Aufgaben:Exercise 3.8Z: Optimal Detection Time for DFE

• Return to book

Table of  $g_d(t)$  samples  (normalized)

As in  "Exercise 3.8",  we consider the bipolar binary system with decision feedback equalization  $\rm (DFE)$.

The pre-equalized basic pulse  $g_d(t)$  at the input of the DFE corresponds to the rectangular response of a Gaussian low-pass filter with the cutoff frequency $f_{\rm G} \cdot T = 0.25$.

In the ideal DFE,  a compensation pulse  $g_w(t)$  is formed which is exactly equal to the input pulse  $g_d(t)$  for all times  $t ≥ T_{\rm D} + T_{\rm V}$,  so that the following applies to the corrected basic pulse:

$$g_k(t) \ = \ g_d(t) - g_w(t) = \ \left\{ \begin{array}{c} g_d(t) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c} t < T_{\rm D} + T_{\rm V}, \\ t \ge T_{\rm D} + T_{\rm V}, \\ \end{array}$$

Here  $T_{\rm D}$  denotes the detection time,  which is a system variable that can be optimized.  $T_{\rm D} = 0$  denotes symbol detection at the pulse midpoint.

• However,  for a system with DFE,  $g_k(t)$  is strongly asymmetric,  so a detection time  $T_{\rm D} < 0$  is more favorable.

• The delay time  $T_{\rm V} = T/2$  indicates that the DFE does not take effect until half a symbol duration after detection.

• However,  $T_{\rm V}$  is not relevant for solving this exercise.

A low-effort realization of the DFE is possible with a delay filter,  where the filter order must be at least  $N = 3$  for the given basic pulse.  The filter coefficients are to be selected as follows:

$$k_1 = g_d(T_{\rm D} + T),\hspace{0.2cm}k_2 = g_d(T_{\rm D} + 2T),\hspace{0.2cm}k_3 = g_d(T_{\rm D} + 3T) \hspace{0.05cm}.$$

Notes:

• The exercise belongs to the chapter  "Decision Feedback".

• Note also that decision feedback is not associated with an increase in noise power,  so that an increase in  (half)  eye opening by a factor of  $K$  simultaneously results in a signal-to-noise ratio gain of  $20 \cdot {\rm lg} \, K$. 

• The pre-equalized basic pulse  $g_d(t)$  at the DFE input corresponds to the rectangular response of a Gaussian low-pass filter with the cutoff frequency  $f_{\rm G} = 0.25/T$.

• The table shows the sample values of  $g_d(t)$  normalized to  $s_0$.  The information section for  "Exercise 3.8"  shows a sketch of  $g_d(t)$. 

Questions

1

Calculate the half eye opening for  $T_{\rm D} = 0$  and ideal DFE.

$100\% \ {\rm DFE} \text{:} \hspace{0.2cm} \ddot{o}(T_{\rm D} = 0)/(2s_0) \ = \ $

2

How must the coefficients of the delay filter be set for this?

$k_1\ = \ $

$k_2\ = \ $

$k_3\ = \ $

3

Let  $T_{\rm D} = 0$. What (half) eye opening results if the DFE compensates the trailers only  $50 \%$? 

$50\% \ {\rm DFE} \text{:} \hspace{0.2cm} \ddot{o}(T_{\rm D} = 0)/(2s_0)\ = \ $

4

Determine the optimal detection time and eye opening with ideal DFE.

$T_{\rm D, \ opt}/T\ = \ $

$100\% \ {\rm DFE} \text{:} \hspace{0.2cm} \ddot{o}(T_\text{D, opt})/(2s_0) \ = \ $

5

How must the coefficients of the delay filter be set for this?

$k_1\ = \ $

$k_2\ = \ $

$k_3\ = \ $

6

How large is the (half) eye opening with  $T_{\rm D, \ opt}$,  if the DFE compensates the trailers only  $50 \%$?  Interpret the result.

$50\% \ {\rm DFE} \text{:} \hspace{0.2cm} \ddot{o}(T_\text{D, opt})/(2s_0)\ = \ $

Solution

Solution

(1)  For detection time  $T_{\rm D} = 0$,  the following holds  (already calculated in Exercise 3.8):

$$\frac{\ddot{o}(T_{\rm D})}{ 2} = g_d(0) - g_d(-T)- g_d(-2T)- g_d(-3T)$$

$$ \Rightarrow \hspace{0.3cm} \frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = 0.470 - 0.235 - 0.029 -0.001 \hspace{0.15cm}\underline {= 0.205} \hspace{0.05cm}.$$

