Optimization of Baseband Transmission Systems

Prerequisites and optimization criterion

The following block diagram applies to this chapter "Optimization of Baseband Transmission Systems". Unless explicitly stated otherwise, the following prerequisites are assumed in the following:

Block diagram of a baseband transmission system

The transmission is binary, bipolar and redundancy-free. The spacing between symbols is $T$ and the (equivalent) bit rate $R = 1/T$. Multilevel and/or redundant systems are not discussed until the main chapter 2: "Coded and Multilevel Transmission" of this book.
The basic transmission pulse $g_s(t)$ is rectangular and has the amplitude $s_0$ and the pulse duration $T_{\rm S} \le T$. If the pulse duration $T_{\rm S}$ coincides with the symbol duration $T$, we speak of NRZ ("non-return-to-zero") rectangular pulses. In the case $T_{\rm S} < T$, the RZ ("return-to-zero") format is present.
The AWGN model with the (one-sided) noise power density $N_0$ is used as the transmission channel, so that $r(t) = s(t) + n(t)$ applies to the received signal. The two-sided noise power density $(N_0/2)$ is more suitable for system-theoretical investigations.
Let the impulse response $h_{\rm E}(t)$ of the receiver filter (German: "Empfangsfilter" ⇒ subscript: "E") also be rectangular, but with width $T_{\rm E}$ and height $1/T_{\rm E}$. The DC transfer factor is therefore $H_{\rm E}(f = 0) = 1$. Only in the special case $T_{\rm E} = T_{\rm S} $ ⇒ $H_{\rm E}(f)$ can be called a "matched filter".
In order to exclude intersymbol interfering, the constraint $T_{\rm S} + T_{\rm E} \le 2T$ must always be observed during optimization. Intersymbol interfering will not be considered until the main chapter 3: "Intersymbol Interfering and Equalization Methods" of this book.
To obtain the sink symbol sequence $〈v_ν〉$ as the best possible estimate for the source symbol sequence $〈q_ν〉$, we use a simple threshold decision with the optimal decision threshold $E = 0$ and optimal detection times $($under the given conditions at $\nu \cdot T)$.

$\text{Definition:}$ By "system optimization" we mean here to determine the parameters $T_{\rm S}$ and $T_{\rm E}$ of the basic transmission pulse $g_s(t)$ and the receiver filter impulse response $h_{\rm E}(t)$ in such a way that the bit error probability $p_{\rm B}$ assumes the smallest possible value.

Power and peak limitation

The optimization of the system variables is decisively influenced by

whether "power limitation"
or "peak limitation"

of the transmitted signal $s(t)$ is required as a constraint of the optimization.

$\text{Definition:}$ Power limitation means that the (average) transmission power (German: "Sendeleistung" ⇒ subscript: "S") $P_{\rm S}$ must not exceed a specified maximum value $P_\text{S, max}$:

$$P_{\rm S}= {\rm E}[s(t)^2] = \overline{s(t)^2} \le P_{\rm S,\hspace{0.05cm} max}\hspace{0.05cm}.$$

In order to achieve the minimum error probability, one will naturally choose the average transmission power $P_{\rm S}$ as large as possible in the allowed range.
Therefore, $P_{\rm S} = P_\text{S, max}$ is always set in the following.

The question of whether power limitation can actually be assumed as a secondary condition of optimization depends on the technical boundary conditions.

This assumption is especially justified for radio transmission systems, among other things
because the impairment of humans and animals known as "electrosmog" depends to a large extent on the (average) radiated power.

Note: Of course, a radio transmission system does not operate in baseband. However, the description variables defined here using baseband transmission as an example are modified in the main chapter 4: "Generalized Description of Digital Modulation Methods" of this book to the effect that they are also applicable to digital carrier frequency systems.

