in Digital Logic
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How do we solve this?
in Digital Logic
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Is it $\sum (0, 1, 3, 6)$ ?
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3 Answers

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$f1$(a, b, c) = $\sum(1, 2 ,3,4)$ and $f2$(a, b, c) = $\sum(0, 2, 4, 6)$

Now, $f1\bigoplus f2$ says that

if $f1$ is $true$ then $f2$ must be $false$.

$and$

if $f2$ is $true$ then $f1$ is $false$.

So, all those minterms where both $f1$ and $f2$ are True should not appear or be True for $f1\bigoplus f2$

Because, $f1\bigoplus f2$ will be true only for those minterms for which either $f1$ or $f2$ (not both) are True.

Hence,

Minterm $f1$ $f2$ $f1 \bigoplus f2$
0 0 1 1
1 1 0 1
2 1 1 0
3 1 0 1
4 1 1 0
6 0 1 1

 

And for all other minterms $f1$ and $f2$ are $False$.

So, $f1 \bigoplus f2$ = $\sum(0, 1, 3, 6)$

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For Solving you need to know that for

  1. AND operation take common number from both f1 and f2
  2. OR operation take union of f1 and f2
  3. Not operation take number not present
  4. for XOR, XNOR, NAND, NOR you can derive from basic operations.

Let the functions 𝑓1 𝑎, 𝑏, 𝑐 = ∑ 1,2,3,4, 𝑓2 𝑎, 𝑏, 𝑐 = ∑ 0,2,4,6 and 𝑓1’ 𝑎, 𝑏, 𝑐 = ∑ 0,5,6,7 and 𝑓2’ 𝑎, 𝑏, 𝑐 = ∑ 1,3,5,7

𝑓1𝑓2’ = ∑ 1,3

𝑓1’𝑓2 = ∑ 0,6

𝑓1 ⊕ 𝑓2 = 𝑓1𝑓2’ + 𝑓1’𝑓2  = ∑ 0,1,3,6

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Method 1)

For the same input, exor gate will produce $0$ whereas for different inputs it will produce $1$ as output. Here is $f_1,f_2$ $2,4$ as common minterm so it will not be in $f_1\oplus f_2$. all the minterm not in common will be in $f_1\oplus f_2$.(Either or)

so $f_1\oplus f_2=\sum(0,1,3,6)$.

Method 2)

Since we know that AND gate takes common minterm, OR gate takes all minterm, and NOT gates take those minterms that are not present.

$f_1\oplus f_2=\bar f_1f_2+f_1\bar f_2$

  • $\bar f_1=(0,5,6,7),f_2=(0,2,4,6)\rightarrow \bar f_1f_2=(0,6)$ (As AND gate takes common minterm)
  • $f_1=(1,2,3,4),\bar f_2=(1,3,5,7)\rightarrow f_1\bar f_2=(1,3)$
  • $\bar f_1f_2+f_1\bar f_2=(0,1,3,6)$ (As OR gate takes all the minterm) 

$\therefore f_1\oplus f_2=\sum(0,1,3,6)$.

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