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Continuous Time Fourier Transform

The Fourier expansion coefficient $X[k]$ ($a_k$ in OWN) of a periodic signal $x_T(t)=x_T(t+T)$ is

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

and the Fourier expansion of the signal is:

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

which can also be written as:

\begin{displaymath}x_T(t)=\frac{1}{T}\sum_{k=-\infty}^{\infty} (T X[k]) e^{jk\om...
...fty}^{\infty} X(k\omega_0) e^{jk\omega_0 t}
\;\;\;\;\;\;\;(a) \end{displaymath}

where $X(k\omega_0)$ is defined as

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

When the period of $x_T(t)$ approaches infinity $T \rightarrow \infty $, the periodic signal $x_T(t)$ 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 (analysis) is:

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 (synthesis) is:

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

Note that $X(\omega)$ is denoted by $X(j\omega)$ in OWN.

Comparing Fourier coefficient of a periodic signal $x_T(t)$ with with Fourier spectrum of a non-periodic signal $x(t)$:

\begin{displaymath}X[k]=\frac{1}{T}\int_T x_T(t)e^{-jk\omega_0 t} dt,\;\;\;\;\;\;
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 $\vert X[k]\vert^2$ represents the energy contained in the kth frequency component of a periodic signal $x_T(t)$, 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 on notations:

The spectrum of a time signal can be denoted by $X(\omega)$ or $X(f)$ to emphasize the fact that the spectrum represents how the energy contained in the signal is distributed as a function of frequency $\omega$ or $f$. Moreover, if $X(f)$ is used, the factor $1/2\pi$ in front of the inverse transform is dropped so that the transform pair takes a more symmetric form. On the other hand, as Fourier transform can be considered as a special case of Laplace transform when the real part $\sigma$ of the complex argument $s=\sigma+j\omega=j\omega$ is zero:

X(s)\vert _{s=j\omega}
=\int_{-\infty}^\infty x(t)e^{-st} dt...}
=\int_{-\infty}^\infty x(t)e^{-j\omega t} dt=X(j\omega)

it is also natural to denote the spectrum of $x(t)$ by $X(j\omega)$ (in OWN).

Example 0:

Consider the unit impulse function:

\begin{displaymath}x(t)=\delta(t) \end{displaymath}

\begin{displaymath}X(j\omega)=\int_{-\infty}^\infty \delta(t) e^{-j\omega t} dt=1 \end{displaymath}

Example 1:

If the spectrum of a signal $x(t)$ is a delta function in frequency domain $X(j\omega)=2\pi\;\delta(\omega)$, the signal can be found to be:

\begin{displaymath}x(t)={\cal F}^{-1}[X(j\omega)]=\frac{1}{2\pi}\int_{-\infty}^{\infty}
2\pi\;\delta(\omega)e^{j\omega t} d\omega=e^0=1 \end{displaymath}


{\cal F}[x(t)]=\int_{-\infty}^\infty e^{-j\omega t} dt=2\pi\;\delta(\omega)

Example 2:

\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(j\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(j\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(j\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 $X(j\omega)=X[k]/\omega$ 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

\int_{-\infty}^\infty e^{-j\omega t} dt
=\int_{-\infty}^\in... 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 3:

\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
$\displaystyle X(j\omega)$ $\textstyle =$ $\displaystyle \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt
[e^{j\omega_0t}+e^{-j\omega_0t}]e^{-j\omega t} dt$  
  $\textstyle =$ $\displaystyle \pi [\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]$  

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 $\omega=\omega_0$.

next up previous
Next: Properties of Fourier Transform Up: handout3 Previous: handout3
Ruye Wang 2009-07-05