Exercise 3.4: Trapezoidal Spectrum and Pulse

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Trapezspektrum & Trapezimpuls

We consider here a trapezoidal spectral function  $X(f)$  according to the upper graph, which is completely described by the three parameters  $X_0$,  $f_1$  and  $f_2$ . For the two corner frequencies,  $f_2 > 0$  and  $0 \leq f_1 \leq f_2$ always apply.

Instead of the corner frequencies  $f_1$  and  $f_2$ , the following two descriptive variables can also be used:

• the  equivalent bandwidth:

$$\Delta f = f_1 + f_2,$$

• the so-called  rolloff factor  (in the frequency domain):

$$r_f = \frac{ {f_2 - f_1 }}{ {f_2 + f_1 }}.$$

With these quantities, the associated time function (see middle graph) is:-

$$x( t ) = X_0 \cdot \Delta f \cdot {\mathop{\rm si}\nolimits} ( { {\rm \pi} \cdot \Delta f \cdot t} ) \cdot {\mathop{\rm si}\nolimits} ( { {\rm \pi} \cdot r_f \cdot \Delta f\cdot t} ).$$

Here  $\text{si}(x) = \text{sin}(x)/x$  is the so-called splitting function.

In this example, the numerical values  $X_0 = 10^{–3}\,\text{V/Hz}$,  $f_1 = 1\,\text{kHz}$  and  $f_2 = 3\,\text{kHz}$  are to be used. The time  $T = 1/\Delta f$  is only used for standardisation purposes. From subtask  (3)  a trapezoidal signal  $y(t)$  is considered, which is identical in shape to the spectrum  $X(f)$ .

The following can be used here as descriptive variables:

• the pulse amplitude  $y_0 = y(t = 0)$,
• the  equivalent pulse duration  (defined via the rectangle with the same area):

$$\Delta t = t_1 + t_2,$$

• the rolloff factor (in the time domain) with comparable definition as  $r_f$:

$$r_t = \frac{ {t_2 - t_1 }}{ {t_2 + t_1 }}.$$

Let  $y_0 = 4\,\text{V}$,  $\Delta t = 1\,\text{ms}$  and  $r_t = 0.5$.

Hints:

• This exercise belongs to the chapter  Fourier Transform Laws.
• Use the  Vertauschungssatz  and the  Similarity Theorem.

• You can check your results using the two interactive applets   Impulses and Spectra  sowie  Frequenzgang und Impulsantwort .

Questions

1

For the given parameters, what is the equivalent bandwidth and rolloff factor of the spectrum  $X(f)$?

$\Delta f \ = \ $  $\text{kHz}$

$r_f \hspace{0.35cm} = \ $

2

What are the signal values of  $x(t)$  at  $t = 0$,  $t = T$  and  $t = T/2$?

$x(t=0)\hspace{0.2cm} = \ $  $\text{V}$

$x(t=T)\ = \ $  $\text{V}$

$x(t=T/2)\ = \ $  $\text{V}$

3

What is the spectrum  $Y(f)$  of the trapezoidal pulse witht  $y_0 = 4\,\text{V}$,  $\Delta t = 1\,\text{ms}$  and  $r_t = 0.5$?
What are the spectral values at the given frequencies?

$Y(f = 0)\hspace{0.2cm} = \ $  $\text{mV/Hz}$

$Y(f = 0.5 \,\text{kHz})\ = \ $  $\text{mV/Hz}$

$Y(f = 1.0 \,\text{kHz})\ = \ $  $\text{mV/Hz}$

4

Which spectral values result with  $y_0 = 8\,\text{V}$,  $\Delta t = 0.5\,\text{ms }$  and  $r_t = 0.5$?

