Exercise 1.5: Cosine-Square Spectrum

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. The inverse Fourier transform of $G(f)$ yields the equation for the time course:

$$g( t )= g_0 \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.05cm}.$$

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 edge 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

$f_{1} \ = \ $

$\ \rm MHz$

$f_{2} \ = \ $

$\ \rm MHz$

$f_{\rm Nyq} \ = \ $

$\ \rm MHz$

$r \ = \ $

$T \ = \ $

$\ \rm µ s$

	$g(t)$ satisfies the first Nyquist criterion because of the $\rm sinc$–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.

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

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 ${\rm sinc}(t/T)$ 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:

$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} \frac {G \left ( f - {k}/{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}= {\rm const.}$$

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:

$$\frac{g( t = T/2)}{g_0} = \ {{\rm sinc}( \frac{t}{T}) \cdot \frac{{\rm d}/{\rm d}t \left [ \cos(\pi \cdot t/T)\right]}{{\rm d}/{\rm d}t\left [ 1 - (2 \cdot t/T)^2\right]}} \bigg |_{t = T/2} = \ {{\rm sinc}( \frac{t}{T}) \cdot \frac{- \pi/T \cdot \sin(\pi \cdot t/T)}{-2 \cdot (2\cdot t/T) \cdot (2/T)}} \bigg |_{t = T/2} = \frac {2}{\pi}\cdot \frac {\pi}{4}\hspace{0.1cm}\underline { = 0.5}\hspace{0.05cm}.$$

A second solution method leads to the expression:

$$\frac{g( t )}{g_0} = {\rm sinc}( \frac{t}{T}) \cdot \frac {\pi}{4} \cdot \big [ {\rm sinc}( t/T + 1/2) + {\rm sinc}(t/T - 1/2)\big] \hspace{0.05cm}.$$

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:

$$\frac{g( t = T/2)}{g_0} = {\rm sinc}( 0.5) \cdot \frac {\pi}{4} \cdot \big [ {\rm sinc}(1 ) + {\rm sinc}(0)\big]= \frac {2}{\pi}\cdot \frac {\pi}{4} = 0.5 \hspace{0.05cm}.$$