Exercise 2.3: Binary Signal and Quaternary Signal
Two redundancy-free transmission systems B and Q each with bipolar amplitude coefficients aν are to be compared. Both systems satisfy the first Nyquist condition. According to the root-root splitting, the spectrum Gd(f) of the basic detection pulse is equal in shape to the power-spectral density Φs(f) of the transmitted signal.
The following properties of the two systems are known:
- From the binary system B, the power-spectral density Φs(f) at the transmitter is known and shown in the graph together with the description parameters.
- The quaternary system Q uses a NRZ rectangular signal with the four possible amplitude values ±s0 and ±s0/3, all with equal probability.
- s02 indicates the maximum instantaneous power that occurs only when one of the two "outer symbols" is transmitted. The descriptive parameters of system 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 (ACF) and power-spectral density (PSD) of a stochastic signal are always related via the Fourier transform.
Questions
Solution
- fNyq=12⋅T⇒T=12⋅fNyq=5ns_.
(2) In the binary system, the bit rate is also the information flow and it holds:
- RB=1/T=200Mbit/s_=2⋅fNyq⋅bit/Hz.
(3) The transmitted power is equal to the integral over Φs(f) and can be calculated as a triangular area:
- PS= ∫+∞−∞Φs(f)df=10−9WHz⋅200MHz=200mW_.
(4) The first two statements are correct:
- The Fourier inverse transform of the power-spectral density Φs(f) gives the sinc2–shaped ACF φs(τ). In general, the following relationship also holds:
- φs(τ)=+∞∑λ=−∞1/T⋅φa(λ)⋅φ∙gs(τ−λ⋅T).
- However, for a redundancy-free binary system, φa(λ=0)=1, while all other discrete ACF values φa(λ≠0) are equal to 0. Thus, the energy ACF also has a sinc2–shaped curve (note: energy ACF and energy PSD are each dotted in this tutorial):
- φ∙gs(τ)=T⋅φs(τ).
- The last statement is not true. For the following reasoning, we assume for simplicity that gs(t) is symmetric and thus Gs(f) is real. Then holds:
- Φs(f)=1/T⋅|Gs(f)|2⇒Gs(f)=√T⋅Φs(f)∙−−−∘gs(t).
- Due to the square root in the above equation, the basic transmission pulse gs(t) is not sinc2–shaped in contrast to the basic detection pulse KORREKTUR: basic transmitter pulsegd(t), which is equal in shape to the energy ACF φ∙gs(τ) and thus sinc2–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}.