CONTINUOUS TIME FOURIER SERIES
A CT fourier series is nothing but the expansion of continuous function $x(t)$ into the complex exponential basis. This is a complicated definition, and below we will unpack it for you.
Consider the set $B$ of complex exponentials:
\begin{eqnarray}
B: \{ \quad \vec{u_{\omega}}=e^{i\omega t} \quad | \quad \omega \in \mathbb{R},\quad t \in \mathbb{R} \quad \}
\end{eqnarray}
B contains uncountably infinite basis vectors. Before expanding any function x(t) into it, we must check if it is orthonormal.
Orthonormality of the basis $B$
\begin{eqnarray}
\vec{u_{\omega_1}}.\vec{u_{\omega_2}} &=& \int_{-\infty}^{\infty} e^{i\omega_1 t} (e^{i\omega_2 t})^* dt\\
&=& \int_{-\infty}^{\infty} e^{i\omega_1 t} e^{-i\omega_2 t} dt \\
&=& \int_{-\infty}^{\infty} e^{i(\omega_1–\omega_2) t} dt \\
&=& \lim_{T \rightarrow \infty} \int_{-T}^{T} e^{i(\omega_1–\omega_2) t} dt \\
&=& \lim_{T \rightarrow \infty} \frac{e^{i(\omega_1–\omega_2) T}-e^{-i(\omega_1–\omega_2) T}}{i(\omega_1–\omega_2)} \\
&=& \lim_{T \rightarrow \infty} 2 \frac{\sin[(\omega_1–\omega_2) T]}{(\omega_1–\omega_2)}\\
&=& 2\pi \delta(\omega_1–\omega_2)
\end{eqnarray}
Because of the extra factor $2\pi$, the vectors in the basis $B$ turn out to be orthogonal but not orthonormal. To make them orthonormal, we modify $B$ as follows:
\begin{eqnarray}
B: \{ \quad \quad \vec{u_{\omega}}=\frac{e^{i\omega t}}{\sqrt{2\pi}} \quad | \quad \omega \in \mathbb{R} \quad \}
\end{eqnarray}
Fourier and the inverse fourier transformations
Now we can expand the expansion of $x(t)$ in the basis $B$ as
\begin{eqnarray}
\vec{x }&=& \int_{-\infty}^{\infty} (\vec{x}.\vec{u_{\omega}}) \vec{u_{\omega}} d\omega\\
\end{eqnarray}
Define $X(\omega)=\vec{v}.\vec{u_{\omega}}$. Substituting
\begin{eqnarray}
\vec{x }&=& \int_{-\infty}^{\infty} X(\omega) \vec{u_{\omega}} d\omega\\
\end{eqnarray}
or,
\begin{eqnarray}
x(t) &=& \int_{-\infty}^{\infty} X(\omega) \frac{e^{i\omega t}}{\sqrt{2\pi}}d\omega \qquad(1)
\end{eqnarray}
where
\begin{eqnarray}
X(\omega) &=& \vec{x }.\vec{u_{\omega}}\\
&=& \int_{-\infty}^{\infty} x(t) \left(\frac{e^{i\omega t}}{\sqrt{2\pi}}\right)^* dt \\
&=& \int_{-\infty}^{\infty} x(t) \frac{e^{-i\omega t}}{\sqrt{2\pi}} dt\qquad(2)\\
\end{eqnarray}
(2) is known as the fourier transformation and (1) is known as the inverse fourier transformation. Note that $X(\omega)$ is the same vector with $x(t)$, just written in a different basis.
More traditionally, Fourier transform – Inverse transform pair is written as follows:
\begin{eqnarray}
x(t) &=& \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i\omega t}dt \omega \qquad(3)\\
X(\omega) &=& \int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt \qquad(4)
\end{eqnarray}
And this is the final form we’ll use them. Note that
- (3) passes from frequency representation $X(\omega)$ to time representation $x(t)$
- (4) passes from time representation $x(t)$ to frequency representation $X(\omega)$
Some examples
1) Find the fourier transform of $e^{-at}U(t)$, for $a \in \mathbb{R}$, $a>0$.
$$ \begin{eqnarray}
\int_{-\infty}^{\infty} e^{-at}U(t) e^{-i \omega t} dt &=& \int_0^{\infty} e^{-at} e^{-i \omega t} dt \\
&=& \int_0^{\infty} e^{-(at+i \omega t)} dt \\
&=& \left. \frac{e^{-(a+i \omega )t}}{-(a+i \omega )} \right|_{t=0}^{\infty}\\
&=& \frac{e^{-a \infty}e^{i \omega \infty}}{-(a+i \omega )} + \frac{1}{(a+i \omega )}
\end{eqnarray}$$
Note that $e^{-a \infty}=0$ for $a<0$, and remember that $|e^{i\theta}|=1$ for all $\theta \in \mathbb{R}$, hence $e^{i \omega \infty}$ is finite. Therefore $e^{-a \infty}e^{i \omega \infty}=0$ and
$$ \begin{eqnarray}
\int_{-\infty}^{\infty} e^{-at}U(t) e^{-i \omega t} dt &=& \frac{1}{(a+i \omega )}
\end{eqnarray}$$
2) Find the fourier transform of $\delta(t)$.
