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I'm struggling to solve the following question. I've solved it partially, but I can't get complete it.

We have the given information about a signal of the form x[n] = Acos(Bn+C)

  1. The signal is even and real.
  2. Its period is N=10.
  3. The eleventh coefficient of its Fourier series is 5
  4. $$\frac{1}{10}\sum_{n=0}^9 |x[n]|^2 = 50$$

Find A, B, C.

So i know that given the signal is even, ak=a-k

Given the time period, it's also clear that a0=a5=a10

Using Perseval's relation, we can also use the fourth equation as $$\frac{1}{2}[\frac{1}{5}\sum_{n=0}^4 |x[n]|^2 + \frac{1}{5}\sum_{n=5}^9 |x[n]|^2] = 50$$ $$\frac{1}{2}[2\times\sum_{k=(N)} |a_k|^2] = 50$$ $$\sum_{k=(N)} |a_k|^2 = 50$$ $$|a_-2|^2 + |a_-1|^2 + |a_0|^2 + |a_1|^2 + |a_2|^2 = 50$$ $$|a_0|^2 + 2\times|a_1|^2 + 2\times |a_2|^2 = 50$$ $$25 + 2\times|a_1|^2 + 2\times |a_2|^2 = 50$$ $$|a_1|^2 + |a_2|^2 = \frac{25}{2}$$

Unfortunately I don't know where to go from here.

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The key to solving this question is to start from $a_{11} = 5 = a_1$.

As such $a_1 = a_{-1} = 5$. So now you have two coefficients. How?

Remember the signal is periodic with $N=10$. So $a_{-1} = a_9 = 5$.

So using Parseval's relation, now you have:

$$|a_0|^2+|a_1|^2 + \sum_{k=2}^{8}|a_k|^2 + |a_9|^2 = 50$$ $$|a_0|^2 + 25 + \sum_{k=2}^{8}|a_k|^2 + 25 = 50$$ $$|a_0|^2 +\sum_{k=2}^{8}|a_k|^2 = 0$$

As such, $a_k = 0$ for $k=0 \; \& \; 2,3,...,8$

Now let's write $x[n]$ using it's DFS coefficients:

\begin{align*} x[n] &= \sum_{k} a_k e^{j2\pi \frac kNn}\\ x[n] &= 5 \cdot (e^{j2\pi \frac {1}{10}n} + \underbrace{e^{j2\pi \frac {9}{10}n}}_{a_9 = a_{-1}})\\ x[n] &= 2\cdot 5 \cdot \underbrace{\frac{(e^{j\pi \frac {n}{5}} + e^{-j\pi \frac{n}{5}})}{2}}_{\cos(\pi\frac n5)}\\ x[n] &= 10\cdot \cos(\pi \frac n5) \end{align*}

As such your final answers are:

\begin{align*} A &= 10\\ B &= \frac{\pi}{5}\\ C &= 0 \end{align*}

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  • $\begingroup$ You're correct. Unfortunately by the eleventh coefficient, they meant a_11, not a_10. I only found out about this after posting the question. but I'll leave it here as it might help someone. $\endgroup$
    – Danialz
    Jun 2, 2023 at 16:35
  • $\begingroup$ Although my question is a bit different from the one specified in the book (which you answered). In the book, we have the period as N=10, but mine was N=5. $\endgroup$
    – Danialz
    Jun 2, 2023 at 16:37
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    $\begingroup$ The question you submitted has period specified as $N=10$ so I solved it as such. I have no idea which book this question is from. To solve it using $N=5$ doesn't change much. $\endgroup$ Jun 2, 2023 at 16:40

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