(5 votes, average: 3.6 out of 5)
Loading ...
Given that we have discussed Binary to Gray code conversion, let us discuss the Gray to BInary conversion.
Conversion from Gray code to natural Binary
Let be the equivalent Gray code for an
bit binary number
with
representing the index of the bit.
1. For ,
i.e, the most significant bit (MSB) of the Gray code is same as the MSB of original binary number.
2. For ,
i.e,
bit of the Binary number is the exclusive-OR (XOR) of
of the bit of the Gray code and
of the bit of the binary number.
Simulation model
Look Up Table based Matlab/Octave mode for Gray to Binary conversion
clear
% binary to gray code conversion
ipBin = [0:15] ; % decimal equivalent for a 4-bit binary
opGray = bitxor(ipBin,floor(ipBin/2))
% Gray to Binary conversion
[tt ind] = sort(opGray); % sorting Gray code elements to form the lookup table
opBin = ind(opGray+1)-1; % picking elements from the array
The Trick from DSP-Guru to do Gray to Binary conversion, thanks to Jerry Avins.
clear
% binary to Gray code
ipBin = [0:15]
opGray = bitxor(ipBin,floor(ipBin/2))
% Gray code to binary
opBin = bitxor(opGray,floor(opGray/2^8)) ;
opBin = bitxor(opBin,floor(opBin/2^4)) ;
opBin = bitxor(opBin,floor(opBin/2^2)) ;
opBin = bitxor(opBin,floor(opBin/2^1)) ;
Concluding thoughts
1. As on now, I do not understand how the DSP Guru trick for Gray code to Binary works. Of course, if I figure out, I will update this post. 
2. Having discussed Binary to Gray code conversion and Gray to Binary conversion, we are now armed to discuss the bit-error rate probabilities for various modulation schemes (Recall: We have been discussing symbol error rate in Additive White Gaussian noise till date).
Thanks, KrishnaRelated posts
- Binary to Gray code for 16QAM
- Binary to Gray code conversion for PSK and PAM
- Bit error rate for 16PSK modulation using Gray mapping
- 16QAM Bit Error Rate (BER) with Gray mapping
- Linear to log conversion
D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN.
{ 11 comments… read them below or add one }
hi krishna,
there is a typo in the article
it shuld be written as
1. For j=n-1;
b[n-1]=g[n-1];
instead of b[n-1]=j[n-1];
i.e, the most significant bit (MSB) of the Gray code is same as the MSB of original binary number.
cheers.
@R.Ramya: Thanks for noticing the typo. I corrected.
I cant understand this line particualr line:
[tt ind] = sort(opGray); % sorting Gray code elements to form the lookup table
here ind refers to ???
@R.Ramya:
The variable ind refers to the index of the elements prior to sorting.
as a quick example,
a = [3 1 4];
[tt ind] = sort(a);
tt =
1 3 4
ind =
2 1 3
Hope this helps.
iam a student of heritage of 2nd yera I.T
Hello!
I need convert 10-bits number from gray to binary. Could anybody help me?
@jerraz: Hope the Lookup Table based Matlab code provided in the post helps.
use this formula
G=(N XOR 2*N)/2
@rayan2: It does the same thing? How does that work?
Since you ask how it works, here is an explanation:
Notation: A single letter represents an 8 bit value
A single letter followed by a digit represents one bit in that 8 bit value.
Bits are numbered from right to left, so bit n has the value 2^n.
Start with the binary value b.
b=b7;b6;b5;b4;b3;b2;b1;b0
Compute its gray code, g = b ^ ( b >> 1 )
g=g7;g6;g5;g4;g3;g2;g1;g0
[0] g7=b7; g6=b7^b6; g5=b6^b5; g4=b5^b4; g3=b4^b3; g2=b3^b2; g1=b2^b1; g0=b1^b0
Now recover the original binary value from g in three steps
[1] x = g ^ (g >> 4)
x=x7;x6;x5;x4;x3;x2;x1;x0
x7=g7; x6=g6; x5=g5; x4=g4 x3=g3^g7 x2=g2^g6 x1=g1^g5; x0=g0^g4
[2] y = x ^ (x >> 2)
y=y7;y6;y5;y4;y3;y2;y1;y0
y7=x7; y6=x6; y5=x5^x7; y4=x4^x6; y3=x3^x5; y2=x2^x4; y1=x1^x3; y0=x0^x2
substitute eqn [1] into eqn [2]
[2a] y7=g7; y6=g6; y5=g5^g7; y4=g4^g6; y3=g3^g7^g5; y2=g2^g6^g4; y1=g1^g5^g3^g7; y0=g0^g4^g2^g6
[3] z = y ^ (y >> 1)
z7=y7; z6=y6^y7; z5=y5^y6; z4=y4^y5; z3=y3^y4; z2=y2^y3; z1=y1^y2; z0=y0^y1
Substitute eqn [2a] into eqn [3]
[3a] z7=g7; z6=g6^g7; z5=g5^g7^g6; z4=g4^g6^g5^g7; z3=g3^g7^g5^g4^g6; z2=g2^g6^g4^g3^g7^g5; z1=g1^g5^g3^g7^g2^g6^g4; z0=g0^g4^g2^g6^g1^g5^g3^g7
Substitute eqn [0] into eqn [3a]
[3b] z7=b7;
z6=b7^b6^b7;
z5=b6^b5^b7^b7^b6;
z4=b5^b4^b7^b6^b6^b5^b7;
z3=b4^b3^b7^b6^b5^b5^b4^b7^b6;
z2=b3^b2^b7^b6^b5^b4^b4^b3^b7^b6^b5;
z1=b2^b1^b6^b5^b4^b3^b7^b3^b2^b7^b6^b5^b4;
z0=b1^b0^b5^b4^b3^b2^b7^b6^b2^b1^b6^b5^b4^b3^b7
Rearrange the bits on the right
[3c] z7=b7;
z6=b6^b7^b7;
z5=b5^b6^b6^b7^b7;
z4=b4^b5^b5^b6^b6^b7^b7;
z3=b3^b4^b4^b5^b5^b6^b6^b7^b7;
z2=b2^b3^b3^b4^b4^b5^b5^b6^b6^b7^b7;
z1=b1^b3^b3^b2^b2^b4^b4^b5^b5^b6^b6^b7^b7;
z0=b0^b1^b1^b2^b2^b3^b3^b4^b4^b5^b5^b6^b6^b7^b7
Now, because x^x=0 and y^0=y, we can just remove each identical pair
[3c] z7=b7;
z6=b6;
z5=b5;
z4=b4;
z3=b3;
z2=b2;
z1=b1;
z0=b0;
And that’s how it works.
@Bob: Thanks much.
I will try to digest and get back, for any clarifications…