Engineering Design a 2 bit multiplier circuit

AI Thread Summary
The discussion focuses on designing a 2-bit multiplier circuit, which requires a truth table, simplified logic expression, logic circuit, and implementation using NAND gates. Participants clarify that the truth table correctly represents the multiplication of two 2-bit numbers, resulting in a 4-bit output. There is confusion about the use of Karnaugh maps for simplifying the logic expressions and how to proceed with the design. The conversation also touches on the general rule that multiplying N-bit and M-bit numbers results in an output of N+M bits, though some participants debate this point. The thread emphasizes the importance of understanding the underlying concepts rather than just seeking direct answers.
soonsoon88
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Design a 2 bit multiplier circuit.
The design should include:
a)truth table
b)simplified logic expression
c)logic circuit
d)implementation of the circuit using NAND gates only.

*Anybody able to help me? Actually that is my quiz on last week, but my lecturer note dint mention about this at all. In short, i really know nothing about it.I hope anyone know the information about 2bit multiplier circuit or anyone know one of the solution or any website can learn about it,share with me about it.THX A LOT !*
 
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soonsoon88 said:
Design a 2 bit multiplier circuit.
The design should include:
a)truth table
b)simplified logic expression
c)logic circuit
d)implementation of the circuit using NAND gates only.

*Anybody able to help me? Actually that is my quiz on last week, but my lecturer note dint mention about this at all. In short, i really know nothing about it.I hope anyone know the information about 2bit multiplier circuit or anyone know one of the solution or any website can learn about it,share with me about it.THX A LOT !*

Well, you should be able to at least start with the truth table, right? Please post the truth table, and tell us about how you would do the Karnaugh Map for it...
 
berkeman said:
Well, you should be able to at least start with the truth table, right? Please post the truth table, and tell us about how you would do the Karnaugh Map for it...

A(1) A(0) B(1) B(0) C(3) C(2) C(1) C(0)

0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 1 0 0 0 0
0 1 0 0 0 0 0 0
0 1 0 1 0 0 0 1
0 1 1 0 0 0 1 0
0 1 1 1 0 0 1 1
1 0 0 0 0 0 0 0
1 0 0 1 0 0 1 0
1 0 1 0 0 1 0 0
1 0 1 1 0 1 1 0
1 1 0 0 0 0 0 0
1 1 0 1 0 0 1 1
1 1 1 0 0 1 1 0
1 1 1 1 1 0 0 1
Am i right? If i am totally wrong, you may let me know.
If it is hard to explain it at here, i may go ask for other first.
Thx for helping first =)
 
Last edited:
soonsoon88 said:
A(1) A(0) B(1) B(0) C(3) C(2) C(1) C(0)

0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 1 0 0 0 0
0 1 0 0 0 0 0 0
0 1 0 1 0 0 0 1
0 1 1 0 0 0 1 0
0 1 1 1 0 0 1 1
1 0 0 0 0 0 0 0
1 0 0 1 0 0 1 0
1 0 1 0 0 1 0 0
1 0 1 1 0 1 1 0
1 1 0 0 0 0 0 0
1 1 0 1 0 0 1 1
1 1 1 0 0 1 1 0
1 1 1 1 1 0 0 1



Am i right? If i am totally wrong, you may let me know.
If it is hard to explain it at here, i may go ask for other first.
Thx for helping first =)

Very good! That is the truth table for A[1:0] * B[1:0] = C[3:0]

Now, draw the 4 Karnaugh maps for the 4 output bits, with the 4 input bits on the sides of each K-map. Remember to use the Grey code ordering of the A and B bits along the sides of the K-maps. If you're unclear about K-maps, you can do a search at wikipedia.org
 
berkeman said:
Very good! That is the truth table for A[1:0] * B[1:0] = C[3:0]

Now, draw the 4 Karnaugh maps for the 4 output bits, with the 4 input bits on the sides of each K-map. Remember to use the Grey code ordering of the A and B bits along the sides of the K-maps. If you're unclear about K-maps, you can do a search at wikipedia.org



Draw ?! >.<
I think it will be a big problem to me since i really know nothing for it xD
However, i will try to search for it.
Anyway, may i know what will be the different if it is 4bit or 3 bit multiplier circuit ??
Actually,i don't really know why i have to do like tis
>> * A(1) A(0) B(1) B(0) C(3) C(2) C(1) C(0) *


Why the input only A and B ? I m vry confuse...sry >.<
 
soonsoon88 said:
Draw ?! >.<
I think it will be a big problem to me since i really know nothing for it xD
However, i will try to search for it.
Anyway, may i know what will be the different if it is 4bit or 3 bit multiplier circuit ??
Actually,i don't really know why i have to do like tis
>> * A(1) A(0) B(1) B(0) C(3) C(2) C(1) C(0) *


Why the input only A and B ? I m vry confuse...sry >.<

As you can see from your truth table, a 2-bit multiplier takes two 2-bit numbers as inputs, and generates a 4-bit result. As you go to more bits at the input, the number of output bits increases.

Quiz Question -- If the two inputs are N-bits and M-bits wide, how many bits will be needed for the output?
 
N+M bits right? I'm a newbie to this as well
 
fizzziks said:
N+M bits right? I'm a newbie to this as well

Correct-o-mundo. Good. How's the K-map going?
 
berkeman said:
Correct-o-mundo. Good. How's the K-map going?

Actually...i don't really know how to carry on already >.<
Even the logic circle..i also don't know how to do via the table >.<
 
  • #10
soonsoon88 said:
Actually...i don't really know how to carry on already >.<
Even the logic circle..i also don't know how to do via the table >.<

Well, we don't really teach you how to use K-maps here (at least I don't). Use your textbook or wikipedia to learn the basics, and post your work here if you still have problems getting it to work. You will not learn to learn if we spoon feed you. Show us the reference links you've explored to learn how to do this, and ask specific questions if you have them.
 
  • #11
fizzziks said:
N+M bits right? I'm a newbie to this as well

berkeman said:
Correct-o-mundo. Good. How's the K-map going?

Reviving an old topic here, but shouldn't it be N*M bits? Because a one bit wide times a one bit wide can be represented with a one bit output, and a 2 bit times a one bit is represented by a 2 bit output (you only need two bits to represent "3")?
 
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