$\text{Definition:}$ Peak limitation is always referred to when the output range of the transmitter is limited. For bipolar signaling, the corresponding condition is:

$$\vert s(t) \vert \le s_0\hspace{0.4cm}{\rm{for} }\hspace{0.15cm}{\rm all}\hspace{0.15cm}t.$$

Of course, "peak limitation" limits also the power, but the "peak power", not the "average power".
Often, instead of "peak limitation", the term "amplitude limitation" is also used, but this does not quite reflect the facts.

The condition "peak limitation" is useful and even necessary, for example, if

the output power range of the transmitter is limited due to nonlinearities of components and power amplifiers, or
the crosstalk noise must not exceed a limit value at any time. This is especially important when communicating over two-wire lines.

$\text{Example 1:}$ We consider three different constellations. Let the basic transmission pulse $g_s(t)$ and the receiver filter impulse response $h_{\rm E}(t)$ each be rectangular and the amplitude $g_0$ of the output pulse $g_d(t)$ always coincide with the amplitude $s_0$ of the input pulse $g_s(t)$ .

$\text{System A}$ $(T_{\rm S} = T, \ T_{\rm E} = T)$:

Impulse response $h_{\rm E}(t)$ and pulses $g_s(t)$ as well as $g_d(t)$ for $\text{System A}$

NRZ basic transmission pulse,
Matched filter, since $T_{\rm E} = T_{\rm S}$,
Basic detection pulse: triangle,
Energy per bit: $E_{\rm B} = s_0^2 \cdot T$,
Noise power: $\sigma_d^2 = N_0/(2T)$,
Best possible constellation
Bit error probability: $p_{\rm B}\hspace{-0.01cm} =\hspace{-0.01cm}{\rm Q} \left( g_0/\sigma_d\right)$

$$\Rightarrow \hspace{0.3cm}p_{\rm B}= {\rm Q} \left( \sqrt{ {2 \cdot s_0^2 \cdot T}/{N_0} }\right) = {\rm Q} \left( \sqrt{ {2 \cdot E_{\rm B} }/{N_0} }\right)\hspace{0.05cm}.$$

$\text{System B}$ $(T_{\rm S} = T, \ T_{\rm E} = T/2)$:

Impulse response $h_{\rm E}(t)$ and pulses $g_s(t)$ as well as $g_d(t)$ for $\text{System B}$

NRZ basic transmission pulse,
No matched filter, since $T_{\rm E} \ne T_{\rm S}$,
Basic detection pulse: trapezoid,
Energy per bit: $E_{\rm B} = s_0^2 \cdot T$,
Noise power: $\sigma_d^2 = N_0/T$,
Always $\text{3 dB}$ worse than $\text{System A}$
Bit error probability: $p_{\rm B}\hspace{-0.01cm} =\hspace{-0.01cm}{\rm Q} \left( g_0/\sigma_d\right)$

$$\Rightarrow \hspace{0.3cm}p_{\rm B}= {\rm Q} \left( \sqrt{ {s_0^2 \cdot T}/{N_0} }\right) = {\rm Q} \left( \sqrt{ E_{\rm B} /{N_0} }\right)\hspace{0.05cm}.$$

$\text{System C}$ $(T_{\rm S} = T/2, \ T_{\rm E} = T/2)$:

Impulse response $h_{\rm E}(t)$ and pulses $g_s(t)$ as well as $g_d(t)$ for $\text{System C}$

RZ basic transmission pulse,
Matched filter, since $T_{\rm E} = T_{\rm S}$,
Basic detection pulse: smaller triangle,
Energy per bit: $E_{\rm B} = 1/2 \cdot s_0^2 \cdot T$,
Noise power: $\sigma_d^2 = N_0/T$,
with power limitation equivalent to $\text{System A}$,
At peak limitation $\text{3 dB}$ worse than $\text{System A}$,
Bit error probability: $p_{\rm B}\hspace{-0.01cm} =\hspace{-0.01cm}{\rm Q} \left( g_0/\sigma_d\right)$

$$\Rightarrow \hspace{0.3cm}p_{\rm B} = {\rm Q} \left( \sqrt{ { s_0^2 \cdot T}/{N_0} }\right)= {\rm Q} \left( \sqrt{2 \cdot {E_{\rm B} }/{N_0} }\right)\hspace{0.05cm}.$$

$\text{Example 2:}$ The same conditions as in $\text{Example 1}$ apply. Graphically shown are the results from $\text{Example 1}$. The diagram shows the bit error probability $p_{\rm B}$ as

System comparison for power and peak limitation

a function of the ratio $E_{\rm B}/N_0$ (left diagram) and
a function of the ratio $s_0^2 \cdot T /N_0$ (right diagram).

