The boolean expression f(x,y,z) in its canonical form for the decoder circuit shown below is

Question

Note : the bubbles are NOT gates

Answer 1

It will be better if we simplifies it as follow 

Now it can be solved easily.

Note that Decoder produces minterms, therefore product of minterms should be zero.

∴ F = F2 ( due to F₁ = 0 )

F2 = ∑m(3,4,6,7).

But note that MUX will be enable if z = 0 only

===> 3 and 7 are invalid, in F2

so,  $F(x,y,z)= \Sigma_m(4,6)$ 

Comments

sir,

1,3,5,7 decoder output will be  "don't care "  why you are taking as 0 . ans will be same beacuse dont care are not minterms or maxterms .

this is new for me may be i am wrong please correct me if i am?

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@Praveen Saini@Vaishali Trivedi Yes, I too have the same doubt as to the Vaishali. Can someone please clarify?

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you need to check the internal structure of a decode with enable bit. E(enable) is connected with all ANDs inside of it, if E is 0, all outputs of decoder remains at 0.

Answer 2

Expression at OR gate would be $\overline{0}+\overline{1}+\overline{2}$

Expression at AND gate would be $\overline{3}*\overline{4}*\overline{6}*\overline{7}$

f(x,y,z)=$\overline{(\overline{0}+\overline{1}+\overline{2}) * (\overline{3}*\overline{4}*\overline{6}*\overline{7})}$

=$=\overline{\overline{0}+\overline{1}+\overline{2}} + \overline{\overline{3}*\overline{4}*\overline{6}*\overline{7}} =0*1*2 + (3+4+6+7)$

Since decoder is active at 1,3,5,7 with output bubbled each will give 0 at D1,D3,D5,D7 and 1 in rest of the outputs.

so function simplifies to 4+6 , Hence the answer $\sum m(4,6)$

Answer 3

option c is right.