Exercise 3.2Z: Optimum Cutoff Frequency for Gaussian Low-pass

From LNTwww

Noiseless and noisy eye diagrams

As in  Exercise 3.2,  a binary bipolar redundancy-free binary system with Gaussian receiver filter  $H_{\rm G}(f)$  is considered.  Its cutoff frequency  $f_{\rm G}$  is to be determined such that the worst-case S/N ratio

$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2 \big]^2}{ \sigma_d^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}} \right)$$

becomes maximum and thus the worst-case error probability  $p_{\rm U}$  becomes minimum.  The thus optimized cutoff frequency  $f_{\rm G, \ opt}$  usually also leads to the minimum mean symbol error probability  $p_{\rm S, \ min}$.

In the above equation,  the following system quantities are used:

  • $\sigma_d^2$  is the detection noise power.  For a Gaussian receiver filter holds:
$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm G}(f)|^2 \,{\rm d} f = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2}}\hspace{0.05cm}.$$
  • $\ddot{o}(T_{\rm D})$  indicates the  "eye opening".  The detection time is always assumed to be  $T_{\rm D} = 0$. 
  • For a Gaussian receiver filter,  the vertical eye opening  $\ddot{o}(T_{\rm D})$  can be expressed solely by the amplitude  $s_0$  of the basic transmission pulse  $($upper boundary line in the noiseless eye diagram$)$ and the maximum value  $g_0$  of the basic detection pulse.  $g_0$ is to be calculated as follows:
$$g_0 = g_d(t = 0) = s_0 \cdot \big [1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\big]\hspace{0.05cm}.$$

The graph shows the eye diagrams of the sought configuration with optimal cutoff frequency.

  • In the upper diagram,  noise is not considered.
  • The lower diagram,  on the other hand,  is valid with AWGN noise for  $10 \cdot {\rm lg} \ E_{\rm B}/N_0 = 10 \ \rm dB$.




Which statements are true for the eye diagram?

The eye opening is calculated without noise.
With Gaussian receiver filter:  $\ddot{o}(T_{\rm D})/2 = s_0 \ – \ g_0$.
With Gaussian receiver filter:  $\ddot{o}(T_{\rm D})/2 = 2 \cdot g_0 \ – \ s_0$.


At what cutoff frequency does a closed eye result?

$f_{\rm G, \ min} \cdot T \ = \ $


Calculate the worst-case SNR for  $10 \cdot {\rm lg} \ E_{\rm B}/N_0 = 10 \ \rm dB$.  What are the values for the cutoff frequencies listed below?

$f_{\rm G} \cdot T = 0.6\text{:} \hspace{0.4cm} 10 \cdot \rm lg \ \rho_{\rm U} \ = \ $

$\ \rm dB$
$f_{\rm G} \cdot T = 0.8\text{:} \hspace{0.4cm} 10 \cdot \rm lg \ \rho_{\rm U}\ = \ $

$\ \rm dB$
$f_{\rm G} \cdot T = 1.0\text{:} \hspace{0.4cm} 10 \cdot \rm lg \ \rho_{\rm U} \ = \ $

$\ \rm dB$


Which statements are true regarding the optimal cutoff frequency?

Optimization with respect to  $p_{\rm U}$  $($or  $\rho_{\rm U})$  yields $f_{\rm G, \ opt} \cdot T \approx 0.8$.
This optimization result is independent of  $E_{\rm B}/N_0$.
Optimization with respect to  $p_{\rm S}$  leads to exactly the same result.


Determine the following quantities for the optimal cutoff frequency  $f_{\rm G, \ opt}$  for  $10 \cdot {\rm lg} \ (E_{\rm B}/N_0) = 10 \ \rm dB$.

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

$\sigma_d/s_0 \ = \ $

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

$\ \rm dB$
$p_{\rm U}\ = \ $

$\ \cdot 10^{\rm -5}$


(1)  The  first and third solutions  are correct:

  • When calculating the vertical eye opening,  the noise component must not be taken into account.  This is captured by the noise rms value  $\sigma_d$.
  • If the eye opening were taken from the lower eye diagram,  the noise component would be captured twice.
  • The upper boundary of the inner eye line results for the symbol sequence  "$\text{ ...} \, \ -\hspace{-0.1cm}1 \ -\hspace{-0.1cm}1, +1, -\hspace{-0.1cm}1, \ -\hspace{-0.1cm}1, \text{ ...} $" .
  • The long  "$-1$"  sequence would lead to the value $-s_0$.
  • In contrast,  the  "worst–case"  sequence leads to the eye line  $-s_0 + 2 \cdot g_d(t)$.
  • Thus,  at detection time  $T_{\rm D} = 0$,  with decision threshold  $E = 0$:
$${\ddot{o}(T_{\rm D})}/{ 2}= 2 \cdot g_0 - s_0 \hspace{0.05cm}.$$