These two diagrams in double logarithmic representation are to be interpreted as follows:

The left diagram compares the systems at the same average power $(P_{\rm S})$ and at constant energy per bit $(E_{\rm B})$, resp. Since the abscissa value is additionally related to $N_0$, the equation $p_{\rm B}(E_{\rm B}/N_0)$ and its graphical representation correctly reflects the situation even for different noise power densities $N_0$.

When power is limited, configurations $\rm A$ and $\rm C$ are equivalent and represent the optimum in each case. As will be shown on the next pages, an optimal system with power limitation always exists if $g_s(t)$ and $h_{\rm E}(t)$ have the same shape $($"matched filter"$)$. The smaller power of system $\rm C$ is compensated by the abscissa chosen here.

In contrast, for system $\rm B$ the matched filter condition is not met $(T_{\rm E} \ne T_{\rm S})$ and the error probability curve is now $\text{3 dB}$ to the right of the boundary curve given by systems $\rm A$ and $\rm C$.

The diagram on the right describes the optimization result with peak limitation, which can be seen from the abscissa labeling. The curve $\rm A$ $($NRZ pulse, matched filter$)$ also indicates here the limit curve, which cannot be undershot by any other system.

Curve $\rm B$ in the diagram on the right has exactly the same shape as in the diagram on the left, since NRZ transmission pulses are again used. The distance of $\text{3 dB}$ from the limit curve is again due to non-compliance with the matched filter condition.

In contrast to the left diagram, the matched filter system $\rm C$ is now also $\text{3 dB}$ to the right of the optimum. The reason for this degradation is that for the same peak value $($same peak power$)$, system $\rm C$ provides only half the average power as system $\rm A$.

System optimization with power limitation

The minimization of the bit error probability $p_{\rm B} = {\rm Q} \left( \sqrt{\rho_d}\right)$ can be attributed to the maximization of the signal–to–noise power ratio $\rho_d$ before the threshold decision $($in short: detection SNR$)$ due to the monotonic function progression of the complementary Gaussian error function $ {\rm Q}(x)$:

$$p_{\rm B} = {\rm Q} \left( \sqrt{\rho_d}\right)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Minimum}\hspace{0.8cm}\Rightarrow \hspace{0.8cm}\rho_d ={g_0^2}/{\sigma_d^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Maximum}\hspace{0.05cm}.$$

Here, $g_0 = g_d(t=0)$ indicates the amplitude of the considered Nyquist pulse and $\sigma_d^2$ denotes the detection noise power for the given receiver filter. At the same time it must be ensured that

the basic detection pulse $g_d(t) = g_s(t) \star h_{\rm E}(t)$ satisfies the first Nyquist criterion, and
the energy of the basic transmission pulse $g_s(t)$ does not exceed a predetermined value $E_{\rm B}$.

