Digital Signal Transmission/Consideration of Channel Distortion and Equalization

1. Digital Signal Transmission
2. Intersymbol Interfering and Equalization Methods
3. Consideration of Channel Distortion and Equalization

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Ideal channel equalizer

For a transmission system whose channel frequency response  $H_{\rm K}(f)$  causes severe distortion,  we assume the following block diagram  (upper graph)  and  equivalent circuit  (lower graph).  To these representations the following is to be noted:

Block diagram and equivalent circuit diagram for consideration of a channel frequency response

• The receiver filter  $H_{\rm E}(f)$  is – at least mentally – composed of an  ideal equalizer  $1/H_{\rm K}(f)$  and a low-pass filter  $H_{\rm G}(f)$.  For the latter we use in this chapter exemplarily a  Gaussian low-pass  with the cutoff frequency  $f_{\rm G}$.
  

• If we now move the ideal equalizer – again purely mentally – to the left side of the noise addition point,  nothing changes with respect to the S/N ratio before the decision device and with respect to the error probability compared to the block diagram drawn above.
  

• From the equivalent diagram below it can be seen that the channel frequency response  $H_{\rm K}(f)$  does not change anything with respect to the signal component  $d_{\rm S}(t)$  of the detection signal if it is fully compensated with  $1/H_{\rm K}(f)$. 

• Thus,  the signal component  $d_{\rm S}(t)$  – originating from the transmitted signal  $s(t)$  – has exactly the same shape as calculated in the chapter  "Error Probability with Intersymbol Interference". 
  

• The degradation due to the frequency response  $H_{\rm K}(f)$  is rather shown by a significant increase of the detection noise power, i.e. the variance of the noise component  $d_{\rm N}(t)$  – originating from the noise signal  $n(t)$  at the receiver input:

$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \,{\rm d} f = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} \frac{1}{|H_{\rm K}(f)|^2}\cdot |H_{\rm G}(f)|^2 \,{\rm d} f \hspace{0.05cm}.$$

• The prerequisite for a finite noise power  $\sigma_d^2$  is that the Gaussian low-pass  $H_{\rm G}(f)$  attenuates the noise  $n(t)$  at  (very)  high frequencies more than it is raised by the ideal equalizer  $1/H_{\rm K}(f)$. 
  
  

Note:   The channel frequency response  $H_{\rm K}(f)$  must be equalized for magnitude and phase,  but only in a limited frequency range specified by  $H_{\rm G}(f)$ .  However,  a complete phase equalization is only possible at the expense of a  (frequency-independent)  delay time,  which will not be considered further in the following.

$\text{Example 1:}$  We consider again a binary system with NRZ rectangular pulses and a Gaussian receiver filter  $H_{\rm E}(f) = H_{\rm G}(f)$  with  (normalized)  cutoff frequency  $f_\text{G, opt} \cdot T = 0.4$.

• The middle graph shows the eye diagram of the signal component  $d_{\rm S}(t)$ of the detection signal for this case – i.e. without taking the noise into account.

• This is identical with the eye diagram shown in the chapter  "Definition and statements of the eye diagram"  in $\text{example 3}$ (right diagram).
  

Binary eye diagrams with intersymbol interferences

The left eye diagram is obtained for ideal channel, i.e., for  $H_{\rm K}(f) = 1$   ⇒   $1/H_{\rm K}(f) = 1$. It takes into account the AWGN noise, but here it was assumed to be very small,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 30 \ \rm dB$.  For this configuration, it was determined by simulation:

$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx 26.8\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}< 10^{-40}\hspace{0.05cm}.$$

In contrast, the right diagram applies to a  "coaxial cable", where the characteristic cable attenuation  $a_\star = 40 \ \rm dB$.  This results in significantly worse system sizes for the same  $E_{\rm B}/N_0$: 

$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx -4.6\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 0.28\hspace{0.05cm}.$$

This result can be interpreted as follows:

• Assuming an ideal channel equalizer  $1/H_{\rm K}(f)$,  the same "eye diagram without noise" (left graph) results for the distorting channel as for the ideal channel  $H_{\rm K}(f) = 1$  (middle graph).
  

