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- Statistics - Discussion

Sign to-commotion proportion (contracted SNR) is a measure utilized as a part of science and designing that analyzes the level of a coveted sign to the level of foundation clamor. It is characterized as the proportion of sign energy to the clamor power, regularly communicated in decibels. A proportion higher than 1:1 (more prominent than 0 dB) shows more flag than clamor. While SNR is regularly cited for electrical signs, it can be connected to any type of sign, (for example, isotope levels in an ice center or biochemical motioning between cells).

Signal-to-noise ratio is defined as the ratio of the power of a signal (meaningful information) and the power of background noise (unwanted signal):

${SNR = \frac{P_{signal}}{P_{noise}}}$

If the variance of the signal and noise are known, and the signal is zero:

${SNR = \frac{\sigma^2_{signal}}{\sigma^2_{noise}}}$

If the signal and the noise are measured across the same impedance, then the SNR can be obtained by calculating the square of the amplitude ratio:

${SNR = \frac{P_{signal}}{P_{noise}} = {(\frac{A_{signal}}{A_{noise}})}^2} $

Where A is root mean square (RMS) amplitude (for example, RMS voltage).

Because many signals have a very wide dynamic range, signals are often expressed using the logarithmic decibel scale. Based upon the definition of decibel, signal and noise may be expressed in decibels (dB) as

${P_{signal,dB} = 10log_{10}(P_{signal})} $

and

${P_{noise,dB} = 10log_{10}(P_{noise})} $

In a similar manner, SNR may be expressed in decibels as

${SNR_{dB} = 10log_{10}(SNR)} $

Using the definition of SNR

${SNR_{dB} = 10log_{10}(\frac{P_{signal}}{P_{noise}})} $

Using the quotient rule for logarithms

${10log_{10}(\frac{P_{signal}}{P_{noise}}) = 10log_{10}(P_{signal}) - 10log_{10}(P_{noise})} $

Substituting the definitions of SNR, signal, and noise in decibels into the above equation results in an important formula for calculating the signal to noise ratio in decibels, when the signal and noise are also in decibels:

${SNR_{dB} = P_{signal,dB} - P_{noise,dB}} $

In the above formula, P is measured in units of power, such as Watts or mill watts, and signal-to-noise ratio is a pure number.

However, when the signal and noise are measured in Volts or Amperes, which are measures of amplitudes, they must be squared to be proportionate to power as shown below:

${SNR_{dB} = 10log_{10}[{(\frac{A_{signal}}{A_{noise}})}^2] \\[7pt]
= 20log_{10}(\frac{A_{signal}}{A_{noise}}) \\[7pt]
= A_{signal,dB} - A_{noise,dB}} $

**Problem Statement:**

Compute the SNR of a 2.5 kHz sinusoid sampled at 48 kHz. Add white noise with standard deviation 0.001. Set the random number generator to the default settings for reproducible results.

**Solution:**

${ F_i = 2500; F_s = 48e3; N = 1024; \\[7pt]
x = sin(2 \times pi \times \frac{F_i}{F_s} \times (1:N)) + 0.001 \times randn(1,N); \\[7pt]
SNR = snr(x,Fs) \\[7pt]
SNR = 57.7103}$

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