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Question:

The fourier series coefficients is given as:

$$c_k= \begin{cases} 1 \qquad & k \ \text{ even} \\ 2 \qquad & k \ \text{ odd} \\ \end{cases}$$

the period of the signal is $T=4$, what is signal $x(t)$?

Attempt: when i am trying to find the signal by applying the general formula at the end i am getting exponential term now i am finding that hard to convert it into impulse again because i know answer of this signal will be a impulse but i am not getting any idea to convert it into impulse again? Please help with this?

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    $\begingroup$ what does your textbook say for $x(t)$ given the information you are given? $\endgroup$ – robert bristow-johnson Jul 15 '17 at 4:02
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    $\begingroup$ i guess i was waiting for you to respond with $$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{j k \frac{2\pi}{T} t } $$ have you seen this equations before? $\endgroup$ – robert bristow-johnson Jul 15 '17 at 4:27
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You should use the synthesis equation of an impulse train with period $T$ (which is easy to derive):

$$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$ That is: the Fourier coefficients for all terms is a constant ($\frac{1}{T}$).

Now assume that there are two impulses with different amplitudes $a$ and $b$ per period, or $$x(t)=a\delta(t)+b\delta(t+2),\ -4< t \le0$$ In such case we have

$$\begin{align} x(t)&=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-kT)+b\cdot\delta(t+\frac{T}{2}-kT)\\&=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-4k)+b\cdot\delta(t+2-4k)\\ &=\sum_{k=-\infty}^{\infty}\frac{a}{4}e^{jk\frac{2\pi}{4} t}+\sum_{k=-\infty}^{\infty}\frac{b}{4}e^{jk\frac{2\pi}{4} (t+2)}\\ &=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+be^{jk\pi}\right)\\ &=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+b(-1)^k\right)\\ &=\begin{cases} \displaystyle\sum_{k=-\infty}^{\infty}\frac{a+b}{4}e^{jk\frac{\pi}{2} t},& k\text{ even}\\[10pt] \displaystyle\sum_{k=-\infty}^{\infty}\frac{a-b}{4}e^{jk\frac{\pi}{2} t},& k\text{ odd} \end{cases} \end{align}$$ Now refering again to $(1)$ and comparing it with the question, we should have $$c_k=\begin{cases} \frac{a+b}{4}=1\\[10pt] \frac{a-b}{4}=2 \end{cases}$$

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