Hot answers tagged

12

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, $...


11

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 ...


11

Some issues here: Your SNR formula only applies to full scale sine waves, your sine wave has -6dB amplitude so your SNR will be 6 dB lower The formula also implies rounding, not truncation, that's another 6 dB You use a frequency that's a small integer divider of the sample rate, that means you are just repeating the same samples over and over again and don'...


9

Because each step in the processing chain is linear we consider a case with only noise and no coherent signal. Denote the noise $\xi(t)$. The $I$ and $Q$ signals are \begin{align}\ I(t) &= \xi(t) \cos(\Omega t) \\ Q(t) &= - \xi(t) \sin(\Omega t) \, . \end{align} We express the effect of the filter as a convolution with the time response function $h$, ...


8

SNR stands for Signal to Noise Ratio. It is a ratio and as such does not have any units, it describes the proportion of signal to undesired noise. There is no single correct measure of SNR, it differs depending on the application. In the equation you have given, the SNR is broken down in the following way: 1) Calculate the power ratio $$\frac{\sigma_s^2}{\...


8

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 ($X$), ...


7

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 ...


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

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.


6

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 ...


6

I was doing quite a bit wrong, but the key thing that I was missing was the fact that the SNR needs to be calculated over the whole Nyquist spectrum instead of only looking at the peaks. This article explains everything very well: Taking the Mystery out of the Infamous Formula, "SNR = 6.02N + 1.76dB," and Why You Should Care. Another issue was that ...


5

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 ...


5

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 ...


5

$\mathsf{SNR}$ (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 ...


5

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 ...


5

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 ...


4

If you define the SNR as the ratio of the signal power and the noise power in dB, you have $$SNR_{dB}=10\log \left(\frac{P_s}{P_w}\right)\tag{1}$$ where $P_s$ is the power of the desired signal and $P_w$ is the noise poiwer. If the noise $w$ has a mean of zero, then $P_w=\sigma^2_w=1$. From (1) (with $P_w=1)$ you get the desired value of $P_s$ for a given ...


4

The multiplication of the two noisy signal gives $$(x_1+n_1)(x_2+n_2)=x_1x_2+x_1n_2+x_2n_1+n_1n_2=x+n\tag{1}$$ with the desired signal $$x=x_1x_2\tag{1}$$ and the noise part $$n=x_1n_2+x_2n_1+n_1n_2\tag{2}$$ Assuming all signals are independent of each other and have zero mean, we get for the signal power $$\sigma^2_{x}=\sigma_{x_1}^2\sigma_{x_2}^2\...


4

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, ...


4

It's often said that pulse compression gives you a gain proportional to the time-bandwidth product (otherwise known as the pulse compression ratio, or $PCR$). This is a really misleading statement, and it had me confused enough to sit down and think about it for awhile. I thought I'd share some of my findings that I pieced together from both reading the ...


4

The noise power continues to be $N_0/2$, independent of bandwidth. The reason is that the noise variance at the output of a filter with frequency response $H(f)$ is $$\sigma_n^2=\int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,df.$$ (This is a direct application of the Wiener-Khinchin theorem). Since you're assuming an orhonormal basis, the noise power is always ...


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 $10\log_{10}(4)$.


4

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

The answer is yes but one has to specify $B_n$ properly to avoide possible confusions. In case if one uses a pulse compression, the bandwidth through which the receiver collects the noise will normally be $B_n = \beta_c$. Then, the "new" signal-to-noise ratio should be written as: $SNR = \dfrac{P_TG_TG_R\lambda^2\sigma{P_g}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \...


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 ...


4

The SNR is simply the mean-square of a demodulated symbol divided by the variance of the signal, or in dB: $$\DeclareMathOperator{\SNR}{SNR}\SNR =10\log_{10}(\mu^2/\sigma^2)$$ $$=20\log_{10}(\mu/\sigma)$$ A typical metric for this in receiver equipment is the Error Vector Magnitude (EVM) where an "error vector" is the Euclidean distance from the ...


3

Your calculation of the SNR is right and the DFT Bin Resolution is also OK. One thing you're missing is the effective resolution due to "Windowing" effect. The DFT of finite number of samples is interpolated by a Dirichlet Kernel (Like Sinc). It means you resolution is also limited by the main lobe width of the Sinc which is proportional to the inverse of ...


3

The SNR is the ratio of the powers. Note that for zero-mean signals, the variance equals the power, so the SNR is the ratio of the variances of signal and noise, but only if both are zero-mean.


3

If you want a complex white noise sample with variance $N_0$, then you should generate two independent noise samples each with variance $\frac{N_0}{2}$. The two samples that you generate make up the real and imaginary parts of the desired complex noise sample. The resulting complex value will have the desired variance $N_0$. You should then sum this with ...


Only top voted, non community-wiki answers of a minimum length are eligible