Test by Bikram | Mock GATE | Test 2 | Question: 43

Question

Match the following:

P. $\sum (1,2,4,7,8,11,13,14)$	1. Full adder sum function
Q. $\sum (0,3,5,6,9,10,12,15)$	2. Full adder carry function
R. $\sum (1,2,4,7)$	3. EX-OR
S. $\sum (3,5,6,7)$	4. EX-NOR

 

 $$\begin{array}{|c|c|c|} \hline P. & \sum(1,2,4,7,8,11,13,14) & 1.& \text{ Full adder sum function} \\ \hline Q.& \sum(0,3,5,6,9,10,12,15) &2.&\text{Full adder carry function} \\ \hline R.& \sum(1,2,4,7) & 3.& \text{EX-OR} \\ \hline S.&\sum(3,5,6,7) & 4. & \text{EX-NOR} \\ \hline  \end{array}$$ 

• A. $P - 1, Q - 2, R - 3, S - 4$
• B. $P - 2, Q - 1, R - 4, S - 3$
• C. $P - 3, Q - 4, R - 1, S - 2$
• D. $P - 4, Q - 3, R - 2, S - 1$

Comments

Table of XOR is this

R has to be XOR i.e. option A

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sir full adder is used for 3- bit summation why we are using it for 4-bit ?

for (0,0,0,0)= sum is 0 and carry is 0 then how Q-2 (which is for carry and taking 0 as minterms)

Answer 1

we have to check the functions from no. $0$ to $15$ .as options include no. till $15$.so the binary representation for  the propostions in truth table should be with $4$ digit input

lets consider the EX-OR option

ABCD       A(EXOR)B    A(EXOR)B (EXOR) C         A(EXOR)B (EXOR) C (EXOR) D

0000              0                  0                                      0

0001              0                 0                                        1

0010              0                 1                                         1

0011               0                  1                                        0

0100              1                     1                                        1

0101               1                      1                                     0

0110                 1                     0                                     0

0111                 1                      0                                     1

1000                1                       1                                   1

1001              1                         1                                    0

1010               1                          0                                   0

1011               1                          0                                  1

1100                0                        0                                       0

1101               0                       0                                         1

1110                 0                       1                                        1

1111                  0                         1                                      0

 $$\begin{array}{|c|c|c|} \hline A & B & C & D & A\oplus B & A\oplus B\oplus C & A\oplus B\oplus C\oplus D \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ \hline 0 & 0 & 1 & 0 & 0 & 1 & 1  \\ \hline 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ \hline 0 & 1 & 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 1 & 1 & 1 & 0 \\ \hline 0 & 1 & 1 & 0 & 1 & 0 & 0   \\ \hline 0 & 1 & 1 & 1 & 1 & 0 & 1 \\ \hline 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 1 & 1 & 1 & 0 \\ \hline 1 & 0 & 1 & 0 & 1 & 0 & 0  \\ \hline 1 & 0 & 1 & 1 & 1 & 0 & 1 \\ \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ \hline 1 & 1 & 1 & 0 & 0 & 1 & 1   \\ \hline 1 & 1 & 1 & 1 & 0 & 1 & 0  \\ \hline \end{array}$$ 

So minimized function for this EXOR function$= \sum(1,2,4,7,8,11,13,14)$

similar way we can verify matching values for other functions.

So the answer should be option C.     

Answer 2

Note: This answer has a lot of extra stuff. Don't read it if you're interested only in the precise answer to the given question.

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0	1	0	1
1	0	1	0
0	1	0	1
1	0	1	0

This is a pretty well known K-map. It results in the XOR of two out of the four variables.

 

1	0	1	0
0	1	0	1
1	0	1	0
0	1	0	1

 This one is the K-map for XNOR of two out of the four variables.

 

Notice the pattern.

Notice that they both complement each other.

 

No need to memorise them forcefully. You can directly solve it, or here's one neat trick.

"0000" ie, the first cell in XOR has to obviously be 0.

$0\oplus 0\oplus 0\oplus 0=0$.

This is how you can immediately differentiate which one is XOR and XNOR.

 

Now, another thing:

• When the variables are even, then XOR and XNOR are the complements of each other. This is evident by the first two K-maps in this very answer.
   
• When given three (or any odd number of) variables, then XOR is equivalent to XNOR. Just make the Truth Table and see for yourself.

 

Hence, $a\oplus b\oplus c= a\odot b\odot c$

So, they will have the same K-map. Which is:-

0	1	0	1
1	0	1	0

This is the K-map of XNOR and XOR both for three variables. As it happens, the Sum in Full Addrer is given by $a\oplus b\oplus c_{in}$

So, this K-map is for:-

1. XOR of three variables
2. XNOR of three variables
3. Sum function of full adder

Because these three are equivalent.

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Now, match the options. Answer is Option C

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0	1	1	0
1	0	0	1
1	0	0	1
0	1	1	0

This one is the K-map for XOR of the remaining two out of four variables — The very first K-map of this answer finds the XOR of say, A and B. This will find XOR of C and D.

 

Similarly

1	0	0	1
0	1	1	0
0	1	1	0
1	0	0	1

This one is the K-map for XNOR of the remaining two out of four variables.

 

You can visualise all K-maps without 0's.

Comments

The above K-maps without 0's.

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	1		1
1		1
	1		1
1		1

XOR

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1		1
	1		1
1		1
	1		1

XNOR

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	1		1
1		1

XOR = XNOR of three variables

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	1	1
1			1
1			1
	1	1

XOR

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1			1
	1	1
	1	1
1			1

XNOR