In the previous sections, it has been mentioned several times that for the AWGN channel with the (one-sided) noise power density $N_0$, the following holds for the optimal system under the constraint of power limitation:

$$p_{\rm B, \hspace{0.08cm}min} = {\rm Q} \left( \sqrt{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} | \hspace{0.05cm} L}}}\right)\hspace{0.5cm}{\rm with}\hspace{0.5cm} \rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} | \hspace{0.05cm} L}}={2 \cdot E_{\rm B}}/{N_0}\hspace{0.05cm}.$$

We use this result for the following definition:

$\text{Definition:}$ The system efficiency under power limitation (German: "Leistungsbegrenzung" ⇒ subscript: "L") of a given configuration is the quotient of the actual and the highest possible signal–to–noise power ratio at the decision point ("detection SNR"):

$$\eta_{\rm L} = \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\vert \hspace{0.05cm} L} } }= \frac{g_0^2 /\sigma_d^2}{2 \cdot E_{\rm B}/N_0}\hspace{0.05cm}.$$

In the following it is proved that

the quantity thus defined actually satisfies the condition $0 \le \eta_{\rm L} \le 1$ and thus can be interpreted as "efficiency",
the value $\eta_{\rm L} = 1$ is obtained when the receiver filter impulse response $h_{\rm E}(t)$ is equal in shape to the basic transmission pulse $g_s(t)$.

$\text{Proof:}$ The proof is done in the frequency domain. For presentation reasons, we normalize the basic transmission pulse:

$$h_{\rm S}(t) = \frac{g_s(t)}{g_0 \cdot T} \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm} H_{\rm S}(f) = \frac{G_s(f)}{g_0 \cdot T} \hspace{0.05cm}.$$

Thus $h_{\rm S}(t)$ has the unit "$\rm 1/s$" and $H_{\rm S}(f)$ is dimensionless. For the individual system quantities it follows:

(1) Due to the first Nyquist criterion, it must hold:

$$ G_d(f) = G_s(f) \cdot H_{\rm E}(f) = G_{\rm Nyq}(f) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_{\rm S}(f) \cdot H_{\rm E}(f)= H_{\rm Nyq}(f)= \frac{G_{\rm Nyq}(f)}{g_0 \cdot T}\hspace{0.05cm}.$$

(2) The amplitude of the basic transmitter pulse is equal to

$$g_d(t=0) = g_0 \cdot T \cdot \int_{-\infty}^{+\infty}H_{\rm Nyq}(f) \,{\rm d} f = g_0\hspace{0.05cm}.$$

(3) The energy of the basic transmission pulse is given as follows:

$$E_{\rm B} = g_0^2 \cdot T^2 \cdot \int_{-\infty}^{+\infty} \vert H_{\rm S}(f)\vert ^2 \,{\rm d} f \hspace{0.05cm}.$$

(4) The detection noise power is:

$$ \sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} \vert H_{\rm E}(f) \vert^2 \,{\rm d} f = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty}\frac {\vert H_{\rm Nyq}(f) \vert^2}{\vert H_{\rm S}(f) \vert^2} \,{\rm d} f\hspace{0.05cm}. $$

(5) Substituting these partial results into the equation for the system efficiency, we obtain:

$$\eta_{\rm L} = \left [ {T \cdot \int_{-\infty}^{+\infty} \vert H_{\rm S}(f) \vert^2 \,{\rm d} f \hspace{0.2cm} \cdot \hspace{0.2cm}T \cdot \int_{-\infty}^{+\infty}\frac { \vert H_{\rm Nyq}(f) \vert ^2}{ \vert H_{\rm S}(f) \vert^2} \,{\rm d} f } \right ]^{-1}\hspace{0.05cm}.$$

(6) We now apply Schwartz's inequality [BSMM15]^[1] to the expression in the parenthesis:

$$\frac{1}{\eta_{\rm L} } = T \cdot \int_{-\infty}^{+\infty} \vert H_{\rm 1}(f) \vert^2 \,{\rm d} f \hspace{0.2cm} \cdot \hspace{0.2cm} T \cdot \int_{-\infty}^{+\infty} \vert H_{\rm 2}(f) \vert^2 \,{\rm d} f \hspace{0.3cm}\ge\hspace{0.3cm} \left [ T \cdot \int_{-\infty}^{+\infty} \vert H_{\rm 1}(f) \cdot H_{\rm 2}(f) \vert \,{\rm d} f \right ]^2$$

