Different values when calculating signal's energy by hand vs MATLAB

Question

Given the following signal:

$$ x(n) = \left\{ \begin{array}{ll} u(0.01n-0.025)2.3704e^{(-0.287682n)}, & n\ge 0 \\ u(-(0.01n+0.025))2.3704e^{(0.287682n)}, & n< 0 \\ \end{array} \right. $$

I'm finding the energy using the summation formula:

$$E_{x(n)}=\sum_{n=-\infty}^\infty |x(n)|^2=\sum_{n=0}^\infty(2.3704e^{(-0.287682n)}u(0.01n-0.025))^2+\sum_{n=-\infty}^0 (2.3704e^{(0.287682n)}u(-(0.01n+0.025)))^2 =\sum_{n=0}^\infty(2.3704e^{(-0.287682n)}u(0.01n-0.025))^2+\sum_{n=0}^\infty(2.3704e^{(-0.287682n)}u(-(0.01n+0.025)))^2 = 2\times2.3704^2\times \sum_{n=0}^\infty (e^{(-0.287682\times 2)}u(0.01n-0.025))^n$$

So due to the sampling frequency and other information given about the problem, the sampling doesn't start until $n=3$ so I adjusted the limits of summation and removed the u(n) function:

$$=2\times2.3704^2\times \sum_{n=3}^\infty (e^{(-0.287682\times 2)})^n$$

And then adjusted the series to begin at $n=1$ so I could use the summation formula:

$$=2\times2.3704^2\times(\sum_{n=1}^\infty (e^{(-0.287682\times 2)})^n - (e^{(-0.287682\times 2)})^1 - (e^{(-0.287682\times 2)})^2)$$

So the series converges since the common ratio $<1$ and the sum of the series is:

$$S_{\infty}=\frac{a_1}{1-r}=\frac{(e^{(-0.287682\times 2)})^1}{1-e^{(-0.287682\times 2)}}=1.2857$$

Replacing the number above in the energy summation equation, I get:

$$E_{x(n)}=2\times 2.3704^2\times(1.2857-0.5625-0.3164)=4.5715$$

However, when I use MATLAB to find the energy of the signal using the following code:

E = 0
for i=1:fs+1
    E = E + abs(x(i))^2
end

where fs is the sampling frequency, the code yields the following result:

$$E=2.2858$$

which doesn't match my mathematical method. However, I noticed that the answer that MATLAB gives me is half the answer of what I got by hand. I also know with certainty that the MATLAB answer is correct.

So how do I go from $4.5715$ to $2.2858$? What's the step/reasoning that I'm missing?

Thank you!

Answer 1

Since I don’t know rest of the given code, I’ve decided to calculate the energy other way round.

The MATLAB code for calculating the given function’s energy is given below:

syms n; % Symbol assignment for summation

Energy_1 = symsum(power((2.3704 * exp(0.287682 * n)), 2), n, -Inf, -3); % Energy for the left-hand side of the sample domain

Energy_2 = symsum(power((2.3704 * exp(-0.287682 * n)), 2), n, 3, Inf ); % Energy for the right-hand side of the sample domain

Energy = Energy_1 + Energy_2 % Overall energy

According to this code above, the resultant overall energy becomes $E = 4.5715$.

As the number of samples increases, the result gets accurate. Also, the coefficient in the exponential’s power affects the accuracy of the result too, but this is not the main case.