Aufgaben:Exercise 4.12: Root-Nyquist Systems

• Return to book

Spectra of fundamental transmission pulse and base detection pulse

In quadrature amplitude modulation systems, the root-Nyquist variant is often chosen instead of a rectangular basic transmission pulse, which gets its name from the spectral range. The reason for this is the significantly smaller bandwidth.

• In this case, base detection pulse  $g_d(t)$  satisfies the  first Nyquist criterion,  since  $G_d(f)$  is point-symmetric about the so-called Nyquist frequency  $f_{\rm Nyq} = 1/T$ .
• The spectral function $G_d(f)$  is a  cosine rolloff spectrum, where the rolloff factor  $r$  can take values from $0$  to  $1$  (including these limits).

Furthermore, the following holds for the Nyquist frequency response:

• When  $|f| < f_1 = f_{\rm Nyq} · (1 – r)$ ,  $G_d(f)$  is constant and equal to  $g_0 · T$.
• At frequencies greater tha  $f_2 = f_{\rm Nyq} · (1 + r)$ , $G_d(f)$  has no components.
• In between, the slope is cosine.

The optimization of digital communication systems requires that the receiver frequency response  $H_{\rm E}(f)$  should be of the same shape as the transmission spectrum $G_s(f)$ .

To obtain dimensionally correct spectral functions for this task and the graph, it is assumed that:

$$G_s(f) = \sqrt{g_0 \cdot T \cdot G_d(f)},\hspace{0.4cm} H_{\rm E}(f) = \frac{1}{g_0 \cdot T}\cdot G_s(f)\hspace{0.05cm}.$$

The top graph shows the transmission spectrum  $G_s(f)$  for the rolloff factors.

• $r = 0$   (green dotted rectangle),
• $r = 0.5$   (blue solid curve),
• $r = 1$   (red dashed curve).

Below, the spectrum   $G_d(f)$  before the decider is shown in the same colors.

• The associated impulse   $g_d(t)$  is a Nyquist impulse pulse for all valid rolloff factors   $(0 ≤ r ≤ 1)$  as opposed to the fundamental transmission pulse   $g_s(t)$.
• For this, the following equation is given in the literature - for example in  [Kam04] :

$$g_s(t) = g_0 \cdot \frac{4 r t/T \cdot \cos \left [\pi \cdot (1+r) \cdot t/T \right ]+ \sin \left [\pi \cdot (1-r) \cdot t/T \right ]}{\left [1- (4 r t/T)^2 \right ] \cdot \pi \cdot t/T}\hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  Quadrature Amplitude Modulation.
• Particular reference is made to the page   Nyquist and root Nyquist systems  in this chapter.
• Further useful informationen can be found in the chapter  Properties of Nyquist Systems  in the book "Digital Signal Transmission".
• [Kam04]  refers to the textbook "Kammeyer, K.D.: Nachrichtenübertragung. Stuttgart: B.G. Teubner, 4. Auflage, 2004".
• Energies are to be specified in $\rm V^2s$  ; they thus refer to the reference resistance  $R = 1 \ \rm \Omega$.

Questions

1

What is the fundamental transmission pulse  $g_s(t)$  for the rolloff factor  $r = 0$? ? What is the signal value at time  $t = 0$?

$g_s(t = 0) \ = \ $ $\ \cdot g_0$

2

What is the fundamental transmission pulse  $g_s(t)$  for the rolloff factor $r = 1$?  What is the signal value at time  $t = 0$?

$g_s(t = 0) \ = \ $ $\ \cdot g_0$

3

Let  $r = 1$.  At what times does  $g_s(t)$  cross the axis?

	At all multiples of the symbol duration  $T$.
	At  $t = ±0.25 T, \ ±0.75 T, \ ±1.25 T, \ ±1.75 T$, ...
	At  $t = ±0.75 T, \ ±1.25 T,\ ±1.75 T$, ...

