Chapter 5, Problem 5.7QAP

Interpretation Introduction

(a)

Interpretation:

The signal-to-noise ratio is to be stated.

Concept introduction:

The signal and noise are two components of instrument measurement. One component is the signal, which carries information that is of interest to scientists and another component is noise, which is the unwanted fluctuations of signals.

Answer to Problem 5.7QAP

The signal-to-noise ratio is $338.74$.

Explanation of Solution

The magnitude of signal is denoted by $S$ and the magnitude of noise is denoted by $N$.

The expression for the mean of the measurements is:

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

Here, the total number of measurement is $N$ and the measurement is $x_i$ where $i=1,2,3,4,\mathrm{...},9$.

Substitute $1.003$ for $x_1$, $1.000$ for $x_2$. $1.001$ for $x_3$, $1.004$ for $x_4$, $1.005$ for $x_5$, $1.006$ for $x_6$, $1.001$ for $x_7$, $0.999$ for $x_8$, $1.007$ for $x_9$, and $9$ for $N$.

$\begin{array}{c}\overline{x}=\frac{\left[\begin{array}{l}1.002+1.000+0.997+1.004+1.003\\ +1.001+1.005+0.999+1.007\end{array}\right]}{9}\\ =\frac{9.018}{9}\\ =1.002\end{array}$

The expression for the standard deviation is:

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

Substitute $1.003$ for $x_1$, $1.000$ for $x_2$. $1.001$ for $x_3$, $1.004$ for $x_4$, $1.005$ for $x_5$, $1.006$ for $x_6$, $1.001$ for $x_7$, $0.999$ for $x_8$, $1.007$ for $x_9$, $1.002$ for $\overline{x}$, and $9$ for $N$.

$\begin{array}{c}s=\sqrt{\frac{\left[\begin{array}{l}{\left(1.003-1.002\right)}^2+{\left(1.000-1.002\right)}^2+{\left(1.001-1.002\right)}^2\\ +{\left(1.004-1.002\right)}^2+{\left(1.005-1.002\right)}^2+{\left(1.001-1.002\right)}^2\\ +{\left(1.006-1.002\right)}^2+{\left(0.999-1.002\right)}^2+{\left(1.007-1.002\right)}^2\end{array}\right]}{9-1}}\\ =\sqrt{\frac{\left[\begin{array}{l}\left({10}^{-6}\right)+\left(4\times {10}^{-6}\right)+\left({10}^{-6}\right)+\left(4\times {10}^{-6}\right)\\ +\left(9\times {10}^{-6}\right)+\left({10}^{-6}\right)+\left(16\times {10}^{-6}\right)+\left(9\times {10}^{-6}\right)+\left(25\times {10}^{-6}\right)\end{array}\right]}{9-1}}\\ =\sqrt{\frac{70\times {10}^{-6}}{8}}\\ =2.958\times {10}^{-3}\end{array}$

The expression for the signal to noise ratio is:

$\frac{S}{N}=\frac{\overline{x}}{s}$ ...... (III)

Substitute $2.5$ for $s$ and $1.002$ for $\overline{x}$ in Equation (III).

$\begin{array}{c}\frac{S}{N}=\frac{1.002}{2.958\times {10}^{-3}}\\ =338.74\end{array}$

Interpretation Introduction

(b)

Interpretation:

The number of measurements required to obtain a $S/N$ of $500$ is to be stated.

Concept introduction:

The signal and noise are two components of instrument measurement. One component is the signal, which carries information that is of interest to scientists and another component is noise, which is the unwanted fluctuations of signals. For $n$ number of repetitions, the signal to noise ratio is proportional to square root of number of repetitions.

Answer to Problem 5.7QAP

The number of measurement required to obtain a $S/N$ of $500$ is $18$.

Explanation of Solution

The expression for $n$ number of repetitions is:

${\left(\frac{S}{N}\right)}_n=\sqrt{n}{\left(\frac{S}{N}\right)}_i$ ...... (IV)

Here, the signal to noise at initial point is ${\left(\frac{S}{N}\right)}_i$.

Substitute $338.74$ for ${\left(\frac{S}{N}\right)}_n$ and $9$ for $N$ in Equation (IV).

$\begin{array}{l}338.74=\sqrt{9}{\left(\frac{S}{N}\right)}_i\\ {\left(\frac{S}{N}\right)}_i=\frac{338.74}{3}\\ {\left(\frac{S}{N}\right)}_i=112.913\end{array}$

Substitute $112.913$ for ${\left(\frac{S}{N}\right)}_i$ and $500$ for ${\left(\frac{S}{N}\right)}_n$ in Equation (IV).

$\begin{array}{l}500=\sqrt{n}\left(112.913\right)\\ n={\left(\frac{500}{112.913}\right)}^2\\ n=19.6\\ n\approx 20\end{array}$