Difference between revisions of "Exercise 2.4: DSL/DMT with IDFT/DFT"

From LNTwww
Line 1: Line 1:
 
   
 
   
{{quiz-Header|Buchseite=Beispiele von Nachrichtensystemen/xDSL als Übertragungstechnik
+
{{quiz-Header|Buchseite=Examples_of_Communication_Systems/xDSL_as_Transmission_Technology
  
 
}}
 
}}
  
[[File:P_ID1972__Bei_A_2_4.png|right|frame|Zeitabtastwerte bei verschiedenen DMT-Spektralbelegungen]]
+
[[File:P_ID1972__Bei_A_2_4.png|right|frame|time samples with different DMT spectral allocations]]
  
Eine  [[Examples_of_Communication_Systems/xDSL_als_Übertragungstechnik#DMT.E2.80.93Realisierung_mit_IDFT.2FDFT|Realisierungsform]]  des  $\rm DMT$–Verfahrens (steht für ''Discrete Multitone Transmission'') basiert auf der ''Inversen Diskreten Fouriertransformation''   $\rm (IDFT)$  beim Sender sowie der ''Diskreten Fouriertransformation''   $\rm (DFT)$  beim Empfänger.
+
A  [[Examples_of_Communication_Systems/xDSL_as_Transmission_Technology#DMT_realization_with_IDFT.2 FDFT|"realization form"]]  of the  $\rm DMT$ method (stands for ''Discrete Multitone Transmission'') is based on the ''Inverse Discrete Fourier Transform''   $\rm (IDFT)$  at the transmitter and the ''Discrete Fourier Transform''   $\rm (DFT)$  at the receiver.
  
Beim Sender werden  $N/2-1$  Nutzer durch die komplexen Spektralkoeffizienten  $D_{k} \ (k = 1,$ ... , $N/2–1)$  den Frequenzen  $f_{k} = k \cdot f_{0}$  zugewiesen. Die Grundfrequenz  $f_{0}$  ist der Kehrwert der Symboldauer  $T$.
+
At the transmitter  $N/2-1$  users are represented by the complex spectral coefficients  $D_{k} \ (k = 1,$ ... , $N/2-1)$  allocated to the frequencies  $f_{k} = k \cdot f_{0}$  . The fundamental frequency  $f_{0}$  is the reciprocal of the symbol duration  $T$.
  
*Es gilt  $D_{k} \in \{ ±1 ± {\rm j} \}$, falls ein Kanal belegt ist, im anderen Fall ist  $D_{k} = 0$.  
+
*It holds  $D_{k} \in \{ ±1 ± {\rm j} \}$ if one channel is occupied, in the other case  $D_{k} = 0$.  
*Die Koeffizienten  $D_{0}$  und  $D_{N/2}$  sind stets Null.  
+
*The coefficients  $D_{0}$  and  $D_{N/2}$  are always zero.  
*Die obersten Koeffizienten werden konjugiert–komplex belegt:
+
*The top coefficients are allocated conjugate-complex:
 
:$$D_k = D_{N-k}^{\star},\hspace{0.2cm}k = N/2 +1,\hspace{0.05cm} \text{...} \hspace{0.05cm}, N-1 \hspace{0.05cm}.$$
 
:$$D_k = D_{N-k}^{\star},\hspace{0.2cm}k = N/2 +1,\hspace{0.05cm} \text{...} \hspace{0.05cm}, N-1 \hspace{0.05cm}.$$
  
Dadurch wird sicher gestellt, dass das Zeitsignal  $s(t)$  stets reell ist. Die Abtastwerte  $s_{0}$, ... , $s_{N–1}$  dieses Signals werden dabei durch die IDFT gebildet, wobei der zeitliche Abstand zweier Abtastwerte
+
This ensures that the time signal  $s(t)$  is always real. The sample values  $s_{0}$, ... , $s_{N-1}$  of this signal are thereby formed by the IDFT, where the temporal distance of two samples is
:$$\Delta t = T/N = 1/(N \cdot f_{0})$$  
+
:$$\Delta t = T/N = 1/(N \cdot f_{0}).$$
beträgt. Durch Tiefpassfilterung erhält man das zeitkontinuierliche Signal.
+
Low-pass filtering is used to obtain the continuous-time signal.
  
