Question

The following data were collected for a voltage measurement, in mV a noisy system: 1.25, 1.74, 1.33, 1.97, 1.10, 1.27, 1.54, 1.62.

(a) Assuming the noise is random, what is the signal-to-noise ratio?

(b) How many measurements would have to be averaged to increase S / N to 10 ?

Short answer

The signal-to-noise ratio is 5.1

Step-by-step solution

Step 1. Given inforamtion

The following data were collected for a voltage measurement, in mV a noisy system: 1.25, 1.74, 1.33, 1.97, 1.10, 1.27, 1.54, 1.62.

Part (a): step 1. Explanation 

The expression for the mean of the measurements is

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^N{x}_l}}{N}$

$\begin{array}{c}\overline{x}=\frac{\left[\begin{array}{l}1.25+1.74+1.33+1.97\\ +1.10+1.27+1.54+1.62\end{array}\right]}{8}\\ =\frac{11.82}{8}\\ =1.47\end{array}$

The expression for the standard deviation is:

$s=\sqrt{\frac{{\displaystyle \sum _{i=1}^N{\left(x_i-\overline{x}\right)}^2}}{N-1}}$

$\begin{array}{c}s=\sqrt{\frac{\left[\begin{array}{l}{\left(1.25-1.47\right)}^2+{\left(1.74-1.47\right)}^2+{\left(1.33-1.47\right)}^2+{\left(1.97-1.47\right)}^2\\ +{\left(1.10-1.47\right)}^2+{\left(1.27-1.47\right)}^2+{\left(1.54-1.47\right)}^2+{\left(1.62-1.47\right)}^2\end{array}\right]}{8-1}}\\ =\sqrt{\frac{\left[\left(0.0484\right)+\left(0.0729\right)+\left(0.0196\right)+\left(0.25\right)+\left(0.137\right)+\left(0.04\right)+\left(0.0049\right)+\left(0.0225\right)\right]}{7}}\\ =\sqrt{\frac{0.595}{7}}\\ =0.291\end{array}$

The expression for the signal to noise ratio is:

$\frac{S}{N}=\frac{\overline{x}}{s}$

$\begin{array}{c}\frac{S}{N}=\frac{1.47}{0.291}\\ =5.05\end{array}$