Exercise 2.2: Binary Bipolar Rectangles

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Binary bipolar rectangular signals

We assume the following signal:

$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm}.$$

The basic transmission pulse  $g_{s}(t)$  is always assumed to be rectangular in this exercise,  with the NRZ format  (blue signal curves in the graph)  as well as the RZ format with duty cycle  $T_{\rm S}/T = 0.5$  (red signal curves)  to be investigated.

The amplitude coefficients have the following properties:

  • They are binary and bipolar:   $a_{\nu} \in \{–1, +1\}$.
  • The symbols within the sequence  $\langle a_{\nu }\rangle$  have no statistical ties.
  • The probabilities for the two possible values  $±1$  are with  $0 < p < 1$:
$${\rm Pr}(a_\nu = +1) \ = \ p,$$
$${\rm Pr}(a_\nu = -1) \ = \ 1 - p \hspace{0.05cm}.$$

The three signal sections shown in the graph are valid for  $p = 0.75$,  $p = 0.50$  and  $p = 0.25$.

Throughout this exercise,  reference is made to the following descriptive quantities:

  • $m_{a} = \E\big[a_{\nu}\big]$  indicates the linear mean  (first order moment)  of the amplitude coefficients.
  • $m_{2a} = \E\big[a_{\nu}^{2}\big]$  is the power  (second order moment).
  • Thus,  the variance  $\sigma_{a}^{2} = m_{2a} - m_{a}^{2}$  can also be calculated.
  • The discrete ACF of the amplitude coefficients is  $\varphi_{a}(\lambda) = \E\big[a_{\nu} \cdot a_{\nu + \lambda} \big]$.  It holds here:
$$\varphi_a(\lambda) = \left\{ \begin{array}{c} m_2 \\ m_1^2 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c}\lambda = 0, \\ \lambda \ne 0 \hspace{0.05cm}.\\ \end{array}$$
  • The energy ACF of the basic transmission pulse is:
$$\varphi^{^{\bullet}}_{g_s}(\tau) = \left\{ \begin{array}{c} s_0^2 \cdot T_{\rm S} \cdot \left( 1 - {|\tau|}/{T_{\rm S}}\right) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c}|\tau| \le T_{\rm S} \\ |\tau| \ge T_{\rm S} \hspace{0.05cm}.\\ \end{array}$$
  • Thus,  for the total ACF of the transmitted signal,  we obtain:
$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{g_s}(\tau - \lambda \cdot T)\hspace{0.05cm}.$$
  • The power-spectral density  ${\it \Phi}_{s}(f)$  is the Fourier transform of the ACF  $\varphi_{s}(\tau)$.

Note:   The exercise belongs to the chapter  "Basics of Coded Transmission".



Which of the three signals shown are redundancy-free?



What is the second order moment  ("power")  $m_{2a}= \E\big[a_{\nu}^{2}\big]$  of the amplitude coefficients as a function of  $p$?

$p = 0.75\text{:} \hspace{0.4cm} m_{2a} \ = \ $

$p = 0.50\text{:} \hspace{0.4cm} m_{2a} \ = \ $

$p = 0.25\text{:} \hspace{0.4cm} m_{2a} \ = \ $


Calculate the first order moment  ("linear mean")  $m_{a}= \E\big[a_{\nu}\big]$  in relation to  $p$.

$p = 0.75\text{:} \hspace{0.4cm} m_{a} \ = \ $

$p = 0.50\text{:} \hspace{0.4cm} m_{a} \ = \ $

$p = 0.25\text{:} \hspace{0.4cm} m_{a} \ = \ $


What is the variance  $\sigma_{a}^{2}$  of the amplitude coefficients?

$p = 0.75\text{:} \hspace{0.4cm} \sigma_{a}^{2} \ = \ $

$p = 0.50\text{:} \hspace{0.4cm} \sigma_{a}^{2} \ = \ $

$p = 0.25\text{:} \hspace{0.4cm} \sigma_{a}^{2} \ = \ $


Let  $p = 0.5$  hold initially.  Sketch the ACF  $\varphi_{s}(\tau)$  for the NRZ and RZ basic transmission pulses and evaluate the following statements:

The ACF is triangular in both cases.
The PSD is  ${\rm sinc}^{2}$–shaped in both cases.
The PSD area is the same in both cases.
In the case of RZ pulses,  ${\it \Phi}_{s}(f)$  involves additional Dirac delta functions.


Let be  $p = 0.75$.  Sketch the ACF for the NRZ basic pulse and evaluate the following statements:

The ACF consists of a triangle and a DC component.
The PSD consists of a  ${\rm sinc}^{2}$ component and a Dirac delta function.
The Dirac delta function has the weight  $s_{0}^{2}$.
With $p = 0.25$,  the same power-spectral density is obtained.


Let be  $p = 0.75$.  Sketch the ACF for the RZ basic pulse and evaluate the following statements:

Again,  the PSD contains a  ${\rm sinc}^{2}$–shaped component.
At the same time,  there are still infinitely many Dirac delta lines in the PSD.


(1)  A digital signal is said to be redundancy-free if

  • the amplitude coefficients do not depend on each other  (this was assumed here),
  • all possible amplitude coefficients are equally probable.

In this sense,  $s_{0.5}(t)$  is a redundancy-free signal   ⇒   solution 2.