$$ \begin{eqnarray}
\int_{-\infty}^{\infty} \delta(t) e^{-i \omega t} dt &=& e^{-i \omega 0} &=& 1
\end{eqnarray}$$
3) Find the fourier transform of $\delta(t-a)$.
$$ \begin{eqnarray}
\int_{-\infty}^{\infty} \delta(t-a) e^{-i \omega t} dt &=& e^{-i \omega a}
\end{eqnarray}$$
4) Find the fourier transform of $\chi_{(-T,T)}(t)$.
$$ \begin{eqnarray}
\int_{-\infty}^{\infty} \chi_{(-T,T)}(t) e^{-i \omega t} dt &=& \int_{-T}^{T} e^{-i \omega t} dt\\
&=& \left. \frac{e^{-i \omega t}}{-i \omega} \right|_{t=-T}^T\\
&=& \frac{e^{i \omega T}-e^{-i \omega T}}{i \omega} \\
&=& 2 \frac{\sin(\omega T)}{\omega}
\end{eqnarray}$$
5) Find the fourier transform of 1.
Note that $\lim_{T \rightarrow \infty} \chi_{(-T,T)}(t) =1$, so this problem is a limiting case of the previous problem.
$$ \begin{eqnarray}
\int_{-\infty}^{\infty}1 e^{-i \omega t} dt &=&\lim_{T \rightarrow \infty} \int_{-T}^{T} e^{-i \omega t} dt\\
&=& \lim_{T \rightarrow \infty} 2 \frac{\sin(\omega T)}{\omega}\\
&=& 2\pi \delta(\omega)
\end{eqnarray}$$
6) Find the fourier transform of $e^{iat}$, $a \in \mathbb{R}$.
$$ \begin{eqnarray}
\int_{-\infty}^{\infty}e^{iat} e^{-i \omega t} dt &=&\lim_{T \rightarrow \infty} \int_{-T}^{T} e^{-i (\omega-a) t} dt\\
&=& \lim_{T \rightarrow \infty} 2 \frac{\sin((\omega-a) T)}{\omega-a}\\
&=& 2\pi \delta(\omega-a)
\end{eqnarray}$$
Note that when $a=0$ this correctly reduces to the above problem.
7) Find the fourier transform of $U(t)$
$$ \begin{eqnarray}
\int_{-\infty}^{\infty}U(t) e^{-i \omega t} dt &=&\lim_{T \rightarrow \infty} \int_{0}^{T} e^{-i \omega t} dt\\
&=& \lim_{T \rightarrow \infty} \left.\frac{e^{-i \omega t} }{-i\omega} \right|_{t=0}^T\\
&=& \frac{1}{i\omega} – \lim_{T \rightarrow \infty} \frac{e^{-i \omega T} }{-i\omega}\\
&=& \frac{1}{i\omega} + \pi \delta(\omega)
\end{eqnarray}$$
The periodic case
Formulas (3) and (4) apply in all cases, periodic or non-periodic. But the periodic case need some special care, as the function do not decay to zero as $t \rightarrow \pm \infty$.
