consequence of overfitting: if we fit train data too well, we'll fit noise too, and thus generalize poorer. "Noise > regularity" can be thought of as low SNR (signal to noise ratio). But here comes a flaw: there need not be noise for overfitting to occur. We can imagine a sine-fitting algorithm where we fed a ...


1

I didn't know the convolution theorem for the DHT before, but it's pretty clear that if it exists, it must be about circular convolution, just like for the DFT. You're comparing that with acyclic convolution, so the results differing is no surprise.


1

If the goal is to get a metric for similarity, consider using direct correlation rather than mapping to the frequency domain to then perform a comparison (which I would then again recommend from that domain to also do as a subsequent direct correlation computation). Assuming no time-shift, the Pearson Correlation Coefficient is ideal for this application in ...


1

10log(165) - 10log(1) is wrong; 1W is reflected; remember, subtraction of logarithm yields the same as division of numbers the logarithm is taken from. By the way, log(1) is always 0, so this should be a giveaway that this operation can't be right. So, 1.3 dB = 1.3489 W no. Definitily not. 1.3 dB is a unitless factor. You need to refresh what a decibel is,...


1

At 8kHz sample rate, an 8192 point FFT would give you about 1 Hz resolution. In order to reduce spectral leakage you may want to go larger than this but this will also make it very slow. Depending on your application, it may be a lot easier to just use a time domain bandpass filter. Since your sample rate is way higher than your frequency range of interest ...


1

One practical and traditionally used approach when the modulated waveform itself does not need to extend to $f=0$ (as in the case of the OP's wanting the transmit $f+f_\Delta$ is to use a PLL, or even an FLL as depicted in the block diagram below, such that the loop bandwidth of the loop is lower than the desired modulation frequency. The modulation signal ...


1

Okay, this function $$x(t)=\sum\limits_{n=-\infty}^{\infty}\operatorname{rect}(t-n)$$ is identically equal to one: $x(t) = 1 \quad \forall t\in\mathbb{R}$ because $$ \operatorname{rect}(x) \triangleq \begin{cases} 1,\quad & |x|<\tfrac12 \\ \tfrac12, \quad & |x|=\tfrac12 \\ 0, \quad & |x|>\tfrac12 \\ \end{cases} $$ Now, if these rectangular ...


1

Many years later I was asked to write some of what I've learned on this. The short answer is that it depends on what the term "subharmonics" means - things are very different if we're looking at bilateral signals or unilateral ones. So this is a partial answer In a very strict sense it's not possible to generate subharmonics using any time-...


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