CONTINUOUS TIME FOURIER SERIES
A CT fourier series is nothing but the expansion of a periodic function into a complex exponential basis. This is a complicated definition, and below we will unpack it for you.
Consider a continuous periodic fuction $x(t)$ periodic with $T$, ie, $x(t)=x(t+T)$ for all $T$. Note that the set $B$ of complex exponentials which are periodic with $T$ is infinite:
\begin{eqnarray}
B: \{ \quad e^{\frac{2\pi k}{T}t} \quad | \quad k \in \mathbb{I} \quad \}
\end{eqnarray}
Note that $e^{\frac{2\pi k}{T}t}$ is periodic with a fundamental period $\frac{T}{k}$. As $k$ is an integer, this means that it is also periodic with period $T$.
Constructing a Vector space over periodic functions with period $T$
Note that periodic functions with period $T$ form a vector space:
- Sum of two periodic function with period $T$ will yield a periodic function with period $T$
- Multiplying a periodic function with period $T$ with a scalar will yield a periodic function with period $T$
Dot product between two periodic functions (or vectors) in this space is defined as
\begin{eqnarray}
x(t).y(t) = \int_{<T>} x(t)y^*(t)dt
\end{eqnarray}
If $x(t)$ and $y(t)$ is real, it is possible to define norms and angles between periodic functions completeley analogously to our previous work. In the complex case, the norm can still be defined.
As we did before, we can pick a basis and express our function (or vector) $x(t)$ in that basis. The basis vectors we are interested in are the set of complex exponentials with period $T$:
\begin{eqnarray}
e_k(t) = e^{\frac{2\pi k}{T}t}
\end{eqnarray}
Note that this basis is orthogonal. For $ k \neq m$:
\begin{eqnarray}
e_k(t).e_m(t) &=& \int_{<T>} e_k(t)e_m^*(t)dt \\
&=& \int_{<T>} e^{\frac{2\pi k}{T}t} (e^{ \frac{2\pi m}{T}t} )^*dt \\
&=& \int_{<T>} e^{\frac{2\pi k}{T}t} e^{ -\frac{2\pi m}{T}t} dt \\
&=& \int_{<T>} e^{\frac{2\pi (k-m)}{T}t} dt \\
&=& \frac{e^{\frac{2\pi (k-m)}{T}t}}{\frac{2\pi (k-m)}{T}} \Big|_{<T>}\\
&=& 0
\end{eqnarray}
For $ k = m$:
\begin{eqnarray}
e_k(t).e_k(t) &=& \int_{<T>} e_k(t)e_k^*(t)dt \\
&=& \int_{<T>} e^{\frac{2\pi k}{T}t} (e^{ \frac{2\pi k}{T}t} )^*dt \\
&=& \int_{<T>} e^{\frac{2\pi k}{T}t} e^{ -\frac{2\pi k}{T}t} dt \\
&=& \int_{<T>} e^{\frac{2\pi (k-k)}{T}t} dt \\
&=& T
\end{eqnarray}
We conclude
\begin{eqnarray}
e_k(t).e_m(t) = T\delta_{km}
\end{eqnarray}
After these basic definitions, everything becomes automatic:
Let us try to express a vector (periodic function) in the basis $e_k(t)$:
\begin{eqnarray}
x(t) &=& \sum_{k=-\infty}^{\infty} a_k e_k(t) \\
&=& \sum_{k=-\infty}^{\infty} a_k e^{ \frac{2\pi i k}{T}t}
\end{eqnarray}
The problem, as usual, is to find the numbers $a_k$. Let us dot product the above equations from right by the vector $e_m(t)$
\begin{eqnarray}
x(t).e_m(t) &=& \left( \sum_{k=-\infty}^{\infty} a_k e_k(t) \right).e_m(t)\\
&=& \sum_{k=-\infty}^{\infty} a_k e_k(t).e_m(t) \\
&=& a_m T
\end{eqnarray}
In the last step we have used the orthogonality property of the vectors $e_k(t)$. Hence we can write
\begin{eqnarray}
a_m = \frac{1}{T} x(t).e_m(t) = \frac{1}{T} \int_{<T>} x(t)e^{ -\frac{2\pi i k}{T}t}dt
\end{eqnarray}
Hence the complete fourier series pair is
\begin{eqnarray}
\boxed{
\begin{aligned}
x(t) = \sum_{k=-\infty}^{\infty} a_k e^{ \frac{2\pi ik}{T}t} \\
a_m = \frac{1}{T} \int_{<T>} x(t)e^{ -\frac{2\pi i k}{T}t}dt
\end{aligned}
}
\end{eqnarray}
Question: 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:
Question: 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 $\frac{T}{2}$.
FOURIER TRANSFORMS
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}
The periodic case
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}
Recall that we had already proven in a previous chapter
$$
\sum_{n=-\infty}^{\infty}e^{{2 \pi i} nt}=\sum_{n=-\infty}^{\infty}\delta(t-n)
$$
In this formula, if we substitute $t \rightarrow \frac{T}{2\pi}\omega$
\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})\\
&=& \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}
where $a[k]$ is the $k$th fourier series coefficient.
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 $U(t)$
$$ \begin{eqnarray}
\int_{-\infty}^{\infty}U(t) e^{-i \omega t} dt &=&\lim_{T \rightarrow \infty} \int_{0}^{T} e^{-i \omega t} dt\\
&=& \lim_{T \rightarrow \infty} \left.\frac{e^{-i \omega t} }{-i\omega} \right|_{t=0}^T\\
&=& \frac{1}{i\omega} – \lim_{T \rightarrow \infty} \frac{e^{-i \omega T} }{-i\omega}\\
&=& \frac{1}{i\omega} + \pi \delta(\omega)
\end{eqnarray}$$