4

What is the fundamental transmission pulse  $g_s(t)$  for the rolloff factor  $r = 0.5$?  Welcher Signalwert ergibt sich zum Zeitpunkt  $t = 0$?

$g_s(t = 0) \ = \ $ $\ \cdot g_0$

5

Which statements are valid for the signal amplitude, independent of  $r$ ?  Solve using the frequency domain.

	The transmit pulse amplitude can take any value in the range   $0 ≤ g_s(t = 0) ≤ g_0$  .
	The transmit pulse amplitude can take any value in the range   $g_0 ≤ g_s(t = 0) ≤ 2 g_0$  .
	The transmit pulse amplitude can take any value in the range   $g_0 ≤ g_s(t = 0) ≤ 4 g_0/π$  .

6

What is the energy $E_{g_s}$  of the fundamental transmission pulse  $g_s(t)$  when  $r = 0$  and  $r = 1$?

$r = 0\text{:} \ \ \ \ E_{g_s} \ = \ $ $\ \cdot g_0^2 \cdot T$

$r = 1\text{:} \ \ \ \ E_{g_s} \ = \ $ $\ \cdot g_0^2 \cdot T$

Solution

Solution

(1)  If we substitute   $r = 0$  into the given equation, the first terms in the numerator and denominator disappear and we get:

$$g_s(t) = g_0 \cdot \frac{\sin \left (\pi \cdot t/T \right )}{\pi \cdot t/T} = g_0 \cdot {\rm si} \left (\pi \cdot {t}/{T} \right )\hspace{0.05cm}.$$

• At time  $t = 0$ ,the   $\rm si$–impulse is equal to  $g_0$:  

$$ g_s(t) \hspace{0.15cm}\underline { = 1.0 } \cdot g_0 \hspace{0.05cm}.$$

(2)  When  $r = 1$ , the given equation simplies as follows:

$$g_s(t) = \frac{4 \cdot g_0}{\pi} \cdot \frac{ \cos \left (2 \pi \cdot t/T \right )}{\left [1- (4 t/T)^2 \right ] }\hspace{0.3cm}\Rightarrow \hspace{0.3cm} g_s(t = 0) = \frac{4 \cdot g_0}{\pi} \hspace{0.15cm}\underline {= 1.273 }\cdot g_0 \hspace{0.05cm}.$$

(3)  The last answer is correct:

• Zero intercepts are only possible for  $r = 1$  if the cosine function in the numerator is zero, that is for all integer values of   $k$:

$$2 \pi \cdot t/T = {\pi}/{2} + k \cdot \pi \hspace{0.3cm}\Rightarrow \hspace{0.3cm} t = \pm 0.25T, \hspace{0.15cm} \pm 0.75T, \hspace{0.15cm}\pm 1.25T, \hspace{0.15cm} ...$$

• However, only the last answer is correct, since the zero values at   $±0.25T$  are cancelled by the zero in the denominator.
• Applying de l'Hospital's rule yields   $g_s(t = ± 0.25T) = g_0$.

(4)  With  $r = 0.5$  and the shortcut  $x = t/T$ , one gets:

$$g_s(x) = \frac{g_0}{\pi} \cdot \frac{2 \cdot x \cdot \cos \left (1.5\pi \cdot x \right )+ \sin \left (0.5\pi \cdot x \right )}{\left (1- 4 \cdot x^2 \right ) \cdot x}\hspace{0.05cm}.$$

• For the calculation at time   $t = 0$ , de l'Hospital's rule must be applied.
• The derivatives of the numerator and denominator give:

$$Z'(x) = 2 \cdot \cos \left (1.5\pi \cdot x \right ) - 3 \pi \cdot x \cdot \sin \left (1.5\pi \cdot x \right ) + 0.5 \pi \cdot \cos \left (0.5\pi \cdot x \right ),$$

Fundamental transmission pulse (root Nyquist) and base detection pulse (Nyquist)

$$N'(x) = \left (1- 4 \cdot x^2 \right ) - 8 \cdot x^2 \hspace{0.05cm}.$$

• The two boundary transitions for  $x → 0$  yield:

$$\lim_{x \rightarrow 0} Z'(x) = 2 +{\pi }/{2},\hspace{0.2cm} \lim_{x \rightarrow 0} N'(x) = 1 \hspace{0.05cm}.$$

