analog signal

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?
 
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?
Please show us what you have tried and exactly where you are stuck.

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


Where?.................
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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