  • Thus,  here the entropy  (the average information content per transmitted binary symbol)  is at most equal to the decision content:
$$H_{\rm max} = {1}/{2}\cdot {\rm log}_2 (2)+{1}/{2}\cdot {\rm log}_2 (2) = 1 \,\,{\rm bit/binary\ symbol} \hspace{0.05cm}.$$
  • In contrast,  the entropies of the other two binary signals are:
$$H = \ \frac{3}{4}\cdot {\rm log}_2 (\frac{4}{3})+ \frac{1}{4}\cdot {\rm log}_2 (4) = \left( \frac{3}{4} + \frac{1}{4}\right)\cdot {\rm log}_2 (4) - \frac{3}{4}\cdot{\rm log}_2 (3) =$$
$$ \hspace{0.5cm} = \ 2 - \frac{3}{4}\cdot{\rm log}_2 (3) = 0.811 \,\,{\rm bit/binary\ symbol} \hspace{0.05cm}.$$
  • From this,  the relative redundancy of these signals is:
$$r = \frac{H_{\rm max} - H}{H_{\rm max}}\hspace{0.15cm} \approx 18.9\%\hspace{0.05cm}.$$

(2)  The second order moment  ("power")  is equal to  $m_{2a} = 1$  independent of  $p$:

$$m_{2a}={\rm E}[a_\nu^2] = p \cdot (+1)^2 + (1-p)\cdot (-1)^2 \hspace{0.15cm}\underline { = 1 \hspace{0.05cm}}.$$

(3)  For the first order moment  ("linear mean")  we get:

$$m_{a}={\rm E}[a_\nu] = p \cdot (+1) + (1-p)\cdot (-1) = 2 p -1 \hspace{0.05cm}.$$
$$\Rightarrow \hspace{0.3cm} p = 0.75\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline {=0.50},\hspace{0.2cm} p = 0.50\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline {=0},\hspace{0.2cm} p = 0.25\text{:} \hspace{0.4cm} m_{a}\hspace{0.15cm}\underline { =-0.50 \hspace{0.05cm}}.$$

(4)  Using the results from  ( (2)  and  (4),  we obtain:

$$p = 0.75\text{:} \hspace{0.4cm} \sigma_{a}^2 \hspace{0.15cm}\underline {=0.75},$$
$$ p = 0.50\text{:} \hspace{0.4cm} \sigma_{a}^2\hspace{0.15cm} \underline { =1.00 \hspace{0.05cm}},$$
$$ p = 0.25\text{:} \hspace{0.4cm} \sigma_{a}^2 \hspace{0.15cm}\underline {=0.75}.$$

ACF with equal symbol probabilities

(5)  Only the  first two statements  are correct:

  • For  $p = 0.5$,   $\varphi_{a}(\lambda = 0) = 1$  and  $\varphi_{a}(\lambda \neq 0) = 0$.  It follows that:
$$\varphi_s(\tau) = \frac{1}{T} \cdot \varphi^{^{\bullet}}_{gs}(\tau )\hspace{0.05cm}.$$
  • This results in a triangular ACF and a  ${\rm sinc}^{2}$–shaped PSD for both the NRZ and RZ basic pulses.
  • The area under the PSD is smaller by a factor of  $T_{\rm S}/T$  for the RZ pulse than for the NRZ pulse, 
    since the ACF values also differ by this factor at  $\tau = 0$.
  • The PSD is continuous in both cases because the ACF does not contain a DC component or periodic components.

ACF with unequal symbol probabilities

(6)  All statements except the third  are correct:

  • For  $p = 0.75$,   the ACF $\varphi_{s}(\tau)$ is composed of infinitely many triangular functions,  all of which have the same height  $s_{0}^{2}/4$  except for the middle triangle around  $\tau = 0$.
  • According to the sketch,  one can combine all these triangle functions into a DC component of height  $m_{a}^{2} \cdot s_{0}^{2} = s_{0}^{2}/4$  and a single triangle around  $\tau = 0$  with height  $\sigma_{a}^{2} \cdot s_{0}^{2} = 3/4 · s_{0}^{2}$.
  • In the PSD,  this leads to a continuous  ${\rm sinc}^{2}$–shaped component and a Dirac delta function at  $f = 0$.  The weight of this Dirac is  $s_{0}^{2}/4$.
  • For  $p = 0.25$  we get the same ACF as with  $p = 0.75$,  since both the second order moment  $m_{2a} = 1$  and  $m_{a}^{2} = 0.25$  coincide.  Thus,  of course,  the power-spectral densities also match.

ACF for RZ rectangular pulses

(7)  Both proposed solutions are correct::

  • With the RZ duty cycle  $T_{\rm S}/T = 0.5$  the sketched ACF is obtained,  which can also be represented by a periodic triangular function of height  $s_{0}^{2}/8$  (red filling)  and a single triangular pulse of height  $3/8 \cdot s_{0}^{2}$  (green filling).
  • This non-periodic component leads to a continuous-valued,  ${\rm sinc}^{2}$–shaped PSD with zeros at multiples of  $2/T$.
  • The periodic triangular ACF causes Dirac delta functions in the PSD at multiples of  $1/T$.
  • However,  due to the antimetry of the periodic component,  the Dirac delta functions at multiples of  $2/T$  each have weight  $0$.
  • The weights of the Dirac delta functions at distance  $1/T$  are proportional to the continuous PSD component.