I write a code for linear feedback shift register.My code is in below:
X=5712;
D1(1)=0;
D2(1)=0;
D3(1)=0;
D4(1)=0;
D5(1)=0;
D6(1)=1;
for i=1:X-1
D6(i+1)=D1(i);
D5(i+1)=xor(D1(i),D6(i));
D4(i+1)=D5(i);
D3(i+1)=D4(i);
D2(i+1)=D3(i);
D1(i+1)=D2(i);
end
In my code i can only use 6 shift register.I know for degree,n=2,3,4,6,7,15,22, the polynomial is x^n+x+1.As the polynomial is same for those degrees so i want to write a common code for all.
Matlab experts Please need your help.
Your problem is that you are making separate vectors for each register. Rather make a single matrix (i.e. D replaces all of your D1, D2, ..., Dn) so that you can loop:
X = 20;
n = 6;
D = zeros(X, n);
D(1,n) = 1;
for ii = 1:X-1
D(ii+1, 1:n-2) = D(ii, 2:n-1);
D(ii+1, n-1) = xor(D(ii,1), D(ii,n));
D(ii+1, n) = D(ii, 1);
end
E = D(:, end:-1:1)
Related
I have the following equation that I want to solve using MATLAB:
X is the unknown variable. I am trying to solve it using MATLAB solve, but I find it hard to code the left part of the equation.
Is it possible to use solve? Are there any other options?
EDIT
Since A and B depends respectively on j and i I have tried to put them in vectors as follows:
A = [A(1) ... A(j) ... A(N)]
B = [B(1) ... B(i) ... B(N)]
I was trying to have something that looks like this:
eqn = sum(A ./ sum(B .* D)) == C;
solve(eqn);
but the whole difficulty is in this part:
because it depends on both i and j.
To write equation you can use this code:
syms x real
C = 1;
beta = 10;
alph = 0.5;
N = 10;
lenA = N;
lenB = N;
A = rand(1,N);
B = rand(1,N);
eq = 0;
for j=2:N
eqaux = 0;
for i=1:N
eqaux = eqaux+B(i)/((alph+beta*x)^(i+j+1));
end
eq = eq+A(j)/eqaux;
end
eq = simplify(eq==C);
If x must be a complex number, delete real of syms x real.
To solve the equation use this code:
sol = solve(eq,x);
sol = vpa(sol);
Of course yu must use your own values of C, alph, beta, A, B and N.
Suppose
x = [x1;x2; ...; xn]
where each xi is a column vector with length l(i). We can set L = sum(l), the total length of x. I would like to generate 2 matrices based on x:
Let's call them A and B. For example, when x only as 2 blocks x1 and x2 then:
A = [x1*x1' zeros(l(1),l(2)); zeros(l(2),l(1)), x2*x2'];
B = [x1 zeros(l(1),1);
zeros(l(2),1), x2];
In the notation of the problem, A is always L by L and B is L by n. I can generate A and B given x using loops but it is tedious. Is there a clever (loop-free) way to generate A and B. I am using MATLAB 2018b but you can assume earlier version of MATLAB if necessary.
I think it is both short and fast:
B = x .* (repelem((1:numel(l)).',l)==(1:numel(l)));
A = B * B.';
If you have large data It is better to use sparse matrix:
B = sparse(1:numel(x), repelem(1:numel(l), l), x);
A = B * B.';
The following should work. In this case I do an inefficient conversion to cell arrays so there may be a more efficient implementation possible.
cuml = [0; cumsum(l(:))];
get_x = #(idx) x((1:l(idx))+cuml(idx));
x_cell = arrayfun(get_x, 1:numel(l), 'UniformOutput', false);
B = blkdiag(x_cell{:});
A = B*B';
Edit
After running some benchmarks I found a direct loop based implementation to be about twice as fast as the cell based approach above.
A = zeros(sum(l));
B = zeros(sum(l), numel(l));
prev = 0;
for idx = 1:numel(l)
xidx = (1:l(idx))+prev;
A(xidx, xidx) = x(xidx,1) * x(xidx,1)';
B(xidx, idx) = x(idx,1);
prev = prev + l(idx);
end
Here's an alternative approach:
s = repelem(1:numel(l), l).';
t = accumarray(s, x, [], #(x){x*x'});
A = blkdiag(t{:});
t = accumarray(s, x, [], #(x){x});
B = blkdiag(t{:});
At first I create a script giving the name multi_002
After I create the function that includes the equation
The script calls the function and reads the array 'a'.
