OR gate gives the output 1 whenever one of its input is 1.
An inverter has a high output when the input is low, and a low output when the input is high.
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.
IC 7402 is different from the other type of IC like 7404,7408,7432 and 7400 when it is connected to have an output desired... if you noticed all i mentioned ic is connected from left to the right. input pin 1 and 2, output pin 3 input pin 4 and 5, output pin 6 input pin 13 and 12, output pin 11 input pin 10 and 9, output pin 8 while in nor gate, to have the desired output it must be connected from right to left... input pin 2 and 3, output pin 1 input pin 5 and 6, output pin 4 input pin 8 and 9, output pin 10 input pin 11 and 12, output pin 13
A nor gate provides an output of 0 when any input is 1.Nor gate provides the opposite of or gate. An or gate provides a 1 or true output when any of the inputs is 1 or true. Therefore the opposite output would be provided by a nor 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.
any instrument gives instanteneous result means no time lag in between the input and output, as you give the input you get the output there and then
An AND gate gives a 1 or True output only and only if all the inputs are True or 1. If even a single output is False or 0, it will not give any output meaning thereby that the output will be False or 0. A NAND gate is quite the reverse of an AND gate and gives an output of 1 or True only if both the inputs are not True or 1. This means that even when both inputs are 0 or False, it still gives a True or 1 output.
One way to make an eight input AND gate out of transistors... Start with one transistor, NPN. Ground the emitter. Connect the collector to Vcc with a resistor. Connect the base to Vcc through two resistors, picked to drive the transistor into saturation. The input of that stage is the junction of the two resistors. If you ground that input, the transistor cuts off, and the collector goes high. Unground the input, and the collector goes low. OK. Now you have an inverter. Build eight of them, but only use one collector resistor, and tie all of the collectors together. If any one input is high, the output is low. If all inputs are low, the output is high. OK. Now you have an 8 input negative logic NAND gate. Follow it up with another inverter stage, reversing the output. OK. Now you still have an 8 input NAND gate, with the output high true. If you want the inputs to be high true also, connect each one to an inverter. This will use 17 transistors, and you will have an 8 input positive logic AND gate.
A 4-input majority logic gate outputs a high signal (1) if the majority of its inputs (at least 3 out of 4) are high (1). The truth table for a 4-input majority gate includes 16 rows, reflecting all possible combinations of the four inputs (A, B, C, D). The output is 1 for the following input combinations: 1110, 1101, 1011, 0111, 1111, and any other combination that has at least three 1s. The output is 0 for combinations with fewer than three 1s.
The trackball would be an input device. Any device that takes information from the outside world into the computer is an input device. Any device that takes information from the computer and gives it to the outside world (such as a monitor or printer) is an output device.
as voltage on the gate increases it will reach a point where any further input will not effect further.
To design an OR using 2:1 mux, we need to tie the "First" input to "Logic 1″ and the "Zeroth" input to the one of the input of the OR Gate. The other input of OR gate would be connected with the select line of the MUX. Now, the output of the MUX would be "1″ when any oth the two inputs would be "1″ otherwise it would be "0″ for all conditions.