Bei ADSL/DMT gilt  $N = 512$  und  $f_{0} = 4.3125 \ \rm kHz$. In dem hier betrachteten Beispiel seien die Parameter zur Vereinfachung wie folgt angenommen:
+
For ADSL/DMT,  $N = 512$  and  $f_{0} = 4.3125 \ \rm kHz$. In the example considered here, let the parameters be assumed as follows for simplicity:
:$$N = 16,\hspace{0.2cm}\Delta t = 10\,{\rm µ s} \hspace{0.05cm}.$$
+
:$$N = 16,\hspace{0.2cm}\delta t = 10\,{\rm µ s} \hspace{0.05cm}.$$
In obiger Tabelle sind für drei verschiedene  $D_{k}$–Belegungen die Abtastwerte  $s_{l} (l = 0$, ... , $15)$  nach der IDFT angegeben. Gesucht sind die zugehörigen Spektralkoeffizienten  $D_{k}\ (k = 0$, ... , $15).$
+
In the above table, for three different  $D_{k}$ occupancies, the sample values  $s_{l} (l = 0$, ... , $15)$  according to the IDFT are given. The corresponding spectral coefficients  $D_{k}\ (k = 0$, ... , $15)$ are sought.
  
  
Line 30: Line 30:
  
  
''Hinweise:''
+
Hints:
*Die Aufgabe gehört zum Kapitel  [[Examples_of_Communication_Systems/xDSL_als_Übertragungstechnik|xDSL als Übertragungstechnik]].  
+
*This exercise belongs to the chapter  [[Examples_of_Communication_Systems/xDSL_as_Transmission_Technology|"xDSL as Transmission Technology"]].  
*Das Sendesignal hat bei DSL die Form
+
*The transmission signal for DSL has the form
 
:$$s(t) = \sum_{k = 1}^{K} \big [ 2 \cdot {\rm Re}\{D_k\} \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}\{D_k\} \cdot \sin(2\pi \cdot k f_0 \cdot t )\big ] \hspace{0.05cm}.$$
 
:$$s(t) = \sum_{k = 1}^{K} \big [ 2 \cdot {\rm Re}\{D_k\} \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}\{D_k\} \cdot \sin(2\pi \cdot k f_0 \cdot t )\big ] \hspace{0.05cm}.$$
*Beachten Sie auch die folgende trigonometrische Beziehung:
+
*Note also the following trigonometric relationship:
 
:$$\cos(2\pi f_0 t + \phi_0) = \cos( \phi_0) \cdot \cos(2\pi f_0 t ) - \sin( \phi_0) \cdot \sin(2\pi f_0 t ) \hspace{0.05cm}.$$
 
:$$\cos(2\pi f_0 t + \phi_0) = \cos( \phi_0) \cdot \cos(2\pi f_0 t ) - \sin( \phi_0) \cdot \sin(2\pi f_0 t ) \hspace{0.05cm}.$$
*Man bezeichnet  das Verhältnis von Maximalwert und Effektivwert als den  ''Crestfaktor''  (oder den Scheitelfaktor) eines Signals.
+
*The ratio of the maximum value and the rms value is called the  ''crest factor'''  (or the crest factor) of a signal.
*Sie können Ihre Lösung mit dem interaktiven Applet  [[Applets:Diskrete_Fouriertransformation_(Applet)|Diskrete Fouriertransformation]] überprüfen.
+
*You can check your solution with the interactive applet  [[Applets:Diskrete_Fouriertransformation_(Applet)|"Discrete Fourier Transform"]].
 
   
 
   
  
Line 44: Line 44:
  
  
===Fragebogen===
+
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{ Wieviele Nutzer&nbsp; $(K)$&nbsp; können mit diesem System versorgt werden?
+
{ How many users&nbsp; $(K)$&nbsp; can be provided with this system?
 
|type="{}"}
 
|type="{}"}
 
$K \ = \ ${ 7 }  
 
$K \ = \ ${ 7 }  
  
{Wie groß ist die Bandbreite&nbsp; $B$&nbsp; des betrachteten DMT–Systems?
+
{What is the bandwidth&nbsp; $B$&nbsp; of the DMT system under consideration?
 
|type="{}"}
 
|type="{}"}
 
$B \ = \ ${ 50 3% } $\ \rm kHz $
 
$B \ = \ ${ 50 3% } $\ \rm kHz $
  
{Wie lauten die Spektralkoeffizienten bei der Belegung&nbsp; $\boldsymbol{\rm A}$?
+
{What are the spectral coefficients at allocation&nbsp; $\boldsymbol{\rm A}$?
 
