Mathematica Equation Evaluation Problem in Mathematica

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The discussion revolves around a Mathematica code issue where the output remains constant despite varying the Signal-to-Noise Ratio (SNR). The user is trying to compute a value, Pout, that should depend on SNR, but it consistently yields the same result. Key points include the need to clear variables to avoid reuse issues, the importance of correctly setting loop indices, and ensuring that mathematical equations are correctly formulated. The user is advised to adjust loop conditions to prevent sums from evaluating to zero. After making adjustments, including using the Refine function and modifying parameters, the user observes that the output changes with SNR, suggesting that previous issues were related to the setup of the loops and parameters. The conversation highlights the necessity of careful variable management and correct mathematical formulation in programming to achieve expected outcomes.
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Hello,

I have the following line in Mathematica:

Code:
Print[Pout = (2^-Q*E^(A/2))/SNR \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(q = 0\), \(Q\)]Binomial[Q, q] \(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(Ne + q\)]
FractionBox[
SuperscriptBox[\((\(-1\))\), \(n\)], \(a[n]\)] Re[
\*FractionBox[\(Meq[\(-
\*FractionBox[\(A + \((2*Pi*I*n)\)\), \(2*SNR\)]\)]\), 
FractionBox[\(A + \((2*Pi*I*n)\)\), \(2*SNR\)]]]\)\)]

But the problem is that for different values of SNR, the result will be the same all the time. Why is that happening? A, Q, and Ne are all constants.

Thanks in advance
 
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What are some examples of what A Q and Ne are so i can try it?
and what is meq?
 
Hepth said:
What are some examples of what A Q and Ne are so i can try it?
and what is meq?

Try these expressions:
Code:
A=23
Q=15
Ne=21
Meq[s_]:=1/(1-s)

Regards
 
and also, the a[n] function or array?

but before that make sure youre clearing any variables youre reusing.
restarting the kernel does that.
 
Hepth said:
and also, the a[n] function or array?

but before that make sure youre clearing any variables youre reusing.
restarting the kernel does that.

Assume
Code:
a[n]=1
How clear all variables? I have many of them.

Regards
 
I have the same problem again in the following code:

Code:
gA = 10;
M = 1;
Ne = 1;
If[M >= 1, m = M, m = 0];
For[SNRdB = 0, SNRdB <= 10, SNRdB++,
 SNR = 10^(SNRdB/10);
 Print[F1 = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(r = m\), \(M\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(M - r\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 0\), \(r + i\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(j*\((Ne - 1)\)\)]
\*SuperscriptBox[\((\(-1\))\), \(i + j\)]*Binomial[M, r]*
        Binomial[M - r, i]*Binomial[r + i, j]*
\*SuperscriptBox[\(E\), 
FractionBox[\(\(-j\)*SNR\), \(gA\)]]*
\*SuperscriptBox[\((
\*FractionBox[\(SNR\), \(g\)])\), \(k\)]\)\)\)\)]]
For[SNRdB = 0, SNRdB <= 10, SNRdB++,
 SNR = 10^(SNRdB/10);
 Print[F2 = 1 - \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(r1 = m\), \(M - 1\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(i1 = 1\), \(M - r1\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(j1 = 1\), \(r1 + i1\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(k1 = 0\), \(j1*\((Ne - 1)\)\)]
\*SuperscriptBox[\((\(-1\))\), \(i1 + j1\)]*Binomial[M, r1]*
         Binomial[M - r1, i1]*Binomial[r1 + i1, j1]*
\*SuperscriptBox[\(E\), 
FractionBox[\(\(-j1\)*SNR\), \(gA\)]]*
\*SuperscriptBox[\((
\*FractionBox[\(SNR\), \(g\)])\), \(k1\)]\)\)\)\)]]

1-1/\[ExponentialE]^(1/10)

1-\[ExponentialE]^-1/10^(9/10)

1-\[ExponentialE]^-1/10^(4/5)

1-\[ExponentialE]^-1/10^(7/10)

1-\[ExponentialE]^-1/10^(3/5)

1-\[ExponentialE]^-1/Sqrt[10]

1-\[ExponentialE]^-1/10^(2/5)

1-\[ExponentialE]^-1/10^(3/10)

1-\[ExponentialE]^-1/10^(1/5)

1-\[ExponentialE]^-1/10^(1/10)

1-1/\[ExponentialE]

1

1

1

1

1

1

1

1

1

1

1

Why is the second part constant, even though it is a dependent on SNR?

