# Exercise 3.4: Trapezoidal Spectrum and Pulse

Trapezoidal spectrum & trapezoidal pulse

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:

$$\Delta f = f_1 + f_2,$$
$$r_{\hspace{-0.05cm}f} = \frac{ {f_2 - f_1 }}{ {f_2 + f_1 }}.$$

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

$$x( t ) = X_0 \cdot \Delta f \cdot {\mathop{\rm si}\nolimits} ( { {\rm \pi} \cdot \Delta f \cdot t} ) \cdot {\mathop{\rm si}\nolimits} ( { {\rm \pi} \cdot r_{\hspace{-0.05cm}f} \cdot \Delta f\cdot t} ).$$

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 normative purpose.

In the 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–in–time with the same area):
$$\Delta t = t_1 + t_2,$$
• the rolloff factor (in the time domain) with comparable definition as  $r_f$:
$$r_{\hspace{-0.05cm}t} = \frac{ {t_2 - t_1 }}{ {t_2 + t_1 }}.$$

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

Hints:

• You can check your results using the two interactive applets
Pulses and Spectra,
Frequency & Impulse Responses.

### Questions

1

What are the equivalent bandwidth and the rolloff factor of the spectrum  $X(f)$  for the given parameters?

 $\Delta f \ = \$  $\text{kHz}$ $r_{\hspace{-0.05cm}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 with  $y_0 = 4\,\text{V}$,  $\Delta t = 1\,\text{ms}$  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:
$${ {r_{\hspace{-0.05cm}f} = }}\frac{ {f_2 - f_1 }}{ {f_2 + f_1 }}\hspace{0.15 cm}\underline{ = 0.5}.$$

(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  $f = 0.5\,\rm{kHz}$ .