$$\Rightarrow \hspace{0.3cm}\frac{1}{\eta_{\rm L} } = T \cdot \int_{-\infty}^{+\infty} \vert H_{\rm S}(f) \vert^2 \,{\rm d} f \hspace{0.1cm} \cdot \hspace{0.1cm} T \cdot \int_{-\infty}^{+\infty}\frac { \vert H_{\rm Nyq}(f) \vert ^2}{ \vert H_{\rm S}(f) \vert ^2} \,{\rm d} f \hspace{0.2cm}\ge\hspace{0.2cm} \left [ T \cdot \int_{-\infty}^{+\infty}\hspace{-0.5cm}H_{\rm Nyq}(f) \,{\rm d} f \right ]^2 = 1. $$

(7) Thus, it is shown that the system efficiency under power limitation indeed satisfies the condition $\eta_{\rm L} \le 1$.

(8) The evaluation shows that for $H_{\rm S, \hspace{0.08cm}opt}(f) = \gamma \cdot \sqrt{H_{\rm Nyq}(f)}$ in the above inequality, the equal sign holds:

$$\gamma^2 \cdot T \cdot \int_{-\infty}^{+\infty}\hspace{-0.3cm} H_{\rm Nyq}(f) \,{\rm d} f \hspace{0.2cm} \cdot \hspace{0.2cm} \frac {1}{\gamma^2} \cdot T \cdot \int_{-\infty}^{+\infty} \hspace{-0.3cm}H_{\rm Nyq}(f) \,{\rm d} f = \left [ T \cdot \int_{-\infty}^{+\infty}\hspace{-0.3cm}H_{\rm Nyq}(f) \,{\rm d} f \right ]^2 = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \eta_{\rm L} = 1 \hspace{0.05cm}.$$

(9) This result is independent of the parameter $\gamma$, which we therefore simplify to $\gamma = 1$: $H_{\rm S, \hspace{0.08cm}opt}(f) = \sqrt{H_{\rm Nyq}(f)}$.

q.e.d.

Root Nyquist systems

The main result of the calculations on the last sections was that for the optimal binary system under the constraint of "power limitation"

the basic detection pulse $g_d(t) = g_s(t) \star h_{\rm E}(t)$ must satisfy the first Nyquist condition, and
the impulse response $h_{\rm E}(t)$ of the receiver filter must be chosen to be equal in shape to the basic transmission pulse $g_s(t)$;
the same applies to the spectral functions $H_{\rm E}(f)$ and $G_s(f)$.

If both $g_s(t)$ and $h_{\rm E}(t)$ are rectangular with $T_{\rm S} = T_{\rm E} \le T$, both conditions are satisfied.

However, the disadvantage of this configuration is the large bandwidth requirement due to the slowly decaying $\rm sinc$–shaped spectral functions $G_s(f)$ and $H_{\rm E}(f)$.
In the diagram below, the spectral function of the rectangular NRZ basic transmission pulse is plotted as a dashed purple curve.

Different transmission spectra for baseband transmission

Assuming the cosine rolloff frequency response ⇒ $H_{\rm E}(f) = H_{\rm CRO}(f)$,

$$G_d(f) = G_s(f) \cdot H_{\rm E}(f) = g_0 \cdot T \cdot {H_{\rm CRO}(f)} $$

$$\Rightarrow \hspace{0.3cm}G_s(f) = g_0 \cdot T \cdot \sqrt{H_{\rm CRO}(f)},\hspace{0.5cm}H_{\rm E}(f)= \sqrt{H_{\rm CRO}(f)}\hspace{0.05cm},$$

then for each rolloff factor $r$, more favorable spectral characteristics and lower bandwidth requirements result.

The diagram shows the normalized transmission spectra $G_s(f)/(g_0 \cdot T)$ in logarithmic representation for the three rolloff factors:

$r = 0$ (green curve),
$r = 0.5$ (blue curve), and
$r = 1$ (red curve).