|type="()"}
 
|type="()"}
- $D_{1} = 1- \rm j, \ alle \ anderen \ 0,$
+
- $D_{1} = 1- \rm j, \ all \ other \ 0,$
+ $D_{1} = 1 + {\rm j}, \ D_{15} = 1 - \rm j, \ alle \ anderen \ 0,$
+
+ $D_{1} = 1 + {\rm j}, \ D_{15} = 1 - \rm j, \ all \ others \ 0,$
- $D_{1} = 1 + {\rm j}, \ D_{15} = 1 + \rm j, \ alle \ anderen \ 0.$
+
- $D_{1} = 1 + {\rm j}, \ D_{15} = 1 + \rm j, \ all \ others \ 0.$
  
{Wie lauten die Spektralkoeffizienten bei der Belegung&nbsp; $\boldsymbol{\rm B}$?
+
{What are the spectral coefficients at allocation&nbsp; $\boldsymbol{\rm B}$?
 
|type="()"}
 
|type="()"}
- $D_{2} = -1 - {\rm j}, \ D_{14} = -1 + \rm j, \ alle \ anderen \ 0$,
+
- $D_{2} = -1 - {\rm j}, \ D_{14} = -1 + \rm j, \ all \ other \ 0$,
- $D_{3} = +1 - {\rm j}, \ D_{13} = +1 + \rm j, \ alle \ anderen \ 0$,
+
- $D_{3} = +1 - {\rm j}, \ D_{13} = +1 + \rm j, \ all \ others \ 0$,
+ $D_{3} = -1 - {\rm j}, \ D_{13} = -1 + \rm j, \ alle \ anderen \ 0$.
+
+ $D_{3} = -1 - {\rm j}, \ D_{13} = -1 + \rm j, \ all \ others \ 0$.
  
{Wie lauten die Spektralkoeffizienten bei der Belegung&nbsp; $\boldsymbol{\rm C}$&nbsp; mit&nbsp; $\boldsymbol{\rm C} = \boldsymbol{\rm A} + \boldsymbol{\rm B}?$
+
{What are the spectral coefficients for the allocation&nbsp; $\boldsymbol{\rm C}$&nbsp; with&nbsp; $\boldsymbol{\rm C} = \boldsymbol{\rm A} + \boldsymbol{\rm B}?$
 
|type="()"}
 
|type="()"}
+ $D_{1} = 1 + {\rm j}, \ D_{3} = -1 -{\rm j}, \ D_{13} = -1 +{\rm j}, \ D_{15} = 1 - {\rm j}$,
+
+ $D_{1} = 1 + {\rm j}, \ D_{3} = -1 -{\rm j}, \ D_{13} = -1 +{\rm j}, \ D_{15} = 1 - {\rm j}$,
- $D_{k} = (-1)^k + {\rm j} \cdot (–1)^{k+1}$.
+
- $D_{k} = (-1)^k + {\rm j} \cdot (-1)^{k+1}$.
  
  
{Wie groß ist der Crestfaktor&nbsp; $(s_{\rm max}/s_{\rm eff})$&nbsp; bei der Belegung&nbsp; $\boldsymbol{\rm C}$?
+
{What is the crest factor&nbsp; $(s_{\rm max}/s_{\rm eff})$&nbsp; at allocation&nbsp; $\boldsymbol{\rm C}$?
 
|type="{}"}
 
|type="{}"}
 
$s_{\rm max}/s_{\rm eff} \ = \ ${ 1.85 3% }  
 
$s_{\rm max}/s_{\rm eff} \ = \ ${ 1.85 3% }  
Line 80: Line 80:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Das System ist für $K = N/2 - 1 \underline{= 7 \ {\rm Nutzer}}$ ausgelegt $(N = 16)$.
+
'''(1)'''&nbsp; The system is designed for $K = N/2 - 1 \underline{= 7 \ {\rm users}}$ $(N = 16)$.
  
