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A priority encoder accepts three input signals $\text{(A, B and C)}$ and produces a two-bit output $(X1, X0 )$ 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.
asked in Digital Logic by Veteran (386k points)
edited by | 978 views
0
+1

This might help ...

0
be aware, in this above lecture, he did some mistake ..!
+1
yes @ankit, he did a mistake, 0's will be put only in 01 rows in second k-map.

1 Answer

+16 votes
Best answer

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

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

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

 

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

answered by Boss (14k points)
edited by
0
Here why u select C as input of MUX?

shouldnot it be signal to see which MUX is active at a time and other deactive at that time?
+1
no we can't do like that, we need both MSB and LSB at any time thats why both the mux should be active all the time
0

https://en.wikipedia.org/wiki/Priority_encoder

why u took o/p 1 for 001 i/p??

0

अनुराग पाण्डेय how you derive expression (K-map??) if yes then how you have taken the inputs please comment.

0
Is this realisation possible if "If none of the inputs are active the output should be 00" - this statement is not present in question?

$\begin{array}{ccccc}

A &B&C&\quad\text{MSB}&\text{LSB}\\
0&0&0&X&X\\
0&0&1&0&0\\
0&1&X&0&1\\
1&X&X&1&0\\\hline
\end{array}$

$MSB=A$

$LSB=A \bar{B}$
0

2018 Given in question "If none of the inputs are active the output should be 00", so lowest priority should start with "01" and not "00"

0

Assume A has the highest priority followed by Band C has the lowest priority.

and wiki says

If two or more inputs are given at the same time, the input having the highest priority will take precedence.[1] An example of a single bit 4 to 2 encoder is shown, where highest-priority inputs are to the left and "x" indicates an irrelevant value - i.e. any input value there yields the same output since it is superseded by higher-priority input. 

0

srestha how the expression for MSB and LSB is derived?

0
for final circuit it is just an assumption

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