CODEWITHSUNDEEP
performancematlabif-statementfor-loopsampling
SolidusVerum
SolidusVerum

Reputation: 123

How can I avoid if else statements within a for loop?

I have a code that yields a solution similar to the desired output, and I don't know how to perfect this.

The code is as follows.

N = 4; % sampling period
for nB = -30:-1; 
  if rem(nB,N)==0
    xnB(abs(nB)) = -(cos(.1*pi*nB)-(4*sin(.2*pi*nB))); 
  else
    xnB(abs(nB)) = 0;
  end
end
for nC = 1:30; 
  if rem(nC,N)==0
    xnC(nC) = cos(.1*pi*nC)-(4*sin(.2*pi*nC)); 
  else
    xnC(nC) = 0;
  end
end
nB = -30:-1; 
nC = 1:30;
nD = 0;
xnD = 0;
plot(nA,xnA,nB,xnB,'r--o',nC,xnC,'r--o',nD,xnD,'r--o')

This produces something that is close, but not close enough for proper data recovery.

I have tried using an index that has the same length but simply starts at 1 but the output was even worse than this, though if that is a viable option please explain thoroughly, how it should be done.

I have tried running this in a single for-loop with one if-statement but there is a problem when the counter passes zero. What is a way around this that would allow me to avoid using two for-loops? (I'm fairly confident that, solving this issue would increase the accuracy of my output enough to successfully recover the signal.)

EDIT/CLARIFICATION/ADD - 1

I do in fact want to evaluate the signal at the index of zero. The if-statement cannot handle an index of zero which is an index that I'd prefer not to skip.

The goal of this code is to be able to sample a signal, and then I will build a code that will put it through a recovery filter.

EDIT/UPDATE - 2

nA = -30:.1:30; % n values for original function
xnA = cos(.1*pi*nA)-(4*sin(.2*pi*nA)); % original function
N = 4; % sampling period
n = -30:30;
xn = zeros(size(n));
xn(rem(n,N)==0) = -(cos(.1*pi*n)-(4*sin(.2*pi*n)));
plot(nA,xnA,n,xn,'r--o')
title('Original seq. x and Sampled seq. xp')
xlabel('n')
ylabel('x(n) and xp(n)')
legend('original','sampled');

This threw an error at the line xn(rem(n,N)==0) = -(cos(.1*pi*n)-(4*sin(.2*pi*n))); which read: In an assignment A(I) = B, the number of elements in B and I must be the same. I have ran into this error before, but my previous encounters were usually the result of faulty looping. Could someone point out why it isn't working this time?

EDIT/Clarification - 3

N = 4; % sampling period
for nB = -30:30;
    if rem(nB,N)==0
        xnB(abs(nB)) = -(cos(.1*pi*nB)-(4*sin(.2*pi*nB)));
    else
        xnB(abs(nB)) = 0; 
    end
end

The error message resulting is as follows: Attempted to access xnB(0); index must be a positive integer or logical.

EDIT/SUCCESS - 4

After taking another look at the answers posted, I realized that the negative sign in front of the cos function wasn't supposed to be in the original coding.

Upvotes: 0

Views: 577

Answers (2)

MrAzzaman
MrAzzaman

Reputation: 4768

You could do something like the following:

nB = -30:1
nC = 1:30
xnB = zeros(size(nB));
remB = rem(nB,N)==0;
xnB(remB) = -cos(.1*pi*nB(remB))-(4*sin(.2*pi*nB(remB));
xnC = zeros(size(nC));
remC = rem(nC,N)==0;
xnC(remC) = cos(.1*pi*nC(remC))-(4*sin(.2*pi*nC(remC)));

This avoids the issue of having for-loops entirely. However, this would produce the exact same output as you had before, so I'm not sure that it would fix your initial problem...

EDIT for your most recent addition:

nB = -30:30;
xnB = zeros(size(nB));
remB = rem(nB,N)==0;
xnB(remB) = -(cos(.1*pi*nB(remB))-(4*sin(.2*pi*nB(remB)));

In your original post you had the sign dependent on the sign of nB - if you wanted to maintain this functionality, you would do the following:

xnB(remB) = sign(nB(remB).*(cos(.1*pi*nB(remB))-(4*sin(.2*pi*nB(remB)));

Upvotes: 3

Roney Michael
Roney Michael

Reputation: 3994

From what I understand, you want to iterate over all integer values in [-30, 30] excluding 0 using a single for loop. this can be easily done as:

for ii = [-30:-1,1:30]
    %Your code
end

Resolution for edit - 2

As per your updated code, try replacing

xn(rem(n,N)==0) = -(cos(.1*pi*n)-(4*sin(.2*pi*n)));

with

xn(rem(n,N)==0) = -(cos(.1*pi*n(rem(n,N)==0))-(4*sin(.2*pi*n(rem(n,N)==0))));

This should fix the dimension mismatch.

Resolution for edit - 3

Try:

N = 4; % sampling period
for nB = -30:30;
    if rem(nB,N)==0
        xnB(nB-(-30)+1) = -(cos(.1*pi*nB)-(4*sin(.2*pi*nB)));
    else
        xnB(nB-(-30)+1) = 0; 
    end
end

Upvotes: 3

Related Questions