Let $x(t)$ be a periodic function with period $T$, ie, $x(t)=x(t+T)$. Taking its Fourier transform
\begin{eqnarray}
X(\omega) &=& \int_{-\infty}^{\infty}x(t)e^{-i\omega t} dt\\
&=& \sum_{n=-\infty}^{\infty} \int_{nT}^{(n+1)T}x(t)e^{-i\omega t} dt\\
&=& \sum_{n=-\infty}^{\infty} \int_{0}^{T}x(t+nT)e^{-i\omega (t+nT)} dt\\
&=& \sum_{n=-\infty}^{\infty} \int_{0}^{T}x(t)e^{-i\omega t}e^{-i\omega nT} dt\\
&=& \left( \int_{0}^{T}x(t)e^{-i\omega t} dt \right) \left( \sum_{n=-\infty}^{\infty} e^{-in \omega T} \right) \\
\end{eqnarray}
Here we have already proven the formula
$$
\sum_{n=-\infty}^{\infty}e^{{2 \pi i} nt}=\sum_{n=-\infty}^{\infty}\delta(t-n)
$$
In this formula, if we substitute $\frac{T}{2\pi}\omega$ in place of $t$, we get
\begin{eqnarray}
\sum_{k=-\infty}^{\infty} e^{in \omega T} &=& \sum_{n=-\infty}^{\infty}\delta(\frac{T}{2\pi}\omega-n) &=& \sum_{n=-\infty}^{\infty}\delta\left( \frac{T}{2\pi} \left( \omega-\frac{2\pi}{T}n \right)\right)
\end{eqnarray}
Note that $\sum_{n=-\infty}^{\infty}e^{-i \omega nT}=\sum_{n=-\infty}^{\infty}e^{i \omega nT}$, and using the scaling property of dirac deltas, we arrive at
\begin{eqnarray}
\sum_{n=-\infty}^{\infty} e^{-in \omega T} &=& \frac{2\pi}{T} \sum_{n=-\infty}^{\infty}\delta\left( \omega-\frac{2\pi}{T}n \right)
\end{eqnarray}
Continuing from the last step of the above derivation
\begin{eqnarray}
X(\omega) &=& \left( \int_{0}^{T}x(t)e^{-i\omega t} dt \right) \left( \sum_{n=-\infty}^{\infty} e^{-in \omega T} \right) \\
&=& \left( \int_{0}^{T}x(t)e^{-i\omega t} dt \right) \left( \frac{2\pi}{T}\sum_{k=-\infty}^{\infty} \delta(\omega – \frac{2 \pi k}{T}) \right) \\
&=& \frac{2\pi}{T}\sum_{k=-\infty}^{\infty} \left( \int_{0}^{T}x(t)e^{-i\omega t} dt \right) \delta(\omega – \frac{2 \pi k}{T}) \\
&=& \frac{2\pi}{T}\sum_{k=-\infty}^{\infty} \left( \int_{0}^{T}x(t)e^{-i\frac{2 \pi k}{T} t} dt \right) \delta(\omega – \frac{2 \pi k}{T}) \qquad (5)
\end{eqnarray}
Let us define $a[k]$ as:
\begin{eqnarray}
a[k] &=& \frac{1}{T}\int_{0}^{T}x(t)e^{-i\frac{2 \pi k}{T} t} dt
\end{eqnarray}
As both $x(t)$ and $e^{-i\frac{2 \pi k}{T} t}$ are periodic with period $T$, the above integral can be taken on any interval of lengh $T$, rather than simply from $0$ to $T$. To indicate this fact, we write it as
\begin{eqnarray}
a[k] &=& \frac{1}{T}\int_{<T>}x(t)e^{-i\frac{2 \pi k}{T} t} dt
\end{eqnarray}
$a[k]$ is the $k$th fourier series coefficient.. Substituting it into (5), we get
\begin{eqnarray}
&=& \frac{2\pi}{T}\sum_{k=-\infty}^{\infty} Ta[k] \delta(\omega – \frac{2 \pi k}{T})\\
&=& 2\pi \sum_{k=-\infty}^{\infty} a[k] \delta(\omega – \frac{2 \pi k}{T})
\end{eqnarray}
which is our first main result.
Can we find $x(t)$ given $X(\omega)$? Evaluating the inverse Fourier transformation
\begin{eqnarray}
x(t)&=& \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i\omega t}dt \\
&=& \frac{1}{2\pi}\int_{-\infty}^{\infty} \left(2\pi \sum_{k=-\infty}^{\infty} a[k] \delta(\omega – \frac{2 \pi k}{T}) \right) e^{i\omega t}dt \\
&=& \sum_{k=-\infty}^{\infty} a[k] \left( \int_{-\infty}^{\infty} \delta(\omega – \frac{2 \pi k}{T}) e^{i\omega t}dt \right) \\
&=& \sum_{k=-\infty}^{\infty} a[k] e^{\frac{2\pi k}{T}t}
\end{eqnarray}
and, this is our second main result.
Putting our results together,
\begin{eqnarray}
\boxed{
\begin{aligned}
x(t) = \sum_{k=-\infty}^{\infty} a[k] e^{ \frac{2\pi ik}{T}t} \\
X(\omega) = 2\pi \sum_{k=-\infty}^{\infty} a[k] \delta(\omega – \frac{2 \pi k}{T})\\
where\\
a[k] = \frac{1}{T} \int_{<T>} x(t)e^{ -\frac{2\pi i k}{T}t}dt
\end{aligned}
}
\end{eqnarray}
which is the special form (3) and (4) take when $x(t)$ is a periodic function with period $T$
Examples for the periodic case
1) Find the fourier coefficients for the function $f(t)$
\begin{eqnarray}
f(t) = \sum_{k=-\infty}^{\infty} \chi_{-1,1}(t-4k)
\end{eqnarray}
and write down its fourier representation.
