analog signal

logistic_guy

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Consider a 4\displaystyle 4-bit digital word D=b3b2b1b0\displaystyle D = b_3b_2b_1b_0 (see Eq. 1.3\displaystyle 1.3) used to represent an analog signal vA\displaystyle v_A that varies between 0 V\displaystyle 0 \ \text{V} and +15 V\displaystyle +15 \ \text{V}.

(a)\displaystyle \bold{(a)} Give D\displaystyle D corresponding to vA=0 V,1 V,2 V,\displaystyle v_A = 0 \ \text{V}, 1 \ \text{V}, 2 \ \text{V}, and 15 V\displaystyle 15 \ \text{V}.
(b)\displaystyle \bold{(b)} What change in vA\displaystyle v_A causes a change from 0\displaystyle 0 to 1\displaystyle 1 in (i) b0,(ii) b1,(iii) b2,\displaystyle (\text{i}) \ b_0, (\text{ii}) \ b_1, (\text{iii}) \ b_2, and (iv) b3\displaystyle (\text{iv}) \ b_3?
(c)\displaystyle \bold{(c)} If vA=5.2 V\displaystyle v_A = 5.2 \ \text{V}, what do you expect D\displaystyle D to be? What is the resulting error in representation?
 
Consider a 4\displaystyle 4-bit digital word D=b3b2b1b0\displaystyle D = b_3b_2b_1b_0 (see Eq. 1.3\displaystyle 1.3) used to represent an analog signal vA\displaystyle v_A that varies between 0 V\displaystyle 0 \ \text{V} and +15 V\displaystyle +15 \ \text{V}.

(a)\displaystyle \bold{(a)} Give D\displaystyle D corresponding to vA=0 V,1 V,2 V,\displaystyle v_A = 0 \ \text{V}, 1 \ \text{V}, 2 \ \text{V}, and 15 V\displaystyle 15 \ \text{V}.
(b)\displaystyle \bold{(b)} What change in vA\displaystyle v_A causes a change from 0\displaystyle 0 to 1\displaystyle 1 in (i) b0,(ii) b1,(iii) b2,\displaystyle (\text{i}) \ b_0, (\text{ii}) \ b_1, (\text{iii}) \ b_2, and (iv) b3\displaystyle (\text{iv}) \ b_3?
(c)\displaystyle \bold{(c)} If vA=5.2 V\displaystyle v_A = 5.2 \ \text{V}, what do you expect D\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. 1.3\displaystyle 1.3 says:

D=b020+b121+b222++bN12N1\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 D\displaystyle D for the digital word, I would change the letter of the equation to avoid confusion.

E=b020+b121+b222++bN12N1\displaystyle E = b_02^0 + b_12^1 + b_22^2 + \cdots + b_{N-1}2^{N-1}

where E=vA\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 #3\displaystyle \#3 except it was in Russian.
Thanks for asking.

Eq. 1.3\displaystyle 1.3 gives us:

vA=b0+b12+b24+b38\displaystyle v_A = b_0 + b_12 + b_24 + b_38
 
(a)\displaystyle \bold{(a)}

vA=0\displaystyle v_A = 0 means:

vA=b0+b12+b24+b38=0\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 0

This also means b0=b1=b2=b3=0\displaystyle b_0 = b_1 = b_2 = b_3 = 0

Then, when vA=0\displaystyle v_A = 0,

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

vA=1\displaystyle v_A = 1 means:

vA=b0+b12+b24+b38=1\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 1

This also means b0=1 and b1=b2=b3=0\displaystyle b_0 = 1 \ \text{and} \ b_1 = b_2 = b_3 = 0

Then, when vA=1\displaystyle v_A = 1,

D=0001\displaystyle D = 0001
 
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