GATE 2010 forms are available from 22nd Sep'09. For more information look at this page
http://examcrazy.com/Engineering/GATE/GATE-Important-dates-how-to-apply-2010.asp
in the mnth of sept may b 2nd week to oct
A+b=
yes (a or b)' = a' and b' a or b = (a' and b')' a or b = a' nand b'
To produce a 3-input OR gate when only 2-input OR gates are available: Use 3 OR gates Inputs to Gate A are input 1 and input 2 Input to Gate B is input 3 (if 2 inputs are necessary, include input...
this shows you everything you need about them Pin Number Description 1 A Input Gate 1 2 B Input Gate 1 3 Y Output Gate 1 4 A Input Gate 2 5 B Input Gate 2 6 Y Output Gate 2 7 Ground 8 Y Output Gate 3 9 B Input Gate 3 10 A Input Gate 3 11 Y Output Gate 4 12 B Input Gate 4 13 A Input Gate 4 14 Positive Supply
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) ) ;
a XOR bis equivalent to: (a AND NOT b) OR (b AND NOT a)
NOR gate = not(A or B) = A nor BAND gate = A and BAND gate = not(not A or not B)AND gate = not(not(A or A) or not(B or B))AND gate = (A nor A) nor (B nor B)Therefore using 2 input NORs to make a 2 input AND you need three NORs. If you wanted something different (e.g. a 5 input AND) the above proof can be modified appropriately to get your answer.
In a circuit board, either an AND gate or an OR gate has two inputs and one output. For example, let's call the inputs A and B--these are wires that, at a given time are either hot (have electric current) or cold (have no current). Let's call the output C; it is either hot or cold depending on the gate and the inputs at the time. If the gate is an AND gete, C will be hot only when both A and B are hot. If the gate is an OR gate, C will be hot if either A or B or both are hot.
1. NAND gate is used to invert the input A (by connecting A to both inputs). 2. NAND gate used to invert B the same way 3. Now put A' and B' into into a third NAND gate. The output will be (A'B')' which is equivalent to A+B.
jab form niklenge.............simple...heeheehahahahaha
You would connect the output of the first AND gate to one input of the second AND gate. You are left with 2 inputs on the first AND gate and 1 input on the second AND gate. The final output is from the second AND gate.