How to calculate sum of all the possible combinations of two vector elements

I have two vectors A and B with different lengths. The B vector always have 4 elements. How can I obtain the vector C containing sum of all the possible combinations of elements of B vector with A.

For example B= [1 2 3 4] and A= [0.1 0.2 0.3 0.4 0.5 0.6]

Then C= [1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 2.5 2.6 3.1 3.2 3.3 3.4 3.5 3.6 4.1 4.2 4.3 4.4 4.5 4.6]

Thank you

[Jon Wieser]

C=[B+A(:)]';

C=reshape(C,1,numel(C));

[Davoud]

It seems it doesn't work for my case of A=[0.0000 + 0.0000i -0.2689+0.6342i 0.2689+0.6342i 0.0000+1.2685i] and B=[0.0000+0.0000i 0.1404 + 0.5011i -0.1404+0.5011i 0.0000+1.0021i]. In each iteration B is otained from a matrix as B=D(:,i).

Answer 1

[Andrei Bobrov]

log(kron(exp(B),exp(A)))

[Ethan Duckworth]

that's kind of amazing

Answer 2

[madhan ravi]

C=reshape((A+B.')',[],1)'

[Stephen23]

Without all those transposes:

reshape(A(:)+B,1,[])

For earlier MATLAB versions:

reshape(bsxfun(@plus,A(:),B),1,[])

[madhan ravi]

Thank you stephen , there is one extra "(" in front of A.

[Stephen23]

@madhan ravi: thank you, fixed.

[Davoud]

It seems it doesn't work for my case of A=[0.0000 + 0.0000i -0.2689+0.6342i 0.2689+0.6342i 0.0000+1.2685i] and B=[0.0000+0.0000i 0.1404 + 0.5011i -0.1404+0.5011i 0.0000+1.0021i]. In each iteration B is otained from a matrix as B=D(:,i).

[Steven Lord]

If you're working with complex numbers, use the transpose operator (.') instead of the complex conjugate transpose operator ('). For real numbers those two operators behave the same, but for complex numbers they don't.

[Andrei Bobrov]

+1

[madhan ravi]

@Andrei Bobrov: Thank you!