1
\$\begingroup\$

enter image description here

enter image description here

I've made an attempt at this question, but the answer was wrong. Could someone please tell me where I was wrong in my calculation?

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ The error might be rounding 14.8 to 15, so the answer should be 14. \$\endgroup\$
    – Klas-Kenny
    Jan 27, 2022 at 7:21
  • \$\begingroup\$ should i not round up to cover every step? \$\endgroup\$ Jan 27, 2022 at 7:27
  • 1
    \$\begingroup\$ Your values are all correct. For an ideal ADC the transfer function, the value is rounded down (if your LSB is 0.5V, 0V=0, 0.1V=0, 0.2V=0, 0.3V=0, 0.4V=0, 0.5V=1), and input of or is offset by half a bit (0V=0, 0.1V=0, 0.2V=0, 0.3V=1, 0.4V=1, 0.5=1). The question must be assuming that the transfer function is not shifted over. Bad question IMO. embedded.com/… \$\endgroup\$ Jan 27, 2022 at 7:30
  • \$\begingroup\$ NB: if voltage reference is 2.56 V and ADC (not shifted) is 4 bits -> voltage resolution is (typo error) 2.56/16=0.16 V. So, the voltage for 1111 is 15*0.16= 2.4 V. \$\endgroup\$
    – Antonio51
    Jan 27, 2022 at 11:04

2 Answers 2

1
\$\begingroup\$

You wrote:

but the answer was wrong

So please provide the correct answer.

In my opinion, when you got 14.82 you cannot change it to 15 which is the nearest; you are to decide the lowest natural number which is 14.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ It depends on which ADC transfer function is used. embedded.com/… \$\endgroup\$ Jan 27, 2022 at 7:31
  • \$\begingroup\$ I think you're using 'Answer' on this Q&A site like 'Reply' is used on discussion sites. Although well-intended, this is a comment as it just requests more information then presents an opinion, not a definite Answer. I appreciate that you don't yet have sufficient Reputation points to have posted this as a comment. \$\endgroup\$
    – TonyM
    Jan 27, 2022 at 8:39
1
\$\begingroup\$

The voltage resolution is wrong — 4 bits gives \$2^k\$ steps = 16, not \$2^k-1\$ (that would be the highest code, 0b1111 = 15).

The binary value sounds correct though, regardless of whether the transfer function is offset by 1/2 LSB or not.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.