a) If we were using a 4-bit adder.

BCD code inputs to provide an output of EX-3. (ignoring the carry out bit)

BCD – in

Constant EX-3

BCD

C-out C-in

0

EX-3 code out

(Remember the EX-3 code was BCD with a little Extra)

b) Using the constant you provided for part (a) demonstrate that the solution did

convert BCD to Excess-3. (ignoring the carry out)

BCD + constant = Excess-3

(BCD 0 = the EX-3 representation of 0) 0000₂ + = 0^{Carry out} 0011

Using a representation of a 4-bit adder chip given below, one of the variable input’s will be excess-3 code with the output of the ADDER being in BCD. Calculate the constant that would be used to ADD together with the excess-3 code that will provide the conversion of the excess-3 code back to BCD using the 4-bit adder chip representation shown below.

Excess-3 code here

4 bit adder chip

C out C in = 0

BCD output here

Given the following Multiplexer chips: Set up as a 16×1 configuration,
1. Label inputs and outputs and label control inputs

8×1

Chip

A

0 0 0 0

0 0 0 1

0 0 1 0

8×1

Chip

B

2×1

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1 1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

Need assistance with this?

Solution