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A ROM is used to store the Truth table for binary multiple units that will multiply two $4-bit$ numbers. The size of the ROM (number of words $\times$ number of bits) that is required to accommodate the Truth table is $M \text{ words}\times N \text{ bits}$. Write the values of $M$ and $N$.
asked in Digital Logic by Veteran (59.5k points)
edited by | 866 views
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ROM stands for Read Only Memory. 
As the name suggest. We can only read from this memory.
So we need to store all the possible outputs of the multiplication of two 4 bits numbers.
Multiplication of two 4 bit numbers can be at most 8 bits long.
With 8 bits - $2^{8}$ words are possible.
And the size of each word is 8 bit.
So memory required is $2^{8} words*8 bits$
So M = $2^{8} = 256$ and N = 8

1 Answer

+13 votes
Best answer

$A$ is $4$ bit binary no $A4A3A2A1$   

$B$ is $4$ bit binary no $B4B3B2B1$

$M$ is result of multiplication $M8M7M6M5M4M3M2M1$     [check biggest no $1111 \times 1111 =11100001$]

$A4$ $A3$ $A2$ $A1$ $B4$ $B3$ $B2$ $B1$ $M8$ $M7$ $M6$ $M5$ $M4$ $M3$ $M2$ $M1$
$0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$
$0$ $0$ $0$ $0$ $0$ $0$ $0$ $1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
$1$ $1$ $1$ $1$ $1$ $1$ $1$ $0$ $1$ $1$ $0$ $1$ $0$ $0$ $1$ $0$
$1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $0$ $0$ $0$ $0$ $1$

So $4$ bit of $A \ 4$ bit of $B$

input will consist of $8$ bit need address $00000000$ to $11111111 = 2^8$ address

output will be of $8$ bits

so memory will be of $2^8 \times 8$

$M = 256 , N = 8$

answered by Veteran (55.1k points)
edited by
+3
Yes. Basically we don't store the input as they are implied. $2^8$ combinations of inputs are possible and hence we store 256 outputs of 8 bits each.
0
Sir, but the question is asking to store truth table, so we should store inputs as well even though they are implied.
0
@ Arjun It is contradicting whatever u told and whatever @Praveen told . Could u clearify by giving some explananations on paper .
+2
Or we can calculate $M$ as:

Inputs A and B are of 4 bits each, so we can represent $2^4\times 2^4=2^8=256$ numbers from them, which is the number of words saved in the ROM.
0
@Arjun sir, why not store input in table ?

Truth table has input and output and size of truth table includes complete truth table.

If we store only output it is partial truth table.Please clarify
0
Inputs will serve as address lines to ROM to get to the desired output.
0
Yes sir.Got it .thanks.


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