• Thus, for the signal amplitude at time   $t = 0$:

$$g_s(t=0) = \frac{g_0}{\pi} \cdot \left ( 2 +{\pi }/{2} \right ) = {g_0} \cdot \left ( 0.5 + {2}/{\pi } \right )\hspace{0.15cm}\underline {= 1.137} \cdot g_0 \hspace{0.05cm}.$$

Here, the graph illustrates the results calculated again:

• The impulse  $g_d(t)$  is a Nyquist pulse, meaning that it has zero crossings at least at all multiples of the symbol duration $T$ (and possibly others depending on the rolloff factor).
• On the other hand, the impulse  $g_s(t)$  does not satisfy the Nyquist criterion.
• Moreover, from this plot one can once again see that for   $r ≠ 0$  the pulse amplitude $g_s(t = 0)$  is always larger than $g_0$ .

(5)  The last answer  is correct $($the first answer is ruled out from the results in questions  (2)  and  (4) $)$.  The validity of the lower bound   $g_0$  and the upper bound   $4g_0/π$  can be proved as follows:

• The pulse amplitude  $g_s(t = 0)$  is generally equal to the area under the spectral function  $G_s(f)$.
• The smallest area is obtained for  $r = 0$.  Here,   $G_s(f) = g_0 · T$  is in the range  $|f| < ±1/(2T)$.  Thus, the area is equal to  $g_0$.
• The largest area is obtained for  $r = 1$. Here   $G_s(f)$  extends to the range   $±1/T$  and has a cosine shape.
• The result  $g_s(t = 0) = 4g_0/π$  was already calculated in question   (3) .  Though it still holds that:

$$g_s(t=0) = 2 \cdot {g_0} \cdot \int_{ 0 }^{1/T} {\cos\left(\frac{\pi }{2}\cdot f \cdot T \right)}\hspace{0.1cm} {\rm d}f = \frac{4 g_0}{\pi} \cdot \int_{ 0 }^{\pi/2} {\cos\left(x \right)}\hspace{0.1cm} {\rm d}x = {4 g_0}/{\pi} \cdot \big[\sin(\pi/2) - \sin(0) \big] = {4 g_0}/{\pi}\hspace{0.05cm}.$$

(6)  The energy of the fundamental transmission pulse   $g_s(t)$  can be found in the time or frequency domain according to Parseval's theorem:

$$E_{g_s} = \int_{ -\infty }^{+\infty} {[g_s(t)]^2}\hspace{0.1cm} {\rm d}t = \int_{ -\infty }^{+\infty} {|G_s(f)|^2}\hspace{0.1cm} {\rm d}f \hspace{0.05cm}.$$

• From the equations and the graph on the exercise page, we can see that   $|G_s(f)|^2$  is the same shape as   $G_d(f)$ , except that the height is now   $(g_0 · T)^2$  instead of   $g_0 · T$ :

$$E_{g_s} = \int_{ -\infty }^{+\infty} {|G_s(f)|^2}\hspace{0.1cm} {\rm d}f = \frac{g_0^2 \cdot T^2}{g_0 \cdot T} \cdot \int_{ -\infty }^{+\infty} {G_d(f)}\hspace{0.1cm} {\rm d}f \hspace{0.05cm}.$$

• Due to the Nyquist form of  $G_d(f)$ , it holds independently of   $r$ that:

$$\int_{ -\infty }^{+\infty} {G_d(f)}\hspace{0.1cm} {\rm d}f = g_0 \hspace{0.05cm}.$$

• Thus, the impulse energy is also independent of   $r$, so it is also valid for   $r = 0$  and  $r = 1$. . In both cases, $E_ {g_s}\hspace{0.15cm}\underline { = 1.0} · g_0^2 · T.$