Achieve Logic Condition

Hi,

Anyone can help me on Logic design?

I only have 2 input and 2 output. The table is showing below. How can we achieve the logic with using simple design. Is it any alternative way to do it?

it is really simple just write karnuoph's map if u cannot get it by sight:
O/p2 = i/p1 AND i/p2
O/p1= i/p1 AND (NOT i/p2)

Hi,

Do you got any ideas... Use 1 chip to achieve ths condition?

In this market we can get one type off IC which can make our own equation but i already forgeted! :|

Thank You.

hello ,
i don't know the exact IC , but u can check here i see both the inverting stage and and gates but in two ics maybe u can check other pages in the site to see if there is an integhrated inverteer along with and gates, but u know buying an inverter isn't that big deal

Hi,

I can't found integration NOT-gate and AND-gate ic.

Do you know how to use Programmable Logic Array(PLA)?

Thank You.

Help,
Use a quad 2-input NOR Gate. To simplify the notation, let
.
A = 1i/p
B = 2i/p
C = 10/p
D = 20/p
' Denotes complement
.
By deMorgan's theorem:
AB = (A' + B')'
AB' = (A' + B)'
.
Use one section of the NOR gate to produce A'
Use the other section of the NOr gat to produce B'
.
The other 2 sections generate the outputs C, D directly.
Regards,
Kral

yes .. U can implement Ur logic in a single IC pack... Convert Ur logic function into NOR logic.

O/p 2 = (not I/P1) nor (not I/P2);
O/P1= (not I/P1) nor (I/P2)

THE TERM (not I/P1) CAN BE REALISED FROM A SINGLE NOR GATE (not IS PERFORMED BY PROVIDING THE SIGNAL TO BOTH INPUTS OF THE nor gate)
THE TERM (not I/P2) CAN BE REALISED FROM A SINGLE NOR GATE
THE REST OF THE TWO NOR LOGICS CAN BE REALISED USING TWO nor gates..
SO IN TOTAL WE NEED 4 nor gates.. THE IC7402 HAS FOUR NOR GATES IN A SINGLE PACKAGE ..

Hi,

Thank You very much!

I can understand the equation and the circuit that you all provided.

But can i know how can we get the equation from my logic-table? Is there any skill to derive the equation?

Thank You

See here - logic gate and Laws of Boolean Algebra

**broken link removed**

See also BOOLE-DEUSTO Logic Minimizer software

Hi,

The BOOLE-DEUSTO Logic Minimizer software is very usefull and very good. Thank You.

I try to play around with the software but this software just given us the standard equation after convert from Karnaugh map: (as safwatonline given)
O/p2 = i/p1 AND i/p2
O/p1= i/p1 AND (NOT i/p2)

Anyone know how to use this software to convert the equation to..
O/p 2 = (not I/P1) nor (not I/P2)
O/P1= (not I/P1) nor (I/P2)


Thank You.

i think it can be solved very easily using k map

hello,
A and B= A' (not and) B'=A' nor B'
simillarly :
A or B= A' (not OR) B' =A' nand B'
this is a general rule.
so in ur case O/p1= i/p1 AND (NOT i/p2)
then O/p1= i/p1' NOR (NOT i/p2)'=i/p1' NOR i/p2

where i used x'=not x
and (not OR) is OR gate with inverters at its input
and (not AND) is AND gate with inverters at its input


in general:
if X=AB+CD then X=(A' nor B') OR (C' nor D')
then X= (A' nor B')' NAND (C' nor D')'=
then X=(A nand B) nand (C nand D)
so any X=AB+CD=(A nand B) nand (C nand D)

similarly
any X=(A+B) nand (C+D)= (A Nor B) nor (C Nor D)

also X'= not X= x nand x = x nor x
regards

Hi,

Thank You for your explaination but i got abit blur!

I know the...
By deMorgan's theorem:
AB = (A' + B')'
AB' = (A' + B)'
is standard and general rule!

Can i know, how can we know AB equation is belong to O/p2 and AB' equation is belong to O/p1 from my logic table.

Thank You.

I think the file I have attached with this will clear Ur doubts

Hi,

Everyone, now then i clear my doubts already!

Thank You very much.

simply u solve by k-maps, the o/p can be expressed by two ways as either sum of products or products of sum.
to do that lets examin the o/p1 for example:
o/p1 is equal one only when A=1 AND B=0 , this means that o/p1=A AND B' cause if u put any thing in this equation other than A=1 And B=0 u will get o/p=0, as u can see i expressed the o/p as sum of products by considering products as the number of times o/p will be equal to one (which is one time in our case) then get expression for each product (by putting the condition and setting A for A=1 and B' for B=0, this shoud be repeated if there is more than one product , i.e. lets say if o/p =1 also at A=0 and B=0 the another expresssion will be o/p=A'B' note products are represented by multiplication of variables and note both A and B are bared(A' and B') cause both are zero) then the last step is to put the final o/p expression as the sum of all products (in this case only AB' so o/p=AB',,, if the example was inluding A'B' then o/p=A'B'+AB')
quick representation to the second o/p2 (only one at A=B=1 the o/p2=AB)

note this can be reversed by representing by products of sums , here u consider when the o/p is equa; to zero (rather than one in the first case) and u put every variable equal to zero as not bared and every vairable equal to one as bared

quick solution to o/p1 by second methode:
o/p1 =0 at (A=0 and B=0) OR (A=1 and B=1) OR(A=0 and B=1) this really is the expression just replace each and by product and each OR by sum also each variable equal one by bared and each variable equal zero by unbared
so o/p1=(A+B)(A'+B')(A+B')
if u simplify the previous expression u should get AB'

note if u solve by k-maps u get simplified expressions from the first time

Added after 1 minutes:

Too late reply , anyway hope this helps a little bit

Hi,

Thank for your good explanation. Now i can understand very well already. Anything i will further with u some more....

Thank for your helping...


Can i ask u 1 quation?

Can we use 1 type of gate to design whatever we want from the equation/K-map? Got any method or tactic to solve it? Like my care just use one type of logic NOR gate 7402IC to implement the design.

Thank You

Easy way to get logic equation is to see wher the o/p is 1
o/p1 is 1 when A=1,B=0 so o/p1=A and B'
o/p2 is 1 when A=1,B=1 so o/p2=A and B.

Help,
By inspection of the logic table, there is only one combination of inputs for which 1/op = "1". This is the state where 1/ip = A = "1" AND 2/ip = B = "0". This is where the 1/op = (1/ip)(2/ip)' = AB' term copmes from.
.
Similarly, there is only one condition under which the 2/op term = "1". This is the state where 1/ip = A = "1" AND 2/ip = B = "1". this is where the 2/op = (1/ip)(2/ip) = AB term comes from.
Regards,
Kral