# Exercise 3.1Z: Spectrum of the Triangular Pulse

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Triangular pulse

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

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:

• You can use the following formulas to solve this task:
$$\sin ^2 \left( \alpha \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right),$$
$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }} + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }.$$

### 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  "slitting 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)  We use the abbreviation  $\omega = 2\pi f$.  Then taking advantage of the above symmetry properties:

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

Normalized spectrum  $X(f)$
$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2 \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
• 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   $\rm si$-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 $f$  holds:  ${X(f)} < 0$  (see sketch).