Aufgaben:Exercise 2.3: Binary Signal and Quaternary Signal

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ACF and PSD of binary signal  $\rm (B)$  and quaternary signal  $\rm (Q)$

Two redundancy-free transmission systems  $\rm B$  and  $\rm Q$  each with bipolar amplitude coefficients  $a_{\nu}$  are to be compared.  Both systems satisfy the first Nyquist condition.  According to the root-root splitting,  the spectrum  $G_{d}(f)$  of the basic detection pulse is equal in shape to the power-spectral density  ${\it \Phi}_{s}(f)$  of the transmitted signal.

The following properties of the two systems are known:

• From the binary system  $\rm B$,  the power-spectral density  ${\it \Phi}_{s}(f)$  at the transmitter is known and shown in the graph together with the description parameters.

• The quaternary system  $\rm Q$  uses a NRZ rectangular signal with the four possible amplitude values  $±s_{0}$  and  $±s_{0}/3$, all with equal probability.

• ${s_{0}}^{2}$  indicates the maximum instantaneous power that occurs only when one of the two  "outer symbols"  is transmitted.  The descriptive parameters of system  $\rm Q$  can be obtained from the triangular ACF in the adjacent graph.

Notes:

• The exercise is part of the chapter  "Basics of Coded Transmission".

• Reference is also made to the chapter  "Redundancy-Free Coding".

• Consider that auto-correlation function  $\rm (ACF)$  and power-spectral density  $\rm (PSD)$  of a stochastic signal are always related via the Fourier transform.

Questions

1

What is the symbol duration  $T$  of the binary system  $\rm B$  with Nyquist property?

$T \ = \ $ $\ \rm ns$

2

What is the (equivalent) bit rate  $R_{\rm B}$  of the binary system  $\rm B$ ?

$R_{\rm B} \ = \ $ $\ \rm Mbit/s$

3

What is the transmitted power of the binary system  $\rm B$?

$P_{\rm S} \ = \ $ $\ \rm mW$

4

Which statements are true regarding the binary system  $\rm B$? 

	The ACF  $\varphi_{s}(\tau)$  of the transmitted signal is  $\rm sinc^{2}$–shaped.
	The energy ACF  $\varphi^{^{\bullet}}_{gs}(\tau)$  of the basic transmission pulse is  $\rm sinc^{2}$–shaped.
	The basic transmission pulse  $g_{s}(t)$  itself is  $\rm sinc^{2}$–shaped.

5

What is the symbol duration  $T$  of the quaternary system  $\rm Q$? 

$T \ = \ $ $\ \rm ns$

6

What is the equivalent bit rate  $R_{\rm B}$  of the quaternary system  $\rm Q$?

$R_{\rm B} \ = \ $ $\ \rm Mbit/s$

7

What is the transmitted power  $P_{\rm S}$  of the quaternary system  $\rm Q$?

$P_{\rm S} \ = \ $ $\ \rm mW$

8

What is the maximum instantaneous transmitted power of the quaternary system  $\rm Q$? 

${s_{0}}^{2} \ = \ $ $\ \rm mW$

Solution

Solution

(1)  The Nyquist frequency  $f_{\rm Nyq} = 100 \ \rm MHz$  can be read from the diagram.  From this follows according to the properties of Nyquist systems:

$$f_{\rm Nyq} = \frac{1 } {2 \cdot T} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} T = \frac{1 } {2 \cdot f_{\rm Nyq}} \hspace{0.15cm}\underline{ =5\,{\rm ns}}\hspace{0.05cm}.$$

(2)  In the binary system,  the bit rate is also the information flow and it holds:

$$R_{\rm B} = {1 }/ { T} \hspace{0.15cm}\underline {= 200\,{\rm Mbit/s}}= 2 \cdot f_{\rm Nyq} \cdot{\rm bit}/{\rm Hz}\hspace{0.05cm}.$$

(3)  The transmitted power is equal to the integral over  ${\it \Phi}_{s}(f)$  and can be calculated as a triangular area:

$$P_{\rm S} = \ \int_{-\infty}^{+\infty} {\it \Phi}_s(f) \,{\rm d} f = 10^{-9} \frac{\rm W}{\rm Hz} \cdot 200\,\,{\rm MHz} \hspace{0.15cm}\underline { = 200\,\,{\rm mW}}.$$

(4)  The  first two statements  are correct:

• The Fourier inverse transform of the power-spectral density  ${\it \Phi}_{s}(f)$  gives the $\rm sinc^{2}$–shaped ACF  $\varphi_{s}(\tau)$.  In general,  the following relationship also holds:

$$ \varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm}.$$

• However,  for a redundancy-free binary system,  $\varphi_{a}(\lambda = 0) = 1$,  while all other discrete ACF values  $\varphi_{a}(\lambda \neq 0)$  are equal to $0$.  Thus,  the energy ACF also has a  $\rm sinc^{2}$–shaped curve   (note:   energy ACF and energy PSD are each dotted in this tutorial):

$$\varphi^{^{\bullet}}_{gs}(\tau ) = T \cdot \varphi_s(\tau) \hspace{0.05cm}.$$

• The last statement is not true.  For the following reasoning,  we assume for simplicity that  $g_{s}(t)$  is symmetric and thus  $G_{s}(f)$  is real.  Then holds:

$${\it \Phi}_{s}(f) = {1 }/ { T} \cdot |G_s(f)|^2\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_s(f) = \sqrt{{ T} \cdot {\it \Phi}_{s}(f)}\hspace{0.4cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}g_s(t) \hspace{0.05cm}.$$

• Due to the square root in the above equation,  the basic transmission pulse  $g_{s}(t)$  is not  $\rm sinc^{2}$–shaped in contrast to the basic detection pulse $g_{d}(t)$,  which is equal in shape to the energy ACF  $\varphi^{^{\bullet}}_{gs}(\tau)$  and thus  $\rm sinc^{2}$–shaped.  At the same time,  $\varphi^{^{\bullet}}_{gs}(\tau) = g_{s}(\tau) ∗ g_{s}(–\tau)$  holds.

(5)  The ACF  $\varphi_{s}(\tau)$  is limited to the range  $|\tau| ≤ T$  when the basic transmission pulse is an NRZ rectangle.  From the graph,  the symbol duration  $T \underline{= 10 \ \rm ns}$.

(6)  For the quaternary signal,  the information flow is the same as for the binary signal above because of the double symbol duration:

$$R_{\rm B} = {{\rm log_2(4)} }/ { T} \hspace{0.15cm}\underline {= 200\,\,{\rm Mbit/s}}\hspace{0.05cm}.$$

(7)  The transmitted power is equal to the ACF value at  $\tau = 0$  and can be read from the graph:

$$P_{\rm S} = \hspace{0.15cm}\underline {100\,\,{\rm mW}}.$$

(8)  For the redundancy-free quaternary signal with NRZ rectangular pulses,  the average transmitted power is:

$$P_{\rm S} = {1}/ { 4} \cdot \left [ (-s_0)^2 + (-s_0/3)^2 + (+s_0/3)^2 +(+s_0)^2 \right ] = {5}/ { 9} \cdot s_0^2$$

$$\Rightarrow \hspace{0.3cm}s_0^2 = {9}/ {5} \cdot P_{\rm S} = {9}/ {5} \cdot 100\,\,{\rm mW}\hspace{0.15cm}\underline { = 180\,\,{\rm mW}}\hspace{0.05cm}.$$