Consideration of Channel Distortion and Equalization

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).

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

To these representations the following is to be noted:

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 at the sink and with respect to the error probability compared to the block diagram drawn above.

From the equivalent circuit below it can be seen that the channel frequency response $H_{\rm K}(f)$ does not change anything with respect to the signal component of the detection signal $d_{\rm S}(t)$ – originating from the transmitted signal $s(t)$ – if it is fully compensated with $1/H_{\rm K}(f)$. Thus, the signal component has exactly the same shape as calculated in the chapter "Error Probability with Intersymbol Interference".

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

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

The prerequisite for a finite noise power $\sigma_d^2$ is that the 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 $H_{\rm K}(f)$ can be equalized according to magnitude and phase, but only in a limited frequency range given by $H_{\rm G}(f)$. However, a complete phase equalization is only possible at the expense of a (frequency-independent) running time, which will not be considered further in the following.

$\text{Example 1:}$ We consider again a binary system with NRZ rectangular pulses and 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 of the detection signal $d_{\rm S}(t)$ 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