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The boolean expression:

y= (a'+b) * (b(c'+d')) = (a'+b) * (bc' + bd')

is given.

Find out the disjunctive normal form to this boolean expression.

My suggestion:

y= a'bc'd' + a'bc'd + a'bcd' + abc'd' + abc'd+abcd'

only when these terms are true the function is 1 and otherwise not. (=0)

Is the DNF right?

Thanks for the help.

Edit: Here ist the truth table for the boolean expression:

enter image description here

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  • \$\begingroup\$ You might notice that a has no effect on the output. \$\endgroup\$ – Finbarr May 5 '19 at 12:01
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That looks good your original equation reduces to (bc' + bd') which is what you have put in your truth table and then extracted in DNF.

Edit to clarify reduction of original equation.

(a'+b) * (bc' + bd')= a'(bc' + bd')+b(bc' + bd')

              = a'(bc' + bd')+(bc' + bd')

              = (bc' + bd')
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  • \$\begingroup\$ the simplified boolean equation would be b(c'+d'), so i could simplify the DNF to y= bc'd' + bc'd + bcd' + bc'd' + bc'd+bcd' , right? \$\endgroup\$ – lightsodium May 5 '19 at 12:17
  • \$\begingroup\$ @lightsodium DNF requires all the variables to be present in all of the terms. \$\endgroup\$ – RoyC May 5 '19 at 12:19
  • \$\begingroup\$ oh,thanks! so it still would be y= a'bc'd' + a'bc'd + a'bcd' + abc'd' + abc'd+abcd' \$\endgroup\$ – lightsodium May 5 '19 at 12:22
  • \$\begingroup\$ "that looks good your original equation reduces to (bc' + bd')": how can I read this just out of the truth table? and not by using laws to reduce the equation \$\endgroup\$ – lightsodium May 5 '19 at 12:42
  • \$\begingroup\$ You can't you have to use the laws of boolean algebra to get to (bc' + bd'). After you put this in your truth table you can read out the DNF form. \$\endgroup\$ – RoyC May 5 '19 at 13:01

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