Exercise 1.6: Root Nyquist System

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Cosine spectrum (transmitter & receiver)

The diagram on the right shows

  • the spectrum  $G_{s}(f)$  of the basic transmission pulse,
  • the frequency response  $H_{\rm E}(f)$  of the receiver filter

of a binary and bipolar transmission system, which are identical in shape to each other:

$$G_s(f) = \left\{ \begin{array}{c} A \cdot \cos \left( \frac {\pi \cdot f}{2 \cdot f_2} \right) \\ \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ \\ \\ \end{array} \begin{array}{*{20}c}|f| \le f_2 \hspace{0.05cm}, \\ \\ {\rm else }\hspace{0.05cm}, \\ \end{array}$$
$$H_{\rm E }(f) = \left\{ \begin{array}{c} 1 \cdot \cos \left( \frac {\pi \cdot f}{2 \cdot f_2} \right) \\ \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\\ \\ \\ \end{array} \begin{array}{*{20}c}|f| \le f_2 \hspace{0.05cm}, \\ \\ {\rm else }\hspace{0.05cm}. \\ \end{array}$$

In the whole exercise  $A = 10^{–6} \ \rm V/Hz$  and  $f_{2} = 1 \ \rm MHz$ are valid.

  • Assuming that the bit rate  $R = 1/T$  is chosen correctly,  the basic transmitter pulse  $g_{d}(t) = g_{s}(t) ∗ h_{\rm E}(t)$  satisfies the first Nyquist criterion.
  • For the associated spectral function  $G_{d}(f)$,  the rolloff thereby occurs cosinusoidally similar to a cosine rolloff spectrum.
  • The rolloff factor  $r$  is to be determined in this exercise.


  • The crest factor is the quotient of the maximum and the rms value of the transmitted signal and thus a measure of the intersymbol interfering at the transmitting end:
$$C_{\rm S} = \frac{s_0}{\sqrt{E_{\rm B}/T}} = \frac{{\rm Max}[s(t)]}{\sqrt{{\rm E}[s^2(t)]}}= {s_0}/{s_{\rm eff}}.$$



Calculate the Nyquist spectrum  $G_{d}(f)$.  What is the Nyquist frequency and the rolloff factor?

$f_{\rm Nyq} \ = \ $

$\ \rm MHz$
$r \ = \ $


What is the bit rate of the Nyquist system at hand?

$R \ = \ $

$\ \rm Mbit/s$


Why is it an optimal system under the constraint  "power limitation"?

The overall system satisfies the Nyquist condition.
The crest factor is  $C_{\rm S} = 1$.
The receiver filter  $H_{\rm E}(f)$  is matched to the basic transmission pulse  $G_{s}(f)$. 


What is the bit error probability if the power density of the AWGN noise is  $N_{0} = 8 \cdot 10^{–8}\ \rm V^{2}/Hz$  $($referenced to  $1 Ω)$? 

$p_{\rm B} \ = \ $

$\ \cdot 10^{-6}$


(1)  With the functions  $G_{s}(f)$  and  $H_{\rm E}(f)$,  the spectrum of the basic detection pulse for  $|f| \leq f_{2}$:

$$G_d(f) = G_s(f) \cdot H_{\rm E}(f) = A \cdot \cos^2 \left( \frac {\pi \cdot f}{2 \cdot f_2} \right).$$
  • According to the general definition of the cosine rolloff spectrum,  the corner frequencies are  $f_{1} = 0$  and  $f_{2} = 1\ \rm MHz$.
  • From this follows for the Nyquist frequency  (symmetry point with respect to the rolloff):
$$f_{\rm Nyq} = \frac{f_1 +f_2 } {2 } \hspace{0.1cm}\underline { = 0.5\,{\rm MHz}}\hspace{0.05cm}.$$
  • The rolloff factor is
$$r = \frac{f_2 -f_1 } {f_2 +f_1 } \hspace{0.1cm}\underline {= 1} \hspace{0.05cm}.$$
  • This means:   $G_{d}(f)$ describes a $\cos^{2}$ spectrum.

(2)  The relationship between Nyquist frequency and symbol duration  $T$  is  $f_{\rm Nyq} = 1/(2T)$.

  • From this it follows for the bit rate  $R = 1/T = 2 \cdot f_{\rm Nyq}\ \underline{= 1 \ \rm Mbit/s}$.
  • Note the different units for frequency and bit rate.

(3)  The  first and third solutions  are correct:

  • This is an optimal binary system under the constraint of power limitation.
  • The crest factor is not important under power limitation.  With the conditions given here,  $C_{\rm S} > 1$ would apply.

(4)  The bit error probability of an optimal system can be calculated as follows:

$$p_{\rm B} = {\rm Q} \left( \sqrt{{2 \cdot E_{\rm B}}/{N_0}}\right)\hspace{0.05cm}.$$
  • In the given example,  we obtain for the average energy per bit:
$$E_{\rm B} = \ \int_{-\infty}^{+\infty}|G_s(f)|^2 \,{\rm d} f = A^2 \cdot \int_{-1/T}^{+1/T} H_{\rm Nyq}(f) \,{\rm d} f = \ \frac {A^2}{T} = \frac {(10^{-6}\,{\rm V/Hz})^2}{10^{-6}\,{\rm s}} = 10^{-6}\,{\rm V^2s}\hspace{0.05cm}.$$
  • With  $N_{0} = 8 \cdot 10^{–8} \ \rm V^{2}/Hz$,  this further gives:
$$p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot 10^{-6}\,{\rm V^2s}}{8 \cdot 10^{-8}\,{\rm V^2/Hz}}}\right)= {\rm Q} \left( \sqrt{25}\right)= {\rm Q} (5) \hspace{0.1cm}\underline {= 0.287 \cdot 10^{-6}}\hspace{0.05cm}.$$