$\rm Notes$:

For baseband transmission, the bandwidth requirement plays only a minor role.
However, the diagram also applies to "carrier frequency systems" when displayed in the equivalent low-pass range.
In these systems, bandwidth plays a very important role. Because: Every additional Hertz of bandwidth can be very expensive.

System optimization with peak limitation

For the AWGN channel with the (one-sided) noise power density $N_0$, the system optimization depends to a large extent on which constraint is specified:

In the case of "power limitation" (German: "Leistungsbegrenzung" ⇒ subscript: "L"), the energy of the basic transmission pulse $g_s(t)$ must not exceed a specified value $E_{\rm B}$. Here, the following applies to the minimum bit error probability and the maximum SNR:

$$p_{\rm B, \hspace{0.08cm}min} = {\rm Q} \left( \sqrt{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} | \hspace{0.05cm} L}}}\right)\hspace{0.5cm}{\rm with}\hspace{0.5cm} \rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} | \hspace{0.05cm} L}}={2 \cdot E_{\rm B}}/{N_0}\hspace{0.05cm}.$$

In the case of "peak limitation" (German: "Spitzenwertbegrenzung" or "Amplitudenbegrenzung" ⇒ subscript: "A"), on the other hand, the modulation range of the transmitter device is limited ⇒ $\vert s(t) \vert \le s_0\hspace{0.4cm}{\rm{for} }\hspace{0.15cm}{\rm all}\hspace{0.15cm}t$. Here, the following applies to the corresponding quantities:

$$p_{\rm B, \hspace{0.08cm}min} = {\rm Q} \left( \sqrt{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} | \hspace{0.05cm}A}}}\right)\hspace{0.5cm}{\rm with}\hspace{0.5cm} \rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm} | \hspace{0.05cm}A}}={2 \cdot s_0^2 \cdot T}/{N_0}\hspace{0.05cm}.$$

For this second case, we define:

$\text{Definition:}$ The system efficiency under peak limitation ("amplitude limitation") is:

$$\eta_{\rm A} = \frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\vert \hspace{0.05cm} A} } }= \frac{g_0^2 /\sigma_d^2}{ 2 \cdot s_0^2 \cdot T/N_0}\hspace{0.05cm}.$$

This system efficiency also satisfies the condition $0 \le \eta_{\rm A} \le 1$.
There is only one system with the result $\eta_{\rm A} = 1$: The NRZ rectangular basic transmission pulse and the receiver filter matched to it..

A comparison with the "system efficiency under power limitation" ⇒ $\eta_{\rm L}$ shows:

$\eta_{\rm A}$ differs from $\eta_{\rm L}$ in that now the denominator contains $s_0^2 \cdot T$ instead of $E_{\rm B}$. The following relationship holds:

$$\eta_{\rm A} = \frac{E_{\rm B}}{s_0^2 \cdot T} \cdot \eta_{\rm L}= \frac{\eta_{\rm L}}{C_{\rm S}^2}\hspace{0.05cm}.$$

Here, the "crest factor" $C_{\rm S}$ denotes the ratio of the maximum value $s_0$ and the rms value $s_{\rm eff}$ of the transmitted signal:

$$C_{\rm S} = \frac{s_0}{\sqrt{E_{\rm B}/T}} = \frac{{\rm Max}[s(t)]}{\sqrt{{\rm E}[s^2(t)]}}= \frac{s_0}{s_{\rm eff}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm eff} = \sqrt {E_{\rm B}/T}.$$

$\text{Example 3:}$ As in "$\text{Example 1}$" we consider three different configurations, each with rectangular functions $g_s(t)$ and $h_{\rm E}(t)$. The system efficiencies are:

