The Continuous Fourier Series is defined as

\begin{nicebox}
\begin{align*}
x(t) &= \sum \limits_{k=-\infty}^{\infty}X_ke^{ik\omega_0t} \quad &\mbox{(synthesis)} \\
X_k &= \frac{1}{p} \int_{\langle p\rangle }x(t)e^{-ik\omega_0t}dt \quad &\mbox{(analysis)}
\end{align*}
\end{nicebox}

\begin{example}
1) Given an inpulse train $x(t)$ of Dirac deltas every $p$ samples, find the spectral coefficients and corresponding frequencies within the signal.


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\begin{align*}
X_k &= \frac{1}{p} \int_{\langle p\rangle }x(t)e^{-ik\omega_0t}dt \\
X_k &= \frac{1}{p} \int_{-p/2}^{p/2}x(t)e^{-ik\omega_0t}dt \\
X_k &= \frac{1}{p} \int_{-p/2}^{p/2}\delta(t)e^{-ik\omega_0t}dt \\
X_k &= \frac{1}{p} \cdot e^{0} \\
X_k &= \frac{1}{p} \\
\end{align*}

Therefore, we can determine that $\forall k$, $X_k = \frac{1}{p}$, and the signal is made up of every harmonic of $\omega_0$, and so we may write the signal in its continuous-time fourier series expansion as

$$ x(t) = \frac{1}{p} \sum \limits_{k=-\infty}^{\infty}e^{ik\omega_0 t} $$

Note that there are infinitely many harmonics of the fundamental frequency, not $p$ harmonics. This is a unique feature of the impulse train in continuous time. How do we know that $e^{ik\frac{\omega_0}{2}t}$ doesn't contibute to the signal? For $k=1$, $e^{i\frac{\omega_0}{2}t}$ has a period $2p$ which contradicts the periodicity of the signal $x$.

\end{example}


\begin{example}
2) Consider a periodic signal in continuous defined by $x(t) = \begin{cases} 1 & -T/2 \leq t \leq T/2  \\ 0 & \mbox{otherwise} \end{cases}$, where $T \in \R$, and $T/2 \langle  p/2$. 


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\rput(5,-0.25){$t$}
\rput(5,1){$x(t)$}

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\begin{align*}
X_k &= \frac{1}{p} \int_{\langle p\rangle }x(t)e^{-ik\omega_0 t}dt \\
X_k &= \frac{1}{p} \int_{-T/2}^{T/2}e^{-ik\omega_0 t}dt \\
X_k &= \frac{1}{p} \lr{ \left. \frac{e^{-ik\omega_0 t}}{-ik\omega_0} \right|_{-T/2}^{T/2}} \\
X_k &= \frac{1}{p} \frac{e^{-ik\omega_0 T/2} - e^{ik\omega_0 T/2}}{-ik\omega_0}  \\
X_k &= \frac{1}{p} \frac{e^{ik\omega_0 T/2} - e^{-ik\omega_0 T/2}}{ik\omega_0}  \\
X_k &= \frac{1}{p} \frac{2i\sin(k\omega_0 T/2)}{ik\omega_0}  \\
X_k &= \frac{1}{k\pi} \sin(k\omega_0 T/2)  \\
\end{align*}

Seems as though we have found a link between the analysis equation and frequency response of the system. 

$$ X_k = \left. X(\omega) \right|_{w=k\omega_0} \Rightarrow X(\omega) = \frac{2}{p}\frac{\sin(\omega T/2)}{\omega} $$

Since $X_k$ has a dependence on $k$, not all frequencies contribute equally. As $k\to \infty$, $\abs{X_k}\to 0$.

\end{example}

FS Examples