#### The Analog-Digital Converter

The input range of a 10 bit ADC varies from +0.1..+4.9V.

a) What input step size has this analog-digital converter?

b) For a full scale sine wave please find the SNR of its quantisation in dB.

c) Find the number of bits for this ADC if the quantization SNR is 90 dB.

d) You have at the input instead $V(t)=0.2*\sin(\omega t)+2.5$ V and you use the inital 10 bit ADC. What is the number of quantization levels?

e) How many effective bits for the previous number of levels?

f) What is the SNR from the previous number of effective bits?

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#### Answer

a) The step of the voltage at the input of this converter is $V(step) = (4.9-0.1)/2^10 =4.6875$ mV

b) The signal to noise ratio is defined as $SNR = 20*log(A(signal)/A(noise))$

where $A(signal) = 4.9-0.1 =4.8 V$ and $A(noise) = V(step) =4.6875 mV$

Therefore the $SNR = 20*log(4.8/4.6875*10^{-3}) =60.2$ dB

c) From the definition of the SNR we have $90 =20*log(4.8/A(noise))$

$A(noise) =4.8/10^{4.5} =0.15179 mV$

Therefore the number of bits required is

$N = log(2) [(4.9-0.1)/0.15179*10^{-3}] =log(2) [31622.3] = 15 bits$

(CHECK $2^{15} =32768$ and $4.8/32768 =0.147$ mV thus is correct!)

d) The total variation of the signal si $2*0.2 =0.4 V$

Thus the number of level the signal is using is

$N = 0.4/4.6875*10^-3 =85.33 =84$ levels.

The number showing the “zero” level of this signal is signal is:

$2.5/4.6875*10^{-3} =373.8 =374$

The levels of the signal begin at the number $N1 =374-84/2 =332$ and end at the number $N2 =374+84/2 =416$

e) The number of bits corresponding to 84 levels is $N =log(2)84 =6.392 bits$

f) $A(signal) =0.2 V$

$A(noise) =4.6875 mV$ (for a total numebr of bits =6.392)

Therefore the signal to noise ratio (SNR) for the previous number of bits is

$SNR = 20*log(0.2 V/4.6875 mV) = 32.6 dB$