Chapter 20, Problem 20.35P

Interpretation Introduction

Interpretation:

The expected detection limits for given an ideal experiment has to be discussed.

Explanation of Solution

The limit for ${\text{O}}_{\text{2}}$ is relative detection air from photoacoustic spectroscopy decreases with increasing signal averaging time.  The more period of times consumed averaging out noise, the lower the detection limit, up to an averaging time of $\sim \text{80 s}$.

The ratio of signal to noise is increased in proportion to the square root of the number of cycles that may be averaged.

$\begin{array}{l}\text{Number}\text{of}\text{cycles=n}\sqrt{\text{n}}\frac{\text{S}}{\text{N}}\text{ratio}\\ \text{1}\text{1}\text{º1}\\ \text{100}\text{10}\text{10}\\ \text{300}\text{17}\text{.32}\text{17}\text{.32}\\ \text{1000}\text{31}\text{.62}\text{31}\text{.62}\\ \end{array}$

We get average of 1000 cycles is observed when the $\frac{\text{S}}{\text{N}}\text{ratio}$ is 60 the calculated the $\frac{\text{S}}{\text{N}}\text{ratio}$ for other experiment can be listed below

$\begin{array}{l}\text{Number}\text{of}\text{cycles=n}\sqrt{\text{n}}\frac{\text{S}}{\text{N}}\text{ratio}\\ \text{1000}\text{60}\text{60}\\ \text{300}\text{60}\left(\frac{\text{17}\text{.32}}{\text{31}\text{.62}}\right)\text{=32}\text{.9}\text{35}\text{.9}\\ \text{100}\text{60}\left(\frac{\text{17}\text{.32}}{\text{31}\text{.62}}\right)\text{=19}\text{20}\text{.9}\\ \text{1}\text{60}\left(\frac{\text{17}\text{.32}}{\text{31}\text{.62}}\right)\text{=}\text{1}\text{.9}\text{1}\text{.95}\\ \end{array}$

The graph is plotted averaging time and relative detection limit.

In this graph measure the absorption of the ${\text{O}}_{\text{2}}$

Figure 1

Conclusion

The expected detection limits for given an ideal experiment has to be discussed.