# Tag Info

10

A discrete signal is often interpreted as an analog voltage signal. The signal x(n) should be considered as a sampled version of x(t) with a unknown sampling frequency. Q1: So when we're talking about the energy of the signal, it is presumed that the voltage signal is obtained in a circuit with an 1Ω resistor in series, so that v(t)^2/R = P. Q2: Signal-...

10

The signal voltage is a sinusoid signal variant from $-V$ to $+V$, thus the RMS value $$V_{rms} = \frac{V}{\sqrt{2}} = \frac{2^N \Delta V / 2}{\sqrt{2}}$$ because $2V = 2^N\Delta V$ where $\Delta V$ is quantization step. Quantization noise is modeled as uniform random variable in $[-\Delta V /2, +\Delta V /2]$ thus its standard deviation is $V_n = \Delta ... 9 SNR is the ratio of the (mean) power of two independent signals, one called "signal" and the other "noise". If the deterministic signal is periodic, then its power is defined as energy per period$E_s / T_s$. SNR is normally expressed in dB:$SNR_{dB} = 10 \log P_s/P_w$. In your particular example, 20 dB means that the signal has 100 times the power of ... 8$E_1=E_{10}sin (\omega t)E_2=E_{20}sin (\omega t + \delta)E_{\theta} = E_1 + E_2 = E_{\theta0}sin (\omega t + \phi )$This can be described as the figure below: Now given the$E_{\theta 0}$, you can rotate$E_1$and$E_2$arbitrarily as long as$E_1$,$E_2$, and$E_{\theta 0}$form the triangular. As a result, your given sine wave can be ... 8 Two exact sine waves of the same frequency but different phases sum to another exact sine wave. That resultant sine wave can be decomposed into an infinite numbers of pairs (or any other greater number) of sine waves of the same frequency, not just the two original ones. Thus more information is required to reduce the possible solution space below infinite.... 7 The common definition of SNR is the power of the wanted signal divided by the noise power. Suppose you have obtained the wanted and the noise signal as arrays, calculation of the SNR in Matlab before noise reduction can be done like this: snr_before = mean( signal .^ 2 ) / mean( noise .^ 2 ); snr_before_db = 10 * log10( snr_before ) % in dB After noise ... 7 If the spectrum of the signal and the spectrum of the noise do not overlap, you might be able to integrate or sum the energy in each of the two frequency ranges, and take the ratio. 7 The quantization error is the quantization noise. It is two different terms for the same thing. 7 When you say that the "information content may remain the same," do you mean the information in the total signal, or the information of the desired signal? Hopefully this will answer both cases. I know Shannon entropy much better than Kolmogorov so I'll use that, but hopefully the logic will translate. Let's say$X = S + N$is your total signal ($X$), ... 6 It can be a little tricky. If your sine wave happens to fall at an FFT bin center, things are a little easier. If your sine wave happens to not fall at a bin center, you have to consider spectral leakage. You'll also need to consider how your window function affects your signal. A simple technique to estimate the signal power would be to sum the ... 6 The measurement that is important to you will depend on the application. If you are looking for an over all measurement of all signal power to noise power then you define signal to be the power in all signal bands and the noise all of the powers in the noise bands. However, if you are trying to use an sub-band adaptive filter to correct for some sort of ... 6 If you can get input signal when it doesn't contain any useful signal (I mean only noise is presented), you can estimate average noise power at first. Simply find a power of such a signal:$P_n=1/N \cdot \displaystyle\sum_{k=0}^{N-1}|s(k)|^2$. Choose$N$in$2^{12}\ldots 2^{15}$range for example. Then you can measure a power of$signal+noise$mixture,$...

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Short answer 10*log(bw/fs) to take into account the oversampling operation because the awgn() function specifies the signal-to-noise ratio per sample, in dB. Longer answer The discrete time AWGN model is $$Y = X+N$$ where X is data from continuous time $X(t)$, N is noise sequence from AWGN process $N(t)$ and Y is receive symbols. If $X(t)$ is ...

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I will show how to calculate the SNR for the case of $N=2$ measurements; it is easy to extend the result to general $N$. Assume a signal $s(t)$ has power $S$, and the noise $n(t)$ has variance $\sigma^2$ and zero mean. Then, the signal $s(t)+n(t)$ has SNR equal to $S/\sigma^2$. Now assume you observe $s(t)$ twice, each time with different, uncorrelated ...

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Let's consider a received signal $$Y(t)=Ap(t)+N(t)\tag{1}$$ where $A$ is the information symbol (modeled as a random variable), $p(t)$ is the transmit pulse, and $N(t)$ is additive white Gaussian noise (AWGN, modeled as a random process). For the sake of simplicity let's assume that all signals are real-valued (baseband case). Filtering with an LTI system ...

