Ace Test Series: Digital Logic - ROM

Question

I am getting (C). But answer given is (B). Where I have gone wrong?

Comments

$1111$ is a $4$ bit binary number, that is, $15$, square of $15 = 225$,  need max 8 bits .

its $16 \times 8$ ROM

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why 16? 4 select lines rt?

[PS: Out of syllabus portion for GATE]

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Yes. 8 bits means 256 addresses with total of 8 bit of size. So answer should be (C).
It is the type of qstn which have been asked 2-3 times previously!

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Yes

$4$ address lines ,

$2^4$ words (addresses) , $8$ bits/address.

$2^4 \times 8 $ bits ROM

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@Tushar You are writing GATE 2016, not GATE 2015 or before :(

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Sorry sir :( I just wanted to clear my confusion

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0

1

0100

1001

10000

.....

11010001 (225)

Totally 16 selections needed and we select each using 4 bits. Each selection can be up to 8 bits. So, answer is B. 

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Ohkay. Got it! Actually I was confused between multiplier and squarer :p

Answer 1

A  $2^{k}\times n$ ROM required $k\times 2^{k}$ decoder and $n$ OR gates

output square of $4$ bit numbers, So, input 4 bit decoder and with this 4 bit we can represent $(1111)_{2}=15$ in decimal

Now, square of $15^{2}=225$ which can be represented by $8$ bits

So, size of ROM will be $2^{4}\times8$

$4$ address lines $8$ data lines