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Exercise 2.3: Binary Signal and Quaternary Signal

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ACF and PSD of binary signal  (B)  and quaternary signal  (Q)

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:

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



Questions

1

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

T = 

 ns

2

What is the (equivalent) bit rate  RB  of the binary system  B ?

RB = 

 Mbit/s

3

What is the transmitted power of the binary system  B?

PS = 

 mW

4

Which statements are true regarding the binary system  B

The ACF  φs(τ)  of the transmitted signal is  sinc2–shaped.
The energy ACF  φgs(τ)  of the basic transmission pulse is  sinc2–shaped.
The basic transmission pulse  gs(t)  itself is  sinc2–shaped.

5

What is the symbol duration  T  of the quaternary system  Q

T = 

 ns

6

What is the equivalent bit rate  RB  of the quaternary system  Q?

RB = 

 Mbit/s

7

What is the transmitted power  PS  of the quaternary system  Q?

PS = 

 mW

8

What is the maximum instantaneous transmitted power of the quaternary system  Q

s02 = 

 mW


Solution

(1)  The Nyquist frequency  fNyq=100 MHz  can be read from the diagram.  From this follows according to the properties of Nyquist systems:

fNyq=12TT=12fNyq=5ns_.


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

RB=1/T=200Mbit/s_=2fNyqbit/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=109WHz200MHz=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/Tφa(λ)φgs(τλT).
  • However,  for a redundancy-free binary system,  φa(λ=0)=1,  while all other discrete ACF values  φa(λ0)  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):
φgs(τ)=Tφs(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)|2Gs(f)=TΦs(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  φgs(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}.