From the Time Point of View#

If we instead take a time point-of-view and let $A=1$

We’ll complete the analysis on the whiteboard in class.

The Sinc Function#

The function, $\sin (\pi x)/\pi x$ crops up again and again in Fourier analysis. The Fourier coefficients $C_k$ are scaled samples of the real continuous normalized sinc function defined as follows:

The function is equal to 1 at $u=0$¹ and has zero crossings at $u=\pm n,n=1,2,3,\dots$ as shown below.

Plot the sinc function#

Plots:

clear all
cd ../matlab
format compact; setappdata(0, "MKernel_plot_format", 'svg')

x = linspace(-5,5,1000);
plot(x,sin(pi.*x)./(pi.*x))
grid
title('Graph of sinc function')
ylabel('sinc(u)')
xlabel('u')

Matlab code to reproduce this plot: plot_sinc.m.

1. Note $sin(\pi u)/(\pi u)$ is undefined at $u=0$. To compute its value we need to take the limit

which can be shown to be 1. I leave the discovery of the proof to you and the internet.

Duty cycle#

• We define the duty cycle $\eta =2t_0/T$ of the rectangular pulse train as the fraction of the time the signal is “on” (equal to 1) over one period.

• The duty cycle is often given as a percentage.

The spectral coefficients expressed using the normalized sinc function and the duty cycle can be written as

Normalize the spectral coefficients#

Let us normalize the spectral coefficients of $\tilde{x}(t)$ by mutiplying them by $T$, and assume $t_0$ is fixed so that the duty cycle $\eta =2t_0/T$ will decrease as we increase $T$:

Then the normalized coefficents $T{C}_k$ of the rectangular wave is a sinc envelope with constant amplitude at the origin equal to $2t_0$, and a zero crossing at fixed frequency $\pi /t_0$ rad/s, both independent of $T$.

Plots follow (use duty_cycle.mlx to reproduce).

open duty_cycle

Durty cycle 50%#

Duty cycle 12.5%#

Comments#

• As the fundamental period increases, we get more spectral lines packed into the lobes of the sinc envelope.

• These normalized spectral coefficients turn out to be samples of the continuous sinc function on the spectrum of $\tilde{x}(t)$

• The two spectra are plotted against the frequency variable $k{\omega }_0$ with units of rad/s rather than index of harmonic component

• The first zeros of each side of the main lobe are at frequencies $\omega =\pm \pi /t_0$ rad/s

• The zero-crossing points of sinc envelope are independent of the period T. They only depend on $t_0$.

Intuition leading to the Fourier Transform#

• An aperiodic signal that has been made periodic by “repeating” its graph every $T$ seconds will have a line spectrum that becomes more and more dense as the fundamental period is made longer and longer.

• The line spectrum has the same continuous envelope.

• As $T$ goes to infinity, the line spectrum will become a continuous function of $\omega$.

• The envelope is this function.

A rectangular pulse#

The aperiodic signal, a single rectangular pulse, can be thought of as being periodic with an infinite fundamental period (we will let $T\to +\mathrm{\infty }$ later).

Since $\tilde{x}(t)=x(t)$ over $t\in [-T/2,T/2]$, the spectral coefficients $C_k$ of the periodic signal $\tilde{x}(t)$ can be written in terms of the aperiodic signal $x(t)$ as follows:

Let us define the envelope $X(j\omega )$ of $T{C}_k$ (we already know this to be the sinc function):

The coefficients $C_k$ are therefore samples of the continuous envelope $X(j\omega )$:

Back to the periodic signal#

$\tilde{x}(t)$ has the Fourier series representation

or equivalently, since ${\mathrm{\Omega }}_0=2\pi /T$:

Now take the limits#

At the limit, as $T\to \mathrm{\infty }$:

We get

• ${\mathrm{\Omega }}_0\to d\omega$ – the fundamental frequency becomes infinitesimally small.

• $k{\mathrm{\Omega }}_0\to \omega$ – harmonic frequencies get so close together they become a continuum.

• The summation becomes an integral.

• $\tilde{x}(t)\to x(t)$ – the periodic signal tends to the aperiodic signal.

Define the Inverse Fourier and Fourier Transform#

Putting these elements together we have an expression for the aperiodic signal in terms of its Fourier transform:

Inverse Fourier Transform:

Similarly, given the expression we have already seen for an arbitrary $x(t)$:

Fourier Transform:

Fourier Transform Pair#

• The two equations on the previous slide are called the Fourier transform pair.

• They are analogous to the Laplace transform pair we have already seen and we can develop tables of properties and transform pairs in the same way.

• Equation $X(j\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j\omega t}dt.$ gives the Fourier transform or the frequency spectrum of the signal $x(t)$.

• Equation $x(t)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X(j\omega )e^{j\omega t}d\omega .$ is the inverse Fourier transform. It gives the function of time that has a given spectrum in the frequency domain.