  
'''(2)'''&nbsp; Die Rahmendauer $T$ ergibt sich zu $N \cdot \Delta t = 0.16 \rm ms$.  
+
'''(2)'''&nbsp; The frame duration $T$ is given by $N \cdot \delta t = 0.16 \rm ms$.  
*Die Grundfrequenz ist hier dementsprechend $f_{0} = 1/T = 6.25 \ \rm kHz$ und die Gesamtbandbreite beträgt $B = 8 \cdot f_{0} \ \underline{= 50 \ \rm kHz}$.  
+
*The fundamental frequency here is accordingly $f_{0} = 1/T = 6.25 \ \rm kHz$ and the total bandwidth is $B = 8 \cdot f_{0} \ \underline{= 50 \ \rm kHz}$.  
*Zum Vergleich: Bei ADSL ergibt sich diese Bandbreite zu $256 \cdot 4.3125 \ \rm kHz= 1104 \ kHz.$
+
*For comparison, for ADSL, this bandwidth results in $256 \cdot 4.3125 \ \rm kHz= 1104 \ kHz.$
  
  
'''(3)'''&nbsp; Richtig ist <u>der zweite Lösungsvorschlag</u>:
+
'''(3)'''&nbsp; Correct is <u>the second proposed solution</u>:
* Aus den $16$ Abtastwerten $s_{l}$ in der ersten Spalte der Tabelle &nbsp;$($Belegung $\boldsymbol{\rm A})$&nbsp; erkennt man, dass $s(t)$ eine harmonische Schwingung mit der Periodendauer $T_{0} = T$ beschreibt (nur eine Schwingung). Die Amplitude ist gleich $2 \cdot \sqrt{2} =2.828$ und die Phase beträgt $\phi_0 = 45^\circ \ (π/4)$.  
+
* From the $16$ samples $s_{l}$ in the first column of the table &nbsp;$($occupancy $\boldsymbol{\rm A})$&nbsp; we see that $s(t)$ describes a harmonic oscillation with period $T_{0} = T$ (only one oscillation). The amplitude is equal to $2 \cdot \sqrt{2} =2.828$ and the phase is $\phi_0 = 45^\circ \ (π/4)$.  
*Damit kann für das zeitkontinuierliche Signal geschrieben werden &nbsp;$($mit  $f_{0} = 1/T)$:
+
*Thus, for the continuous-time signal, we can write &nbsp;$($with $f_{0} = 1/T)$:
 
:$$s(t) = 2 \cdot \sqrt{2}\cdot \cos(2\pi f_0 t + \pi /4) \hspace{0.05cm}.$$
 
:$$s(t) = 2 \cdot \sqrt{2}\cdot \cos(2\pi f_0 t + \pi /4) \hspace{0.05cm}.$$
*Mit der angegebenen trigonometrischen Umformung und ${\rm cos} \ (π/4) \ = \ {\rm sin} \ (π/4) \ = \ \sqrt{2}$ gilt weiterhin:
+
*With the given trigonometric transformation and ${\rm cos} \ (π/4) \ = \ {\rm sin} \ (π/4) \ = \ \sqrt{2}$ still holds:
 
:$$s(t) = 2 \cdot \cos(2\pi f_0 t ) - 2 \cdot \sin(2\pi f_0 t ) \hspace{0.05cm}.$$
 
:$$s(t) = 2 \cdot \cos(2\pi f_0 t ) - 2 \cdot \sin(2\pi f_0 t ) \hspace{0.05cm}.$$
*Ein Koeffizientenvergleich mit der weiteren Gleichung
+
*A coefficient comparison with the further equation.
 
:$$s(t) = \sum_{k = 1}^{K} \left [ 2 \cdot {\rm Re}[D_k] \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}[D_k] \cdot \sin(2\pi \cdot k f_0 \cdot t )\right ] \hspace{0.05cm}$$
 
:$$s(t) = \sum_{k = 1}^{K} \left [ 2 \cdot {\rm Re}[D_k] \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}[D_k] \cdot \sin(2\pi \cdot k f_0 \cdot t )\right ] \hspace{0.05cm}$$
:liefert das Ergebnis:
+
:returns the result:
 
:$$2 \cdot {\rm Re}[D_1] = 2 \hspace{0.3cm} \ \Rightarrow \ \hspace{0.3cm} {\rm Re}[D_1] = 1\hspace{0.05cm},$$  
 
:$$2 \cdot {\rm Re}[D_1] = 2 \hspace{0.3cm} \ \Rightarrow \ \hspace{0.3cm} {\rm Re}[D_1] = 1\hspace{0.05cm},$$  
 
:$$2 \cdot {\rm Im}[D_1] = 2 \hspace{0.3cm} \ \Rightarrow \ \hspace{0.3cm} {\rm Im}[D_1] = 1\hspace{0.05cm}.$$
 
