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Asked by Samuel Hammarberg on 18 May 2013

Hello!

I would appreaciate any help vectorizing the for loop below. The problem i cannot solve is how to take out the constants from wdrive and multiply it with the rest, according to the loop below. I want to become better at vectorizing my MATLAB code so I would really appreciate any help! Thanks alot!

%S and R are square matrices (404*404). %fc is a column vector

wdrive=0:1:800; a2=zeros(2*dofs,(length(wdrive))); b2=zeros(2*dofs,(length(wdrive))); Amp=zeros(1,length(wdrive));

SRS=S*(R\S); SRfc=S*(R\fc);

for i=1:length(wdrive) a2(:,i)=(wdrive(i)^2*SRS+R)\(fs-wdrive(i)*SRfc); b2(:,i)=(R)\(fc+wdrive(i)*S*a2(:,i)); Amp(i)=sqrt(a2(dofs-1,i)^2+b2(dofs-1,i)^2); end

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Answer by Matt J on 18 May 2013

Edited by Matt J on 18 May 2013

Accepted answer

I don't think you can vectorize everything, but you can cut down on the operations done inside the loop,

Rfc=R\fc; SRS=S*(R\S); SRfc=S*(Rfc);

rhs=bsxfun(@minus,fs,SRfc*wdrive);

for i=1:length(wdrive) a2(:,i)=(wdrive(i)^2*SRS+R)\rhs(:,i); end

b2=Rfc+R\S*bsxfun(@times, wdrive, a2); Amp=abs(a2(dofs-1,:)+i*b2(dofs-1,:));

Show 2 older comments

Matt J on 19 May 2013

**Samuel Hammarberg commented:**

It would still be sweet if you, somehow, could get rid of the for loop altogether. So if some one could crack that, that would be great!

Thank you very much!

Matt J on 19 May 2013

The for-loop isn't slowing you down significantly. You just have a lot of matrix inversions

(wdrive(i)^2*SRS+R)

to do. There's no simplification to this that I can see, unless there is some special structure in S and R that you haven't mentioned.

Answer by Samuel Hammarberg on 18 May 2013

Edited by Matt J on 19 May 2013

**Relocated to Comment by Matt J**

## 2 Comments

Direct link to this comment:http://mathworks.com/matlabcentral/answers/76241#comment_149633

What is "fs"?

Direct link to this comment:http://mathworks.com/matlabcentral/answers/76241#comment_149656

Ops! Forgot to specify that. Fc is a column vector with the same size as Fs.