How to become

Question

I have this project, listen below, and I'm not sure where to start, maybe someone can give me some pointers or maybe point me in the right direction to start this? Thanks!!

Input: A, B = octal digits (see below); Cin = binary digit

Output: S = octal digit (see below); Cout = binary digit

Task: Using binary FAs, create a circuit that acts like an octal FA . More specifically, this scheme will introduce two octal digits A, B, convert them to binary numbers, add they use only binary FAs, convert the binary result back to octal and display the sum as an octal digit and a binary operation.

binary representation of octal digit input / output

Each octal digit will be represented using the following 8-bit binary representation:

octal 8-bit input lines:

Digit: 0 1 2 3 4 5 6 7
0 1 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0
2 0 0 1 0 0 0 0 0
3 0 0 0 1 0 0 0 0
4 0 0 0 0 1 0 0 0
5 & ​​nbsp; 0 0 0 0 0 1 0 0
6 0 0 0 0 0 0 1 0
7 0 0 0 0 0 0 0 1

You must design the circuit in a structured way.

Answer 1

Well, that's why essentially you are asked to design an 8-in-3 encoder and a decoder from 3 to 8. Because you are given the FA to work with it for other purposes.

First we need to determine how the encoder and decoder work. So, we will build a truth table:

encoder:

Input | Output 01234567 | 421 ----------------- 10000000 | 000 01000000 | 001 00100000 | 010 00010000 | 011 00001000 | 100 00000100 | 101 00000010 | 110 00000001 | 111 

and the decoder is just the opposite.

Next, how do we build our encoder? Well, we can just attack it one bit at a time.

So, for the 1st digit we have, if bit 1, 3, 5 or 7 is entered, then it is 1, otherwise it is 0. Therefore, we just need a giant OR with 4 inputs connected to 1, 3, 5 and 7.

For the 2s digit, we need an OR gate connected to 2, 3, 6, 7. Finally, for the 4th shutter, connect them to 4, 5, 6, 7. This does not do any error checking to make sure the extra bits are not set . Although, the behavior in this case looks undefined by specification, so it’s probably OK.

Then you take three lines and feed them to your adders. It’s easy, so I won’t go into it.

Finally, you need a decoder, it's a little more complicated than an encoder.

Look at the decoder truth table:

 Input | Output 421 | 01234567 ---------------- 000 | 10000000 001 | 01000000 010 | 00100000 011 | 00010000 100 | 00001000 101 | 00000100 110 | 00000010 111 | 00000001 

This time we can’t just use 3 or the gate and call it day.

Write this in a C-like code:

 if (!input[0] && !input[1] && !input[2]) output[0] = 1 if (input[0] && !input[1] && !input[2]) output[1] = 1 if (!input[0] && input[1] && !input[2]) output[2] = 1 if (input[0] && input[1] && !input[2]) output[3] = 1 if (!input[0] && !input[1] && input[2]) output[4] = 1 if (input[0] && !input[1] && input[2]) output[5] = 1 if (!input[0] && input[1] && input[2]) output[6] = 1 if (input[0] && input[1] && input[2]) output[7] = 1 

So, it looks like we will be using 8 3 input I-gates and three NOT-gates!

This is a bit more complicated, so I made an example implementation:

Answer 2

If the conversion must be done manually in the class, you can try the following.

Convert Octal to binary:

To convert octal to binary, replace each octal digit with its binary representation. Example: Converting 518 to binary: 58 = 1012 18 = 0012 Therefore, 518 = 101 0012.

Convert binary to octal:

The process is the reverse of the previous algorithm. Binary digits are grouped in triples, starting with the decimal point (if any) or the last digit and go left and right. Add leading 0s (or trailing zeros to the right of the decimal point) to fill in the last group of three, if necessary. Then replace each trio with an equivalent octal digit.

Example: convert binary code 1010111100 to octal: (Adding two leading zeros, number - 001010111100) 001 = 1, 010 = 2, 111 = 7, 100 = 4 Therefore 1010111100 = 1274

Answer 3

To convert to and from octal, you can use a pair of encoder and decoder ( http://www.asic-world.com/digital/combo3.html ). A 3-bit adder can be made from 3 FA chain.