:$$2 \cdot {\rm Im}[D_1] = 2 \hspace{0.3cm} \ \Rightarrow \ \hspace{0.3cm} {\rm Im}[D_1] = 1\hspace{0.05cm}.$$
*Weiterhin ist zu beachten, dass der Koeffizient $D_{15}$ mit dem konjugiert–komplexen Wert zu belegen ist:
+
*Further note that the coefficient $D_{15}$ is to be allocated the conjugate complex value:
 
:$$D_{15} = D_{1}^{\star} = 1 - {\rm j}\hspace{0.05cm}.$$
 
:$$D_{15} = D_{1}^{\star} = 1 - {\rm j}\hspace{0.05cm}.$$
  
  
Zum gleichen Ergebnis wäre man durch Auswertung der (zeitkontinuierlichen) Fouriertransformierten von $s(t)$ gekommen:
+
The same result would have been obtained by evaluating the (continuous-time) Fourier transform of $s(t)$:
 
:$$S(f) = (1 + {\rm j}) \cdot \delta (f - f_0) + (1 - {\rm j}) \cdot \delta (f + f_0)\hspace{0.05cm}.$$
 
:$$S(f) = (1 + {\rm j}) \cdot \delta (f - f_0) + (1 - {\rm j}) \cdot \delta (f + f_0)\hspace{0.05cm}.$$
Der Koeffizient $D_1$ beschreibt das Gewicht bei der ersten Diracfunktion (also bei $f = f_0$), der Koeffizient $D_{15} = D_{-1}$ das Gewicht der Diracfunktion bei $f = -f_0$. Hierbei ist die implizite periodische Fortsetzung bei der DFT (bzw. IDFT) zu beachten.
+
The coefficient $D_1$ describes the weight at the first Dirac function (i.e., at $f = f_0$), and the coefficient $D_{15} = D_{-1}$ describes the weight of the Dirac function at $f = -f_0$. Here, the implicit periodic continuation in the DFT (or IDFT) should be noted.
  
  
  
'''(4)'''&nbsp; Richtig ist <u>der Lösungsvorschlag 3</u>, wobei nun $D_{13} = D_{3}^∗$ zu berücksichtigen ist.
+
'''(4)'''&nbsp; Correct is <u>the proposed solution 3</u>, where now $D_{13} = D_{3}^∗$ has to be considered.
*Zeichnet man die Abtastwerte $s_l$ auf, so erkennt man nun die 3–fache Frequenz. Zum Beispiel  ergibt sich aus dem Vergleich von $s_2$ und $s_{10}$:
+
*If one plots the samples $s_l$, one now recognizes the 3-fold frequency. For example, comparing $s_2$ and $s_{10}$ gives:
 
:$$8 \cdot \Delta t ={T}/{2} = 1.5 \cdot T_0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T_0 = {T}/{3}\hspace{0.05cm}.$$
 
:$$8 \cdot \Delta t ={T}/{2} = 1.5 \cdot T_0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T_0 = {T}/{3}\hspace{0.05cm}.$$
*Die Amplitude ist gegenüber der Belegung $\boldsymbol{\rm A}$ unverändert. Die Phase $\phi_0$ erkennt man aus dem ersten Maximum bei $l = 2$:
+
*The amplitude is unchanged compared to the allocation $\boldsymbol{\rm A}$. The phase $\phi_0$ can be recognized from the first maximum at $l = 2$:
:$$ s(t) \ = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot ( t - 2 \cdot \Delta t)) = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot t + \phi_0), \hspace{0.3cm} \phi_0 = 12 \pi \cdot \frac{\Delta t}{T} = \frac{3 \pi}{4} \hspace{0.05cm}.$$
+
:$$ s(t) \ = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot ( t - 2 \cdot \delta t)) = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot t + \phi_0), \hspace{0.3cm} \phi_0 = 12 \pi \cdot \frac{\delta t}{T} = \frac{3 \pi}{4} \hspace{0.05cm}.$$
*Nach gleicher Vorgehensweise wie bei Aufgabe '''(3)''' erhält man nun mit $ {\rm cos}(3π/4) \ = \sin(3π/4) = \sqrt{2}/2$:
+
*Following the same procedure as in exercise '''(3)''', we now obtain $ {\rm cos}(3π/4) \ = \sin(3π/4) = -\sqrt{2}/2$:
 
