03-21-2013 11:47 PM
how to implement following equation in labview
y (n+1) = x(n+1) if n(mod T) = 0 and ∏ Q(n-iT) =1 where ∏ denotes
multiplication, its lower limit is i=0, upper limit is L
x(n) otherwise
where T is length of comparison intervals and L is number of comparison.
03-22-2013 01:24 AM
03-22-2013 01:50 AM
Hi....
I have added a picture to try and see what you are saying.
is the picture correct ?
Do you have the X array and you are creating the Y array ?
Do you have initial values for one of the arrays ?
can you give a reference to the article you are using ?
03-22-2013 03:41 AM
yes, the limit applied are correct but the limits are of Q(n-iT) , n(mod T) =0
i just want to know how to apply limits and how to give two conditions
03-22-2013 03:44 AM
X(n) is vector which change its value at every instant.
n is number of samples. n(mod T) = 0 means modulus of length of comparison interval is zero
03-22-2013 08:28 AM
Do you have any code of what you have tried? I'm also wondering where X and Q are coming from. Things that might help you is the modulus can be calculated with the Quotient & Remainder function (the remainder is the modulus) and there is a Multiply Array Elements.
03-22-2013 09:20 AM - edited 03-22-2013 09:46 AM
OK.....
more questions and I will try to create an example.
1. What is Q ?
2. What are ranges of the indexes, do we have X0? do we have X1 ?
(notice that in order to know Y0 we should have X(-1)
3. the multiplication is also (mod T) ?
03-22-2013 10:07 AM
@rashi wrote:
i just want to know how to apply limits and how to give two conditions
OK... without knowing too many details, about what is Q and if there is Y0 or X0 ,
here is an example that will show you "how to apply limits and how to give two conditions"
tell me if it help in any way
03-22-2013 10:29 PM
i am sending you detail of equation in attachment
03-22-2013 10:31 PM
my labview version is 2010, so VI sent by you could not be opened so kindly send me in 2010 version