Exercise 1.1: Basic Transmission Pulses

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Considered basic transmission pulses

In this exercise,  we examine the two transmitted signals  $s_{\rm R}(t)$  and  $s_{\rm C}(t)$  with rectangular resp. cosine–square basic transmission pulse,  shown in the diagram.

In particular,  the following characteristics are to be calculated for the respective basic transmission pulses  $g_s(t)$: 

  • the equivalent pulse duration of  $g_s(t)$:
$$\Delta t_{\rm S} = \frac {\int ^{+\infty} _{-\infty} \hspace{0.15cm} g_s(t)\,{\rm d}t}{{\rm Max} \hspace{0.05cm}[g_s(t)]} \hspace{0.05cm},$$
  • the energy of  $g_s(t)$:
$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t \hspace{0.05cm},$$
  • the power of the transmitted signal  $s(t)$:
$$P_{\rm S} = \lim_{T_{\rm M} \to \infty} \frac{1}{+T_{\rm M}} \cdot \int^{+T_{\rm M}/2} _{-T_{\rm M}/2} s^2(t)\,{\rm d}t \hspace{0.05cm}.$$

Always assume in your calculations that the two possible amplitude coefficients are equally likely and that the distance between adjacent symbols is  $T = 1 \ \rm µ s$.  This corresponds to a bit rate of  $R = 1 \ \rm Mbit/s$.

  • The  (positive)  maximum value of the transmitted signal is the same in both cases:
$$s_0 = \sqrt{0.5\, {\rm W}} \hspace{0.05cm}.$$
  • Assuming that the transmitter is terminated with a  $50\ \rm Ω$  resistor,  this corresponds to the following voltage value:
$$s_0^2 = 0.5\, {\rm W} \cdot 50\, {\rm \Omega} = 25\, {\rm V}^2 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} s_0 =5\, {\rm V} \hspace{0.05cm}.$$


$$\int \cos^4(a x)\,{\rm d}x = \frac{3}{8} \cdot x + \frac{1}{4a} \cdot \sin(2 a x)+ \frac{1}{32a} \cdot \sin(4 a x)\hspace{0.05cm}.$$



Are $s_{\rm R}(t)$ and $s_{\rm C}(t)$ unipolar or bipolar signals?

$s_{\rm R}(t)$  is a bipolar signal and  $s_{\rm C}(t)$  is a unipolar signal.
$s_{\rm C}(t)$  is a bipolar signal and  $s_{\rm R}(t)$  is a unipolar signal.


What is the equivalent pulse duration  $\Delta t_{\rm S}$,  normalized to the symbol duration  $T$?

$\text{For the signal}\ \ s_{\rm R}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $

$\text{For the signal}\ \ s_{\rm C}(t) \text{:} \ \ \Delta t_{\rm S}/T \ = \ $


What is the energy of the rectangular basic transmission pulse  $g_s(t)$?

$E_g \ = \ $

$\ \cdot 10^{-6}\ \rm Ws$


What is the power of the rectangular transmitted signal  $s_{\rm R}(t)$?

$P_{\rm S} \ = \ $

$\ \rm W$


What is the energy of the cosine–square basic transmission pulse  $g_s(t)$?

$E_g \ = \ $

$\ \cdot 10^{-6}\ \rm Ws$


What is the power of the transmitted signal  $s_{\rm C}(t)$?

$P_{\rm S} \ = \ $

$\ \rm W$


(1)  Solution 2  is correct:

  • In both cases,  the transmitted signal can be represented in the following form:
$$s(t) = \sum_{(\nu)} a_\nu \cdot g_s ( t - \nu \cdot T)$$
  • For the signal $s_{\rm R}(t)$,  the amplitude coefficients  $a_ν$  are either  $0$  or  $1$.  Thus,  a unipolar signal is present.
  • In contrast,  for the bipolar signal $s_{\rm R}(t)$   ⇒   $a_ν ∈ \{–1, +1\}$ holds.

