Design a 4 bit Grey to Binary code converter.

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written 7.9 years ago by

ak.amitkhare.ak • 560

Step1: Write the truth table The truth table of Gray code and its equivalent binary code is as shown in table:

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Step2: Write K – ma for each binary output and get simplified expression The K – map for various binary outputs and the corresponding simplified expression are given below: For output B_3

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Simplified expression $B_3= G_3$

For output B_2

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$B_(2 )=G_(3 ) (G_2 ) ̅ + G_(2 ) (G_3 ) ̅ = G_(3 )⨁G_(2 )$

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${B_1} = {G_1}\overline {{G_2}{G_3}} + {G_2}\overline {{G_3}{G_1}} + {G_3}{G_2}{G_1} + {G_3}\overline {{G_1}{G_2}} = \overline {{G_1}} ({G_3} \oplus {G_2}) + {G_1}(\overline {{G_3} \oplus {G_2}} )$

$ = \overline {{G_1}} X + {G_1}\overline X $ where $X = {G_3} \oplus {G_2}$

$ = {G_1} \oplus X = {G_1} \oplus {G_2} \oplus {G_3}$

For output $B_0$

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${B_0} = {G_0}\overline {{G_1}{G_2}{G_3}} + {G_2}\overline {{G_3}{G_1}{G_0}} + {G_1}\overline {{G_3}{G_2}{G_0}} + {G_1}\overline {{G_3}{G_2}{G_0}} + {G_2}\overline {{G_3}} {G_1}{G_0} + {G_2}\overline {{G_1}} {G_3}{G_0} + {G_2}\overline {{G_0}} {G_3}{G_1} + {G_3}\overline {{G_1}{G_2}{G_0}} + {G_0}\overline {{G_2}} {G_3}{G_1}$

$ = \overline {{G_1}{G_0}} ({G_3} \oplus {G_2}) + {G_0}\overline {{G_1}} (\overline {{G_3} \oplus {G_2}} ) + {G_0}\overline {{G_1}} (\overline {{G_3} \oplus {G_2}} ) + {G_1}\overline {{G_0}} (\overline {{G_3} \oplus {G_2}} )$

$ = ({G_3} \oplus {G_2}) + (\overline {{G_1} \oplus {G_0}} ) + ({G_1} \oplus {G_0}) + (\overline {{G_3} \oplus {G_2}} )$

$ = \overline Y X + Y\overline X $ where $X = {G_3} \oplus {G_2}$ and $Y = {G_1} \oplus {G_0}$

$ = ({G_2} \oplus {G_3}) \oplus ({G_1} \oplus {G_0})$

$ = {G_2} \oplus {G_3} \oplus {G_1} \oplus {G_0}$

Step3: Realization

The Gray to Binary code converter is as shown in figure 7.

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