Aufgaben:Exercise 3.1Z: Spectrum of the Triangular Pulse

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Dreieckimpuls

A triangular pulses  ${x(t)}$, is considered, which is described in the range  $–T ≤ t ≤ T$  by the following equation:

$$x(t) = A \cdot \left( {1 - {\left| \hspace{0.05cm}t \hspace{0.05cm}\right|}/{T}} \right).$$

Let the pulse amplitude be  $A = 1\, \text{V}$, the time parameter  $T = 1 \text{ ms}$. For all times  $|\hspace{0.05cm} t \hspace{0.05cm} | > T$  ist  ${x(t)} = 0$.

To calculate the spectral function  ${X(f)}$  you can exploit the following properties:

• The time function is even and thus the spectral function is real:

$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{ \infty } {x(t)} \cdot \cos \left( {2\pi ft} \right){\rm d}t.$$

• For  $|\hspace{0.05cm} t \hspace{0.05cm} | > T$, ${x(t)}$  has no components:

$$X\left( f \right) = 2 \cdot \int_0^T {x(t)} \cdot \cos \left( {2\pi ft} \right){\rm d}t.$$

Hints:

• This exercise belongs to the chapter  Fourier Transform and Its Inverse.
• Further information on this topic can be found in the learning video  Kontinuierliche und diskrete Spektren.

• Zur Lösung dieser Aufgabe können Sie auf die folgenden Formeln zurückgreifen:

$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }} + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }, \hspace{0.5cm} \sin ^2 \left( \alpha \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right).$$

Questions

1 Calculate the spectral function  ${X(f)}$. What spectral value results at the frequency   $f = 500 \,\text{Hz}$?

$X(f = 500 \,\text{Hz}) \ = \ $  $\text{mV/Hz}$

2 Give the spectral function  ${X(f)}$  using the slit function  $\text{si}(x) = \sin(x)/x$ . What value results for  $f = 0$?

$X(f = 0) \ = \ $  $\text{mV/Hz}$

3 At what frequency  $f = f_0$  does the spectrum  ${X(f)}$  have the first zero?

$f_0 \ = \ $  $\text{kHz}$

4 Which of the two statements is true?

	At all multiples of  $f_0$  the spectrum has zeros.
	At the frequency  $f = 1.5 \cdot f_0$  the spectral function is negative.

Solution

Solution

(1)  Taking advantage of the above symmetry properties, the abbreviation  $\omega = 2\pi f$ holds:

$$X(f) = 2A \cdot \int_0^T {\left( {1 -{t}/{T}} \right)} \cdot \cos \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t.$$

• This integral is composed of two parts:

$$X_1 (f) = 2A \cdot \int_0^T {\cos } \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t = \frac{2A}{\omega } \cdot \sin \left( {\omega T} \right),$$

$$X_2 (f) = - \frac{2A}{T} \cdot \int_0^T {t \cdot \cos } \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t = - \frac{2A}{T} \cdot \left. {\left[ {\frac{{\cos \left( {\omega t} \right)}}{\omega ^2 } + \frac{{t \cdot \sin \left( {\omega t} \right)}}{\omega }} \right]} \right|_0^T .$$

• Taking into account the upper and lower limits, one obtains:

$$X_2 \left( f \right) = - \frac{2A}{T} \cdot \left[ {\frac{{\cos \left( {\omega T} \right)}}{\omega ^2 } - \frac{1}{\omega ^2 } + \frac{{T \cdot \sin \left( {\omega T} \right)}}{\omega }} \right].$$

• Adding the two parts, we get:

$$X(f) = \frac{2A}{\omega ^2 \cdot T}\cdot \big[ {1 - \cos \left( {\omega T} \right)} \big] = \frac{A}{2\pi ^2 f^2 T} \cdot \big[ {1 - \cos \left( {2\pi fT} \right)} \big].$$

• At frequency  $f = 1/(2T) = 500 \,\text{Hz}$  the argument of the cosine function is equal to  $\pi$  and the cosine function itself is equal to  $-1$. It follows:

$$X( {f ={1}/{2T} = 500\;{\rm Hz}} ) = \frac{4}{\pi^2} \cdot A \cdot T = \frac{4}{\pi^2} \cdot 1\;{\rm V} \cdot 10^{ - 3}\;{\rm s}\hspace{0.15 cm}\underline{= 0.405 \,{\rm mV/Hz}}.$$

(2)  With the trigonometric transformation  ${1}/{2} \cdot (1 - \cos (2 \alpha)) = \sin^2(\alpha)$  one obtains for the spectral function:

$\rm si$-Quadrat-Spektrum

$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2 \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$

• At the frequency  $f = 0$  the  $\rm si$-function is equal to  $1$. From this follows:

$$X( {f = 0} ) = A \cdot T \hspace{0.15 cm}\underline{= 1\,{\rm mV/Hz}}.$$

(3)  The first zero occurs when the argument of the si function is equal to  $\pi$ .

• From this follows  $f_0 \cdot T = 1$  bzw.  $f_0 = 1/T \hspace{0.15 cm}\underline{= 1 \ \text{kHz}}$.

(4)  The first statement is correct:

• The spectrum  ${X(f)}$  is equal to  ${\rm si}^2(n \cdot \pi) = 0$ at multiples of  $f_0$  $(f = n \cdot f_0)$ .
• The second statement is false:  At no frequency  $f$  is  ${X(f)} < 0$  (see sketch).