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I need to multiply 2 3bit numbers. I tried that:

enter image description here

but it does not seem to work.

In this example, the output should be 49. so that is that 77?

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Are the blocks labelled "FA" meant to be full adders? –  The Photon Nov 21 '12 at 22:21
    
Look at the Wiki article referred to by @clabacchio. Specifically, the binary multiplication section. You'll see that your project is 2 shifts, 2 adds, and a handful of AND gates. –  Tony Ennis Nov 22 '12 at 1:31

1 Answer 1

I'm having a hard time trying to understand the schematic...however: there are several different architectures for binary multipliers, but all of them are based on simple considerations.

1- The multiplication in binary logic is made by the AND operator. Therefore you will need a battery of AND gates that multiply each bit of one factor for each bit of the other.

To make it simpler, let's call the factors A and B, of M and N bits respectively. The AND product of the two numbers will give N vectors of M bits, with increasing weight. Now you have to sum up all the vectors considering the weight of each one, to obtain the result.

Long story short: I see the full adders, but you first miss the AND operators.

Some more info on wiki.

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Thanks. but I didn't understand .. where to use the AND gates? –  Billie Nov 22 '12 at 14:43
    
@user1798362 did you try to google it? –  clabacchio Nov 22 '12 at 15:09
    
Of course. I ask here just after a lot of search on Google without answer. –  Billie Nov 22 '12 at 15:30
    
@user1798362 I find it strange, because I found it in the first results :) asic-world.com/images/digital/binary_multiplier.gif –  clabacchio Nov 22 '12 at 16:12
    
Oh , Well , I didn't think about search images. thanks. I'll try it. –  Billie Nov 22 '12 at 21:44

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