Each array include one equation a(i) = x - i
I want to minimize its equation of the 4 lines array 'a'.
I suspect that something doesn't work right. I notice that
Optimtool works but f1,f2,f3,f4 doesn't go to zero. With another words
there isn't convergence.
FitnessFunction = #array_002;
numberOfVariables = 1;
[x,fval] = gamultiobj(FitnessFunction,numberOfVariables);
function [a,y,c]= array_002(x)
size = 4; n = 4;
y = zeros(size,1);
c = zeros(size,1);
for i = 1:n
y(i) = x;
c(i) = i;
a = y - c;
end
Why my genetic algorithm doesn't converge? Is there any idea?
Thank you in advance!
I need to do function that works like this :
N1 = size(X,1);
N2 = size(Xtrain,1);
Dist = zeros(N1,N2);
for i=1:N1
for j=1:N2
Dist(i,j)=D-sum(X(i,:)==Xtrain(j,:));
end
end
(X and Xtrain are sparse logical matrixes)
It works fine and passes the tests, but I believe it's not very optimal and well-written solution.
How can I improve that function using some built Matlab functions? I'm absolutely new to Matlab, so I don't know if there really is an opportunity to make it better somehow.
You wanted to learn about vectorization, here some code to study comparing different implementations of this pair-wise distance.
First we build two binary matrices as input (where each row is an instance):
m = 5;
n = 4;
p = 3;
A = double(rand(m,p) > 0.5);
B = double(rand(n,p) > 0.5);
1. double-loop over each pair of instances
D0 = zeros(m,n);
for i=1:m
for j=1:n
D0(i,j) = sum(A(i,:) ~= B(j,:)) / p;
end
end
2. PDIST2
D1 = pdist2(A, B, 'hamming');
3. single-loop over each instance against all other instances
D2 = zeros(m,n);
for i=1:n
D2(:,i) = sum(bsxfun(#ne, A, B(i,:)), 2) ./ p;
end
4. vectorized with grid indexing, all against all
D3 = zeros(m,n);
[x,y] = ndgrid(1:m,1:n);
D3(:) = sum(A(x(:),:) ~= B(y(:),:), 2) ./ p;
5. vectorized in third dimension, all against all
D4 = sum(bsxfun(#ne, A, reshape(B.',[1 p n])), 2) ./ p;
D4 = permute(D4, [1 3 2]);
Finally we compare all methods are equal
assert(isequal(D0,D1,D2,D3,D4))
Hello I am relatively new to MATLAB and have received and assignment in which we could use any programming language. I would like to continue MATLAB and have decided to use it for this assignment. The questions has to do with the following formula:
x(t) = A[1+a1*E(t)]*sin{w[1+a2*E(t)]*t+y}(+/-)a3*E(t)
The first question we have is to develop an appropriate discretization of x(t) with a time step h. I think i understand how to do this using step but because there is a +/- in the end I am running into errors. Here is what I have (I have simplified the equation by assigning arbitrary values to each variable):
A = 1;
E = 1;
a1 = 1;
a2 = 2;
a3 = 3;
w = 1;
y = 0;
% ts = .1;
% t = 0:ts:10;
t = 1:1:10;
x1(t) = A*(1+a1*E)*sin(w*(1+a2*E)*t+y);
x2(t) = a3*E;
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
plot(y)
The problem is I keep getting the following error because of the +/-:
In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in Try1 (line 21)
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
Any help?? Thanks!
You can remove the (t) from the left-hand side of all three assignments.
y = [x1+x2, x1-x2]
MATLAB knows what to do with vectors and matrices.
Or, if you want to write it out the long way, tell MATLAB there will be two columns:
y(t, 1:2) = [x1(t)'+x2(t)', x1(t)'-x2(t)']
or two rows:
y(1:2, t) = [x1(t)+x2(t); x1(t)-x2(t)]
But this won't work when you have fractional values of t. The value in parentheses is required to be the index, not a dependent variable. If you want the whole vector, just leave it out.