Question

Q5.3 3 Points Show an implementation of f using the minimum number of 2-1 multiplexers and inverters, i.e., you can only use 2-1 MUXs and inverters to implement f. Please select file(s) Q5.4 3 Points Show an implementation of f using a single 4-1 MUX and inverters only. You must use Input ' c ' as the MSB of the selection line and Input ' a ' as the LSB of the selection line. Please select file(s)

Answer 1

To implement f using the minimum number of 2-1 multiplexers and inverters, we can use the equations for Junction b and Junction e. Since the equations are equivalent, we only need to keep one of them.

Step-by-step explanation:

We can use the equations for Junction b and Junction e to implement f with the fewest 2-1 multiplexers and inverters possible. Only one of the equations needs to be kept because they are equivalent. Now let us look at the equation for Junction b: I₁ = I₂ + I₃.

We can connect I₃ as the input to the multiplexers' data inputs and I₃ as the input to the selectors of the multiplexers in order to implement this equation using 2-1 multiplexers. The values of I₁ in the equation will be the multiplexers' outputs.

For example, if we have two 2-1 multiplexers, we can connect I₂ to the select line of one multiplexer, I₃ to the select line of the other multiplexer, and I₁ to the data inputs of both multiplexers. The outputs of the two multiplexers will be the values of I₁ = I₂ + I₃.