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+21 votes

The amount of ROM needed to implement a $4-bit$ multiplier is

  1. $64$ bits
  2. $128$ bits
  3. $1$ Kbits
  4. $2$ Kbits
asked in Digital Logic by Veteran (19.7k points)
retagged by | 2.7k views

2 Answers

+26 votes
Best answer
A ROM cannot be written. So, to implement a 4-bit multiplier we must store all the possible combinations of $2^4 * 2^4$ inputs and their corresponding $8$ output bits giving a total of $ 2^4 * 2^4* 8$ bits $= 2048$ bits. So, (D) is the answer.

PS: We are not storing the input bits explicitly -- those are considered in order while accessing the output 8 bits. In this way, by storing all the possible outputs in order we can avoid storing the input combinations.
answered by Veteran (347k points)
edited by
Arjun Suresh sir- Can you tell the book where this topic is covered. It will be great if you can provide some link.
I guess Morris Mano book covers this. I don't have the book now to verify. This ppt also gives a very good detail:
total possible cases $ = 2^4 * 2^4$ and output of each case $= 8\;bit$
How is the input mapped to.output in the multiplier?

this image clarifies the ROM thing here..

@arjun sir ,please update the link.. it is dead.
@Arjun Suresh Sir, I have a doubt here... With 4 bits, 16 numbers (multiplicands) are possibe. Each of this number takes 4 bits in memory.. Similarly each multiplier will also take 4 bits then why dont we multiply the result u found with 4*4... 2^4 is just number of multiplicands.. It is not number of bits used to store those many multiplicands... Please clarify
@Anchit see now.
+5 votes
Since we want to multiply 4bit two number .

Then number of combination of input for multiplication is=16*16

Number of output lines =8

Size of ROM=16*16*8=2048bits=2Kbits
answered by Boss (8.4k points)
Please update the link..Link is not working..

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