Aufgaben:Exercise 1.5: Cosine-Square Spectrum

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Cosine-square Nyquist spectrum

The spectrum  $G(f)$  with  $\cos^{2}$–shaped course is considered according to the sketch. This satisfies the first Nyquist criterion:

$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) = {\rm const.}$$

Accordingly, the associated pulse  $g(t)$  has zero crossings at multiples of  $T$, where  $T$  remains to be determined. Fourier inverse transformation of  $G(f)$  yields the equation for the time course:

$$g( t )= g_0 \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot

t/T)^2}\cdot {\rm si}(\pi \cdot {t}/{T})\hspace{0.05cm}.$$ The questions for this exercise refer to the following properties:

• The spectral function  $G(f)$  is a special case of the cosine rolloff spectrum, which is point symmetric about the Nyquist frequency  $f_{\rm Nyq}$. 
• The cosine rolloff spectrum is completely characterized by the corner frequencies  $f_{1}$  and  $f_{2}$. 
• For  $| f | < f_{1}$,   $G(f) = g_{0} \cdot T = \rm const.$, while the spectrum for  $| f | > f_{2}$  has no components.
• The relation between the Nyquist frequency and the corner frequencies is:

$$f_{\rm Nyq}= \frac{f_1 +f_2 }

{2 }\hspace{0.05cm}.$$

• The transition steepness is characterized by the so-called rolloff factor:

$$r = \frac{f_2 -f_1 }

{f_2 +f_1 }\hspace{0.2cm}(0 \le r \le 1) \hspace{0.05cm}.$$

Note:

• The exercise belongs to the chapter  Properties of Nyquist Systems.

Questions

1 What are the corner frequencies of this cosine rolloff spectrum?

$f_{1} \ = \ $ $\ \rm MHz$

$f_{2} \ = \ $ $\ \rm MHz$

2 What are the Nyquist frequency and the rolloff factor?

$f_{\rm Nyq} \ = \ $ $\ \rm MHz$

$r \ = \ $

3 At what time interval  $T$  does  $g(t)$  have zero crossings?

$T \ = \ $ $\ \rm µ s$

4 Which of the following statements is true?

	$g(t)$  satisfies the first Nyquist criterion because of the  $\rm si$ term.
	$g(t)$  has further zero crossings at  $\pm 0.5T,  \pm 1.5T,  \pm 2.5 T, \text{...}$
	The  $\cos^{2}$ spectrum also satisfies the second Nyquist criterion.

5 What is the (normalized) value of the pulse at time  $t = T/2$?

$g(t = T/2)/g_{0} \ = \ $

Solution

Solution

(1)  The upper corner frequency can be read from the diagram:   $f_{2} \underline{= 2 \ \rm MHz}$. Since the spectrum is not constant in any range, $f_{1} \underline {= 0}$.

(2)  From the given equations we obtain:

$$f_{\rm Nyq} = \ \frac{f_1 +f_2 }

{2 }\hspace{0.1cm}\underline { = 1\,{\rm MHz}}\hspace{0.05cm},\hspace{0.5cm} r = \ \frac{f_2 -f_1 } {f_2 +f_1 }\hspace{0.1cm}\underline { = 1 }\hspace{0.05cm}.$$

(3)  The spacing of equidistant zero crossings is directly related to the Nyquist frequency:

$$f_{\rm Nyq}= \frac{1}{2T} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T= \frac{1}{2f_{\rm Nyq}}\hspace{0.1cm}\underline { = 0.5\,{\rm µ s}}\hspace{0.05cm}.$$

(4)  Statements 1 and 3 are correct:

• The first statement is correct:   The function $si(π · t/T)$ leads to zero crossings at $\nu T (\nu \neq 0)$.
• The last statement is also true:  Because of $g(t) = 0$ for $t =\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$ the second Nyquist criterion is also fulfilled.
• On the other hand, the middle statement is false, since $g(t = T/2) \neq 0$.

