Difference between revisions of "Aufgaben:Exercise 3.8: Delay Filter DFE Realization"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Decision_Feedback}} |
− | [[File:P_ID1453__Dig_A_3_8.png|right|frame| | + | [[File:P_ID1453__Dig_A_3_8.png|right|frame|Decision feedback with delay filter]] |
− | + | We consider a bipolar binary system with decision feedback equalization $\rm (DFE)$. | |
− | + | The pre-equalized basic pulse $g_d(t)$ can be calculated as the rectangular response of a Gaussian low-pass filter with the cutoff frequency $f_{\rm G} \cdot T = 0.25$ and is shown in red in the diagram. In [[Aufgaben:Exercise_3.8Z:_Optimal_Detection_Time_for_DFE|Exercise 3.8Z]] the samples of $g_d(t)$ are tabulated in the distance $T/10$. | |
− | + | With ideal decision feedback – dimensioned for the detection time $T_{\rm D} = 0$ – applies: | |
+ | *A compensation pulse $g_w(t)$ is formed, which is equal to $g_d(t)$ for $t ≥ T_{\rm V} = T/2$ and identical to zero for $t < T_{\rm V}$ (blue filled area). | ||
− | + | *The corrected basic pulse $g_k(t) = g_d(t) - g_w(t)$ is thus always zero for $t > T/2$. | |
− | :$$\rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ | + | |
+ | |||
+ | By simulation, the worst-case S/N ratio at the decision and from this the worst–case error probability were determined for this system with ideal DFE and with detection at time $T_{\rm D} = 0$. The result was as follows: | ||
+ | :$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2\big]^2}{ | ||
\sigma_d^2} = 25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | \sigma_d^2} = 25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \approx 14\,{\rm dB} | 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \approx 14\,{\rm dB} | ||
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\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | + | A low-effort realization of the DFE is possible with a delay filter. In the graph, the compensation pulse $g_w(t)$ for such a delay filter with order $N = 2$ and coefficients $k_1 = 0.2$ and $k_2 = 0.05$ is plotted (blue curve). | |
− | |||
− | |||
− | === | + | |
+ | Note: The exercise belongs to the chapter [[Digital_Signal_Transmission/Decision_Feedback|"Decision Feedback"]]. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {How large is the (normalized) half eye opening at ideal DFE? Also for the further exercises $T_{\rm D} = 0$ is valid. |
|type="{}"} | |type="{}"} | ||
− | $\ddot{o}(T_{\rm D} | + | $\ddot{o}(T_{\rm D})/(2s_0) \ = \ $ { 0.205 3% } |
− | { | + | {Calculate the (normalized) noise rms value from the given SNR. |
|type="{}"} | |type="{}"} | ||
− | $\sigma_d/s_0$ | + | $\sigma_d/s_0\ = \ $ { 0.041 3% } |
− | { | + | {Calculate the half normalized eye opening and the signal-to-noise ratio if the DFE is implemented by a delay filter with $N = 2$, $k_1 = 0.2$ and $k_2 = 0.05$. |
|type="{}"} | |type="{}"} | ||
− | $\ddot{o}(T_{\rm D})/(2s_0)$ | + | $\ddot{o}(T_{\rm D})/(2s_0) \ = \ $ { 0.148 3% } |
− | $10 \cdot {\rm lg} \, \rho_{\rm U}$ | + | $10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 11.1 3% } $\ \rm dB$ |
− | { | + | {Calculate the half normalized eye opening and the signal-to-noise ratio with $N = 2$ when the coefficients $k_1$ and $k_2$ are chosen as best as possible? |
|type="{}"} | |type="{}"} | ||
− | $\ddot{o}(T_{\rm D})/(2s_0)$ | + | $\ddot{o}(T_{\rm D})/(2s_0) \ = \ $ { 0.204 3% } |
− | $10 \cdot {\rm lg} \, \rho_{\rm U}$ | + | $10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 13.94 3% } $\ \rm dB$ |
− | { | + | {Which of the following statements are true in the present case? |
|type="[]"} | |type="[]"} | ||
− | + | + | + For a receiver without DFE, the eye is closed. |
− | + | + | + A disadvantage of the DFE is the propagation of uncertainty. |
− | - | + | - The DFE improves each symbol decision. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' With ideal DFE, all trailers are compensated. Thus one obtains for the half eye opening under the condition $T_{\rm D} = 0$: |
