Aufgaben:Exercise 3.8: Delay Filter DFE Realization

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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  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

1

How large is the (normalized) half eye opening at ideal DFE?  Also for the further exercises  $T_{\rm D} = 0$ is valid.

$\ddot{o}(T_{\rm D})/(2s_0) \ = \ $

2

Calculate the (normalized) noise rms value from the given SNR.

$\sigma_d/s_0\ = \ $

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  $k_2 = 0.05$. 

$\ddot{o}(T_{\rm D})/(2s_0) \ = \ $

$10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ $\ \rm dB$

4

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?

$\ddot{o}(T_{\rm D})/(2s_0) \ = \ $

$10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ $\ \rm dB$

5

Which of the following statements are true in the present case?

	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.

Solution

Solution

(1)  With ideal DFE,  all trailers are compensated.  Thus one obtains for the half eye opening under the condition  $T_{\rm D} = 0$:

$$\frac{\ddot{o}(T_{\rm D})}{ 2} = g_d(0) - g_d(-T)- g_d(-2T)- g_d(-3T)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = 0.470 - 0.235 - 0.029 -0.001\hspace{0.15cm}\underline { = 0.205} \hspace{0.05cm}.$$

(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.