Difference between revisions of "Exercise 2.4: DSL/DMT with IDFT/DFT"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Examples_of_Communication_Systems/xDSL_as_Transmission_Technology |
}} | }} | ||
− | [[File:P_ID1972__Bei_A_2_4.png|right|frame| | + | [[File:P_ID1972__Bei_A_2_4.png|right|frame|time samples with different DMT spectral allocations]] |
− | + | 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. | |
− | + | 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: |
:$$D_k = D_{N-k}^{\star},\hspace{0.2cm}k = N/2 +1,\hspace{0.05cm} \text{...} \hspace{0.05cm}, N-1 \hspace{0.05cm}.$$ | :$$D_k = D_{N-k}^{\star},\hspace{0.2cm}k = N/2 +1,\hspace{0.05cm} \text{...} \hspace{0.05cm}, N-1 \hspace{0.05cm}.$$ | ||
− | + | 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 | |
− | :$$\Delta t = T/N = 1/(N \cdot f_{0})$$ | + | :$$\Delta t = T/N = 1/(N \cdot f_{0}).$$ |
− | + | 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: | |
− | :$$N = 16,\hspace{0.2cm}\ | + | :$$N = 16,\hspace{0.2cm}\delta t = 10\,{\rm µ s} \hspace{0.05cm}.$$ |
− | In | + | In the above table, for three different $D_{k}$ occupancies, the sample values $s_{l} (l = 0$, ... , $15)$ according to the IDFT are given. The corresponding spectral coefficients $D_{k}\ (k = 0$, ... , $15)$ are sought. |
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− | + | Hints: | |
− | * | + | *This exercise belongs to the chapter [[Examples_of_Communication_Systems/xDSL_as_Transmission_Technology|"xDSL as Transmission Technology"]]. |
− | * | + | *The transmission signal for DSL has the form |
:$$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}.$$ | :$$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}.$$ | :$$\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 [[Applets:Diskrete_Fouriertransformation_(Applet)|"Discrete Fourier Transform"]]. |
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− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | { How many users $(K)$ can be provided with this system? |
|type="{}"} | |type="{}"} | ||
$K \ = \ ${ 7 } | $K \ = \ ${ 7 } | ||
− | { | + | {What is the bandwidth $B$ of the DMT system under consideration? |
|type="{}"} | |type="{}"} | ||
$B \ = \ ${ 50 3% } $\ \rm kHz $ | $B \ = \ ${ 50 3% } $\ \rm kHz $ | ||
− | { | + | {What are the spectral coefficients at allocation $\boldsymbol{\rm A}$? |
|type="()"} | |type="()"} | ||
− | - $D_{1} = 1- \rm j, \ | + | - $D_{1} = 1- \rm j, \ all \ other \ 0,$ |
− | + $D_{1} = 1 + {\rm j}, \ D_{15} = 1 - \rm j, \ | + | + $D_{1} = 1 + {\rm j}, \ D_{15} = 1 - \rm j, \ all \ others \ 0,$ |
− | - $D_{1} = 1 + {\rm j}, \ D_{15} = 1 + \rm j, \ | + | - $D_{1} = 1 + {\rm j}, \ D_{15} = 1 + \rm j, \ all \ others \ 0.$ |
− | { | + | {What are the spectral coefficients at allocation $\boldsymbol{\rm B}$? |
|type="()"} | |type="()"} | ||
− | - $D_{2} = -1 - {\rm j}, \ D_{14} = -1 + \rm j, \ | + | - $D_{2} = -1 - {\rm j}, \ D_{14} = -1 + \rm j, \ all \ other \ 0$, |
− | - $D_{3} = +1 - {\rm j}, \ D_{13} = +1 + \rm j, \ | + | - $D_{3} = +1 - {\rm j}, \ D_{13} = +1 + \rm j, \ all \ others \ 0$, |
− | + $D_{3} = -1 - {\rm j}, \ D_{13} = -1 + \rm j, \ | + | + $D_{3} = -1 - {\rm j}, \ D_{13} = -1 + \rm j, \ all \ others \ 0$. |
− | { | + | {What are the spectral coefficients for the allocation $\boldsymbol{\rm C}$ with $\boldsymbol{\rm C} = \boldsymbol{\rm A} + \boldsymbol{\rm B}?$ |
|type="()"} | |type="()"} | ||
− | + $D_{1} = 1 + {\rm j}, \ D_{3} = -1 -{\rm | + | + $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 ( | + | - $D_{k} = (-1)^k + {\rm j} \cdot (-1)^{k+1}$. |
− | { | + | {What is the crest factor $(s_{\rm max}/s_{\rm eff})$ at allocation $\boldsymbol{\rm C}$? |
|type="{}"} | |type="{}"} | ||
