Implement decoder using simplified SOP

Question

Implement Z using single 2x4 decoder with 1-enable and active low outputs without logic gate and no complemented literals

Z = A'BC'DE + (D'+B)' + BDE

My attempt at Simplified SOP

Z = A'BC'DE + (D'+B)' + BDE
Z = A'BC'DE + DB' + BDE
Z = BDE(A'C+1) + DB'
Z = BDE + DB'

Answer:

enter image description here

I have the answer but i can't seem to figure out how to use the values to construct the decoder

• Yet Another Michael: This question is confusing... you say you need to implement a decoder with logic, but it seems the solution wants no logic and only the decoder. Where do A and C come from? If I were to guess, the question asked to implement the logic function: Z = D and B and not(E), which is implemented by the decoder in the answer.
• Mason: @Michael A and C come from Z. Updated the question thank you!

Answer 1

From what you have reached for the simplified SOP, you can see that "D" is present in both terms. So that can hint you to use it as the Enable input for the decoder, as the enable input would be AND ed with each minterm. Using D as the enable input leaves only B and E to be used as the inputs to S1 and S0; so you can assign them to the inputs arbitrarily(either (S1 = B, S0 = E) or (S0 = E, S1 = B). Then you just have to build Z using the outputs of the decoder(based on the S1 and S0 you have chosen).

Expanding what you have reached for the SOP, you get:(Oi is the i-th output of the decoder)

Z = D(BE + B') = D(BE + B'E + B'E') = O0 + O1 + O3

It appears that the Z marked on the image in your question is the complement of the Z in the equation.