Ace Test Series 2018: Digital Logic - Rom

Question

Comments

Is it 5 address bits??

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yes
post your solution.

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1024 bits = 2⁵ * 2⁵ bits [ according to given condtion]

now rows = 2⁵ [ represent number of address bits] = 5bits

column = 2⁵ [ represent size of each entry] = 5bits 

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Rom can be represent using ROM matrix, which has rows and columns and intersection between them is called as links.

A n*2ⁿ decoder and a MUX is used for this matrix which has m select lines hence 2^m address lines.

now, links = 1024 = 2^m * 2ⁿ , now given that rows and columns are equal, hence 1024 = 2⁵ * 2⁵

Hence, rows = 2⁵ and columns = 2⁵, bits for rows = 5

Answer 1

NxM ROM with equal rows and columns, since inputs are lgN (Log base 2 N) and Output is M

For equal rows and columnns lgN = M and lgN*M = 1024  

M= 32 , lgN=32 , N= 5

5 bits are required

Answer 2

Total memory = no of rows *no of cols = 1024

we get no of rows =32

henece to identify row we need 5 bits .