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These stuffs $E_s/N_0, E_b/N_0 \textrm{ and SNR}$ are convertible. \begin{align} E_s/N_0 &= E_b/N_0 + 10\log_{10}(k) \\ E_s/N_0 &= 10\log_{10}(T_{sym}/T_{samp}) + \mathrm{SNR} \end{align} where $k$ is the number of information bits per symbol, $T_{sym}$ is the signal's symbol period and $T_{samp}$ is the signal's sampling period. More details can ...

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I am not really sure what you want to do so please comment and I will modify my answer. The case that $x$ and $u$ seems statistically identical is that $x$ is standard Gaussian random variable and $u$ has $\pm 1$ values. Use your notations, $x$ is Gaussian random variable with zero mean and unit variance (deducted from the fact that you used the standard ...

4

So far you seemed to identify snoring through the presence of a strong periodic pattern in the audio. If you tell me that there might be other sources with the same property, it's time to move on and focus on a property of the signal more specific of snoring ; and I would suggest to look more precisely into the timbre of the recorded sound itself. Given that ...

4

It's not practical to come up with a comprehensive list of definitions of signal-to-noise ratio, because different measures are relevant in different applications. Here are a few common measures that I've come across and/or used in the past (you'll find that they are bent toward communications applications): "Pure" SNR: I call this "pure" because it is a ...

4

Question 1) Yes, you must know what the right answer is before-hand in order to judge the receiver's performance. I am doing this right now on a project. I generate data, scramble it, modulate it, add noise and carrier offset, etc., then I give it to the receiver. The bits I started out with are my "yes/no" labels. Question 2) All such measurements are ...

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No, the data in the USB cable is digital. Namely there is either an error in the data or the data is the same at any place along the cable.

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So, let's not forget what SNR is: it's a relation of powers present. The thing that improves SNR is a propoer low-pass: it leaves the signal power alone and reduces the power of the noise. An ideal low-pass filter will leave zero noise outside its specified bandwidth – so it doesn't matter whether you "cut off" these bandwidths using decimation or not, ...

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$\mathsf{SNR}$ (signal-to-noise ratio) is a generic term whose value can be defined in different ways by different people, and as long as one states clearly what is meant by $\mathsf{SNR}$ in a particular document, there is no confusion. Thus, there is no "arriving" at the formula $\mathsf{SNR} = \frac{E[y^2]}{\sigma_w^2}$ at all: it is the definition of ...

4

This is to take into account the oversampling operation. Symbol time = sample time implies no oversampling. See AWGN model for more details about the conversion among EbN0, EsN0 and SNR. For example, if a complex baseband signal is oversampled by a factor of 4, then EsNo exceeds the corresponding SNR by $10\log_{10}(4)$.

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Let's start by fixing a symbol rate $R_s$ symbols per second. To modulate $R_s$ symbols per second without ISI, Nyquist says that we need a bandwidth at least $BW_0=1/R_s$ Hz. With spread spectrum, we use more bandwidth, say $M$ times, than what Nyquist advised. The used bandwidth is $BW=M\times BW_0=M/R_s$. In LoRa specifications, $BW=125, 250, 500 \... 4 Decimating a signal (selecting every Dth sample and discarding the rest) does not distort the signal within the passband in any way other than to cause aliases from higher frequencies to fold into the signal bandwidth. Depending on how we model the system the phase may be effected since$z^{-n}$is replaced with$z^{-n/D}$, but the phase will still be ... 4 The phase of the sinusoid does not matter: A phase shift of a sinusoid is equivalent to shifting it in time, which results in a time shift of both the quantized sinusoid and the quantization error. The power spectrum is invariant to time shifts. We choose to work with sinusoid$A\cos(x)$. Optimality by def. 1 Equivalently to maximizing signal-to-noise ... 3 The noise seems to be irrelevant to this problem since you say that it is a frequency band far removed from the sinusoids of interest. Thus, with a relatively simple bandpass filter, you should be able to recover a signal $$s(t) = C\cos(2\pi f_c t + \theta) = A\cos(2\pi f_ct + \theta_1) + B\cos(2\pi f_ct + \theta_2), \tag{1}$$ where$C$and$\theta$are ... 3 There are many ways to add two sine waves of the same frequency and different phase and get back a sine wave. These cant be separated without more information. 3 Why it is plotting against only 500hz and what information I can get on this?? Plotting against$500Hz$: Because that is the frequency of the sampled signal. We usually sample a signal at least two times of its max frequency in order to avoid aliasing. In your case$fs=1000Hz$(sampling frequency), so max frequency is (no higher than)$500Hz\$. The ...

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