Please help me answer these 6 questions. Thank you. An analog-to-digital convert

Question

Please help me answer these 6 questions.  Thank you.

An analog-to-digital converter (ADC) has an input range from +0.1V to +4.9V. It is a 10-bit ADC. What is the voltage step size (volts/level) of this ADC? What would the quantization SNR in dB be for this ADC for a full-scale sine wave input? How many bits would be required if we wished for a quantization SNR of at least 90dB for a full-scalc sine wave input? (Round UP to the next integer value.) Suppose that the input is NOT a full-scale sine wave but is instead Vi(t) = 0.2sin( omega t) +2.5 volts. (Note the 2.5V DC offset simply places the sine wave within the converter's input range and does not affect the SNR.) Suppose that the original 10-bit ADC is being used. Approximately how many of the quantization levels are being exercised by this signal? Convert this effective number of levels to an effective number of bits. (You are allowed to have a non-integer number of bits for this answer.) What is the resulting effective quantization SNR? (Again, you may use your non-integer number of bits here.)

Explanation / Answer

a) step voltage is V(step) = (4.9-0.1)/2^10 =4.6875 mV

b) SNR = 20*log(Asignal/Anoise)

Asignal = 4.9-0.1 =4.8 V

Anoise = Vstep =4.6875 mV

SNR = 20*log(4.8/4.6875*10^-3) =60.2 dB

c) 90 =20*log(4.8/Anoise)

Anoise =4.8/10^4.5 =0.15179 mV

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 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

Anoise =4.6875 mV (for a total numebr of bits =6.392)

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

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.