(2)  The coefficients should be chosen such that  $g_k(t)$  fully compensates for the trailer of  $g_d(t)$:

$$k_1 = g_d( T)\hspace{0.15cm}\underline {= 0.235},\hspace{0.2cm}k_2 = g_d(2T)\hspace{0.15cm}\underline {= 0.029},\hspace{0.2cm}k_3 = g_d(3T)\hspace{0.15cm}\underline {= 0.001} \hspace{0.05cm}.$$

(3)  Based on the result of subtask  (1),  we obtain:

$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = 0.205 - 0.5 \cdot (0.235 + 0.029 + 0.001)\hspace{0.15cm}\underline { = 0.072} \hspace{0.05cm}.$$

(4)  Optimizing  $T_{\rm D}$  according to the entries in the table yields:

$$T_{\rm D}/T = 0: \hspace{0.5cm} \ddot{o}(T_{\rm D})/(2 \, s_0) = 0.470 – 0.235 – 0.029 – 0.001 = 0.205,$$

$$T_{\rm D}/T = \ –0.1: \hspace{0.2cm} \ddot{o}(T_{\rm D})/(2 \, s_0) = 0.466 \ – \ 0.204 \ – \ 0.022 \ – \ 0.001 = 0.240,$$

$$T_{\rm D}/T = \ –0.2: \hspace{0.2cm} \ddot{o}(T_{\rm D})/(2 \, s_0) = 0.456 \ – \ 0.174 \ – \ 0.016 \ – \ 0.001 = 0.266,$$

$$T_{\rm D}/T = \ –0.3: \hspace{0.2cm} \ddot{o}(T_{\rm D})/(2 \, s_0) = 0.441 \ – \ 0.146 \ – \ 0.012 \ – \ 0.001 = 0.283,$$

$${\bf {\it T}_{\rm D}/{\it T} = \ –0.4: \hspace{0.2cm} \ddot{o}({\it T}_{\rm D})/(2 \, {\it s}_0) = 0.420 \ – \ 0.121 \ – \ 0.008 \ – \ 0.001 = 0.291,}$$

$$T_{\rm D}/T = \ –0.5: \hspace{0.2cm} \ddot{o}(T_{\rm D})/(2 \, s_0) = 0.395 \ – \ 0.099 \ – \ 0.006 \ – \ 0.001 = 0.290,$$

$$T_{\rm D}/T = \ –0.6: \hspace{0.2cm} \ddot{o}(T_{\rm D})/(2 \, s_0) = 0.366 \ – \ 0.080 \ – \ 0.004 \ – \ 0.001 = 0.282,$$

• Thus,  the optimal detection time is  $T_{\rm D, \ opt} \ \underline {= \ –0.4T}$  (probably slightly larger).

• For this,  the maximum value $(\underline{0.291})$  was determined for the half eye opening.

(5)  With  $T_{\rm D} = \ –0.4 \ T$,  the filter coefficients are:

$$k_1 = g_d(0.6 T)\hspace{0.15cm}\underline {= 0.366},\hspace{0.2cm}k_2 = g_d(1.6T)\hspace{0.15cm}\underline {= 0.080},\hspace{0.2cm}k_3 = g_d(2.6T)\hspace{0.15cm}\underline {= 0.004} \hspace{0.05cm}.$$

(6)  Using the same procedure as in subtask  (3),  we obtain here:

$$\frac{\ddot{o}(T_{\rm D,\hspace{0.05cm} opt})}{ 2 \cdot s_0} = 0.291 - 0.5 \cdot (0.366 + 0.080 + 0.004) \hspace{0.15cm}\underline {= 0.066} \hspace{0.05cm}.$$

The results of this exercise can be summarized as follows:

1. Optimizing the detection timing ideally increases the eye opening by a factor of  $0.291/0.205 = 1.42$,  which corresponds to the signal-to-noise ratio gain of  $20 \cdot {\rm lg} \, 1.42 \approx 3 \ \rm dB$.
2. However,  if the DFE functions only  $50\%$  due to realization inaccuracies,  then with  $T_{\rm D} = \ –0.4T$  there is a degradation by the amplitude factor  $0.291/0.066 \approx 4.4$  compared to the ideal DFE.  For  $T_{\rm D} = 0$,  this factor is much smaller with  $2.05/0.072 \approx 3$.
3. In fact,  the actually worse system  $($with  $T_{\rm D} = 0)$  is superior to the actually better system  $($with  $T_{\rm D} = \ –0.4T)$,  if the decision feedback works only  $50\%$.  Then there is a SNR loss of  $20 \cdot {\rm lg} \, (0.072/0.066) \approx 0.75 \ \rm dB$.
4. One can generalize these statements:   The larger the improvement by system optimization  (here:  the optimization of the detection time)  is in the ideal case,  the larger is also the degradation at non-ideal conditions,  e.g.,  at tolerance-bounded realization.