Aufgaben:Exercise 3.3Z: Rectangular Pulse and Dirac Delta

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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

1

Which of the following statements are true regarding the spectrum  $X_1(f)$?

	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  $4 \,\text{kHz}$.

2

Which of the following statements are true regarding the spectrum  $X_2(f)$?

	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  $4 \,\text{kHz}$.

3

Let  $k = 10$.  Calculate the frequency  $f_{10}$  of the first zero and the spectral value at  $f = 2 \,\text{kHz}$.

$f_{10} \ = \ $  $\text{kHz}$

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

4

What is the spectral value at  $f = 2 \,\text{kHz}$  in the limiting case  $k \rightarrow \infty$?  Interpret the result.

$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   $f = 2 \,\text{kHz}$ .