$Y(f=0)\hspace{0.2cm}= \ $  $\text{mV/Hz}$

$Y(f=1.0 \,\text{kHz})\ = \ $  $\text{mV/Hz}$

Solution

Solution

(1)  The equivalent bandwidth is by definition equal to the width of the equal-area rectangle:

$$\Delta f = f_1 + f_2 \hspace{0.15 cm}\underline{= 4\;{\rm{kHz}}}{\rm{.}}$$

• For the rolloff factor holds:

$${ {r_f = }}\frac{ {f_2 - f_1 }}{ {f_2 + f_1 }}\hspace{0.15 cm}\underline{ = 0.5}.$$

(2)  The maximum value of the pulse  $x(t)$  occurs at time  $t = 0$ :

$$x_0 = x(t = 0) = X_0 \cdot \Delta f \hspace{0.15 cm}\underline{= 4\, \text{V}}.$$

• At time  $t = T = 1/\Delta f$  applies due to  $\text{si}(\pi) = 0$:

$$x( {t = T} ) = x_0 \cdot {\mathop{\rm si}\nolimits} ( {\rm{\pi }} ) \cdot {\mathop{\rm si}\nolimits} ( { { {\rm{\pi }}}/{2}} )\hspace{0.15 cm}\underline{ = 0}.$$

• Also at all multiples of  $T$    $x(t)$  exhibits zero crossings. At time  $t = T/2$  holds:

$$x( {t = T/2} ) = x_0 \cdot {\mathop{\rm si}\nolimits} ( { { {\rm{\pi }}}/{2}} ) \cdot {\mathop{\rm si}\nolimits}( { { {\rm{\pi }}}/{4}} ) = x_0 \cdot \frac{ { 1 \cdot \sqrt 2 /2}}{ { {\rm{\pi /}}2 \cdot {\rm{\pi /4}}}} = x_0 \cdot \frac{ {4 \cdot \sqrt 2 }}{ { {\rm{\pi }}^{\rm{2}} }} \hspace{0.15 cm}\underline{= 2.293\;{\rm{V}}}{\rm{.}}$$

(3)  The time function associated with the trapezoidal spectrum  $X(f)$  is according to the specification:

$$x( t ) = X_0 \cdot \Delta f \cdot {\mathop{\rm si}\nolimits} ( { {\rm{\pi }} \cdot \Delta f \cdot t} ) \cdot {\mathop{\rm si}\nolimits} ( { {\rm{\pi }} \cdot r_f \cdot \Delta f \cdot t} ).$$

• Since both  $X(f)$  and  $x(t)$  are real and, moreover,  $y(t)$  is of the same form as  $X(f)$  , we obtain, taking into account all equivalences for the spectral function of the trapezoidal pulse:

$$Y( f ) = y_0 \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( { {\rm{\pi }} \cdot \Delta t \cdot f} ) \cdot {\mathop{\rm si}\nolimits} ( { {\rm{\pi }} \cdot r_t \cdot \Delta t \cdot f} ).$$

• In particular, holds:

$$Y( {f = 0} ) = y_0 \cdot \Delta t \hspace{0.15 cm}\underline{= 4 \;{\rm{mV/Hz}}}{\rm{,}}$$

$$Y( {f = 0.5\;{\rm{kHz}}} ) = y_0 \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} ) \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{4}} ) \hspace{0.15 cm}\underline{= 2.293 \;{\rm{mV/Hz}}}{\rm{,}}$$

$$Y( {f = 1\;{\rm{kHz}}} ) = y_0 \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {\rm{\pi }} ) \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} )\hspace{0.15 cm}\underline{ = 0}\;{\rm{.}}$$

(4)  The spectral value at frequency  $f = 0$  is not changed:  

$$Y_0 = y_0 \cdot \Delta t \hspace{0.15 cm}\underline{= 4 \,\rm{mV/Hz}}.$$

• But since the time function is now only half as wide, the spectrum widens by a factor of  $2$:

$$Y( {f = 1\;{\rm{kHz}}} ) = Y_0 \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} ) \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{4}} ) \hspace{0.15 cm}\underline{= 2.293\,{\rm{mV/Hz}}}{\rm{.}}$$

• In subtask  (3)  this spectral value occurred at the frequency  $f = 0.5\,\rm{kHz}$  aufgetreten.