Sir can you please explain this step?

17 votes

Consider three $4$-variable functions $f_1, f_2$, and $f_3$, which are expressed in sum-of-minterms as

$f_1=\Sigma(0,2,5,8,14),$

$f_2=\Sigma(2,3,6,8,14,15),$

$f_3=\Sigma (2,7,11,14)$

For the following circuit with one AND gate and one XOR gate the output function $f$ can be expressed as:

- $\Sigma(7,8,11)$
- $\Sigma (2,7,8,11,14)$
- $\Sigma (2,14)$
- $\Sigma (0,2,3,5,6,7,8,11,14,15)$

29 votes

Best answer

Perform $f_1 \cdot f_2$ first, then with the result perform $XOR$ with $f_3$.

$f1 \cdot f2$ means just take common minterms in $f_1$ and $f_2$ (WHY? due to AND gate present, the minterm should be present in both functions.)

$f_1.f_2 = \Sigma (0,2,5,8,14) \cdot \Sigma (2,3,6,8,14,15) = \Sigma (2,8,14)$

$\Sigma (2,8,14) \oplus \Sigma (2,7,11,14) = \Sigma (7,8,11)$

$f1 \cdot f2$ means just take common minterms in $f_1$ and $f_2$ (WHY? due to AND gate present, the minterm should be present in both functions.)

$f_1.f_2 = \Sigma (0,2,5,8,14) \cdot \Sigma (2,3,6,8,14,15) = \Sigma (2,8,14)$

$\Sigma (2,8,14) \oplus \Sigma (2,7,11,14) = \Sigma (7,8,11)$

12

ex-or means, odd no.of times that minterm exist !

2 appeared even number of times, so it can't be in the result !

8 appeared odd number of times, so it should be in the result !

continue this procedure...

2 appeared even number of times, so it can't be in the result !

8 appeared odd number of times, so it should be in the result !

continue this procedure...

10

XOR i.e. $\bigoplus$ is related to Symmetric Difference. Let two sets $A=\{2,8,14\}$ and $B=\{2,7,11,14\}$

$A\bigoplus B=A\cup B-A\cap B=\{2,7,8,11,14\}-\{2,14\}=\{7,8,11\}$

0

* Ex-OR* is

Can you prove that, * how it works on Number of minterms*?

1

XOR i.e. ⨁⨁ is related to Symmetric Difference. Let two sets A={2,8,14}A={2,8,14} and B={2,7,11,14}B={2,7,11,14}

A⨁B=A∪B−A∩B={2,7,8,11,14}−{2,14}={7,8,11}

This is a good point but i don't think its the fundamental property of $\mathbf{\oplus}$, right?

0

ayushsomani that’s because if say x is present 1 time and is in f1 then it will occur in f1.f2’ . If x is in f2 then it will generate from term f2.f1’ in result . (remember **.** results in common terms )

7 votes

f1 and f2 = ∑(0,2,5,8,14) and ∑(2,3,6,8,14,15) = ∑(2,8,14) (common minterms)

(f1 and f2) xor f3 = ∑(2,8,14) xor ∑(2,7,11,14)

we know a ⊕ b = a'b + ab', so

∑(2,8,14) xor ∑(2,7,11,14)

= (∑(0,1,3,4,5,6,7,9,10,11,12,13,15) and ∑(2,7,11,14)) or (∑(2,8,14) and ∑(0,1,3,4,5,6,8,9,10,12,13,15))

= ∑(7,11) or ∑(8)

= ∑(7,8,11) (union)

(f1 and f2) xor f3 = ∑(2,8,14) xor ∑(2,7,11,14)

we know a ⊕ b = a'b + ab', so

∑(2,8,14) xor ∑(2,7,11,14)

= (∑(0,1,3,4,5,6,7,9,10,11,12,13,15) and ∑(2,7,11,14)) or (∑(2,8,14) and ∑(0,1,3,4,5,6,8,9,10,12,13,15))

= ∑(7,11) or ∑(8)

= ∑(7,8,11) (union)

4 votes

From the given diagram we can write like

$f=(f_{1}.f_{2})\oplus f_{3}$

We know that $A\oplus B=A\overline{B}+\overline{A}B$

`So ,we can expend it and get`

$f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{(f_{1}.f_{2})}.f_{3}$

Apply** **De-Morgan's** **laws

$\overline{(A.B)}=\overline{A}+\overline{B}$

$\overline{(A+B)}=\overline{A}.\overline{B}$

Now $f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{(f_{1}}+\overline{f_{2})}.f_{3}$

`Apply distributive rule`

$f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{f_{1}}.f_{3}+\overline{f_{2}}.f_{3}$ -------------$>(1)$

Given that

$f_1=\Sigma(0,2,5,8,14)$

$\overline{f_{1}}=\Sigma(1,3,4,6,7,9,10,11,12,13,15)$

$f_2=\Sigma(2,3,6,8,14,15)$

$\overline{f_{2}}=\Sigma(0,1,4,5,7,9,10,11,12,13)$

$f_3=\Sigma (2,7,11,14)$

$\overline{f_{3}}=\Sigma(0,1,3,4,5,6,8,9,10,12,13,15)$

Now

$f_{1}.f_{2}=\Sigma(0,2,5,8,14).\Sigma(2,3,6,8,14,15)$

`Perform Intersection and get it`

$f_{1}.f_{2}=\Sigma(2,8,14)$

$(f_{1}.f_{2}).\overline{f_{3}}=\Sigma(2,8,14).\Sigma(0,1,3,4,5,6,8,9,10,12,13,15)$

`Perform Intersection and get it`

$(f_{1}.f_{2}).\overline{f_{3}}=\Sigma(8)$

and $\overline{f_{1}}.f_{3}=\Sigma(1,3,4,6,7,9,10,11,12,13,15).\Sigma (2,7,11,14)$

`Perform Intersection and get it`

$\overline{f_{1}}.f_{3}=\Sigma(7,11)$

and last one

$\overline{f_{2}}.f_{3}=\Sigma(0,1,4,5,7,9,10,11,12,13).\Sigma (2,7,11,14)$

`Perform Intersection and get it`

$\overline{f_{2}}.f_{3}=\Sigma(7,11)$

Now put the values in the equation $(1)$ and get

$f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{f_{1}}.f_{3}+\overline{f_{2}}.f_{3}$

$f=\Sigma(8)+\Sigma(7,11)+\Sigma(7,11)$

`Perform Union and get it`

$f=\Sigma(7,8,11)$

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This is better way to analyze,how `GATE`

works.

2 votes

**Option:A**

$(f1.f2)\bigoplus f3$

$(f1.f2)'.f3+(f1.f2).f3'$

$(2,8,14)'.(2,7,11,14)+(2,8,14).(2,7,11,14)'$

$(0,1,3,4,5,6,7,9,10,11,12,13,15).(2,7,11,14)+(2,8,14).(0,1,3,4,5,6,8,9,10,12,13,15)$

$(7,11)+(8)$

$(7,8,11)$

0 votes