analog signal

logistic_guy

logistic_guy

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Consider a \(\displaystyle 4\)-bit digital word \(\displaystyle D = b_3b_2b_1b_0\) (see Eq. \(\displaystyle 1.3\)) used to represent an analog signal \(\displaystyle v_A\) that varies between \(\displaystyle 0 \ \text{V}\) and \(\displaystyle +15 \ \text{V}\).

\(\displaystyle \bold{(a)}\) Give \(\displaystyle D\) corresponding to \(\displaystyle v_A = 0 \ \text{V}, 1 \ \text{V}, 2 \ \text{V},\) and \(\displaystyle 15 \ \text{V}\).
\(\displaystyle \bold{(b)}\) What change in \(\displaystyle v_A\) causes a change from \(\displaystyle 0\) to \(\displaystyle 1\) in \(\displaystyle (\text{i}) \ b_0, (\text{ii}) \ b_1, (\text{iii}) \ b_2,\) and \(\displaystyle (\text{iv}) \ b_3\)?
\(\displaystyle \bold{(c)}\) If \(\displaystyle v_A = 5.2 \ \text{V}\), what do you expect \(\displaystyle D\) to be? What is the resulting error in representation?
 
logistic_guy said:
Consider a \(\displaystyle 4\)-bit digital word \(\displaystyle D = b_3b_2b_1b_0\) (see Eq. \(\displaystyle 1.3\)) used to represent an analog signal \(\displaystyle v_A\) that varies between \(\displaystyle 0 \ \text{V}\) and \(\displaystyle +15 \ \text{V}\).

\(\displaystyle \bold{(a)}\) Give \(\displaystyle D\) corresponding to \(\displaystyle v_A = 0 \ \text{V}, 1 \ \text{V}, 2 \ \text{V},\) and \(\displaystyle 15 \ \text{V}\).
\(\displaystyle \bold{(b)}\) What change in \(\displaystyle v_A\) causes a change from \(\displaystyle 0\) to \(\displaystyle 1\) in \(\displaystyle (\text{i}) \ b_0, (\text{ii}) \ b_1, (\text{iii}) \ b_2,\) and \(\displaystyle (\text{iv}) \ b_3\)?
\(\displaystyle \bold{(c)}\) If \(\displaystyle v_A = 5.2 \ \text{V}\), what do you expect \(\displaystyle D\) to be? What is the resulting error in representation?
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
Eq. \(\displaystyle 1.3\) says:

\(\displaystyle D = b_02^0 + b_12^1 + b_22^2 + \cdots + b_{N-1}2^{N-1}\)

But since the question used the letter \(\displaystyle D\) for the digital word, I would change the letter of the equation to avoid confusion.

\(\displaystyle E = b_02^0 + b_12^1 + b_22^2 + \cdots + b_{N-1}2^{N-1}\)

where \(\displaystyle E = v_A\)
 
khansaheb said:
where is equation 1.3 first shown?


Where?.................
Click to expand...
I don't remember the precise location, but it was the same as post \(\displaystyle \#3\) except it was in Russian.
Thanks for asking.

Eq. \(\displaystyle 1.3\) gives us:

\(\displaystyle v_A = b_0 + b_12 + b_24 + b_38\)
 
\(\displaystyle \bold{(a)}\)

\(\displaystyle v_A = 0\) means:

\(\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 0\)

This also means \(\displaystyle b_0 = b_1 = b_2 = b_3 = 0\)

Then, when \(\displaystyle v_A = 0\),

\(\displaystyle D = 0000\)
 
\(\displaystyle \bold{(a)}\)

\(\displaystyle v_A = 1\) means:

\(\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 1\)

This also means \(\displaystyle b_0 = 1 \ \text{and} \ b_1 = b_2 = b_3 = 0\)

Then, when \(\displaystyle v_A = 1\),

\(\displaystyle D = 0001\)
 
\(\displaystyle \bold{(c)}\)

\(\displaystyle v_A = 2\) means:

\(\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 2\)

This also means \(\displaystyle b_1 = 1 \ \text{and} \ b_0 = b_2 = b_3 = 0\)

Then, when \(\displaystyle v_A = 2\),

\(\displaystyle D = 0010\)
 
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