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Consider the signal

$$x(t)=\cos(2\pi t)$$

Since $x(t)$ is periodic with a fundamental period of $1$, it is also periodic with a period of $N$, where $N$ is any positive integer. What are the Fourier series coefficients of $x(t)$ of we regard it as a periodic signal with period $N=3$?

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    $\begingroup$ This looks a lot like homework, so it is appreciated if you add your own thoughts. We can help you if we know what it is that you don't understand, but we can't just present you a solution to a homework problem. $\endgroup$ – Matt L. Feb 18 '18 at 16:36
  • $\begingroup$ "This student was looking for the Fourier Series of a Cosine, and what he found might SHOCK you!" ahem, what are the coefficients that you have to add to a sum of sines and cosines to get a cosine? $\endgroup$ – Marcus Müller Feb 18 '18 at 16:43
  • $\begingroup$ The infinitude of periodicities of a signal does not affect its Fourier series coefficients. It is the non-orthogonality of the signal with the Fourier basis functions of $\cos(2\pi nt)$ and $\sin(2\pi nt)$ that determine the coefficients. Your pure cosine signal will only be non-orthogonal to one Fourier basis function; so it will have only 1 Fourier series coefficient no matter what periodicity you pretend it has. $\endgroup$ – Andy Walls Feb 18 '18 at 19:27
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    $\begingroup$ @AndyWalls: The index of the respective Fourier coefficient does change with the chosen period. $\endgroup$ – Matt L. Feb 18 '18 at 19:28
  • $\begingroup$ @MattL. True. And that is really the answer. It's just a single coefficient whose index depends on what one defines as the base period of the basis functions. $\endgroup$ – Andy Walls Feb 18 '18 at 19:34
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HINT: Compare the given signal $x(t)$ with the general Fourier series representation of an $N$-periodic signal:

$$a_0+\sum_{k=1}^{\infty}a_k\cos\left(\frac{2\pi k}{N}t\right)+\sum_{k=1}^{\infty}b_k\sin\left(\frac{2\pi k}{N}t\right)$$

Now it should be easy to figure out which of the coefficients $a_k$ and $b_k$ are non-zero and what their value is.

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  • $\begingroup$ This is my doubt Matt. If we regard x(t) as a periodic signal with period=1, then fundamental frequency=2*pi, but if we regard it as a signal with period=3, then frequency=2*pi/3. I.e now coefficients at index 3 and -3 are equal to 1/2 while previously coefficients at 1 and -1 we're equal to 1/2. Right? $\endgroup$ – YOGENDRA SINGH Feb 19 '18 at 0:23
  • $\begingroup$ But we look at the time scaling property of the Fourier series, it dawns on us that the Fourier series coefficients remain same regardless of the time period or frequency. So, I am a bit confused about the two cases. $\endgroup$ – YOGENDRA SINGH Feb 19 '18 at 0:42
  • $\begingroup$ @YOGENDRASINGH: Yes to your first comment. That's it, the coefficient remains the same but its index changes, simply because the index is related to a different fundamental frequency. $\endgroup$ – Matt L. Feb 19 '18 at 8:02

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