:$$\frac{\ddot{o}(T_{\rm D})}{ | :$$\frac{\ddot{o}(T_{\rm D})}{ | ||
− | 2} = g_d(0) - g_d(-T)- g_d(-2T)- g_d(-3T) | + | 2} = g_d(0) - g_d(-T)- g_d(-2T)- g_d(-3T)\hspace{0.3cm} |
− | + | \Rightarrow \hspace{0.3cm} \frac{\ddot{o}(T_{\rm D})}{ | |
2 \cdot s_0} = 0.470 - 0.235 - 0.029 -0.001\hspace{0.15cm}\underline { = 0.205} | 2 \cdot s_0} = 0.470 - 0.235 - 0.029 -0.001\hspace{0.15cm}\underline { = 0.205} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | '''(2)''' | + | '''(2)''' From the given worst-case S/N ratio $\rho_{\rm U} = 25$, it follows: |
:$$\rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ | :$$\rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ | ||
\sigma_d^2} = 25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | \sigma_d^2} = 25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
(\sigma_d/s_0)^2 = \frac{[\ddot{o}(T_{\rm D})/(2s_0)]^2}{ | (\sigma_d/s_0)^2 = \frac{[\ddot{o}(T_{\rm D})/(2s_0)]^2}{ | ||
− | 25} | + | 25}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} |
− | |||
\sigma_d/s_0 = \frac{\ddot{o}(T_{\rm D})/(2s_0)}{ | \sigma_d/s_0 = \frac{\ddot{o}(T_{\rm D})/(2s_0)}{ | ||
5} \hspace{0.15cm}\underline {= 0.041} \hspace{0.05cm}.$$ | 5} \hspace{0.15cm}\underline {= 0.041} \hspace{0.05cm}.$$ | ||
− | '''(3)''' | + | '''(3)''' By this filter, the first two trailers are only partially compensated and the third trailer is not compensated at all. |
+ | *From this follows with the result of subtask '''(1)''': | ||
:$$\frac{\ddot{o}(T_{\rm D})}{ | :$$\frac{\ddot{o}(T_{\rm D})}{ | ||
− | 2 \cdot s_0} \ = \ 0.205 - | 0.235 - 0.2 | - |0.029 -0.05 | -0.001 = | + | 2 \cdot s_0} \ = \ 0.205 - | 0.235 - 0.2 | - |0.029 -0.05 | -0.001 = 0.205 - 0.035 - 0.021 -0.001 \hspace{0.15cm}\underline {= 0.148}$$ |
− | |||
:$$\Rightarrow \hspace{0.3cm} \rho_{\rm U} =\frac{0.148^2}{ | :$$\Rightarrow \hspace{0.3cm} \rho_{\rm U} =\frac{0.148^2}{ | ||
0.041^2} \approx 13 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | 0.041^2} \approx 13 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
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− | '''(4)''' | + | '''(4)''' Almost the same results are obtained as for ideal DFE (only the third trailer is not compensated): |
:$$\frac{\ddot{o}(T_{\rm D})}{ | :$$\frac{\ddot{o}(T_{\rm D})}{ | ||
2 \cdot s_0}\hspace{0.15cm}\underline { = 0.204} | 2 \cdot s_0}\hspace{0.15cm}\underline { = 0.204} | ||
\hspace{0.3cm}\Rightarrow \hspace{0.3cm} | \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
− | 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 13.94 \,{\rm dB}}\hspace{0.3cm}{\rm ( | + | 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 13.94 \,{\rm dB}}\hspace{0.3cm}{\rm (Ideal \hspace{0.15cm}DFE\hspace{-0.15cm}:}\hspace{0.15cm}{13.98 \,{\rm dB)}} |
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | '''(5)''' | + | '''(5)''' The <u>first two statements</u> are correct: |
+ | * Without DFE, the half eye opening is: | ||
:$$\frac{\ddot{o}(T_{\rm D})}{ | :$$\frac{\ddot{o}(T_{\rm D})}{ | ||
2 \cdot s_0} = 0.470 - 2 \cdot 0.235 - 2 \cdot 0.029 - 2 \cdot | 2 \cdot s_0} = 0.470 - 2 \cdot 0.235 - 2 \cdot 0.029 - 2 \cdot | ||
0.001 < 0 | 0.001 < 0 | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
+ | *If at some point a wrong decision is made due to a too large noise component, the falsification probability of the subsequent symbols is significantly increased. However, there are always symbol combinations in each sequence that interrupt this propagation of uncertainty. | ||
− | + | *The last statement is wrong. It is rather true: Small distances from the decision threshold are increased, large distances, on the other hand, are decreased and their falsification probabilities are consequently increased. On average, however, this leads to a smaller error probability. | |
− | |||
− | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Digital Signal Transmission: Exercises|^3.6 Decision Feedback Equalization^]] |
Latest revision as of 14:55, 27 June 2022
We consider a bipolar binary system with decision feedback equalization $\rm (DFE)$.