• Channel equalization  $1/H_{\rm K}(f)$  extremely amplifies the noise component. In the right-hand example, equally strong equalization is required over a wide frequency range because of the strong distortion.
• The noise power  $\sigma_d^2$  here is larger by a factor of  $1300$  than in the left constellation (no distortion   ⇒   no equalization). Thus, the error probability results in  $p_{\rm S}\approx p_{\rm U}\approx 50 \%$.
  

• An acceptable error probability results only with smaller noise power density  $N_0$. For example, with  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 50 \ \rm dB$  $($instead of $30 \ \rm dB)$  the following result is obtained:

$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = -4.6 +20 \approx 15.4\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 2 \cdot 10^{-9} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm S} \ge p_{\rm U}/4 \approx 0.5 \cdot 10^{-9}\hspace{0.05cm}.$$

Increase of the noise power by linear equalization

The eye diagrams on the last section impressively document the increase of the noise power  $\sigma_d^2$  with unchanged vertical eye opening, if one compensates the channel frequency response  $H_{\rm K}(f)$  on the receiving side by its inverse. This result shall now be interpreted in terms of the noise power density  ${\it \Phi}_{d{\rm N}}(f)$  after the receiver filter (before the decision), with the following settings:

• Let the channel be a  "coaxial cable"  with the magnitude frequency response

$$|H_{\rm K}(f)| = {\rm exp}\left [- a_{\star}\cdot \sqrt{2 f T}\hspace{0.05cm} \right ]\hspace{0.2cm}{\rm with}\hspace{0.2cm} a_{\star} = 1.7\,\,{\rm Np}\hspace{0.2cm} ({\rm corresponding to} \hspace{0.2cm} 15\,\,{\rm dB}) \hspace{0.05cm}.$$

• The  "ideal channel equalizer"  $1/H_{\rm K}(f)$  compensates the channel frequency response completely. No statement is made here about the realization of the attenuation and phase equalization.
  

• For noise power limitation a  "Gaussian low-pass filter"  is used:

$$|H_{\rm G}(f)| = {\rm exp}\left [- \pi \cdot \left (\frac{f }{2 f_{\rm G}}\right )^2 \right ]\hspace{0.2cm}{\rm with}\hspace{0.2cm} f_{\rm G} = 0.8/T \hspace{0.2cm} {\rm and} \hspace{0.2cm} f_{\rm G} = 0.4/T \hspace{0.05cm}.$$

Thus, the noise power density before the decision is:

$${\it \Phi}_{d{\rm N}}(f) = \frac{N_0}{2} \cdot \frac{|H_{\rm G }(f)|^2}{|H_{\rm K}(f)|^2} = \frac{N_0}{2} \cdot {\rm exp}\left [2 \cdot a_{\star}\cdot \sqrt{2 f T} - {\pi}/{2} \cdot \left ({f }/{f_{\rm G}}\right )^2 \right ] \hspace{0.05cm}.$$

Noise rise due to distorting channel

This curve is shown on the left for the two (normalized) cutoff frequencies

• $f_\text{G} \cdot T = 0.8$ (left) and
• $f_\text{G} \cdot T = 0.4$ (right)

Note that here, for reasons of presentation, the characteristic cable attenuation of  $a_\star = 15 \ \rm dB$   $($corresponding to  $1.7 \ \rm Np)$  is chosen to be significantly smaller than in the right eye diagram in  "$\text{Example 1}$"  in the last section  $($valid for  $a_\star = 40 \ \rm dB)$.
Let us first consider the left graph for the (normalized) cutoff frequency  $f_\text{G} \cdot T = 0.8$ which, according to the calculations in the  "last chapter"  represents the optimum for the ideal channel   ⇒   $H_{\rm K}(f) = 1$. 

• The constant noise power density  $N_0/2$  at the receiver input is highlighted in yellow. With an ideal channel, this is limited by the Gaussian receive filter  $H_{\rm E}(f) = H_{\rm G}(f)$  and results in the detection noise power  $\sigma_d^2$  (indicated by the blue area in the graph).
  

