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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 $\text{M words}\times \text{ N bits}$. Write the values of $\text{M}$ and $\text{N}$.
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25 votes
$A$ is $4$ bit binary number: $A_{4}A_{3}A_{2}A_{1}$
$B$ is $4$ bit binary number: $B_{4}B_{3}B_{2}B_{1}$

$M$ is result of multiplication $M_{8}M_{7}M_{6}M_{5}M_{4}M_{3}M_{2}M_{1}$     [check biggest no $1111 \times 1111 =11100001$]
$$\small{\begin{array}{|cccc|cccc|cccccccc|} \hline \textbf{A4} & \textbf {A3} & \textbf {A2} &  \textbf{A1 }&\textbf{B4} & \textbf {B3} & \textbf {B2} &  \textbf{B1 }&\textbf{M8} & \textbf {M7} & \textbf {M6} &  \textbf{M5 }&\textbf{M4} & \textbf {M3} & \textbf {M2} &  \textbf{M1 }\\\hline 0&0&0&0&0&0&0&0&0&0&0&0&0&0 &0&0\\
0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\
.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.\\
.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.\\
1&1&1&1&1&1&1&0&1&1&0&1&0&0&1&0 \\
1&1&1&1&1&1&1&1&1&1&1&0&0&0&0&1 \\\hline \end{array}}$$
$4$ bits of $A$ and $4$ bits of $B$ mean input will consist of $8$ bits and need address $00000000$ to $11111111 = 2^8$ address

The output will be of $8$ bits

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

$M = 256 , N = 8$
edited by
9 votes
9 votes

In ROM, we don't compute anything, we just have to store the computed result. So just see how many address lines are needed to store all possible combination of result. here 4 bits need to multiply by 4 bits. so 8 address lines are
required to address 2^8=256 lines and in these lines, different value(result) will be stored in the form of
links. There are 8 data lines are required to receive the result because of max. value will be 1111 x 1111 =11100001.

Here 1st line will store the multiplication result of 0000 x 0000, and 256th line will store the result of 1111 x 1111 in the form of links between horizontal and vertical lines. Note that 0001 x 0000 & 0000 x 0001 will store the same result at 2 different places.

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