) Show how you would build a 4:1 multiplexer out of 2:1 multiplexers. (b) Show how you would build an 8:1 multiplexer out of 4:1 and 2:1 multiplexers. Use as few muxes as possible in your solution.

a. The answer to this question is attached. the number first attached picture.

Explanation:

a. You will need four inverters (NOT gates), four 5

input AND gates (well, they don't come as 5-

input devices, so either use cascaded 2-input or

3-input or four 8-input AND gates) and a 4-input

OR gate. The inputs to the decoder will

represent the MUX select lines. Let's call the

outputs from the decoder A, B, C, and D and lets

call the inputs to the MUX, INO, IN1, IN2 and IN3,

and lets call the MUX output OUT. So, by

implementing the following logic equation, we

will perform the function of a 4:1 MUX.

OUT (NOTA.NOTB.NOTC.D.INO)+

(NOTA.NOTB.C.NOTD.IN1)+

(NOTA.B.NOTC.NOTD.IN2)+

(A.NOTB.NOTC.NOTD.IN3)

Here NOTA = inverted A etc. and ""=AND and

"+"=OR

b. So, we require two 4x1 Multiplexers in first stage in order to get the 8 data inputs. Since, each 4x1 Multiplexer produces one output, we require a 2x1 Multiplexer in second stage by considering the outputs of first stage as inputs and to produce the final output.

Let the 8x1 Multiplexer has eight data inputs I7 to I0, three selection lines s2, s1 & s0 and one output Y. The Truth table of 8x1 Multiplexer is shown on the last picture attached.

The same selection lines, s1 & s0 are applied to both 4x1 Multiplexers. The data inputs of upper 4x1 Multiplexer are I7 to I4 and the data inputs of lower 4x1 Multiplexer are I3 to I0. Therefore, each 4x1 Multiplexer produces an output based on the values of selection lines, s1 & s0.

The outputs of first stage 4x1 Multiplexers are applied as inputs of 2x1 Multiplexer that is present in second stage. The other selection line, s2 is applied to 2x1 Multiplexer.

If s2 is zero, then the output of 2x1 Multiplexer will be one of the 4 inputs I3 to I0 based on the values of selection lines s1 & s0.

If s2 is one, then the output of 2x1 Multiplexer will be one of the 4 inputs I7 to I4 based on the values of selection lines s1 & s0.

Therefore, the overall combination of two 4x1 Multiplexers and one 2x1 Multiplexer performs as one 8x1 Multiplexer.

the answer is e ; )(: