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From Fourier Series to Fourier Transform

The Fourier expansion of a periodic signal xT(t)=xT(t+T) is

\begin{displaymath}x_T(t)={\cal F}^{-1}[X[k]]=\sum_{k=-\infty}^{\infty} X[k] e^{jk\omega_0 t}

where X[k] is the Fourier coefficient

\begin{displaymath}X[k]={\cal F}[x_T(t)]=\frac{1}{T}\int_T x_T(t)e^{-jk\omega_0 t} dt
\;\;\;\;\;(k=0, \pm 1, \pm 2, \cdots)

If we define

\begin{displaymath}X(k\omega_0)\stackrel{\triangle}{=}TX[k]=\int_T x_T(t)e^{-jk\omega_0t}dt \end{displaymath}

the Fourier expansion becomes

\begin{displaymath}x_T(t)=\frac{1}{T}\sum_{k=-\infty}^{\infty} T X[k] e^{jk2\pi ...
...um_{k=-\infty}^{\infty} X(k\omega_0) e^{jk\omega_0 t} \omega_0 \end{displaymath}

When the period of xT(t) approaches infinity $T \rightarrow \infty $, the periodic signal xT(t) becomes a non-periodic signal x(t) and the following will result:

In summary, when the signal is non-periodic $x(t)=\lim_{T\rightarrow \infty}x_T(t)$, the Fourier expansion becomes Fourier transform. The forward transform is

\begin{displaymath}X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t} dt \;\;\;...
\;\;\;\;X(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft} dt

and the inverse transform is

\begin{displaymath}x(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega
=\int_{-\infty}^{\infty} X(f) e^{j2\pi f t} df

Comparing Fourier coefficient of a periodic signal xT(t)

\begin{displaymath}X[k]=\frac{1}{T}\int_T x_T(t)e^{-jk\omega_0 t} dt \end{displaymath}

with Fourier spectrum of a non-periodic signal x(t)

\begin{displaymath}X(\omega)=\int_{-\infty}^\infty x(t)e^{-j\omega t} dt \end{displaymath}

we see that the dimension of $X(\omega)$ is different from that of X[k]:

\begin{displaymath}[X(\omega)]=[X[k]][t]=\frac{[X[k]]}{[\omega]} \end{displaymath}

If |X[k]|2 represents the energy contained in the kth frequency component of a periodic signal xT(t), then $\vert X(\omega)\vert^2$ represents the energy density of a non-periodic signal x(t) distributed along the frequency axis. We can only speak of the energy contained in a particular frequency band $\omega_1 < \omega < \omega_2$:

\begin{displaymath}\mbox{Energy contained in band $\omega_1 < \omega_2$ }=\int_{\omega_1}^{\omega_2}
\vert X(\omega)\vert^2 d\omega \end{displaymath}

Note: The argument $\omega$ of a continuous time signal's spectrum

\begin{displaymath}X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t} dt
=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft} dt \end{displaymath}

always appears in the form of $e^{j\omega t}=e^{j2\pi ft}$, therefore $X(\omega)$can also be expressed as X(f), $X(j\omega)$, or $X(e^{j\omega})$. The expressions $X(\omega)$ or X(f) emphasizes the fact that this is the spectrum of the signal representing how the energy contained in the signal is distributed as a function of frequency $\omega$ or f (instead of $j\omega$ or $e^{j\omega}$). Moreover, if X(f) is used (instead of $X(\omega)$), the factor $1/2\pi$ in front of the inverse transform is dropped so that the transform pair looks more symmetric. However, as Fourier transform can be considered as a special case of Laplace transform when $s=j\omega$ (i.e., the real part of s is zero, $\sigma=0$):

\begin{displaymath}{\cal F}[x(t)]={\cal L}[x(t)]\vert _{s=j\omega}
=X(s)\vert _{s=j\omega}=X(j\omega) \end{displaymath}

it is also natural to write Fourier transform of x(t) as $X(j\omega)$.

Example 1:

\begin{displaymath}x(t)=\left\{ \begin{array}{ll} 1 & \vert t\vert<a \\ 0 & \mbox{else} \end{array} \right. \end{displaymath}

The spectrum is

\begin{displaymath}X(\omega)=\int_{-a}^a e^{-j\omega t} dt =\frac{1}{-j\omega}e^{-j\omega t}\vert _{-a}^a
=\frac{2}{\omega} sin(a\omega)

This is the sinc function with a parameter a, as shown in the figure.


Note that the height of the main peak is 2a and it gets taller and narrower as a gets larger. Also note

\begin{displaymath}\int_{-\infty}^\infty X(\omega) d\omega
=2\int_{-\infty}^\infty \frac{sin(a\omega)}{\omega} d\omega=2\pi \end{displaymath}

When a approaches infinity, x(t)=1 for all t, and the spectrum becomes

\begin{displaymath}X(\omega)=\int_{-\infty}^\infty e^{-j\omega t} dt =
...\frac{2}{\omega} sin(a\omega)] = 2\pi \delta(\omega)=\delta(f) \end{displaymath}

Recall that the Fourier coefficient of x(t)=1 is

\begin{displaymath}X[k]=\delta[k]=\left\{ \begin{array}{ll} 1 & k=0 \\ 0 & \mbox{else}
\end{array} \right. \end{displaymath}

which represents the energy contained in the signal at k=0 (DC component at zero frequency), and the spectrum $X(\omega)=X[k]/\omega$ is the energy density or distribution which is infinity at zero frequency.

The integral in the above transform is an important formula to be used frequently later:

\begin{displaymath}\int_{-\infty}^\infty e^{-j\omega t} dt=2\pi \delta(\omega)
\int_{-\infty}^\infty e^{-j2\pi ft} dt=\delta(f) \end{displaymath}

which can also be written as

\begin{displaymath}\int_{-\infty}^\infty e^{-j\omega t} dt
=\int_{-\infty}^\inf... t)] dt
=\int_{-\infty}^\infty cos(\omega t) dt = \delta(f) \end{displaymath}

Switching t and f in the equation above, we also have

\begin{displaymath}\int_{-\infty}^\infty e^{-j2\pi ft} df
=\int_{-\infty}^\infty cos(2\pi ft) df=\delta(t) \end{displaymath}

representing a superposition of an infinite number of cosine functions of all frequencies, which cancel each other any where along the time axis except at t=0 where they add up to infinity, an impulse.

Example 2:

\begin{displaymath}x(t)=cos(\omega_0 t)=\frac{1}{2}[e^{j\omega_0t}+e^{-j\omega_0t}] \end{displaymath}

The spectrum of the cosine function is
$\displaystyle X(\omega)$ = $\displaystyle \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt
[e^{j\omega_0t}+e^{-j\omega_0t}]e^{-j\omega t} dt$  
  = $\displaystyle \pi [\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]$  

The spectrum of the sine function

\begin{displaymath}x(t)=sin(\omega_0 t)=\frac{1}{2j}[e^{j\omega_0t}-e^{-j\omega_0t}] \end{displaymath}

can be similarly obtained to be

=-j\pi [\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]

Again, these spectra represent the energy density distribution of the sinusoids, while the corresponding Fourier coefficients

\begin{displaymath}X[k]={\cal F}[cos(\omega_0 t)]=\frac{1}{2}[\delta[k-1]+\delta[k+1]] \end{displaymath}


\begin{displaymath}X[k]={\cal F}[sin(\omega_0 t)]=\frac{1}{2j}[\delta[k-1]-\delta[k+1]] \end{displaymath}

represent the energy contained at frequency $\omega=\omega_0$.

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Next: Fourier Transform of Periodic Up: No Title Previous: No Title
Ruye Wang