GATE1992-04-b

Question

A priority encoder accepts three input signals $\text{(A, B and C)}$ and produces a two-bit output $(X_1, X_0 )$ corresponding to the highest priority active input signal. Assume $A$ has the highest priority followed by $B$ and $C$ has the lowest priority. If none of the inputs are active the output should be $00$, design the priority encoder using $4:1$ multiplexers as the main components.

Comments

be aware, in this above lecture, he did some mistake ..!

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yes @ankit, he did a mistake, 0's will be put only in 01 rows in second k-map.

Answer 1

$\text{MSB} - \text{Most Significant Bit}$
$\text{LSB} - \text{Least Significant Bit}$

$\begin{array}{ccccc}
\rlap{\text{TRUTH TABLE}} \\
\rlap{\text{Inputs}: A,B,C}\\
\rlap{\text{Outputs}: \text{MSB},\text{LSB}}\\
A &B&C&\quad\text{MSB}&\text{LSB}\\
0&0&0&0&0\\
0&0&1&0&1\\
0&1&X&1&0\\
1&X&X&1&1\\\hline
\end{array}$

$\text{MSB} = A+B$
$\text{LSB} =A + \bar B C$ 

 

It can be implemented using two $4 \times 1$ Multiplexers.

Comments

I think that the positions of (A - S0, B - S1) as the selection inputs should be interchanged to (A - S1, B - S0) in the diagram, as A is more significant than B.

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For the second mux,(in my previous comment) the equation is as follows:

$\overline{AB}C + A\overline{B} + AB$

$= \overline{AB}C + A$

$= (\overline{A}+A) (\overline{B}+A) (C+A)$

$= (\overline{B}+A) (C+A)$

$= \overline{B}C + \overline{B}A + AC + A$

$= \overline{B}C + A(\overline{B}+C+1)$

$= \overline{B}C + A$

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How LSB and MSB are mapped? What I mean is why 11 is maped with ABC=1XX only and not with any other input?

Answer 2

We can implement the priority encoder using two $4:1$. Multiplexers as main components and $1$ NOT gate and $1$ OR gate
Using following truth table (Priority encoder with highest priority to $A$)$$\begin{array}{cccc|cc}  \textbf{A} & \textbf {B} &\textbf {C} & \textbf {X} &  \textbf{$X_1$ } & \textbf{$X_0$}\\ \hline 0& 0& 0  & \text{X} &0 &0\\ 0& 0& 1  & \text{X} &0 &1\\  0& 1& \text{X} & \text{X} &1 &0 \\ 1& \text{X}& \text{X}  & \text{X} &1 &1 \\ \end{array}$$

Comments

Can you please explain logic behind second mux? I'm getting the answer here : https://gateoverflow.in/17407/gate1992_04_b

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i know the priority encoders but not able to understand this above concept plz clear me????

Answer 3

we know the truth table of priority encoder .

now we have a multiplexer. we need to do selection in a abnormal way. means we have to play with select lines . and we have 2 output from the above function. relize them and then give input to the select line of muxs.

y1= i3+i2

y0=i3+ i2' i1

now make y1 as s1 select line .and y0 as s0.

Comments

I have understood that here we have made selection i/p(s0 and s1) a bit complex in order to select any of the (x3,x2,or x1). But this encoder o/p should be 2 bits. How are we acheiving that thing?

Answer 4

$X_0 = \bar{A} \bar{B} C + A$

$X_1 = \bar{A} B  + A$

 

Comments

Input to the first MUX ($X_1$) should be 0111 then characteristic equation:

$X_1=\ A'B'(0)+A'B(1)+AB'(1)+AB(1) = A'B+A(B'+B) = A'B+A = A+B$.

Input to second MUX ($X_0$) should be C011 then characteristic equation:

$X_0=\ A'B'C+A'B(0)+AB'(1)+AB(1) = A'B'C+0+A(B'+B) = A'B'C+ A$.

Answer 5

 

Suggestions are welcome.

Comments

How to know what inputs to be given to multiplexers??

Please explain anyone