Loading [MathJax]/jax/element/mml/optable/Latin1Supplement.js

Difference between revisions of "Aufgaben:Exercise 4.7: Spectra of ASK and BPSK"

From LNTwww
 
(20 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulationsverfahren
+
{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 
}}
 
}}
  
[[File:P_ID1701__Mod_A_4_6.png|right|]]
+
[[File:P_ID1701__Mod_A_4_6.png|right|frame|Power-spectral densities of  q(t)  and  s(t)   – valid for ASK and BPSK]]
Die Sendesignale von ASK (Amplitude Shift Keying) und BPSK (Binary Phase Shift Keying) können beide in der Form s(t)=q(t)·z(t) dargestellt werden, wobei z(t) eine harmonische Schwingung mit der Frequenz $f_T$ und der Amplitude 1 darstellt. Die Trägerphase $ϕ_T$ ist für die hier betrachteten Leistungsdichtespektren nicht von Bedeutung.
+
The transmitted signals of  $\rm ASK$  ("Amplitude Shift Keying")  and  $\rm BPSK$  ("Binary Phase Shift Keying")  can both be expressed in the form
 +
:$$s(t) = q(t) · z(t),$$
 +
where the carrier  z(t)  represents a harmonic oscillation with frequency  fT  and amplitude  $1$.  The carrier phase  $ϕ_{\rm T}$  is not important for the power-spectral densities considered here.
  
Bei ASK sind unipolare Amplitudenkoeffizienten – das heißt: aν0,1 – des Quellensignals
+
*In each case,  the source is redundancy-free,  which means that the two possible symbols are equally probable and the symbols are statistically independent of each other.
q(t)=+ν=aνgq(tνT)
+
*For ASK,  the unipolar amplitude coefficients – that is:  $a_ν ∈ \{0, 1\}$  – of the source signal are
anzusetzen, während im Fall der BPSK $a_ν$ ∈ {–1, +1} zu berücksichtigen ist. Die Quelle ist jeweils redundanzfrei, was bedeutet, dass die beiden möglichen Symbole ±1 gleichwahrscheinlich sind und die Symbole statistisch voneinander unabhängig.
+
:q(t)=+ν=aνgq(tνT)
 +
:while in the case of BPSK  $a_ν ∈ \{-1, +1\}$  has to be considered.
  
In der Grafik sind die Leistungsdichtespektren Φq(f) und Φs(f) von Quellensignal und Sendesignal angegeben, die sich bei einem NRZ–Rechteckimpuls gq(t) mit der Amplitude s0=2V und der Dauer T=1μs ergeben. Damit lautet die Spektralfunktion:
 
Gq(f)=s0Tsi(πfT).
 
Zu bestimmen sind in dieser Aufgabe die Konstanten A, B, C und D für ASK und BPSK.
 
  
'''Hinweis:''' Die Aufgabe bezieht sich auf das [http://en.lntwww.de/Modulationsverfahren/Lineare_digitale_Modulationsverfahren Kapitel 4.2] dieses Buches sowie auf das Kapitel 2.1 im Buch „Digitalsignalübertragung”.
+
In the diagram,  the power-spectral densities  Φq(f)  and  Φs(f)  of the source signal and the transmitted signal are given,  respectively,  for an NRZ rectangular pulse  gq(t)  with amplitude  s0=2 V  and duration  T = 1 \ \rm µ s.  Thus the spectral function is:
===Fragebogen===
+
:$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$
 +
The constants  A,  B,  C  and  D  for the  ASK  and  BPSK modulation methods are to be determined.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Notes:
 +
*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]].
 +
*Reference is also made to the chapter   [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]]   in the book  "Digital Signal Transmission".
 +
*The powers are to be specified in   $\rm V^2;  they thus refer to the reference resistance  R = 1 \ \rm \Omega$.
 +
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß sind der Parameter $A = Φ_q(f = 0)$ und das Diracgewicht B bei ASK?
+
{Which values result with ASK for the parameters &nbsp;$A = {\it Φ}_q(f = 0)$&nbsp; and &nbsp;$B&nbsp;(Dirac weight at &nbsp;f = 0)$?
 
|type="{}"}
 
|type="{}"}
$ASK: A$ = { 1 3% } $10^{-6}V^2$
+
$A \ = \ { 1 3% }\ \cdot 10^{-6} \  \rm V^2/Hz$
B = { 1 35 } V2  
+
$B \ = \ $ { 1 3% } $\ \rm V^2$  
  
