View source for Exercise 2.4: DSL/DMT with IDFT/DFT

 
{{quiz-Header|Buchseite=Examples_of_Communication_Systems/xDSL_as_Transmission_Technology

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[[File:P_ID1972__Bei_A_2_4.png|right|frame|time samples with different DMT spectral allocations]]

A&nbsp; [[Examples_of_Communication_Systems/xDSL_as_Transmission_Technology#DMT_realization_with_IDFT.2 FDFT|"realization form"]]&nbsp; of the&nbsp; $\rm DMT$ method (stands for ''Discrete Multitone Transmission'') is based on the ''Inverse Discrete Fourier Transform'' &nbsp; $\rm (IDFT)$&nbsp; at the transmitter and the ''Discrete Fourier Transform'' &nbsp; $\rm (DFT)$&nbsp; at the receiver.

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

*It holds&nbsp; $D_{k} \in \{ ±1 ± {\rm j} \}$ if one channel is allocated , in the other case&nbsp; $D_{k} = 0$. 
*The coefficients&nbsp; $D_{0}$&nbsp; and&nbsp; $D_{N/2}$&nbsp; 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&nbsp; $s(t)$&nbsp; is always real. The sample values&nbsp; $s_{0}$, ... , $s_{N-1}$&nbsp; 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,&nbsp; $N = 512$&nbsp; and&nbsp; $f_{0} = 4.3125 \ \rm kHz$. In the example considered here, let the parameters be assumed as follows for simplicity:
:$$N = 16,\hspace{0.2cm}\delta t = 10\,{\rm &micro; s} \hspace{0.05cm}.$$
In the above table, for three different&nbsp; $D_{k}$ allocations, the sample values&nbsp; $s_{l} (l = 0$, ... , $15)$&nbsp; according to the IDFT are given. The corresponding spectral coefficients&nbsp; $D_{k}\ (k = 0$, ... , $15)$ are sought.








Hints:
*This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/xDSL_as_Transmission_Technology|"xDSL as Transmission Technology"]]. 
*The transmission signal for DSL has the form
:$$s(t) = \sum_{k = 1}^{K} \big [ 2 \cdot {\rm Re}\{D_k\} \cdot \cos(2\pi \cdot k f_0 \cdot t ) - 2 \cdot {\rm Im}\{D_k\} \cdot \sin(2\pi \cdot k f_0 \cdot t )\big ] \hspace{0.05cm}.$$
*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&nbsp; ''crest factor'''&nbsp; (or the crest factor) of a signal.
*You can check your solution with the interactive applet&nbsp; [[Applets:Diskrete_Fouriertransformation_(Applet)|"Discrete Fourier Transform"]].
 





===Questions===
<quiz display=simple>
{ How many users&nbsp; $(K)$&nbsp; can be provided with this system?
|type="{}"}
$K \ = \ ${ 7 } 

{What is the bandwidth&nbsp; $B$&nbsp; of the DMT system under consideration?
|type="{}"}
$B \ = \ ${ 50 3% } $\ \rm kHz $

{What are the spectral coefficients at allocation&nbsp; $\boldsymbol{\rm A}$?
|type="()"}
- $D_{1} = 1- \rm j, \ all \ other \ 0,$
+ $D_{1} = 1 + {\rm j}, \ D_{15} = 1 - \rm j, \ all \ others \ 0,$
- $D_{1} = 1 + {\rm j}, \ D_{15} = 1 + \rm j, \ all \ others \ 0.$

{What are the spectral coefficients at allocation&nbsp; $\boldsymbol{\rm B}$?
|type="()"}
- $D_{2} = -1 - {\rm j}, \ D_{14} = -1 + \rm j, \ all \ other \ 0$,
- $D_{3} = +1 - {\rm j}, \ D_{13} = +1 + \rm j, \ all \ others \ 0$,
+ $D_{3} = -1 - {\rm j}, \ D_{13} = -1 + \rm j, \ all \ others \ 0$.

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


{What is the crest factor&nbsp; $(s_{\rm max}/s_{\rm eff})$&nbsp; at allocation&nbsp; $\boldsymbol{\rm C}$?
|type="{}"}
$s_{\rm max}/s_{\rm eff} \ = \ ${ 1.85 3% } 



</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; The system is designed for $K = N/2 - 1 \underline{= 7 \ {\rm users}}$ $(N = 16)$.


'''(2)'''&nbsp; 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)'''&nbsp; Correct is <u>the second proposed solution</u>:
* From the $16$ samples $s_{l}$ in the first column of the table &nbsp;$($allocation $\boldsymbol{\rm A})$&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 $\phi_0 = 45^\circ \ (π/4)$. 
*Thus, for the continuous-time signal, we can write &nbsp;$($with $f_{0} = 1/T)$:
:$$s(t) = 2 \cdot \sqrt{2}\cdot \cos(2\pi f_0 t + \pi /4) \hspace{0.05cm}.$$
*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)'''&nbsp; Correct is <u>the proposed solution 3</u>, 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)'''&nbsp; The correct solution here is <u>the first proposed solution</u>:
* 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; 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.


{{ML-Fuß}}



[[Category:Examples of Communication Systems: Exercises|^2.3 xDSL Transmission Technology
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