How to realise F = ~A*B + ~B*C using 4-input multiplexer...

[student2005]

Would someone pls show me how the function F = ~A*B + ~B*C could be realised using a 4-input multiplexer and some logic gates? Many thanks.

 

[lokeyh]

Re: How to realise F = ~A*B + ~B*C using 4-input multiplexer

Check out the part on Multiplexers as Universal Combinational Module at https://homepages.ius.edu/JFDOYLE/c421/html/Chapter9.htm

 

[student2005]

Do you think we can implement the function F with 2-input muliplexer? Thanks.

I have read the article, but still could not get the clue. Possible to show how a 4-input multiplexer could be used to implement the function F? Many thanks.

 

[lokeyh]

Re: How to realise F = ~A*B + ~B*C using 4-input multiplexer

F = ~A*B + ~B*C

If B=0, ~B=1, which means that ~A*B is always 0 and ~B*C depends on C. Hence F will only depend on C.
If B=1, ~B=0, which means that ~B*C is always 0 and ~A*B depends on ~A. Hence F will only depend on ~A.

So B is the select of the multiplexer and ~A and C are the inputs of the multiplexer. You can just use 2-input multiplexer.

 

[student2005]

F = ~A*B + ~B*C

If B=0, ~B=1, which means that ~A*B is always 0 and ~B*C depends on C. Hence F will only depend on C.
If B=1, ~B=0, which means that ~B*C is always 0 and ~A*B depends on ~A. Hence F will only depend on ~A.

So B is the select of the multiplexer and ~A and C are the inputs of the multiplexer. You can just use 2-input multiplexer.

Since 2-input multiplexer can do the job, so I think 4-input multiplexer sure will do the job as well, right?

If it's true, how am I going to configure the 4-input multiplexer in order to implement the function F mentioned above?

Many thanks...

Added after 4 hours 48 minutes:

Pls advise whether the circuit configuration shown in the figure could implement the function F or not.

In my opinion, there are many ways to implement function F with ONE 4-to-1 MUX and additional logic gates. Am I correct? Pls advise. Thanks.

 

[timer]

Re: How to realise F = ~A*B + ~B*C using 4-input multiplexer

dear student2005, i'm afraid that the config you've used for the 4:1 mux would not yield the desired function.
according to your design, when B=0 (and thus S1S0=00), IO is selected to give the output as (~A), while according to the function, the output must be C.
Similarly, when B=1, I1 is selected to give the output as (~C), while acording to the function, the output must be (~A).
So just connect I0 to C and I1 to ~A, with S1 grounded & S0 tied to B. That will give you the desired function.

There are many ways to implement a function using MUXs. An alternative way would be to choose A & B as the select lines, tying I0 & I2 to C, I1 to logic HIGH and I3 to GND. All this becomes clear when you write the truth table.

Hope this helps.

 

[sp]

dear student2005, i hav read ur PM...reply u here...

my suggestion is tht u do the truth table for the boolean equation for the function:

F = ~A*B + ~B*C equivalent to 
F = ((NOT A) AND B) OR ((NOT B) AND C)

it is hard to use words to describe the behaviour of the equation... it is misleading n sometimes u will miss out something,,, so truth table please...and u will see clearly how the output respond to the input...

and then u can choose which to b ur select line & which to b ur inputs..

actually dont let the physical MUX output wth equation form confuse u... when u know wad type of outputs truth table is w.r.t ur inputs.... u can just output(verb here the desire output value according to ur truth table...

and again... we dont do homework for u here... u hav to help urself so tht we can help u...

warm regards,
sp

 

[fidjouss100]

Re: How to realise F = ~A*B + ~B*C using 4-input multiplexer

you can also go this way

control bits [B,C] b:Most significant

I0=0 (Gnd)
I1=1 (Vdd)
I2=I3=~A

 

[student2005]

Don't worry, it's not a homework. I'm not in school anymore. When I have this problem, I have no one to ask and therefore I ask in this forum. Thank you guys...