Digital-Grand Test -Q4

+1 vote

Assume you need a 64-to-1 multiplexer controlled by 6 select signals A through F, i.e., a binary number ABCDEF specifies the index of the input to be sent to the output. However, as we assume a logic gate can not have more than 8 inputs, such a MUX is not directly available. So you build this 64-to-1 MUX using 4-to-1 mux, such that minimum number of 4-to-1 mux are used.  We build this 64-to-1 MUX using three levels of 4-to-1 MUXs where level 3 has one 4-to-1 mux and its output is the output of the original 64-to-1 mux. 

The realization of this 64-to-1 mux using three levels of 4-to-1 muxes is such that Level i (i = 1,2,3) is controlled by X_iY_i select signals i.e. In level i, the most significant select line of each 4-to-1 mux is connected to X_i and the least significant control line of each 4-to-1 mux is connected to Y_i.  

To select input 27 i.e. (ABCDEF = 011011) and send it to the output, the value of X_1Y_1X_2Y_2X_3Y_3 should be

(A). 011011

(B). 110110

(C). 111001

(D). 100111

asked Jun 10 in Digital by gbeditor (44,560 points)
reshown Jun 12 by gbeditor

1 Answer

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Best answer

Use three levels of 4x1 MUXs. 16 in first level controlled by EF to receive 64 inputs and generate 16 outputs, 4 in 2nd level controlled by CD to receive 16 outputs from first level and generate 4 outputs, and one in last level controlled by AB to select one of the four outputs from the 2nd level as the final output.

So, X_1Y_1 \equiv EF ; X_2Y_2 \equiv CD; X_3Y_3 \equiv AB

Hence, answer is Option C.

NOTE : A common mistake student might do is that they might assign AB to the select lines of first level. This will not give desried behavious of Mux.

answered Jun 25 by deepak-gatebook (112,760 points)