• If – as usual for conducted transmission üblich – higher frequencies are strongly attenuated,  $|H_{\rm E}(f)| = |H_{\rm G}(f)|/|H_{\rm K}(f)|$  increases very strongly due to the ideal channel equalizer before the attenuating influence of the Gaussian filter becomes effective for  $f \cdot T \ge 0.6$  $($only valid for  $a_\star = 15 \ \rm dB$  and  $f_\text{G} \cdot T = 0.8)$. 
  

• The noise power  $\sigma_d^2$  is now equal to the area under the red curve, which is about a factor of  $28$  larger than the blue area. The effects of these different noise powers can also be seen in the eye diagrams on the last section, but for  $a_\star = 40 \ \rm dB$.
  
  

The right graph shows the noise power density  ${\it \Phi}_{d{\rm N}}(f)$  for the normalized cutoff frequency  $f_\text{G} \cdot T = 0.4$. Here the noise power is only increased by a factor of  $9$  by the ideal channel equalizer (ratio between the area under the red curve and the blue area).

$\text{Conclusion:}$  From the above graph and the previous explanations it is already clear that with distorting channel   ⇒   $H_{\rm K}(f) \ne 1$ the cutoff frequency  $f_\text{G} \cdot T = 0.8$  of the Gaussian low-pass filter $H_{\rm G}(f)$ will no longer be optimal after the ideal channel equalizer  $1/H_{\rm K}(f)$. 

Optimization of the cutoff frequency

The graph shows the signal-to-noise ratios as a function of the cutoff frequency  $f_{\rm G}$  of the overall Gaussian frequency response  $H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$. This picture is valid for

• a  "coaxial transmission channel"  with characteristic cable attenuation  $a_\star = 15 \ \rm dB$,
  
• AWGN noise with  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$, where  $E_{\rm B} = s_0^2 \cdot T$  is to be set   ⇒   NRZ rectangular pulses.
  
  

Optimal cut-off frequency of the GTP with distorting channel  $(a_\star = 15 \ \rm dB)$

Notes:

• The circles show the dB values for  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_d$   ⇒   "mean" detection SNR (measure of mean error probability  $p_{\rm S})$.
• The squares show the dB values for  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U}$   ⇒   "worst-case" SNR  $($measure of worst-case error probability  $p_{\rm U})$.

One can see from this plot and by comparison with the  "corresponding plot"  in the last chapter, which was valid for  $H_{\rm K}(f) = 1$  and  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$: 

• Even with a highly distorting channel,  $\rho_{\rm U}$  is a suitable lower bound for  $\rho_d$   ⇒   $\rho_{d} \ge \rho_{\rm U}$. Correspondingly,  $p_{\rm U} \ge p_{\rm S} $  is also a reasonable upper bound for  $p_{\rm S}$.

• For the considered cable attenuation  $a_\star = 15 \ \rm dB$,  the cutoff frequency  $f_\text{G} \cdot T \approx 0.55$  is optimal and  $\ddot{o}/s_0 \approx 1.327$  and  $\sigma_d/s_0 \approx 0.106$ hold.
• This gives the (worst-case) signal-to-noise ratio  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$  and the worst–case error probability  $p_{\rm U} \approx 2 \cdot 10^{-9}.$
  

• A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$  holds:

$$\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx 14.1\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$$

• If the cutoff frequency  $f_\text{G}$  is too large, the noise is limited less effectively. For example, the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:

$$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx 14.2\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$$

• The optimal values of  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for ideal channel.

When comparing the signal-to-noise ratios, however, it must be taken into account that  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$  is the basis here; in contrast,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$  was assumed in the  "corresponding graph"  for the ideal channel.

Optimal cut-off frequency as a function of cable attenuation

Optimal cutoff frequency and system efficiency as a function of characteristic cable attenuation. In particular:
$\hspace{0.8cm} 10 · \lg \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ dB;$   $\hspace{0.8cm} 10 · \lg \eta\hspace{0.05cm}(a_\star = 80 \ \rm dB) = -78.2 \ dB;$

We further consider

• a binary system with NRZ transmission pulses   ⇒   $E_{\rm B} = s_0^2 \cdot T$,
  
• a coaxial cable $H_{\rm K}(f)$, characteristic attenuation  $a_\star$,
  
• a Gaussian total frequency response  $H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.

