## FANDOM

133 Pages

Real world signals are analogue signals. Often we want to work with them in the digital domain, and then output the results back to the real world. To do this, we firstly need a method of creating a digital representation of the analogue signal we are interested in.

Analogue to Digital (A/D) conversion is the process of creating a digital representation of an analogue signal. There is a problem with this: analogue signals can take on any value, but digital signals are limited in the number of values they can take. Specifically, if a digital signal is represented by $N$ bits then it can only take on $2^{N}$ values - i.e. digital signals are quantised.

Electrical engineers get around this by sampling the analogue signal, and then rounding it to the nearest digital 'step', where the step size is given by

$\Delta V=\frac{V_{pp}}{2^{N}}$

where $V_{pp}$ is the peak-to-peak voltage of the desired signal. Evidently this results in a maximum quantisation error of half the step size:

$|e_{max}|=\frac{\Delta V}{2}$

It is generally assumed that $e$ is uniformly distributed on either side of each step (i.e. it can take on any value in this interval with equal probability $\frac{1}{\Delta V}$ ).

$e\sim U\left(-\frac{\Delta V}{2},\,+\frac{\Delta V}{2}\right)$

We can then calculate the variance of this error as

$\sigma_{e}^{2}=\frac{1}{\Delta V}\int\limits _{-\frac{\Delta V}{2}}^{\frac{\Delta V}{2}}e^{2}de=\frac{(\Delta V)^{2}}{12}=P_{noise}$

This variance is also known as the quantisation noise, because it gives an idea of how much noise is present in the signal due to the rounding errors. We define the signal to quantisation noise power ratio of an A/D converter as

$SQNR=10\log\left(\frac{P_{signal}}{P_{noise}}\right)$

Where $P_{signal}$ is the power of the analogue signal being converted, and $SQNR$ is the signal to quantisation noise power ratio measured in dB. For a simple sine wave with amplitude A (i.e. $V_{pp}=2A$ , $P_{signal}=\frac{A^{2}}{2}$ )

$SQNR=10\log\left(\frac{\frac{A^{2}}{2}}{\frac{(\Delta V)^{2}}{12}}\right) =10\log\left(\frac{\frac{A^{2}}{2}}{\frac{(2A/2^{N})^{2}}{12}}\right) =10\log\left(\frac{3\times2^{N}}{2}\right)$

i.e. for a simple sine wave,

Failed to parse (unknown function\boxed): \boxed{SQNR\approx6.02N+1.76\, dB}

where N is again the number of bits used to represent our digital signal.

The SQNR tells us the relative composition of our digital signal - how much of it is the analogue signal we're actually interested in, and how much of it is noise. Evidently a higher SQNR is better because it means more of the signal is getting through. To achieve this we simply represent our digital signal with more bits, thereby giving a higher resolution and minimising the quantisation error.

Credits to the lecture notes of Prof. E. Ambikairajah.