:$${\rm Re}\{D_3\} = -1, \hspace{0.2cm} {\rm Im}\{D_3\} = -1\hspace{0.05cm}.$$
 
:$${\rm Re}\{D_3\} = -1, \hspace{0.2cm} {\rm Im}\{D_3\} = -1\hspace{0.05cm}.$$
  
  
  
'''(5)'''&nbsp; Richtig ist hier <u>der erste Lösungsvorschlag</u>:
+
'''(5)'''&nbsp; The correct solution here is <u>the first proposed solution</u>:
* Aufgrund der Linearität der IDFT ergeben sich die Koeffizienten $D_1$, $D_3$, $D_{13}$ und $D_{15}$ entsprechend den Ergebnissen der Teilaufgaben '''(4)''' und '''(5)'''.
+
* Due to the linearity of the IDFT, the coefficients $D_1$, $D_3$, $D_{13}$ and $D_{15}$ are obtained according to the results of the subtasks '''(4)''' and '''(5)'''.
  
  
  
  
'''(6)'''&nbsp; Die Belegung $\boldsymbol{\rm C}$ führt zu der Summe zweier harmonischer Schwingungen (mit $f_0$ bzw. $3f_0$), jeweils mit gleicher Amplitude $A$. Somit ergibt sich für die mittlere Signalleistung:
+
'''(6)'''&nbsp; The allocation $\boldsymbol{\rm C}$ leads to the sum of two harmonic oscillations (with $f_0$ and $3f_0$, respectively), each with the same amplitude $A$. Thus, the average signal power is given by:
 
:$$P_{\rm S} = 2 \cdot \frac{A^2}{2} = A^2 = 8\hspace{0.05cm}.$$
 
:$$P_{\rm S} = 2 \cdot \frac{A^2}{2} = A^2 = 8\hspace{0.05cm}.$$
Der Effektivwert ist gleich der Wurzel aus der Sendeleistung $P_{\rm S}$:
+
The rms value is equal to the square root of the transmitted power $P_{\rm S}$:
 
:$$s_{\rm eff} = \sqrt{P_{\rm S}} = A = 2.828\hspace{0.05cm}.$$
 
:$$s_{\rm eff} = \sqrt{P_{\rm S}} = A = 2.828\hspace{0.05cm}.$$
Der Maximalwert ist aus der Tabelle ablesbar:
+
The maximum value can be read from the table:
 
:$$s_{\rm max} = 5.226\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm max}/s_{\rm eff} = \frac{5.226}{2.828} \hspace{0.15cm} \underline{\approx 1.85 \hspace{0.05cm}}.$$
 
:$$s_{\rm max} = 5.226\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm max}/s_{\rm eff} = \frac{5.226}{2.828} \hspace{0.15cm} \underline{\approx 1.85 \hspace{0.05cm}}.$$
Dagegen würde bei den beiden Belegungen $\boldsymbol{\rm A}$ und $\boldsymbol{\rm B}$ jeweils $s_{\rm max}/s_{\rm eff}= \sqrt{2} = 1.414$ gelten.
+
In contrast, $s_{\rm max}/s_{\rm eff}= \sqrt{2} = 1.414$ would hold for both $\boldsymbol{\rm A}$ and $\boldsymbol{\rm B}$ allocations.
 
 
  
  

Revision as of 17:30, 7 March 2023

time samples with different DMT spectral allocations

"realization form"  of the  $\rm DMT$ method (stands for Discrete Multitone Transmission) is based on the Inverse Discrete Fourier Transform   $\rm (IDFT)$  at the transmitter and the Discrete Fourier Transform   $\rm (DFT)$  at the receiver.

At the transmitter  $N/2-1$  users are represented by the complex spectral coefficients  $D_{k} \ (k = 1,$ ... , $N/2-1)$  allocated to the frequencies  $f_{k} = k \cdot f_{0}$  . The fundamental frequency  $f_{0}$  is the reciprocal of the symbol duration  $T$.

  • It holds  $D_{k} \in \{ ±1 ± {\rm j} \}$ if one channel is occupied, in the other case  $D_{k} = 0$.
  • The coefficients  $D_{0}$  and  $D_{N/2}$  are always zero.
  • The top coefficients are allocated conjugate-complex:
$$D_k = D_{N-k}^{\star},\hspace{0.2cm}k = N/2 +1,\hspace{0.05cm} \text{...} \hspace{0.05cm}, N-1 \hspace{0.05cm}.$$

This ensures that the time signal  $s(t)$  is always real. The sample values  $s_{0}$, ... , $s_{N-1}$  of this signal are thereby formed by the IDFT, where the temporal distance of two samples is

$$\Delta t = T/N = 1/(N \cdot f_{0}).$$

Low-pass filtering is used to obtain the continuous-time signal.