(2)  The signal  $s_{\rm R}(t)$  is NRZ  ("non-return-to-zero")  rectangular.

  • Accordingly,  both the absolute pulse duration  $T_{\rm S}$  and the equivalent pulse duration  $\Delta t_{\rm S}$  are equal to the symbol duration  $T$:
$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 1 }\hspace{0.05cm}.$$
  • The basic transmission pulse for the signal  $s_{\rm C}(t)$  is:
$$g_s(t) = \left\{ \begin{array}{c} s_0 \cdot \cos^2(\pi \cdot \frac{t}{T}) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} -T/2 \le t \le +T/2 \hspace{0.05cm}, \\ {\rm else} \hspace{0.05cm}. \\ \end{array}$$
  • From the diagram on the information section,  we can see that the following values apply to the cosine–square pulse:
$$T_{\rm S} / T = 1\hspace{0.05cm},\hspace{0.3cm}\Delta t_{\rm S} / T \hspace{0.1cm}\underline{ = 0.5} \hspace{0.05cm}.$$

(3)  For the energy of the rectangular pulse holds:

$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = s_0^2 \cdot T = 0.5\, {\rm W} \cdot 1\, {\rm µ s} \hspace{0.1cm}\underline{= 0.5 \cdot 10^{-6}\, {\rm Ws}}\hspace{0.05cm}.$$

(4)  For a bipolar rectangular signal,  the following would apply:

$$s_{\rm R}^2(t)= s_0^2 = {\rm const.} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_s = s_0^2 \cdot \lim_{T_{\rm M} \to \infty} \frac{1}{T_{\rm M}} \cdot \int ^{T_{\rm M}/2} _{-T_{\rm M}/2} \,{\rm d}t = s_0^2 \hspace{0.05cm}.$$
  • However,  since the signal  $s_{\rm R}(t)$ is  unipolar here,  in half the time $s_{\rm R}(t)= 0$.  Thus,  we get:
$$P_{\rm S} = {1}/{2} \cdot s_0^2 \hspace{0.1cm}\underline{= 0.25 \,{\rm W}} \hspace{0.05cm}.$$

(5)  For the energy of the cosine–square pulse holds:

$$E_g = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = 2 \cdot s_0^2 \cdot \int^{T/2} _{0} \cos^4(\pi \cdot {t}/{T})\,{\rm d}t \hspace{0.05cm}.$$
  • Here,  the formula derived in subtask  (3)  and the symmetry of  $g_s(t)$  about time  $t = 0$  are considered.
  • The integral is given in the task description,  where $a = π/T$  is to be set:
$$E_g = 2 \cdot s_0^2 \cdot \left [ \frac{3}{8} \cdot t + \frac{T}{4\pi} \cdot \sin(2 \pi \frac{t}{T})+ \frac{T}{32\pi} \cdot \sin(4 \pi \frac{t}{T})\right ]_{0}^{T/2}\hspace{0.05cm}.$$
  • The lower bound  $t = 0$  always yields the result  $0$.   With respect to the upper bound,  only the first term yields a result different from  $0$.  Thus:
$$E_g = 2 \cdot s_0^2 \cdot \frac{3}{8} \cdot \frac{T}{2} = \frac{3}{8} \cdot 5 \cdot 10^{-7}\, {\rm Ws} \hspace{0.1cm}\underline{ = 0.1875 \cdot 10^{-6}\, {\rm Ws}}\hspace{0.05cm}.$$

(6)  The following relationship holds for the bipolar signal $s_{\rm C}(t)$:

$$P_{\rm S} = \frac{ E_g}{T} = \frac{ 1.875 \cdot 10^{-7}\, {\rm Ws}}{10^{-6}\, {\rm s}}\hspace{0.1cm}\underline{ = 0.1875 \,{\rm W}} \hspace{0.05cm}.$$