$\text{System A:}$ $\rho_d = {2 \cdot E_{\rm B} }/{N_0} = { 2 \cdot s_0^2 \cdot T}/{N_0}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\eta_{\rm L} = 1.0,\hspace{0.3cm}\eta_{\rm A} = 1.0\hspace{0.05cm}.$
$\text{System B:}$ $\rho_d = {E_{\rm B} }/{N_0} ={ s_0^2 \cdot T}/{N_0}\hspace{1.35cm}\Rightarrow \hspace{0.3cm}\eta_{\rm L} = 0.5,\hspace{0.3cm}\eta_{\rm A} = 0.5\hspace{0.05cm}.$
$\text{System C:}$ $\rho_d = {2 \cdot E_{\rm B} }/{N_0} = { s_0^2 \cdot T}/{N_0}\hspace{0.8cm}\Rightarrow \hspace{0.3cm}\eta_{\rm L} = 1.0,\hspace{0.3cm}\eta_{\rm A} = 0.5\hspace{0.05cm}.$

It can be seen:

For $\text{System A}$ both system efficiencies are at most equal $1$.
For $\text{System B}$, both system efficiencies are only $0.5$ each due to the lack of matching $(T_{\rm E} \ne T_{\rm S})$.
For $\text{System C}$ the system efficiency $\eta_{\rm L}$ has the maximum value $\eta_{\rm L} = 1$ because of $T_{\rm E} = T_{\rm S}$. In contrast, $\eta_{\rm A} = 0.5$ because the RZ pulse does not have the maximum energy that would be allowed due with the peak constraint. The crest factor here has the value $C_{\rm S} = \sqrt{2}$.

$\text{Example 4:}$ We now consider a "root–Nyquist–configuration" with cosine–rolloff total frequency response:

$$H_{\rm S}(f)= \sqrt{H_{\rm CRO}(f)}, \hspace{0.5cm}H_{\rm E}(f)= \sqrt{H_{\rm CRO}(f)} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_{\rm S}(f) \cdot H_{\rm E}(f) = {H_{\rm CRO}(f)}\hspace{0.05cm}.$$

Here, the frequency responses $H_{\rm S}(f)$ and $H_{\rm CRO}(f)$ give the normalized spectral functions of the basic transmission pulse and the basic detection pulse. It holds:

$$G_s(f) = g_0 \cdot T \cdot \sqrt{H_{\rm S}(f)},\hspace{0.5cm} G_d(f) = g_0 \cdot T \cdot {H_{\rm CRO}(f)} = G_{\rm Nyq}(f)\hspace{0.05cm}.$$

The graphic shows the eye diagrams at the transmitter (top) resp. at the receiver (bottom) for the rolloff–factors $r = 0.25$, $r = 0.5$ and $r = 0. 1$. It should be recalled, that such a configuration is optimal under the constraint of power limitation independent of the rolloff–factor $r$. In front of the decision device there is always a fully open eye (see lower row of figures) and it holds for the system efficiency: $\eta_{\rm L} = 1$.

Eye diagrams for root Nyquist configurations

One can see from this plot:

The basic transmission pulse $g_s(t)$ does not satisfy the Nyquist condition: The eye at the transmitter (upper row of images) is not fully open and the maximum value of of the transmitted signal is greater than its rms value.

The crest factor $C_{\rm S} = s_0/s_{\rm eff}$ is always greater than $1$ and thus the efficiency $\eta_{\rm A}<1 $. For $r = 0.5$: $C_{\rm S} \approx 1.45$ ⇒ $\eta_{\rm A} \approx 0.47$. The detection–SNR is then reduced by $10 \cdot \lg \ \eta_{\rm A} \approx 3.2 \ \rm dB$ than in the rectangle–rectangle–configuration.

In the limiting case $r = 0$ even $C_{\rm S} \to \infty$ and $\eta_{\rm A} \to 0$. Here, the basic transmission pulse $g_s(t)$ falls here even more slowly than with $1/t$ and it holds:

$$\max_t\{ s(t) \} = \max_t \hspace{0.15cm}\left [ \sum_{(\nu)} a_\nu \cdot g_s ( t - \nu \cdot T)\ \right ]\rightarrow \infty\hspace{0.05cm}.$$

Limiting the transmitted signal $s(t)$ to a finite maximum value $s_0$ by a weighting factor approaching zero, leads to a closed eye in front of the decision device.

