Draw the diagram for the decoder question below ($($Digital System)$)$:b.$.$ Decoders 3-
to−8: 

 

Combine two 3$-$to−8$<br>$ decoders to create a 4$-$to−16$<br>$ decoder.

Connect inputs A and B to the select lines of the first decoder.

Connect inputs C and D to the select lines of the second decoder.

Enable only the outputs corresponding to the minterms (0,2,4,6,8,10) of the function.

OR the enabled outputs together to produce F($($A,$,$ B,$,$ C,$,$ D).$).$

Here's a text−$-$based description of the diagram for implementing F($($A,$,$ B,$,$ C,$,$ D)$)$ using 3$-$to−8 decoders:

Diagram:

Inputs: A,$,$ B,$,$ C,$,$ D

Output: F($($A,$,$ B,$,$ C,$,$ D)$)$

Components:

Two 3-to−8$<br>$ decoders

Six OR gates

Connections:

Connect inputs A and B to the select lines ($($S0$0$ and S1)$1)$ of the first decoder.

Connect inputs C and D to the select lines ($($S0$0$ and S1)$1)$ of the second decoder.

Enable only the following outputs of the decoders using AND gates:

Output 0$0$ of both decoders (0000)$(0000)$

Output 2$2$ of both decoders (0010)$(0010)$

Output 4$4$ of both decoders (0100)$(0100)$

Output 6$6$ of both decoders (0110)$(0110)$

Output 8$8$ of both decoders (1000)$(1000)$

Output 10$10$ of both decoders (1010)$(1010)$

Connect these enabled outputs to a 6$-$input OR gate.

Connect the output of the OR gate to the output F($($A,$,$ B,$,$ C,$,$ D).$).$

Explanation:

The decoders create 16 possible output combinations based on the 4$4$ inputs A,$,$ B,$,$ C,$,$ and D.$.$

By enabling only the outputs corresponding to the minterms of the function (0,2,4,6,8,10), we ensure that the output F($($A,$,$ B,$,$ C,$,$ D)$)$ is only 1 when one of these minterms is true.

The OR gate combines these enabled outputs, producing a final output of 1$1$ only when one or more of the minterms are true, correctly implementing the function.

Answer rating: 100% (QA)

Here is the diagram for implementing FA B C D using two 3to8 decoders to create a 4to16 decoder Inpu... View the full answer