The blue circles (left axis labels) mark the optimal cutoff frequencies  $f_\text{G, opt}$  for the respective cable attenuation  $a_\star$.

In addition, the graph shows with red squares the  "system efficiency"  (at peak limitation)  $\eta$,  which is the ratio of the SNR  $\rho_{d}$  achievable with the considered configuration to the maximum possible S/N ratio  $\rho_{d, \ {\rm max}}$. 

Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:

$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}|\hspace{0.05cm} A}}}= \frac{\rho_d}{2 \cdot E_{\rm B}/N_0}\approx \frac{\rho_{\rm U}}{2 \cdot E_{\rm B}/N_0}.$$

One can see from the arrangement of the blue circles:

• The optimal cutoff frequency  $f_\text{G, opt}$  depends significantly on the strength of the distortions of the coaxial cable, more precisely:   exclusively on the characteristic cable attenuation  $a_\star$  at half the bit rate.
• The larger the cable attenuation  $a_\star$  and thus the noise influence, the lower the optimal cutoff frequency  $f_\text{G, opt}$.
  
• However, always  $f_\text{G, opt} > 0.27/T$. Otherwise, the eye would be closed, equivalent to the "worst–case" error probability  $p_{\rm U} = 0.5$.

Let us now discuss the dependence of the system efficiency  $\eta$  (red squares) on the characteristic cable attenuation  $a_\star$. The right ordinate starts at the top at  $0 \ \rm dB$  and extends downward to  $-100 \ \rm dB$.

As will now be illustrated by some numerical examples, the representation  $\eta = \eta\hspace{0.05cm}(a_\star)$ avoids some problems arising from the wide range of values of S/N ratios:

• The ordinate value  $10 \cdot {\rm lg}\hspace{0.1cm} \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ \rm dB$  states that the best possible Gaussian low-pass with cutoff frequency  $f_\text{G} \cdot T = 0.8$  at ideal channel is  $1.4 \ \rm dB$  worse than the optimal (matched filter) receiver.
  

• If we assume ideal channel  $(a_\star = 0 \ \rm dB)$  and  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 10 \ \rm dB$,  the above equation also states that this configuration will lead to the following (worst-case) error probability:

$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 10 \cdot {\rm lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} + 10 \cdot {\rm lg}\hspace{0.1cm}(2) + 10 \cdot {\rm lg}\hspace{0.1cm}(\eta) \approx \approx 10\,{\rm dB} \hspace{0.1cm}+\hspace{0.1cm}3\,{\rm dB} \hspace{0.1cm}-\hspace{0.1cm}1.4\, {\rm dB}= 11.6\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 7 \cdot 10^{-5}\hspace{0.05cm}.$$

• Accordingly, if this (worst-case) error probability  $p_{\rm U} = 7 \cdot 10^{-5}$   ⇒   $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} = 11.6 \ \rm dB$  is not to be exceeded for the channel with characteristic cable attenuation  $a_\star = 80 \ \rm dB$,  then the following must apply to the  $E_{\rm B}/N_0$  ratio:

\[10 \cdot {\rm lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} \ge 11.6\,{\rm dB} \hspace{0.1cm}-3\,{\rm dB} \hspace{0.1cm}-\hspace{0.1cm}(-78.2)\,{\rm dB}= 86.8\,{\rm dB} \hspace{0.2cm} \Rightarrow \hspace{0.2cm}{E_{\rm B}}/{N_0}\approx 5 \cdot 10^{8}\hspace{0.05cm}.\]

• To achieve this, however, the cutoff frequency of the Gaussian low-pass filter must be lowered to  $f_{\rm G}= 0.33/T$  according to the blue circles in the graph.
  

Exercises for the chapter

Exercise 3.3: Noise at Channel Equalization

Exercise 3.3Z: Optimization of a Coaxial Cable System