analog signal

[logistic_guy]

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?

 

[khansaheb]

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

 

[logistic_guy]

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]

Eq. 1.3\displaystyle 1.31.3 says:

where is equation 1.3 first shown?

(see Eq. 1.3\displaystyle 1.31.3)

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

 

[logistic_guy]

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:

$\displaystyle v_A = b_0 + b_12 + b_24 + b_38$

 

[logistic_guy]

$\displaystyle \bold{(a)}$

$\displaystyle v_A = 0$ 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 $\displaystyle v_A = 0$,

$\displaystyle D = 0000$

 

[logistic_guy]

$\displaystyle \bold{(a)}$

$\displaystyle v_A = 1$ 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 $\displaystyle v_A = 1$,

$\displaystyle D = 0001$

 

[logistic_guy]

$\displaystyle \bold{(c)}$

$\displaystyle v_A = 2$ 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 $\displaystyle v_A = 2$,

$\displaystyle D = 0010$

 

[logistic_guy]

$\displaystyle \bold{(a)}$

$\displaystyle v_A = 15$ means:

$\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 15$

This also means $\displaystyle \ b_0 = b_1 = b_2 = b_3 = 1$

Then, when $\displaystyle v_A = 15$,

$\displaystyle D = 1111$

 

[logistic_guy]

$\displaystyle \bold{(b)}$

Let us assume that $\displaystyle v_A = 0$.

This means that $\displaystyle D = 0000$ and we know that $\displaystyle D = b_3b_2b_1b_0$.

Let us change $\displaystyle b_0$ to $\displaystyle 1$.

Now $\displaystyle D = 0001$ and $\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 1 + (0)2 + (0)4 + (0)8 = 1 \ \text{V}$

Therefore,

If we change $\displaystyle b_0$ from $\displaystyle 0$ to $\displaystyle 1$, then $\displaystyle v_A$ wil change from $\displaystyle 0$ to $\displaystyle 1$.

 

[logistic_guy]

$\displaystyle \bold{(b)}$

Let us assume that $\displaystyle v_A = 0$.

This means that $\displaystyle D = 0000$ and we know that $\displaystyle D = b_3b_2b_1b_0$.

Let us change $\displaystyle b_1$ to $\displaystyle 1$.

Now $\displaystyle D = 0010$ and $\displaystyle v_A = b_0 + b_12 + b_24 + b_38 = 0 + (1)2 + (0)4 + (0)8 = 2 \ \text{V}$

Therefore,

If we change $\displaystyle b_1$ from $\displaystyle 0$ to $\displaystyle 1$, then $\displaystyle v_A$ will change from $\displaystyle 0$ to $\displaystyle 2$.