Anyone please help me in this un-coded code.

What's not correct in this code that I'm unable to get curve.

close all, clc, clear all; 
p = 0:0.01:0.1

p = 1×11

0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000

for a = 1:length(p)
z = randi([0 1],1,10) % Random matrix
nz = bsc(z,0.001); % Binary symmetric channel
numerrs(a) = biterr(z,nz) % Number of errors
end 

z = 1×10

1 1 1 0 1 0 0 0 1 0

numerrs = 0

z = 1×10

0 0 0 0 1 1 1 1 1 1

numerrs = 1×2

0 0

z = 1×10

1 0 0 0 0 1 0 0 1 0

numerrs = 1×3

0 0 0

z = 1×10

0 0 0 0 1 0 0 0 1 0

numerrs = 1×4

0 0 0 0

z = 1×10

1 0 0 1 0 1 0 0 0 0

numerrs = 1×5

0 0 0 0 0

z = 1×10

1 1 0 1 0 1 1 1 0 1

numerrs = 1×6

0 0 0 0 0 0

z = 1×10

0 1 1 0 1 0 1 0 0 0

numerrs = 1×7

0 0 0 0 0 0 0

z = 1×10

1 1 0 1 0 0 1 1 0 0

numerrs = 1×8

0 0 0 0 0 0 0 0

z = 1×10

1 0 0 0 0 1 1 1 0 1

numerrs = 1×9

0 0 0 0 0 0 0 0 0

z = 1×10

1 1 0 0 1 0 0 1 1 1

numerrs = 1×10

0 0 0 0 0 0 0 0 0 0

z = 1×10

1 0 0 0 0 0 0 0 1 0

numerrs = 1×11

0 0 0 0 0 0 0 0 0 0 0

semilogy(z,nz,'b')
xlim([0 0.1]) 
ylim([0.0001 1])
legend('Uncoded')
xlabel('p crossover probability for the bsc')
ylabel('Probability of error')
title('Un-coded')
grid on

Answer 1

[Image Analyst]

Well, what is the value of nz? Is it a 10 element vector, or a scalar. Did you mean to give nz an index? You're overwriting z and nz on every loop iteration and then just plotting after the loop has finished, which means you'll just be plotting the last thing - last pair of z and nz.

[Walter Roberson]

You are using z as your independent coordinate, but your z variable contains all 0 and 1, so if it worked at all it would be drawing back and forth between the two edges of your plot.

[Reema Memon]

Actually I've to plot a curve for specific value of p. May you kindly let me know then what can I do now to get the curve?

[Reema Memon]

Can you help me that where I should make changes?

[Walter Roberson]

Your code does not use the value of p, just the size of p.