  
{Bestimmen Sie die Parameter $C = Φ_s(f = f_T)$ und D des ASK–Sendesignals.
+
{Determine the parameters &nbsp;$C = {\it Φ}_s(f = f_{\rm T})$&nbsp; and &nbsp;$D&nbsp;(Diracweightatf = f_{\rm T})$ for the ASK transmitted signal.
 
|type="{}"}
 
|type="{}"}
$ASK:  C$ = { 0.25 3% } $10^{-6}V^2/Hz$  
+
$C \ = \ { 0.25 3% }\ \cdot 10^{-6} \  \rm V^2/Hz$
D = { 0.25 3% } V2  
+
$D \ = \ { 0.25 3% }\ \rm V^2$  
  
{Wie groß sind die Parameter $A = Φ_q(f = 0)$ und B bei BPSK?
+
{What are the values for parameters &nbsp;A&nbsp; and &nbsp;$B$&nbsp; for BPSK?
 
|type="{}"}
 
|type="{}"}
$BPSK:  A$ = { 4 3% }  $10^{-6}V^2/Hz$  
+
$A \ = \ { 4 3% }\ \cdot 10^{-6} \  \rm V^2/Hz$
B = { 0 3% } V2  
+
$B \ = \ $ { 0. } $\ \rm V^2$  
  
{Bestimmen Sie die Parameter $C = Φ_s(f = f_T)$ und D des BPSK–Sendedsignals.
+
{What are the values of parameters &nbsp;C&nbsp; and &nbsp;$D$&nbsp; for BPSK?
 
|type="{}"}
 
|type="{}"}
$BPSK:  C$ = { 1 3% } $10^{-6}V^2/Hz$  
+
$C \ = \ { 1 3% }\ \cdot 10^{-6} \  \rm V^2/Hz$  
D = { 0 3% V2  
+
$D \ = \ $ { 0. }  $\ \rm V^2$
  
{Welche Aussagen treffen zu, auch wenn gq(t) kein NRZ–Rechteckimpuls ist?
+
{Which statements are always true  for BPSK even if &nbsp;gq(t)&nbsp; is not an NRZ rectangular pulse?
|type="{}"}
+
|type="[]"}
+ Der kontinuierliche Anteil von $Φ_q(f)$ ist formgleich mit $|Gq(f)|^2$.
+
+ The continuous part of &nbsp;$ {\it Φ}_q(f)$&nbsp; is equal in form to &nbsp;$|G_q(f)|^2$.
- $Φ_q(f)$ beinhaltet bei ASK genau eine Diraclinie bei f=0.
+
- ${\it Φ}_q(f)$&nbsp; contains a single Dirac delta line at ASK&nbsp; for BPSK (at $f = 0)$.
- $Φ_q(f)$ beinhaltet bei BPSK genau eine Diraclinie bei f=0.
+
- ${\it Φ}_q(f)$ contains a single Dirac delta line at BPSK&nbsp; for BPSK (at $f = 0)$.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.''' Der Gleichanteil des unipolaren redundanzfreien Quellensignals beträgt mq=s0/2. Das Diracgewicht ist somit B=m2q=s20/4=1V2. Ohne diesen Gleichanteil ergibt sich das stochastische Rechtecksignal $q(t) m_q$ ∈ {$+s_0/2, –s_0/2$}. Dieses gleichsignalfreie Signal besitzt den kontinuierlichen LDS–Anteil $(s_0/2)^2 · T · si^2(πfT)$, woraus der gesuchte Wert bei der Frequenz f = 0 ermittelt werden kann:
+
'''(1)'''&nbsp; The DC component of the unipolar redundancy-free source signal is&nbsp; mq=s0/2.&nbsp; Thus,&nbsp; the Dirac weight is&nbsp; $B = m_q^2 = s_0^2/4\hspace{0.15cm}\underline{ = 1 \ \rm V^2}$.  
$$A = \frac{s_0^2 \cdot T}{4} = \frac{(2V)^2 \cdot 10^{-6} s}{4} = 10^{-6} V^2/Hz$$
+
 
