Exercise 2.2: Binary Bipolar Rectangles

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".

Questions

Solution

(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.

	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.

	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.

	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.

	$s_{0.75}(t)$,
	$s_{0.50}(t)$,
	$s_{0.25}(t)$,