$s_{\rm max}/s_{\rm eff} \ = \ ${ 1.85 3% } | $s_{\rm max}/s_{\rm eff} \ = \ ${ 1.85 3% } | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' The system is designed for $K = N/2 - 1 \underline{= 7 \ {\rm users}}$ $(N = 16)$. |
− | '''(2)''' | + | '''(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)''' | + | '''(3)''' Correct is <u>the second proposed solution</u>: |
− | * | + | * 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}.$$ | :$$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}.$$ | :$$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}$$ | :$$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 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}.$$ | ||
− | * | + | *Further note that the coefficient $D_{15}$ is to be allocated the conjugate complex value: |
:$$D_{15} = D_{1}^{\star} = 1 - {\rm j}\hspace{0.05cm}.$$ | :$$D_{15} = D_{1}^{\star} = 1 - {\rm j}\hspace{0.05cm}.$$ | ||
− | + | The same result would have been obtained by evaluating the (continuous-time) Fourier transform of $s(t)$: | |
:$$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}.$$ | ||
− | + | The coefficient $D_1$ describes the weight at the first Dirac function (i.e., at $f = f_0$), and the coefficient $D_{15} = D_{-1}$ describes the weight of the Dirac function at $f = -f_0$. Here, the implicit periodic continuation in the DFT (or IDFT) should be noted. | |
− | '''(4)''' | + | '''(4)''' Correct is <u>the proposed solution 3</u>, where now $D_{13} = D_{3}^∗$ has to be considered. |
− | * | + | *If one plots the samples $s_l$, one now recognizes the 3-fold frequency. For example, comparing $s_2$ and $s_{10}$ gives: |
:$$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}.$$ | :$$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 \ | + | :$$ s(t) \ = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot ( t - 2 \cdot \delta t)) = \ 2 \cdot \sqrt{2}\cdot \cos(2\pi \cdot 3 f_0 \cdot t + \phi_0), \hspace{0.3cm} \phi_0 = 12 \pi \cdot \frac{\delta t}{T} = \frac{3 \pi}{4} \hspace{0.05cm}.$$ |
− | * | + | *Following the same procedure as in exercise '''(3)''', we now obtain $ {\rm cos}(3π/4) \ = \sin(3π/4) = -\sqrt{2}/2$: |
:$${\rm Re}\{D_3\} = -1, \hspace{0.2cm} {\rm Im}\{D_3\} = -1\hspace{0.05cm}.$$ | :$${\rm Re}\{D_3\} = -1, \hspace{0.2cm} {\rm Im}\{D_3\} = -1\hspace{0.05cm}.$$ | ||
− | '''(5)''' | + | '''(5)''' 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)''' | + | '''(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: |
:$$P_{\rm S} = 2 \cdot \frac{A^2}{2} = A^2 = 8\hspace{0.05cm}.$$ | :$$P_{\rm S} = 2 \cdot \frac{A^2}{2} = A^2 = 8\hspace{0.05cm}.$$ | ||
− | + | The rms value is equal to the square root of the transmitted power $P_{\rm S}$: | |
:$$s_{\rm eff} = \sqrt{P_{\rm S}} = A = 2.828\hspace{0.05cm}.$$ | :$$s_{\rm eff} = \sqrt{P_{\rm S}} = A = 2.828\hspace{0.05cm}.$$ | ||
− | + | The maximum value can be read from the table: | |
:$$s_{\rm max} = 5.226\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm max}/s_{\rm eff} = \frac{5.226}{2.828} \hspace{0.15cm} \underline{\approx 1.85 \hspace{0.05cm}}.$$ | :$$s_{\rm max} = 5.226\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm max}/s_{\rm eff} = \frac{5.226}{2.828} \hspace{0.15cm} \underline{\approx 1.85 \hspace{0.05cm}}.$$ | ||
− | + | In contrast, $s_{\rm max}/s_{\rm eff}= \sqrt{2} = 1.414$ would hold for both $\boldsymbol{\rm A}$ and $\boldsymbol{\rm B}$ allocations. | |
− | |||
Revision as of 17:30, 7 March 2023
A "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:
- This exercise belongs to the chapter "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 crest factor' (or the crest factor) of a signal.
- You can check your solution with the interactive applet "Discrete Fourier Transform".
Questions
Solution
(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.