Exercise 3.3Z: Rectangular Pulse and Dirac Delta

Various rectangular pulses

We consider here a multitude of symmetrical rectangular functions $x_k(t)$. The individual rectangles differ in amplitudes (heights)

$$A_k = k \cdot A$$

and different pulse durations (widths)

$$T_k = T/k.$$

Let $k$ be any positive value.

The rectangular pulse $x_1(t)$ shown in red has the amplitude $A_1 = {A} = 2 \,\text{V}$ and the duration $T_1 = {T} = 500 \,µ\text{s}$.
The pulse $x_2(t)$ shown in blue is half as wide ⇒ $T_2 =250 \,µ\text{s}$, but twice as high ⇒ $A_2 = 4 \text{ V}$.

Hints:

This task belongs to the chapter Special Cases of Pulses.
Use one of the functions $\text{si}(x) = \sin(x)/x$ or $\text{sinc}(x) = \sin(\pi x)/(\pi x)$.

You can check your results using the two interactive applets

Pulses and Spectra,

Frequency & Impulse Responses.

Questions

	The spectral value $X_1(f = 0)$ is equal to $10^{–3} \,\text{V/Hz}$.
	$X_1(f)$ has zeros at the interval of $2 \,\text{kHz}$.
	$X_1(f)$ has zeros at the interval of $4 \,\text{kHz}$.

	The spectral value is $X_2(f = 0)$ is equal to $10^{–3} \,\text{V/Hz}$.
	$X_2(f)$ has zeros at the interval of $2\, \text{kHz}$.
	$X_2(f)$ has zeros at the interval of $4 \,\text{kHz}$.

$f_{10} \ = \ $

$\text{kHz}$

$X_{10}(f = 2 \text{kHz})\ = \ $

$\text{mV/Hz}$

$X_{\infty}(f = 2 \,\text{kHz})\ = \ $

$\text{mV/Hz}$

Solutions

Solution

(1) The proposed solutions 1 and 2 are correct:

The spectral value at frequency $f = 0$ is always equal to the area under the time function according to the first Fourier integral :

$$X( f ) = \int_{ - \infty }^{ + \infty } {x( t )} \cdot {\rm{e}}^{ - {\rm{j2\pi }}ft} \hspace{0.1cm} {\rm d}t \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \;X( {f = 0} ) = \int_{ - \infty }^{ + \infty } {x( t )}\hspace{0.1cm} {\rm d}t.$$

In the present case, the pulse area is always $A \cdot T = 10^{–3} \,\text{Vs} = 1\, \text{mV/Hz}$.
Because of $T_1 = 500 \,µ\text{s}$ the spectrum $X_1(f)$ has zero crossings at the interval $f_1 = 1/T_1 = 2 \,\text{kHz}$ .

(2) The proposed solutions 1 and 3 are correct:

Due to equal pulse areas, the spectral value is not changed at the frequency $f = 0$ .
The equidistant zero crossings now occur at the interval $f_2 = 1/T_2 = 4 \,\text{kHz}$.

(3) Zero crossings occur at multiples of $f_{10} = 1/T_{10} = 20 \,\text{kHz}$, and the spectral function is:

$$X_{10} ( f ) = X_0 \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}f/f_{10} } ).$$

At frequency $f = 2 \,\text{kHz}$ the argument of the $\rm si$-function is equal to $\pi/10$ $($or $18^{\circ})$:

$$X_{10} ( {f = 2\;{\rm{kHz}}}) = 10^{ - 3} \;{\rm{V/Hz}} \cdot \frac{{\sin ( {18^\circ } )}}{{{\rm{\pi /10}}}} \hspace{0.15 cm}\underline{= 0.984 \;{\rm{mV/Hz}}}{\rm{.}}$$

(4) In the limiting case $k \rightarrow \infty$ the then infinitely high and infinitely narrow Rectangular pulse changes into the Dirac delta impulse.

Its spectrum is constant for all frequencies.
Thus the spectral value $X_{\infty}(f = 2 \,\text{kHz})\hspace{0.15 cm}\underline{=1 \text{ mV/Hz}}$ also applies at the frequency $f = 2 \,\text{kHz}$ .