For ADSL/DMT,  $N = 512$  and  $f_{0} = 4.3125 \ \rm kHz$. In the example considered here, let the parameters be assumed as follows for simplicity:

$$N = 16,\hspace{0.2cm}\delta t = 10\,{\rm µ s} \hspace{0.05cm}.$$

In the above table, for three different  $D_{k}$ occupancies, the sample values  $s_{l} (l = 0$, ... , $15)$  according to the IDFT are given. The corresponding spectral coefficients  $D_{k}\ (k = 0$, ... , $15)$ are sought.





Hints:

$$s(t) = \sum_{k = 1}^{K} \big [ 2 \cdot {\rm Re}\{D_k\} \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}\{D_k\} \cdot \sin(2\pi \cdot k f_0 \cdot t )\big ] \hspace{0.05cm}.$$
  • Note also the following trigonometric relationship:
$$\cos(2\pi f_0 t + \phi_0) = \cos( \phi_0) \cdot \cos(2\pi f_0 t ) - \sin( \phi_0) \cdot \sin(2\pi f_0 t ) \hspace{0.05cm}.$$
  • The ratio of the maximum value and the rms value is called the  crest factor'  (or the crest factor) of a signal.
  • You can check your solution with the interactive applet  "Discrete Fourier Transform".




Questions

1

How many users  $(K)$  can be provided with this system?

$K \ = \ $

2

What is the bandwidth  $B$  of the DMT system under consideration?

$B \ = \ $

$\ \rm kHz $

3

What are the spectral coefficients at allocation  $\boldsymbol{\rm A}$?

$D_{1} = 1- \rm j, \ all \ other \ 0,$
$D_{1} = 1 + {\rm j}, \ D_{15} = 1 - \rm j, \ all \ others \ 0,$
$D_{1} = 1 + {\rm j}, \ D_{15} = 1 + \rm j, \ all \ others \ 0.$

4

What are the spectral coefficients at allocation  $\boldsymbol{\rm B}$?

$D_{2} = -1 - {\rm j}, \ D_{14} = -1 + \rm j, \ all \ other \ 0$,
$D_{3} = +1 - {\rm j}, \ D_{13} = +1 + \rm j, \ all \ others \ 0$,
$D_{3} = -1 - {\rm j}, \ D_{13} = -1 + \rm j, \ all \ others \ 0$.

5

What are the spectral coefficients for the allocation  $\boldsymbol{\rm C}$  with  $\boldsymbol{\rm C} = \boldsymbol{\rm A} + \boldsymbol{\rm B}?$

$D_{1} = 1 + {\rm j}, \ D_{3} = -1 -{\rm j}, \ D_{13} = -1 +{\rm j}, \ D_{15} = 1 - {\rm j}$,
$D_{k} = (-1)^k + {\rm j} \cdot (-1)^{k+1}$.

6

What is the crest factor  $(s_{\rm max}/s_{\rm eff})$  at allocation  $\boldsymbol{\rm C}$?

$s_{\rm max}/s_{\rm eff} \ = \ $


Solution

(1)  The system is designed for $K = N/2 - 1 \underline{= 7 \ {\rm users}}$ $(N = 16)$.


(2)  The frame duration $T$ is given by $N \cdot \delta t = 0.16 \rm ms$.

  • The fundamental frequency here is accordingly $f_{0} = 1/T = 6.25 \ \rm kHz$ and the total bandwidth is $B = 8 \cdot f_{0} \ \underline{= 50 \ \rm kHz}$.
  • For comparison, for ADSL, this bandwidth results in $256 \cdot 4.3125 \ \rm kHz= 1104 \ kHz.$


(3)  Correct is the second proposed solution:

  • From the $16$ samples $s_{l}$ in the first column of the table  $($occupancy $\boldsymbol{\rm A})$  we see that $s(t)$ describes a harmonic oscillation with period $T_{0} = T$ (only one oscillation). The amplitude is equal to $2 \cdot \sqrt{2} =2.828$ and the phase is $\phi_0 = 45^\circ \ (π/4)$.
  • Thus, for the continuous-time signal, we can write  $($with $f_{0} = 1/T)$:
$$s(t) = 2 \cdot \sqrt{2}\cdot \cos(2\pi f_0 t + \pi /4) \hspace{0.05cm}.$$
  • With the given trigonometric transformation and ${\rm cos} \ (π/4) \ = \ {\rm sin} \ (π/4) \ = \ \sqrt{2}$ still holds:
$$s(t) = 2 \cdot \cos(2\pi f_0 t ) - 2 \cdot \sin(2\pi f_0 t ) \hspace{0.05cm}.$$
  • A coefficient comparison with the further equation.
$$s(t) = \sum_{k = 1}^{K} \left [ 2 \cdot {\rm Re}[D_k] \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}[D_k] \cdot \sin(2\pi \cdot k f_0 \cdot t )\right ] \hspace{0.05cm}$$
returns the result:
$$2 \cdot {\rm Re}[D_1] = 2 \hspace{0.3cm} \ \Rightarrow \ \hspace{0.3cm} {\rm Re}[D_1] = 1\hspace{0.05cm},$$
$$2 \cdot {\rm Im}[D_1] = 2 \hspace{0.3cm} \ \Rightarrow \ \hspace{0.3cm} {\rm Im}[D_1] = 1\hspace{0.05cm}.$$
  • Further note that the coefficient $D_{15}$ is to be allocated the conjugate complex value:
$$D_{15} = D_{1}^{\star} = 1 - {\rm j}\hspace{0.05cm}.$$


The same result would have been obtained by evaluating the (continuous-time) Fourier transform of $s(t)$:

$$S(f) = (1 + {\rm j}) \cdot \delta (f - f_0) + (1 - {\rm j}) \cdot \delta (f + f_0)\hspace{0.05cm}.$$

The coefficient $D_1$ describes the weight at the first Dirac function (i.e., at $f = f_0$), and the coefficient $D_{15} = D_{-1}$ describes the weight of the Dirac function at $f = -f_0$. Here, the implicit periodic continuation in the DFT (or IDFT) should be noted.


(4)  Correct is the proposed solution 3, where now $D_{13} = D_{3}^∗$ has to be considered.

  • If one plots the samples $s_l$, one now recognizes the 3-fold frequency. For example, comparing $s_2$ and $s_{10}$ gives:
$$8 \cdot \Delta t ={T}/{2} = 1.5 \cdot T_0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T_0 = {T}/{3}\hspace{0.05cm}.$$
  • The amplitude is unchanged compared to the allocation $\boldsymbol{\rm A}$. The phase $\phi_0$ can be recognized from the first maximum at $l = 2$:
$$ s(t) \ = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot ( t - 2 \cdot \delta t)) = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot t + \phi_0), \hspace{0.3cm} \phi_0 = 12 \pi \cdot \frac{\delta t}{T} = \frac{3 \pi}{4} \hspace{0.05cm}.$$
  • Following the same procedure as in exercise (3), we now obtain $ {\rm cos}(3π/4) \ = \sin(3π/4) = -\sqrt{2}/2$:
$${\rm Re}\{D_3\} = -1, \hspace{0.2cm} {\rm Im}\{D_3\} = -1\hspace{0.05cm}.$$


(5)  The correct solution here is the first proposed solution:

  • Due to the linearity of the IDFT, the coefficients $D_1$, $D_3$, $D_{13}$ and $D_{15}$ are obtained according to the results of the subtasks (4) and (5).



(6)  The allocation $\boldsymbol{\rm C}$ leads to the sum of two harmonic oscillations (with $f_0$ and $3f_0$, respectively), each with the same amplitude $A$. Thus, the average signal power is given by:

$$P_{\rm S} = 2 \cdot \frac{A^2}{2} = A^2 = 8\hspace{0.05cm}.$$

The rms value is equal to the square root of the transmitted power $P_{\rm S}$:

$$s_{\rm eff} = \sqrt{P_{\rm S}} = A = 2.828\hspace{0.05cm}.$$

The maximum value can be read from the table:

$$s_{\rm max} = 5.226\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm max}/s_{\rm eff} = \frac{5.226}{2.828} \hspace{0.15cm} \underline{\approx 1.85 \hspace{0.05cm}}.$$

In contrast, $s_{\rm max}/s_{\rm eff}= \sqrt{2} = 1.414$ would hold for both $\boldsymbol{\rm A}$ and $\boldsymbol{\rm B}$ allocations.