• The condition for the second Nyquist criterion is in the frequency domain:

$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} \frac {G \left ( f -

\frac{k}{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}= {\rm const.}$$

• The condition is indeed fulfilled for the cos$^{2}$ spectrum, as can be shown after a longer calculation. We restrict ourselves here to the frequency range $| f · T | \leq 1$ and set $g_{0} \cdot T = 1$ for simplicity:

$$G_{\rm Per}(f) = \frac {\cos^2 \left [\pi/2 \cdot ( f_{\rm Nyq}

- f) \cdot T \right ]}{\cos \left [\pi \cdot ( f_{\rm Nyq} - f) \cdot T \right ]}+\frac {\cos^2 \left [\pi/2 \cdot ( f_{\rm Nyq} + f) \cdot T \right ]}{\cos \left [\pi \cdot ( f_{\rm Nyq} + f) \cdot T \right ]}\hspace{0.05cm}.$$

• Further holds:

$$\frac {\cos^2 (x)}{\cos(2x)} = {1}/{2} \cdot \frac

{1+\cos(2x)}{\cos(2x)}= {1}/{2} \cdot \left [1+ \frac {1}{\cos(2x)}\right ]$$

$$\Rightarrow \hspace{0.3cm} G_{\rm Per}(f) = {1}/{2}

\cdot \left [1+ \frac {1}{\cos \left [\pi \cdot ( f_{\rm Nyq} - f) \cdot T \right ]} +1- \frac {1}{\cos \left [\pi \cdot ( f_{\rm Nyq} + f) \cdot T \right ]}\right ]\hspace{0.05cm}.$$

• Because of $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos

\left ( {\pi}/{2} \pm \pi f T \right) = \sin \left ( \pm \pi f T \right)\text{:}$

$$\Rightarrow \hspace{0.3cm} G_{\rm Per}(f) = 2 - \frac {1}{\sin

(\pi f T)} + \frac {1}{\sin (\pi f T)} = 2 = {\rm const}\hspace{0.05cm}.$$

(5)  For $t = T/2$, the given equation yields an indeterminate value (0 divided by 0), which can be determined using l'Hospital's rule.

• To do this, form the derivatives of the numerator and denominator and insert the desired time $t = T/2$ into the result:

$$\frac{g( t = T/2)}{g_0} = \ {{\rm si}(\pi \cdot \frac{t}{T})

\cdot \frac{{\rm d}/{\rm d}t \left [ \cos(\pi \cdot t/T)\right]}{{\rm d}/{\rm d}t\left [ 1 - (2 \cdot t/T)^2\right]}} \bigg |_{t = T/2} = \ {{\rm si}(\pi \cdot \frac{t}{T}) \cdot \frac{- \pi/T \cdot

\sin(\pi \cdot t/T)}{-2 \cdot (2\cdot t/T) \cdot (2/T)}} \bigg |_{t = T/2} = \frac {2}{\pi}\cdot

\frac {\pi}{4}\hspace{0.1cm}\underline { = 0.5}\hspace{0.05cm}.$$

• A second solution method leads to the expression:

$$\frac{g( t )}{g_0} = {\rm si}(\pi \cdot \frac{t}{T}) \cdot

\frac {\pi}{4} \cdot \big [ {\rm si}(\pi \cdot (t/T + 1/2)) + {\rm si}(\pi \cdot (t/T - 1/2))\big] \hspace{0.05cm}.$$

• The second bracket expression can be transformed as follows:

$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}

\bigg ] = \ \frac {\pi}{4} \cdot \left [ \frac {{\rm sin}(\pi \cdot t/T + \pi/2)}{\pi \cdot t/T + \pi/2} + \frac {{\rm sin}(\pi \cdot t/T - \pi/2)}{\pi \cdot t/T - \pi/2}\right] = \ \frac {1}{2} \cdot {\rm cos}(\pi \cdot t/T )\cdot \left [ \frac {1}{2 \cdot t/T + 1} - \frac {1}{ 2 \cdot t/T - 1}\right] $$

$$\Rightarrow \hspace{0.3cm} \frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}

\bigg ] = \ \frac {1}{2} \cdot {\rm cos}(\pi \cdot t/T )\cdot \frac{1- 2 \cdot t/T + 1+ 2 \cdot t/T}{(1+ 2 \cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot t/T)^2}\hspace{0.05cm}.$$

• It follows that both expressions are actually equal. Thus, for time $t = T/2$, the following is still true:

$$\frac{g( t = T/2)}{g_0} = {\rm si}( \frac{\pi}{2}) \cdot \frac

{\pi}{4} \cdot \left [ {\rm si}(\pi ) + {\rm si}(0)\right]= \frac {2}{\pi}\cdot \frac {\pi}{4} = 0.5 \hspace{0.05cm}.$$