Charles daEngineer

Base Number Conversion

Published on Thu Oct 20 2022

Introduction

This post is a derivation of equation 2 at wolfram and the equation at the bottom of page 2 in this paper. I have only seen these equations used for determining nth digits of $π$ . I would like to share the more general derivation, for explicitly determining any nth digit of any integer base number.

Proof

The equation for a base numbering system is: $x_{b} = \sum_{k = M}^{N} a_{k} b^{k}$ Where $M$ is the position of the digit to the furthest right side of the number and $N$ is the position of the digit to the furthest left side of the number.

A desired $a_{i}$ can be found with the following steps: $\frac{x_{b}}{b^{i + 1}} = \sum_{k = i + 2}^{N} a_{k} b^{k - i - 1} + a_{i + 1} + a_{i} b^{- 1} + \sum_{k = M}^{i - 1} a_{k} b^{k - i - 1}$ ${\frac{x_{b}}{b^{i + 1}}} = a_{i} b^{- 1} + \sum_{k = M}^{i - 1} a_{k} b^{k - i - 1}$ $b {\frac{x_{b}}{b^{i + 1}}} = a_{i} + \sum_{k = M}^{i - 1} a_{k} b^{k - i}$ $\begin{matrix} (1) & ⌊ b {\frac{x_{b}}{b^{i + 1}}} ⌋ = a_{i} \end{matrix}$ Where ${\frac{x_{b}}{b^{i + 1}}}$ is the fractional part of $\frac{x_{b}}{b^{i + 1}}$ and $⌊ \sum_{k = M}^{i - 1} a_{k} b^{k - i} ⌋ = 0$ , this occurs because the maximum value of $a_{k} = b - 1$ so: $\sum_{k = M}^{i - 1} a_{k} b^{k - i} \leq (b - 1) \sum_{k = M}^{i - 1} (b^{k - i})$ and: $(b - 1) \sum_{k = M}^{i - 1} (b^{k - i}) = 1 - b^{M - i}$ If the number I am trying to convert is not something like $0. \overset{―}{9}$ and for measuring real world signals this is okay so I can say $⌊ 1 - b^{M - i} ⌋ = 0$

Conclusion

One example of equation $1$ is going from a base 10 to a base 2 number. The procedure is to divide the base 10 number by 2 $i$ times then take the floor, if it's even then $a_{i} = 0$ if it's odd then $a_{i} = 1$ .

An easy example is converting 10.5. 10 is even so $a_{0} = 0$ . $10.5 / 2 = 5.25$ 5 is odd so $a_{1} = 1$ . $10.5 / 2^{2} = 2.625$ 2 is even so $a_{2} = 0$ . $10.5 / 2^{3} = 1.3125$ 1 is odd so $a_{3} = 1$ continuing to divide by 2 and taking the floor always results in 0. Almost done; $10.5 * 2 = 21$ ; 21 is odd so $a_{- 1} = 1$ . Now continuing to multiply by two gets us an even number everytime. So we have: ${10.5}_{10} = {1010.1}_{2}$

Applied to an ADC

Using these results to implement an ADC is now straight forward using spice or eesim.dev:

ADC test

.func frac(x)=x-floor(x)
.func A2D(ith)=floor(2*frac(V(1)/2^(ith+1)))

B0 2 0 V= A2D(0)
B1 3 0 V= A2D(1)
B2 4 0 V= A2D(2)
B3 5 0 V= A2D(3)
B4 6 0 V= A2D(4)
Bsum 99 0 V = V(2)*2^0+V(3)*2^1+V(4)*2^2+V(5)*2^3+V(6)*2^4
* DC source for voltage reference
Vs    1 0       dc 0

* DC test
.dc Vs 0 50 0.1

.end

Here I have implemented a 5 bit adc using ngspice's nonlinear element "B"; it allows me to define the voltage which I set to the function in equation $1$ . "Bsum" converts the digital signal to an analog one; using both Vs and Bsum I can see how accurate a 5bit ADC is.

Lastly, notice that this is a direct conversion from a base 10 number to a base 2 number. However; in most ADC applications we really want to "bin" values that are between a range. So, lets do this experiment again where the values all land between V_max and V_min and $V_{m a x} = 5 V$ and $V_{m i n} = 0$ . This can be done by using the binary values that represent less than 1V so:

ADC test
.param Vmax=5 Vmin=0 nth=5
.func frac(x)=x-floor(x)
.func A2D(ith)=floor(2*frac(V(1)/Vmax/2^(ith+1)))
B0 2 0 V= A2D(0-nth)
B1 3 0 V= A2D(1-nth)
B2 4 0 V= A2D(2-nth)
B3 5 0 V= A2D(3-nth)
B4 6 0 V= A2D(4-nth)
Bsum 99 0 V = Vmax*(V(2)*2^(0-nth)+V(3)*2^(1-nth)+V(4)*2^(2-nth)+V(5)*2^(3-nth)+V(6)*2^(4-nth))
* DC source for voltage reference
Vs    1 0       dc 0

* DC test
.dc Vs {Vmin-1} {Vmax+1} {(Vmax-Vmin)/(nth)/100}

.end