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Can someone please help me with this Fourier series question:

Determine the Fourier series coefficients of $x(t)$ given as $x(t) = > \cos4t +\sin8t+3$?

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    $\begingroup$ For homework questions it is very much appreciated if you show your own work and explain where you're stuck, instead of just copying the original question. $\endgroup$ – Matt L. Dec 10 '14 at 8:03
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$$x(t) = \cos4t +\sin8t+3$$

Knowing that $$\cos(\theta) = \frac{1}{2}[\exp(j\theta) + \exp(-j\theta)]$$ and that $$\sin(\theta) = \frac{1}{2j}[\exp(j\theta) - \exp(-j\theta)]$$

you can rewrite $x(t)$ as follows:

$$x(t) = \frac{1}{2}\exp(j4t) + \frac{1}{2}\exp(-j4t) + \frac{1}{2j}\exp(j8t) - \frac{1}{2j}\exp(-j8t) + 3$$

  • The fundamental frequency is the greatest common factor between the components' (4 and 8), so 4

  • The line spectrum (which plots frequency vs the magnitudes of the c coefficients) defines components for $c(0)$, $c(1)$ and $c(-1)$, $c(2)$ and $c(-2)$.

  • $c(0)$ is the magnitude of the DC component, which is 3

  • $c(1), c(-1)$ correspond to the coefficients multiplying the terms that are 1 times the fundamental frequency, so $\dfrac{1}{2}$
  • $c(2), c(-2)$ correspond to the coefficients multiplying the terms that are 2 times the fundamental frequency, so $\dfrac{1}{2j}$

The signal is then equivalent to $c(k) = \dfrac{1}{T}$ (multiplied by the integral over a period of $x(t)\exp(-jkw_ot)dt$, and you'll have five terms.

Hope that helps!

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