Answer: This function is periodic with period 4. As we need to integrate over a single period, we can take the integral from $-2$ to $2$.
\begin{eqnarray}
a_k &=& \frac{1}{4} \int_{-2}^{2} \chi_{-1,1}(t) e^{ -\frac{2\pi ki}{4}t}dt \\
&=& \frac{1}{4} \int_{-1}^{1} e^{ -\frac{\pi i k}{2}t}dt \\
&=& \frac{e^{ -\frac{\pi i k}{2}1}-e^{ -\frac{\pi i k}{2}(-1)}}{ -\frac{\pi i k}{2}} \\
&=& { \frac{4}{\pi k}} \frac{e^{ \frac{\pi i k}{2}}-e^{ \frac{\pi i k}{2}}}{2i} \\
&=& \frac{4}{\pi k} \sin \left( \frac{\pi k}{2} \right)
\end{eqnarray}
$a_0$ requires special care, as it gives a $0/0$. There are two ways to resolve this situation:
2) Find the fourier coefficients and the fourier representation for
\begin{eqnarray}
f(t) = \sum_{k=-\infty}^{\infty} \delta(t-Tk)
\end{eqnarray}
Answer: This function is periodic with period $T$. As we need to integrate over a single period, we can take the integral from $-\frac{T}{2}$ to $\frac{T}{2}$.
Appendix: Deriving Fourier transforms from Fourier series
(to students: you are not responsible from this part)
We want to expand the definition of fourier series into non-periodic signals. One way of doing this is to take a periodic signal and let the period go to infinity, ie, $T\rightarrow \infty$. In the limit the periodic signal will convert into a non-periodic signal. Note that what we are doing is to treat the non-periodic signals as a special case of periodic signals..
What will happen to FS equations when $T\rightarrow \infty$? The first step of seeing this is to make a notational change: We will index $a$ not with integers $k$ but with real numbers $\omega_0 k$. In other words, we will do the substitution $a[k] \Rightarrow a(\omega_0 k)$. Let me emphasize again that this is just a notational change. If $\omega_0=0.3$, instead of writing $\ldots, a[-3], a[-2], a[-1], a[0], a[1], a[2], a[3], \ldots $ we will just write $\ldots, a(-0.9), a(-0.6), a(-0.3), a(0), a(0.3), a(0.6), a(0.9), \ldots $. Note that $a(\omega_0 k)$ is technically not a discrete signal, as it is indexed by a real number, not an integer. Hence we replaced the corner brackets with parantheses.
After the substitution $a[k] \Rightarrow a(\omega_0 k)$, the FS equations become
\begin{eqnarray}
x(t) &=& \sum_{k=-\infty}^{\infty} a(\omega_0 k ) e^{i\omega_0 k t}\\
a(\omega_0 k ) &=& \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) e^{-i\omega_0 k t} dt
\end{eqnarray}
Now we will do a second notational change and define the new signal $X(\omega_0 k )=Ta(\omega_0 k )$. Substituting this, and remembering that $\omega_0=\frac{2\pi}{T}$, we get
\begin{eqnarray}
x(t) &=& \sum_{k=-\infty}^{\infty} \frac{X(\omega_0 k )}{T} e^{i\omega_0 k t}&=& \frac{1}{2\pi} \sum_{k=-\infty}^{\infty} X(\omega_0 k ) e^{i\omega_0 k t} \omega_0\\
X(\omega_0 k ) &=& \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) e^{-i\omega_0 k t} dt
\end{eqnarray}
Now we can take the limit $T\rightarrow \infty$. Or, remembering that $\omega_0=\frac{2\pi}{T}$, we can equivalently take the limit $\omega_0\rightarrow 0$. In this limit, $k \omega_0, k \in \mathbb{I}$ will become a continuous variable. Let us denote this as $k \omega_0 \Rightarrow \omega$, $\omega_0 \Rightarrow d\omega$. Taking the limits in the above equations
\begin{eqnarray}
x(t) &=& \lim_{\omega_0\rightarrow 0 } \frac{1}{2\pi} \sum_{k=-\infty}^{\infty} X(\omega_0 k ) e^{i\omega_0 k t} \omega_0\\
\lim_{\omega_0\rightarrow 0 } X(\omega_0 k ) &=& \lim_{\omega_0\rightarrow 0 } \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) e^{-i\omega_0 k t} dt
\end{eqnarray}
Remembering the definition of integral, we get
\begin{eqnarray}
\boxed{
\begin{aligned}
x(t) &=& \frac{1}{2\pi} \int_{\omega=-\infty}^{\infty} X(\omega ) e^{i\omega t} d\omega\\
X(\omega ) &=& \int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt
\end{aligned}
}
\end{eqnarray}