Regards
 
Because of your sums and their indices. To fix this, add:

If[M > 1, m = M, m = 0];

before the second For loop. It changes the greater than equal to to just a greater than. That way the second sum in the second for loop doesn't go from 1 to 0 (1 to M-r1 == M-m==0)
 
As for your first question, I DO get something different each time I change SNR.
 
Hepth said:
As for your first question, I DO get something different each time I change SNR.

Really? How is that? Try to put the SNR in a For[] loop, and tell me what will happen. Because I am using a For[] loop actually.

Regarding your previous post, I have doubts that I have something wrong in the mathematical equations. So, I will double check them and see what happen then.

Thank you
 
  • #10
FOR:
Code:
A = 23;
Q = 15;
Ne = 21;
Meq[s_] := 1/(1 - s)
a[n_] = 1;
For[SNRdB = 0, SNRdB <= 10, SNRdB++,
  SNR = 10^(SNRdB/10);
Print[
   Pout =
          Refine[(1.0) (2^-Q*E^(A/2))/SNR
<br /> \left.\left.\left.\sum _{q=0}^Q \text{Binomial}[Q,q]\sum _{n=0}^{\text{Ne}+q} \frac{(-1)^n}{a[n]}\text{Re}\left[\frac{\text{Meq}\left[-\frac{A+(2*\text{Pi}*I*n)}{2*\text{SNR}}\right]}{\frac{A+(2*\text{Pi}*I*n)}{2*\text{SNR}}}\right]\right]\right]\right]<br />

The changes I made were adding the Refine to simplify the complex stuff, and multiplying by 1.0 to give me a real value.
I also added the SNR changing in the for loop. Looks like it changes when SNR does. Or did I do something wrong?

EDIT: oops, here's my output:
343.991

424.207

520.658

635.431

770.336

926.63

1104.7

1303.75

1521.56

1754.43

1997.27
 
Last edited:
  • #11
S_David said:
Regarding your previous post, I have doubts that I have something wrong in the mathematical equations. So, I will double check them and see what happen then.

Thank you

What I mean is that for your given values, you have a sum over
<br /> \sum _{\text{r1}=m}^{M-1} \sum _{\text{i1}=1}^{M-\text{r1}} \sum _{\text{j1}=1}^{\text{r1}+\text{i1}} \sum _{\text{k1}=0}^{\text{j1}*(\text{Ne}-1)} <br />
but you have defined :
If[M >= 1, m = M, m = 0];
and M IS 1, so m=M=1;
Then in your sum
<br /> \sum _{\text{r1}=m}^{M-1} \sum _{\text{i1}=1}^{M-\text{r1}} <br />
you have:
r1 from {m to M-1} which is r1 from {1 to (1-1)} or {1 to 0}
then you have i1 from {1 to M-r1} which is {1 to (1-1)} or {1 to 0}

So it doesn't sum anything.
see :
<br /> M=1;m=1;\sum _{\text{r1}=m}^{M-1} \sum _{\text{i1}=1}^{M-\text{r1}} 1==0<br />
 
  • #12
Yes, now the fake code is working. I said fake because I gave you fake parameters, so the values you got are not the expected one, because there are not in the range between 0 and 1 as it must be by definition. When I did a slight change toward the real parameters I got the expected results as following:

Code:
A = 23;
Q = 15;
Ne = 21;
Meq[s_] := 1/(1 - 0.5 s)^4 ;
a[n_] = If[n == 0, 2, 1];
For[SNRdB = 0, SNRdB <= 10, SNRdB++,
 SNR = 10^(SNRdB/10);
 Print[Pout = Refine[(1.0) (2^-Q*E^(A/2))/SNR \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(q = 0\), \(Q\)]Binomial[Q, q] \(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(Ne + q\)]
FractionBox[
SuperscriptBox[\((\(-1\))\), \(n\)], \(a[n]\)] Re[
\*FractionBox[\(Meq[\(-
\*FractionBox[\(A + \((2*Pi*I*n)\)\), \(2*SNR\)]\)]\), 
FractionBox[\(A + \((2*Pi*I*n)\)\), \(2*SNR\)]]]\)\)]]]

0.142877

0.246246

0.390748

0.564677

0.738295

0.875494

0.956523

0.989842

0.998584

0.999898

0.999997

But when I turn my attention to my real, relatively long code, I faced with the same problem again, although I used the same procedure as you described. I don't know why.

Best Regards
 

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