# PLC Program to Convert Gray Code to Binary

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This is a PLC Program to Implement Gray Code to Binary Conversion.

Problem Description

Implementing Gray Code to Binary conversion in PLC using Ladder Diagram.

Problem Solution
• Writing truth table showing the relation between Gray Code as input and Binary as output.
• The MSB of Gray and Binary are same, so pass the same bit directly.
• Add Binary MSB to the next bit of Gray code. Record the result and ignore the carries.
• Similarly continue until the LSB is reached.
• To obtain these equations, Karnaugh-Map method is again used.
• For each Binary output D3, D2, D1 and D0, write Karnaugh-Map.
• From the K-Map, obtaining a simplified expression for each Binary output in terms of Gray Code inputs.
• Realize the code converter using the Logic Gates.
• By following the process to convert Gray Code into Binary, Truth Table can be written as given below.

Truth Table relating Binary to BCD

```Decimal	Gray Code Input	                Binary Output
D3	D2	D2	D0	B3	B2	B1	B0
0	0	0	0	0	0	0	0	0
1	0	0	0	1	0	0	0	1
3	0	0	1	1	0	0	1	0
2	0	0	1	0	0	0	1	1
6	0	1	1	0	0	1	0	0
7	0	1	1	1	0	1	0	1
5	0	1	0	1	0	1	1	0
4	0	1	0	0	0	1	1	1
12	1	1	0	0	1	0	0	0
13	1	1	0	1	1	0	0	1
15	1	1	1	1	1	0	1	0
14	1	1	1	0	1	0	1	1
10	1	0	1	0	1	1	0	0
11	1	0	1	1	1	1	0	1
9	1	0	0	1	1	1	1	0
8	1	0	0	0	1	1	1	1```

Boolean expression for each BCD bits can be written as

``` B3= m(8, 9, 10, 11, 12, 13, 14, 15)
B2= m(4, 5, 6, 7, 8, 9, 10, 11)
B1= m(2, 3, 4, 5, 8, 9, 14, 15)
B0= m(1, 2, 4, 7, 8, 11, 13, 14)```

Karnaugh-Map for each output

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Realizing code conversion using Logic Gates

PLC Program

Here is PLC program to Implement Gray Code to Binary Conversion, along with program explanation and run time test cases.

```List of Inputs and Outputs
D3=		I:1/0	(Input)
D2=		I:1/1	(Input)
D1=		I:1/2	(Input)
D0=		I:1/3	(Input)
B3=		O:2/0	(Output)
B2=		O:2/1	(Output)
B1=		O:2/2	(Output)
B0=		O:2/3	(Output)```

Ladder Diagram to obtain Binary output

Program Description
• Simply by looking at the logic circuit, ladder diagram is made.
• RUNG000 for output bit B3 (O:2/0) is as described earlier that the MSB of Gray and Binary are same, so passed the same D3 (I:1/0) bit directly.
• By simplifying Boolean expressions, we can see that the output of RUNG001 is fed to input of RUNG002 and EX-ORed with D1 (I:1/2) bit.
• Similarly in RUNG003 comprises of RUNG002’s output as input and EX-ORed with bit D0 (I:1/3).
Runtime Test Cases
```Decimal	Gray Code Input	                Binary Output
D3	D2	D2	D0	B3	B2	B1	B0
0	0	0	0	0	0	0	0	0
1	0	0	0	1	0	0	0	1
3	0	0	1	1	0	0	1	0
2	0	0	1	0	0	0	1	1
6	0	1	1	0	0	1	0	0
7	0	1	1	1	0	1	0	1
5	0	1	0	1	0	1	1	0
4	0	1	0	0	0	1	1	1
12	1	1	0	0	1	0	0	0
13	1	1	0	1	1	0	0	1
15	1	1	1	1	1	0	1	0
14	1	1	1	0	1	0	1	1
10	1	0	1	0	1	1	0	0
11	1	0	1	1	1	1	0	1
9	1	0	0	1	1	1	1	0
8	1	0	0	0	1	1	1	1```

Sanfoundry Global Education & Learning Series – PLC Algorithms.
To practice all PLC programs, here is complete set of 100+ PLC Problems and Solutions.

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