The pre-equalized basic pulse $g_d(t)$ can be calculated as the rectangular response of a Gaussian low-pass filter with the cutoff frequency $f_{\rm G} \cdot T = 0.25$ and is shown in red in the diagram. In Exercise 3.8Z the samples of $g_d(t)$ are tabulated in the distance $T/10$.
With ideal decision feedback – dimensioned for the detection time $T_{\rm D} = 0$ – applies:
- A compensation pulse $g_w(t)$ is formed, which is equal to $g_d(t)$ for $t ≥ T_{\rm V} = T/2$ and identical to zero for $t < T_{\rm V}$ (blue filled area).
- The corrected basic pulse $g_k(t) = g_d(t) - g_w(t)$ is thus always zero for $t > T/2$.
By simulation, the worst-case S/N ratio at the decision and from this the worst–case error probability were determined for this system with ideal DFE and with detection at time $T_{\rm D} = 0$. The result was as follows:
- $$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2\big]^2}{ \sigma_d^2} = 25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \approx 14\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}(\sqrt{\rho_{\rm U}}) \approx 2.9 \cdot 10^{-7} \hspace{0.05cm}.$$
A low-effort realization of the DFE is possible with a delay filter. In the graph, the compensation pulse $g_w(t)$ for such a delay filter with order $N = 2$ and coefficients $k_1 = 0.2$ and $k_2 = 0.05$ is plotted (blue curve).
Note: The exercise belongs to the chapter "Decision Feedback".
Questions
Solution
- $$\frac{\ddot{o}(T_{\rm D})}{ 2} = g_d(0) - g_d(-T)- g_d(-2T)- g_d(-3T)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = 0.470 - 0.235 - 0.029 -0.001\hspace{0.15cm}\underline { = 0.205} \hspace{0.05cm}.$$
(2) From the given worst-case S/N ratio $\rho_{\rm U} = 25$, it follows:
- $$\rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ \sigma_d^2} = 25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} (\sigma_d/s_0)^2 = \frac{[\ddot{o}(T_{\rm D})/(2s_0)]^2}{ 25}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma_d/s_0 = \frac{\ddot{o}(T_{\rm D})/(2s_0)}{ 5} \hspace{0.15cm}\underline {= 0.041} \hspace{0.05cm}.$$
(3) By this filter, the first two trailers are only partially compensated and the third trailer is not compensated at all.
- From this follows with the result of subtask (1):
- $$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} \ = \ 0.205 - | 0.235 - 0.2 | - |0.029 -0.05 | -0.001 = 0.205 - 0.035 - 0.021 -0.001 \hspace{0.15cm}\underline {= 0.148}$$
- $$\Rightarrow \hspace{0.3cm} \rho_{\rm U} =\frac{0.148^2}{ 0.041^2} \approx 13 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\hspace{0.15cm}\underline { \approx 11.1\,{\rm dB}} \hspace{0.05cm}.$$
(4) Almost the same results are obtained as for ideal DFE (only the third trailer is not compensated):
- $$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0}\hspace{0.15cm}\underline { = 0.204} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 13.94 \,{\rm dB}}\hspace{0.3cm}{\rm (Ideal \hspace{0.15cm}DFE\hspace{-0.15cm}:}\hspace{0.15cm}{13.98 \,{\rm dB)}} \hspace{0.05cm}.$$
(5) The first two statements are correct:
- Without DFE, the half eye opening is:
- $$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = 0.470 - 2 \cdot 0.235 - 2 \cdot 0.029 - 2 \cdot 0.001 < 0 \hspace{0.05cm}.$$
- If at some point a wrong decision is made due to a too large noise component, the falsification probability of the subsequent symbols is significantly increased. However, there are always symbol combinations in each sequence that interrupt this propagation of uncertainty.
- The last statement is wrong. It is rather true: Small distances from the decision threshold are increased, large distances, on the other hand, are decreased and their falsification probabilities are consequently increased. On average, however, this leads to a smaller error probability.