 +
*Without this DC component,&nbsp; the stochastic  signal&nbsp; $q(t) - m_q ∈ \{+s_0/2, -s_0/2\}$ would be obtained.
 +
*This DC-free signal has the continuous PSD component&nbsp; $(s_0/2)^2 · T · {\rm sinc}^2(fT)$.
 +
*From this,&nbsp; the value we are looking for at frequency&nbsp; $f = 0$&nbsp; can be determined:
 +
:$$A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot
 +
10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.$$
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; The spectrum&nbsp; Z(f)&nbsp; of a cosine signal&nbsp; z(t)&nbsp; consists of two Dirac delta functions at&nbsp; ±fT,&nbsp; each with weight&nbsp; 1/2.
 +
*The power-spectral density&nbsp; Φz(f)&nbsp; also consists of the two Dirac delta functions,&nbsp; but now with respective weights&nbsp; 1/4.
 +
*The convolution&nbsp; Φq(f)Φz(f)&nbsp; gives the power-spectral density&nbsp; Φs(f)&nbsp; of the transmitted signal.&nbsp; It follows that:
 +
:$$C =  {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm
 +
V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.$$
 +
 
 +
Note: &nbsp; The power per bit is obtained as the integral over&nbsp; Φs(f):
 +
:$$P_{\rm S}  = \int_{ - \infty }^\infty \hspace{-0.3cm}  {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f
 +
= 2 \cdot \int_{ 0 }^\infty \hspace{-0.3cm}  {\left [ C \cdot {\rm sinc}^2(f T) + D \cdot \delta (f - f_{\rm T}]\right ]}\hspace{0.1cm} {\rm
 +
d}f= 2 \cdot \left [ \frac{C}{T}  + D \right ] =
 +
2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm
 +
V^{2}/Hz}}{10^{-6} \,{\rm s}}  + 0.25 \,{\rm V^{2}} \right ] \hspace{0.15cm}\underline {= 1
 +
\,{\rm V^{2}}}.$$
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp; For BPSK, the source signal&nbsp; q(t)&nbsp; is to be bipolar.
 +
*Therefore,&nbsp; the Dirac delta line is missing in the power-spectral density&nbsp;  ⇒  &nbsp; B=0_.
 +
*  The continuous PSD component is four times larger than with ASK:
 +
:$$A =  {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.$$
 +
 
 +
 
  
'''2.''' Das Spektrum Z(f) eines Cosinussignals z(t) besteht aus zwei Diracfunktionen bei ±fT, jeweils mit dem Gewicht 1/2. Das Leistungsdichtespektrum Φz(f) besteht ebenfalls aus den beiden Diracfunktionen, nun aber mit jeweiligem Gewicht 1/4. Die Faltung Φq(f) ∗ Φz(f) ergibt das Leistungsdichtespektrum Φs(f) des Sendesignals. Daraus folgt:  
+
'''(4)'''&nbsp; For the PSD parameters of the BPSK transmitted signal,&nbsp; the following applies analogously to the ASK:
$$C = \frac{A}{4} = 0.25 \cdot 10^{-6} V^2/Hz, D = \frac{B}{4} = 0.25 V^2$$
+
:$$C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D =
 +
\frac {B}{4} \hspace{0.15cm}\underline {= 0}.$$
  
  
'''3.'''
 
  
'''4.'''  
+
'''(5)'''&nbsp; Only the&nbsp; <u>first statement</u>&nbsp; is correct:
 +
* For BPSK,&nbsp;  Φq(f)&nbsp; does not contain a single Dirac delta line even if&nbsp; gq(t)&nbsp; deviates from the rectangular form&nbsp; (assuming equally probable symbols).
 +
*In contrast, the unipolar ASK source signal contains infinitely many Dirac delta lines at all multiples of&nbsp; 1/T.
 +
*For more information on this topic, see the page&nbsp;  "ACF and PSD for unipolar binary signals" in the book "Digital Signal Transmission".
  
'''5.'''
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^4.2 Lineare digitale Modulationsverfahren^]]
+
[[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]]

Latest revision as of 14:31, 18 January 2023

Power-spectral densities of  q(t)  and  s(t)  – valid for ASK and BPSK

The transmitted signals of  ASK  ("Amplitude Shift Keying")  and  BPSK  ("Binary Phase Shift Keying")  can both be expressed in the form

s(t)=q(t)·z(t),

where the carrier  z(t)  represents a harmonic oscillation with frequency  fT  and amplitude  1.  The carrier phase  ϕT  is not important for the power-spectral densities considered here.

  • In each case,  the source is redundancy-free,  which means that the two possible symbols are equally probable and the symbols are statistically independent of each other.
  • For ASK,  the unipolar amplitude coefficients – that is:  aν{0,1}  – of the source signal are
q(t)=+ν=aνgq(tνT)
while in the case of BPSK  aν{1,+1}  has to be considered.


In the diagram,  the power-spectral densities  Φq(f)  and  Φs(f)  of the source signal and the transmitted signal are given,  respectively,  for an NRZ rectangular pulse  gq(t)  with amplitude  s0=2 V  and duration  T = 1 \ \rm µ s.  Thus the spectral function is:

G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.

