An XOR (exclusive OR gate) has two inputs and one output. If only one of the inputs is at level 1, then the output is 1 otherwise the output is 0. The truth table looks like this: A B Out
0 0 0
0 1 1
1 0 1
1 1 0 Exclusive OR represents in logic what "or" means in English; for example, if asked if you want tea or coffee it's usually meant that you can have one or the other - not both.
An XOR gate is a logic gate performing a Boolean logic XOR operation.
Three 2-input XOR gates and one 3-input NOR gate will do the work. Connect each output of each XOR gate to one input of the 3-input NOR gate and apply the two 3-bit words to the inputs of the XOR gates. If X (X2X1X0) and Y(Y2Y1Y0) are two 3-bit words, X2 and Y2 will connect to one XOR gate, X1 and Y1 to the next XOR gate and X0 and Y0 to the last XOR gate. You could see the result of the operation on a LED connected to the output of the NOR gate. Other implementations are also possible of course. The solution above is absolutely correct, but includes a 3 input gate. If the task is to use only two input gates, then a small change will be needed. Take the outputs from any two XOR gates into a 2 input OR gate. Then take the output of the OR gate and the output of the third XOR gate into a 2 input NOR gate. The operation remains identical to the first solution but adheres to the brief of using gates with 2 inputs. In the real world, there is probably no reason to impose such a limitation on a design so the first solution would normally be the preferred route to take.
Out= A'B'C+AB'C'+AC'A'+ABC
An AND gate is a logic gate performing a Boolean logic AND operation.
You use the first HS with entries of x0 and y0 . then you get D0=X0 XOR Y0 and B0=X0 NOT*Y0.(*=AND) Then you use the sacond HS with entries of D0(from the first HS) and B-1(borrow from the previous level). then you get D1=X0 XOR Y0 XOR B-1(exactly like D in FS) and B1=D0 NOT*B-1. Then you use the OR gate with entries B0(from the first HS) and B1(from the second HS) and you get B=B0 OR B-1 AND D0 NOT. If you check the options you can get their similar to the borrow of FS
NAND gates are universal gates and can be used to construct any of the logic gates (AND, OR, NOT, NOR, XOR, XNOR). The easiest way to figure this out is to use basic Boolean Laws. For instance, to create a NOT gate (A'), tie one of the NAND gate's input to logic high: (A+1)' = A'. To create an AND gate (AxB), use two NANDs in series, with the second one configured as an inverter: (AxB) = ((AxB)')'
xor and xnor gates are derived from not gate
yes... xor is derived gate from primary gates
No, XOR gate is a not a universal gate. There are basically two universal gates NAND and NOR.
its not
XOR (Exclusive OR) gate is exclusively for either. OR allows 10, 01, 11. XOR allows 10 or 01, but not 11.
XOR (Exclusive OR) gate is exclusively for either. OR allows 10, 01, 11. XOR allows 10 or 01, but not 11.
1 gate.
Three 2-input XOR gates and one 3-input NOR gate will do the work. Connect each output of each XOR gate to one input of the 3-input NOR gate and apply the two 3-bit words to the inputs of the XOR gates. If X (X2X1X0) and Y(Y2Y1Y0) are two 3-bit words, X2 and Y2 will connect to one XOR gate, X1 and Y1 to the next XOR gate and X0 and Y0 to the last XOR gate. You could see the result of the operation on a LED connected to the output of the NOR gate. Other implementations are also possible of course. The solution above is absolutely correct, but includes a 3 input gate. If the task is to use only two input gates, then a small change will be needed. Take the outputs from any two XOR gates into a 2 input OR gate. Then take the output of the OR gate and the output of the third XOR gate into a 2 input NOR gate. The operation remains identical to the first solution but adheres to the brief of using gates with 2 inputs. In the real world, there is probably no reason to impose such a limitation on a design so the first solution would normally be the preferred route to take.
yes
For 2-input EX-OR gate, if one input is A, the other input is B, and the output is Y. Then the Boolean expression for EX-OR (XOR) function (gate) is Y=A⊕B The output Y is true if either input A or if input B is true, but not both.Y= ( (A and NOT B) or (NOT A and B) ) ;
XOR
Anot + Bnot +Cnot= y not meaning one single line on top of each letter.