draw logical diagram 2 to 4 line decoder with nor gates only
You need 9 3-to-8 decoders. 8 decoders for selecting one of 64 lines. 1 decoder for enabling 1 decoder out of 8 decoder.
for a two input gate to represent as an n-input gate excatly n-1 two input gates are required. this implies that for a two input OR gate to represent a four input OR gate exactly three two input OR gates are required let F is =a+b+c+d =(((a+b)+c)+d) =((a+b)+(c+d)) in both the above cases + is used three times so three two input OR gates make a four input OR gates. This discussion doesnot hold good for NAND gates an example can illlustrate the reson:- take F=(a.b.c.d)'=a'+b'+c'+d' --------------------------->(1) (this is obtained by a four input NAND gate) let us take this in the manner we did it for an OR gate and we will then verify the result. =((a.b)'(c.d)')' =((a'+b').(c'+d'))' =(a'+b')'+(c'+d')' =ab+cd <------------------------(2) (1)is not equal to (2) so we can say that a NAND gate cannot be replaced in the manner as OR gate is replaced
Design and draw a combinational circuit using AND-OR-NOT gates that accepts 4 input bits, and produces two bit output; the first of the two bits is set to 1 if the number of 1's in the input is even; and the second of the output bit is set to 1 if the input have 3 or more (all four) 1 bits.
It is a data selector. There are 16 digital input lines, 4-bit decoder, strobe and output pin. So you put a 4-bit binary number from 0-15 into ABCD bits and the corresponding input value is found on output at the strobe time.
Amplifiers comes in IC packages diagram is triangle with +/- input by adding a resistor r1 in series to the - input the gain can be found by adding a resistor from output to the - input The + terminal must be satisfied with the proper resistor
draw the logic diagram of 2 to 4 line decoder decoder using nor gates include enable input
You need 9 3-to-8 decoders. 8 decoders for selecting one of 64 lines. 1 decoder for enabling 1 decoder out of 8 decoder.
the basic difference is that there is 1 input line for the demux whereas a decoder has no input lines
A 4-input decoder can produce (2^n) outputs, where (n) is the number of inputs. For a 4-input decoder, (n = 4), so the number of possible outputs is (2^4 = 16). Therefore, a 4-input decoder can generate 16 distinct output lines based on the 4 input combinations.
An n-to-2^n decoder has n input lines. Each combination of the n input lines corresponds to one of the 2^n output lines being activated. Thus, for a decoder to function correctly, it requires exactly n input lines to decode the binary input into a specific output line.
In a decoder circuit, only one output is activated at a time because it is designed to represent a unique binary input combination. The logic gates within the decoder ensure that each specific input corresponds to a single output line being high (active), while all other outputs remain low (inactive). This one-hot encoding scheme prevents ambiguity and ensures clear and distinct representation of the input state, which is crucial for accurate digital processing and control.
To create a decoder with symbols, you need to assign a unique symbol to each input combination of the decoder. For example, in a 2-to-4 decoder, you could use symbols like A, B, C, and D to represent the output signals corresponding to the input combinations. By using symbols that are easy to understand and differentiate, you can effectively represent the decoder's logic and functionality.
An 8 to 1 multiplexer (MUX) is a digital switch that selects one of eight input lines and forwards the selected input to a single output line based on three selection lines. The logic diagram consists of eight input lines (I0 to I7), three selection lines (S0, S1, S2), and one output line (Y). The selection lines determine which input is connected to the output, with each combination of the selection lines corresponding to one of the input lines. The logic gates used in the diagram typically include AND, OR, and NOT gates to implement the necessary connections and selection logic.
connect S1 and S0 to a 2x4 active high decoder then connect each output of the decoder with the enable of each three-state buffer. Each of with has an input of D0,D1,D2,D3 respectively. Connect the outputs of each with an OR gate (since only one can be active at a time, whichever's active will be the output).
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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...