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Asked by Satya Narayaan Rao
on 10 Jul 2013

How to get rid of this error!! this is my code below for

for i = 18:-1:13

test=ff(i) [B,A] = oct3dsgn(ff(i),Fs,N); if narg y = filter(B,A,x); test=sum(y) P(i) = sum(y.^2)/m;

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Answer by Tom Lane
on 11 Jul 2013

If the variable y is not a vector (if it is a matrix with more than 1 row and column), then sum(y.^2) is not a scalar. It looks like you are trying to assign it into a single element of P.

Satya Narayaan Rao
on 11 Jul 2013

Ya y is a row matrix. it contains more than one row or shall i double p

## 3 Comments

## Jan Simon

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/81633#comment_159143

Seeing the code, which produces an error, does not allow to fix it as long as the intention is not clear.

The dimensions of the used variables matter, so please post all required details.

## Satya Narayaan Rao

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/81633#comment_159457

Ya Y is a complex number a+Bi so does it occupy more than one cell after performing the operation and Y is a row matrix.

i am squaring all the elements and then summing them so do need to assign different memory allocation different

## Satya Narayaan Rao

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/81633#comment_159467

pi = 3.14159265358979;

Fs = 25600; % Sampling Frequency

N = 3; % Order of analysis filters.

F = [ 50 100 125 160, 200 250 315, 400 500 630, 800 1000 1250, ... 1600 2000 2500, 3150 4000 5000 ]; % Preferred labeling freq.

ff = (1000).*((2^(1/3)).^[-10:7]); % Exact center freq. P = zeros(1,18);

P=double(P);

if narg

m = length(x);

end;

% Design filters and compute RMS powers in 1/3-oct. bands % 5000 Hz band to 1600 Hz band, direct implementation of filters. BB=[];

AA=[];

for i = 18:-1:13

[B,A] = oct3dsgn(ff(i),Fs,N);

if narg y = filter(B,A,x);