I need to have output of qfunc as a handle function

all,

I need to have output of qfunc as a handle function but I face following error:

error:  """Error using qfunc 
The argument of the Q function must be a real array."""

so does anybody know how can I use handle function in qfunction??????

 myf = @(ebn0)BER32APSK(c,M,ebn0)-p;

and inside of BER32APSK I would have:

fsum = fsum + qfunc(sqrt(((d2)*(log2(M))*ebn0)/2));

Answer 1

[Walter Roberson]

One of d2 or log2(M) or ebn0 are not real-valued.

In particular check for the possibility that M is negative, as log of a negative number would give imaginary values.

[farah arabian]

Walter, d and M are real but ebn0 is actually the variable of a handle function, what do you think?

[Walter Roberson]

You need to invoke ebn0 with the appropriate argument. For example,

fsum = fsum + qfunc(sqrt(((d2)*(log2(M))*ebn0(d2)/2));

[farah arabian]

kindly notice d2 is not index here is my m.file

[Walter Roberson]

Change your line

      EbN02 = fzero(myf,x0);

to

      EbN02 = fzero(myf,[0 100]);

where 100 should be an appropriate upper bound.

You were using an unconstrained parameter, which was becoming negative, which was leading to sqrt() of a negative number.

[farah arabian]

Walter thanks for your response,

mmm.... when I changed interval to the mentioned one or [0 50] then I face following error: "The function values at the interval endpoints must differ in sign." I understand that if function doesn't have sign changing then fzero cannot find the answer but on the other hand ebno range is suppose to be a positive something like 4 or 5 6

[Walter Roberson]

What are some example x values?

[farah arabian]

like 4.7, 7.1, 5.6, 6.1

[Walter Roberson]

I was expecting pairs -- I need to know what inputs to use to the cool* function.

[Walter Roberson]

For the input pair [4.7, 7.1], the solution is approximately 232.76669.

[farah arabian]

o, sorry walter, input pairs are actually two numbers which each are between 1 to let's say 7. but I need to find the answer for all the possible pairs during mentioned interval

[Walter Roberson]

There is no closed formula for that, and it is a quite long expression involving multiple erfc() calls. You need to solve it numerically -- but you have to take into account that the solutions for the code as given involve some numbers a lot bigger than you expect.

[Walter Roberson]

I have gone through the formula in more detail. It has the following properties:

• it is the sum of 442 terms of erfc(), then minus 1/10000000000
• if x(1) == 1 exactly or x(2) == 1 exactly then there a few terms that have a limit of +1/40 as EBN0 goes to inf. As even one +1/40 is greater than -1/10000000000 in the sum, the sum is guaranteed positive for that case -- so there are no solutions if either x is exactly 1
• if x(1) == x(2) exactly, then there is one term that has a limit of +1/40 as EBN0 goes to inf. As even one +1/40 is greater than -1/10000000000 in the sum, the sum is guaranteed positive for that case -- so there are no solutions if the x are exactly equal
• in all other cases, x not 1 exactly and the x not exactly equal, the limit of all of the erfc() terms as EBN0 goes to inf is 0, so the limit as EBN0 goes to inf of the overall sum is -1/10000000000. As the terms start out positive for smaller EBN0 (30999999999/10000000000 for EBN0 = 0), the question involved becomes how far out do you have to go along EBN0 before the sum of those 442 terms happens to get small enough to just balance the -1/10000000000 . This is generally going to be a lot higher than the roughly 4 that you were expecting.

If you are expecting EBN0 solutions of roughly 4, then that would suggest that your equations may be wrong.

[farah arabian]

hey Walter....thanks for your comment body...the problem solved by taking x0=[1 200] and I was wrong for the range because my answer is 137.7241 dB.

[Walter Roberson]

I saw values over 1200 for sure, so I would recommend you use at least 1500 as the upper bound.