Optimization of the rolloff factor with peak limitation

For this chapter, the following assumptions are made:

Let the basic transmission pulse $g_s(t)$ be NRZ rectangular; with peak limitation this is optimal.
The overall frequency response $H_{\rm S}(f) \cdot H_{\rm E}(f) ={H_{\rm Nyq}(f)}$ satisfy the Nyquist condition.
The Nyquist frequency response is realized by a cosine rolloff low-pass: $H_{\rm Nyq}(f) = H_{\rm CRO}(f)$.
Since the pulse amplitude $g_0$ is independent of the rolloff factor $r$, the SNR maximization can be attributed to the minimization of the noise power at the decision:

$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty}|H_{\rm E}(f)|^2 \,{\rm d} f\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Minimum,} \hspace{0.5cm}{\rm where}\hspace{0.5cm} H_{\rm E}(f) =\frac {H_{\rm CRO}(f)}{{\rm sinc}(f T)}\hspace{0.05cm}.$$

The diagram shows the power transmission function $|H_{\rm E}(f)|^2$ for three different rolloff factors. The areas under each of these curves are a measure of the noise power $\sigma_d^2$ before the decision. One can see from this plot:

For optimization of the rolloff factor with peak limitation

The rolloff factor $r = 0$ (rectangular frequency response) leads only to efficiency $\eta_{\rm A} \approx 0.65$, despite the very narrowband receiver filter, since $H_{\rm E}(f)$ increases with increasing $f$ because of the $\rm sinc$-function in the denominator.
Although $r = 1$ causes a spectrum twice as wide, but it does not lead to any noise enhancement. Since the area under the red curve is smaller than that under the green curve, the result is a better value: $\eta_{\rm A} \approx 0.88$.
The highest system efficiency results for $r \approx 0.8$ (flat maximum) with $\eta_{\rm A} \approx 0.89$. For this one achieves the best possible compromise between bandwidth and boost.
By comparison with the optimal frequency response $H_{\rm E}(f) = {\rm sinc}(f T)$ with peak limitation, which leads to the result $\sigma_d^2 = N_0/(2T)$ ⇒ $\eta_{\rm A}= 1$, we obtain for the system efficiency:

$$\eta_{\rm A} = \left [T \cdot \int_{-\infty}^{+\infty}\hspace{-0.15cm} |H_{\rm E}(f)|^2 \,{\rm d} f \right ]^{-1} \hspace{0.05cm}.$$

$\text{Conclusion:}$

The absolute optimum at peak limitation ⇒ $\eta_{\rm A}= 1$ results only with a rectangular basic transmission pulse $g_s(t)$ and a likewise rectangular receiver filter pulse response $h_{\rm E}(t)$ of the same width $T$.
The best cosine rolloff Nyquist spectrum with $r = 0.8$ (blue curve) is about $0.5 \ \rm dB$ worse compared to the matched filter (violet-dashed curve), because the area under the blue curve is about $12\%$ larger than the area under the violet curve.
The so-called "root–root configuration" ⇒ $H_{\rm S}(f) = H_{\rm E}(f) =\sqrt{H_{\rm CRO}(f)}$ thus only makes sense if one assumes power limitation.

Exercises for the chapter

Exercise 1.6: Root Nyquist System

Exercise 1.6Z: Two Optimal Systems

Exercise 1.7: System Efficiencies

References

↑ Bronshtein, I.N.; Semendyayew, K.A.; Musiol, G.; Mühlig, H.: Handbook of Mathematics. 6. Edition. Heidelberg, New York, Dorderecht, London: Springer Verlag, 2015. ISBN: 978-3-662-46220-1

[1] Bronshtein, I.N.; Semendyayew, K.A.; Musiol, G.; Mühlig, H.: Handbook of Mathematics. 6. Edition. Heidelberg, New York, Dorderecht, London: Springer Verlag, 2015. ISBN: 978-3-662-46220-1

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