The constants  A,  B,  C  and  D  for the  \rm ASK  and  \rm BPSK modulation methods are to be determined.




Notes:

  • The exercise belongs to the chapter  Linear Digital Modulation.
  • Reference is also made to the chapter  Basics of Coded Transmission  in the book  "Digital Signal Transmission".
  • The powers are to be specified in  \rm V^2;  they thus refer to the reference resistance  R = 1 \ \rm \Omega.



Questions

1

Which values result with ASK for the parameters  A = {\it Φ}_q(f = 0)  and  B  (Dirac weight at  f = 0)?

A \ = \

\ \cdot 10^{-6} \ \rm V^2/Hz
B \ = \

\ \rm V^2

2

Determine the parameters  C = {\it Φ}_s(f = f_{\rm T})  and  D  (Dirac weight at f = f_{\rm T}) for the ASK transmitted signal.

C \ = \

\ \cdot 10^{-6} \ \rm V^2/Hz
D \ = \

\ \rm V^2

3

What are the values for parameters  A  and  B  for BPSK?

A \ = \

\ \cdot 10^{-6} \ \rm V^2/Hz
B \ = \

\ \rm V^2

4

What are the values of parameters  C  and  D  for BPSK?

C \ = \

\ \cdot 10^{-6} \ \rm V^2/Hz
D \ = \

\ \rm V^2

5

Which statements are always true for BPSK even if  g_q(t)  is not an NRZ rectangular pulse?

The continuous part of   {\it Φ}_q(f)  is equal in form to  |G_q(f)|^2.
{\it Φ}_q(f)  contains a single Dirac delta line at ASK  for BPSK (at f = 0).
{\it Φ}_q(f) contains a single Dirac delta line at BPSK  for BPSK (at f = 0).


Solution

(1)  The DC component of the unipolar redundancy-free source signal is  m_q = s_0/2.  Thus,  the Dirac weight is  B = m_q^2 = s_0^2/4\hspace{0.15cm}\underline{ = 1 \ \rm V^2}.

  • Without this DC component,  the stochastic signal  q(t) - m_q ∈ \{+s_0/2, -s_0/2\} would be obtained.
  • This DC-free signal has the continuous PSD component  (s_0/2)^2 · T · {\rm sinc}^2(fT).
  • From this,  the value we are looking for at frequency  f = 0  can be determined:
A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.


(2)  The spectrum  Z(f)  of a cosine signal  z(t)  consists of two Dirac delta functions at  \pm f_{\rm T},  each with weight  1/2.

  • The power-spectral density  {\it Φ}_z(f)  also consists of the two Dirac delta functions,  but now with respective weights  1/4.
  • The convolution  {\it Φ}_q(f) ∗ {\it Φ}_z(f)  gives the power-spectral density  {\it Φ}_s(f)  of the transmitted signal.  It follows that:
C = {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.

Note:   The power per bit is obtained as the integral over  {\it Φ}_s(f):

P_{\rm S} = \int_{ - \infty }^\infty \hspace{-0.3cm} {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f = 2 \cdot \int_{ 0 }^\infty \hspace{-0.3cm} {\left [ C \cdot {\rm sinc}^2(f T) + D \cdot \delta (f - f_{\rm T}]\right ]}\hspace{0.1cm} {\rm d}f= 2 \cdot \left [ \frac{C}{T} + D \right ] = 2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm V^{2}/Hz}}{10^{-6} \,{\rm s}} + 0.25 \,{\rm V^{2}} \right ] \hspace{0.15cm}\underline {= 1 \,{\rm V^{2}}}.


(3)  For BPSK, the source signal  q(t)  is to be bipolar.

  • Therefore,  the Dirac delta line is missing in the power-spectral density  ⇒   \underline{B = 0}.
  • The continuous PSD component is four times larger than with ASK:
A = {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.


(4)  For the PSD parameters of the BPSK transmitted signal,  the following applies analogously to the ASK:

C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = \frac {B}{4} \hspace{0.15cm}\underline {= 0}.


(5)  Only the  first statement  is correct:

  • For BPSK,  {\it Φ}_q(f)  does not contain a single Dirac delta line even if  g_q(t)  deviates from the rectangular form  (assuming equally probable symbols).
  • In contrast, the unipolar ASK source signal contains infinitely many Dirac delta lines at all multiples of  1/T.
  • For more information on this topic, see the page  "ACF and PSD for unipolar binary signals" in the book "Digital Signal Transmission".