(2)  For the half vertical eye opening holds:

$${\ddot{o}(T_{\rm D})}/{ 2} \ = \ 2 \cdot g_0 - s_0 = 2 \cdot s_0 \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right] - s_0 = s_0 \cdot\left [ 1- 4 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right] \hspace{0.05cm}.$$
  • A closed eye results according to the given applet for
$${\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right) \ge 0.25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sqrt{2\pi} \cdot f_{\rm G} \cdot T< 0.675\hspace{0.3cm}\Rightarrow \hspace{0.3cm} f_{\rm G, min} \cdot T \approx \frac{0.675}{2.5}\hspace{0.15cm}\underline { \approx 0.27} \hspace{0.05cm}.$$

(3)  Using the equations on the information section and the previous calculations,  we obtain

$\rho_{\rm U}$ as a function of (normalized) cutoff frequency
$$\rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ \sigma_d^2} = \frac{s_0^2 \cdot\left [ 1- 4 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right]^2}{ N_0 \cdot f_{\rm G} / \sqrt{2}}$$
  • With the specification  $E_{\rm B}/N_0 = 10 \ \rm dB $,  the following determining equation is obtained:
$$10 \cdot {\rm lg}\hspace{0.1cm} {E_{\rm B}}/{ N_0} = 10 \, {\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {E_{\rm B}}/{ N_0} = {s_0^2 \cdot T}/{ N_0} = 10$$
$$\Rightarrow \hspace{0.3cm} \rho_{\rm U} = 10 \cdot \sqrt{2} \cdot \frac{ \left [ 1- 4 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right]^2}{ f_{\rm G} \cdot T}\hspace{0.05cm}.$$
  • The figure shows this function plot as a function of the  (normalized)  cutoff frequency.  For the given cutoff frequencies holds:
  • $f_{\rm G} \cdot T = 0.6\text{:} \hspace{0.4cm} \rho_{\rm U} \approx 12.7 \Rightarrow 10 \cdot \rm lg \ \rho_{\rm U} \ \underline {\approx \ 11.04 \ \rm dB},$
  • $f_{\rm G} \cdot T = 0.8\text{:} \hspace{0.4cm} \rho_{\rm U} \approx 14.7 \Rightarrow 10 \cdot \rm lg \ \rho_{\rm U} \ \underline {\approx \ 11.66 \ \rm dB},$
  • $f_{\rm G} \cdot T = 1.0\text{:} \hspace{0.4cm} \rho_{\rm U} \approx 13.5 \Rightarrow 10 \cdot \rm lg \ \rho_{\rm U} \ \underline {\approx \ 11.30 \ \rm dB}.$
  • From the above graph,  one can also see the minimum cutoff frequency   ⇒   subtask (2).

(4)  The  first two solutions  are correct:

  • The validity of the first statement is evident from the above graph.
  • Since in  $\rho_{\rm U}$  equation the ratio  $E_{\rm B}/N_0$  occurs only a factor,  the optimization  (setting derivative to zero)  always leads to the same result independent of  $E_{\rm B}/N_0$.
  • The optimal cutoff frequency with respect to  $p_{\rm U}$  is approximately optimal with respect to  $p_{\rm S}$  as well,  but not exactly.
  • For very large values of  $E_{\rm B}/N_0$  (small noise),  this approximation is very correct and  $p_{\rm S} \ \approx \ p_{\rm U}/4$  holds.
  • In contrast,  for large noise e.g.  $10 \cdot {\rm lg} \ E_{\rm B}/N_0 = 0 \ \rm dB$,  a smaller optimal cutoff frequency results when the optimization is based on  $p_{\rm S}$:
$f_{\rm G} \cdot T = 0.8\text{:} \hspace{0.4cm} p_{\rm U} = 0.113,\hspace{0.4cm} p_{\rm S} = 0.102,$
$f_{\rm G} \cdot T = 0.6\text{:} \hspace{0.4cm} p_{\rm U} = 0.129,\hspace{0.4cm} p_{\rm S} = 0.094.$
  • However,  the error probabilities are then so large that these results are not practically relevant.

(5)  With the result of subtask  (2)   ⇒   $E_{\rm B}/N_0 = 10$ and $f_{\rm G} \cdot T = 0.8$  holds:

$${\ddot{o}(T_{\rm D})}/{ s_0} = 2 \cdot \left [ 1- 4 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot 0.8 \right)\right] = 2 \cdot \left [ 1- 4 \cdot 0.022\right]\hspace{0.15cm}\underline { = 1.824} \hspace{0.05cm},$$
$${\sigma_d^2}/{ s_0^2} = \frac{N_0 \cdot f_{\rm G} }{\sqrt{2}\cdot s_0^2}= \frac{N_0 }{s_0^2 \cdot T} \cdot \frac{f_{\rm G} \cdot T}{\sqrt{2}} = 0.1 \cdot \frac{0.8}{\sqrt{2}} \approx 0.0566 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\sigma_d}/{ s_0}\hspace{0.15cm}\underline { \approx 0.238} \hspace{0.05cm},$$
$$\rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})]^2}{ 4 \cdot \sigma_d^2} = \frac{1.824^2}{ 4 \cdot 0.0566}\approx 14.7 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U}\hspace{0.15cm}\underline {\approx 11.66\, {\rm dB}} \hspace{0.05cm}.$$
$$p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}} \right) = {\rm Q} \left( \sqrt{14.7} \right) \hspace{0.15cm}\underline { \approx 6.4 \cdot 10^{-5}}\hspace{0.05cm}.$$