# Some doubts regarding truth table for the function $(cd+b'c+bd')(b+d)$?

$$(cd+b'c+bd')(b+d)$$

Expanding we have $$(cd+b'c+bd')(b+d)=bcd+bd'+cd+b'cd$$

Express each function as sum of Minterm and Product of Maxterm.

If you have n-variables then you will have $$2^n$$ products. Following n=3 we have 8 minterm. But my book suggest that there are 16 terms.

I need some clarification on this. Help please.

Another question is why don't we simplified the terms using Boolean Algebra like we normally do?

• This is electrical forum you can ask this type of questions here math.stackexchange.com – Hazem Dec 22 '17 at 4:49
• @Hazem Digital logic and boolean algebra are just fine here at EE.SE. – Shamtam Dec 22 '17 at 4:50
• Perhaps they were assuming there was another variable "a", whose value is irrelevant? – user253751 Dec 22 '17 at 4:56
• It makes sense but the variable is not mentioned in the question. – Crazy Dec 22 '17 at 4:59
• I am working out the Digital Design by Morris. – Crazy Dec 22 '17 at 5:00

## 1 Answer

If your variables are $b$, $c$, and $d$, then your eight minterms are:

\begin{align} & bcd \\ & bcd' \\ & bc'd \\ & bc'd' \\ & b'cd \\ & b'cd' \\ & b'c'd \\ & b'c'd' \\ \end{align}

Your formula was:

$$bcd + bd′ + cd + b′cd$$

To express this in terms of minterms, you have to fill in the missing variables. Let's look at $bd'$. The $c$ is missing, which means it has no effect on $bd'$. So we can write this as:

$$bd' = bcd' + bc'd'$$

You could simplify the equation first:

$$bcd + bd' + cd + b'cd = bd' + cd$$

but if you need the minterms, you'll just have to expand it again.