Digital Logic

Question

The total number of comparisons performed in a 8-bit magnitude comparator consist of inputs A[A₄, A₃, A₂, A₁] and B[B₄, B₃, B₂, B₁] then condition for A>B is :

A 255 x 2⁶

B 255 x 2⁷

C 255 x 2⁸​​​​​

D 255 x 2⁹

Answer 1

for A > B 

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=  (2 (^ (2^n) )  – 2^n )/2    (n = number of bits )
=  (2 (^ (2^8) )  – 2^8 )/2 
=  (2 (^ (16) )  – 2^8 )/2 

= ((2^8) * (2^8) – 1)  )/2 
= 128(255)
= 255 x 2^7

Comments

Can you explain more ?

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Like what exactly question wants to ask . And what is formula you have used.

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Say you have 1 bit comparator and making comparation between A and B 
 

A	B	A>B	A<B	A=B
0	0	0	0	1
0	1	0	1	0
1	0	1	0	0
1	1	0	0	1

Now see.. 

with 1 bit we get 4 encodings...2^2n

for A = B we get 2 encodings… which is 2^n

A > B and A< B will always have an equal number of encodings..observe the table above.

therefore just divide everything by 2.

All encodings after removing A = B ….

now encodings for A<B or A>B will be  → (2^2n  –  2^n)/ 2 

put 8 in place of n… (2^2*8  – 2^8) / 2

2^8(2^8 - 1)  / 2

2^7(255) – > 255 x 2^7 

 

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Thank you so much bro.

But here it is given 8 bit but in question it have 4 bits in A and 4 bits in B . It means it is 4 bit comparator ???