Generate a random signal in Matlab

I need to generate a transmit signal of 10,000 random bits in MATLAB with values of -1 or +1, with both having a equal probabilty of occuring and then verify the probability.

I am using y=randi([-1 1],10000)

But I keep getting 0's and do not know how to take them out and am unsure if my function is even correct.

I tried verifying the probabilty by doing 2 different matrices and adding them together and should be equal to 0, but bc i have zeros this did not work.

Any suggestions?

Answer 1

[Daniel M]

y = (rand(10000,1) > 0.5)*2 -1;

[Walter Roberson]

Alternately,

y = randi([0 1], 10000, 1)*2 - 1;

[Walter Roberson]

Note: do not expect an exact balance of -1 and +1: the balance is statistical, not exact.

[Chase Murphy]

Thank you all! How would I go about verifying afterwards, that it is about =0? Like trying to add them together, as I mentioned I could do 2 different vectors and add together, but unsure how to do this

[Adam Danz]

To determine the proportion of 1s in y,

p = sum(y==1)/numel(y); 

I'll add a 3rd option that requires the Machine Learning Toolbox:

y = randsample([-1,1],10000,true); 

Each of these methods have about the same probability of choosing 1. This verification produces 10k values of [-1,1] and it repeats that process 10000 times. It then returns the mean proportions of the value 1 in each iteration.

% number of reps 
n = 10000; 

prob = [0,0,0]; 

fprintf('Testing first method...\n')
tempProb = zeros(1,n); 
for i = 1:n
    y = (rand(10000,1) > 0.5)*2 -1;
    tempProb(i) = sum(y==1)/10000; 
end
prob(1) = mean(tempProb); 
% prob(1) = sum(tempProb==.5);  % alternative

fprintf('Testing second method...\n')
tempProb = zeros(1,n); 
for i = 1:n
    y = randi([0 1], 10000, 1)*2 - 1;
    tempProb(i) = sum(y==1)/10000; 
end
prob(2) = mean(tempProb); 
% prob(2) = sum(tempProb==.5);  % alternative

fprintf('Testing third method...\n')
tempProb = zeros(1,n); 
for i = 1:n
    y = randsample([-1,1],10000,true); 
    tempProb(i) = sum(y==1)/10000; 
end
prob(3) = mean(tempProb); 
% prob(3) = sum(tempProb==.5);  % alternative

%display results
disp(prob)

I was currious about the speed of these methods so I ran each of the (rand, randi, randsamp) 100,000 times. Below are the median execution times and the 95% confidence intervals (percentile method).

[Daniel M]

For a binomial distribution with n number of trials and p probability of success:

mean = n*p

variance = n*p*(1-p)

So in this example the mean would be 10000*0.5 = 5000. The variance is 2500, and the standard deviation is 50. Therefore, you can check if your values for y fall within 1 standard deviation. This is a simplification that makes some assumptions, but should be fine for your situation.

y = randi([0 1], 10000, 1)*2 - 1;
y1 = sum(y==1) % ans = 5025, but will be different every time

The value of y1 should be within 5000 +/- 50 about 68% of the time.

[Chase Murphy]

How would you transmit this signal over a channel z[n] of 1 at n=0 and .5 at n=+/-1

[Walter Roberson]

p = sum(y==1)/numel(y); 

for vector y that can be abbreviated to

p = mean(y==1)

for non-vector toss in a (:) after the y or use mean2() instead of mean()

[Steven Lord]

for non-vector toss in a (:) after the y or use mean2() instead of mean()

You can specify 'all' as the dimension input to mean if you're using release R2018b or newer of MATLAB.

>> A = magic(5);
>> mean(A, 'all')
ans =
    13

>> mean(A(:))
ans =
    13

>> mean(A)
ans =
    13    13    13    13    13