Not able to figure out the way to deal with large vectors

Hi ,

I was doing problem and faced a issue how to deal with large vectors .

I have a function

x(n) = e0.1n [u (n + 5) – u (n − 10)]

and want to fetch the even and odd components of it to perform some operations over them .The issue i am computing them by iterating through the since to get access of the values and checking if it is odd or even and perform operation right away there.But it might be too silly to use when 'n' value goes very high .

can anyone suggest any good way for that ? Here is the Pseduo code

n=-10:1:10;

for loop

%%cheking the conditions if the current x(n)

%%value is odd or even and performing operations there only .

end

%%plot the parts

Answer 1

[Harsh Kumar]

Hi Bhaskar ,

I understand that you are looking to compute the odd and even parts of a function optimistically in MATLAB .

To achieve this, you can use the vector definition of function and use the theoretical definition of odd/even function to compute them optimistically .

Odd= (func(x)-func(-x))/2

Even= (func(x)+func(-x))/2

Refer to the below code snippet for better understanding .

n=-50:1:50;
func1=exp(0.1*n).*((n>=-5)-(n>=10)); % orginal function 
func2=exp(-1*0.1*n).*((n<=5)-(n<=-10)); %conjugate  function 
odd=(func1-func2)/2; %odd function calculated using given formula 
even=(func1+func2)/2; %even function calculated using given formula
stem(n,odd);

figure();
stem(n,even);

Refer to the documentation for more details on vector operations :https://in.mathworks.com/help/stateflow/ug/operations-for-vectors-and-matrices.html

Answer 2

[KSSV]

n=-10:1:10;
c1 = 0 ; c2 = 0 ; 
E = zeros([],1) ;
O = zeros([],1) ; 
for i = 1:length(n)
    %% Get your x(n)
    if mod(x(n))
        c1 = c1+1 ; 
        % Odd do your calculations
        % save the value 
        O(c1)=val ; 
    else
        c2 = c2+1 ; 
        % even do your calculation
        % save the value 
        E(c2) = val ;
    end
end
%%plot the parts
plot(E,'r')
hold on
plot(O,b) 

Alternatively, you can calculate x in the loop and later find it is even or odd using mod. This would be vectorised.

Answer 3

[Walter Roberson]

mask = mod(x,2) == 1;

Now do the "odd" operations on x(mask) and do the "even" operations on x(~mask).