Aufgaben:Exercise 1.5: Cosine-Square Spectrum - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]		[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]
	[[de:Aufgaben:1.~~5_Cosinus~~-Quadrat-Spektrum]]		[[de:Aufgaben:Aufgabe 1.5: Cosinus-Quadrat-Spektrum]]

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

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

	*A second solution method leads to the expression:		*A second solution method leads to the expression:
	:$$\frac{g( t )}{g_0} = {\rm sinc}( \frac{t}{T}) \cdot\frac {\pi}{4} \cdot \big [ {\rm sinc}( t/T + 1/2) +{\rm sinc}(t/T - 1/2)\big] \hspace{0.05cm}.$$		:$$\frac{g( t )}{g_0} = {\rm sinc}( \frac{t}{T}) \cdot\frac {\pi}{4} \cdot \big [ {\rm sinc}( t/T + 1/2) +{\rm sinc}(t/T - 1/2)\big] \hspace{0.05cm}.$$
	*The second bracket expression can be transformed as follows:		*The second bracket expression can be transformed as follows:
	:$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}		:$$\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] $$
	\bigg ] = \ \frac {\pi}{4} \cdot \left [ \frac {{\rm sin}(\pi
	\cdot t/T + \pi/2)}{\pi \cdot t/T + \pi/2} + \frac {{\rm sin}(\pi
	\cdot t/T - \pi/2)}{\pi \cdot t/T - \pi/2}\right] = \ \frac {1}{2} \cdot {\rm cos}(\pi
	\cdot t/T )\cdot \left [ \frac {1}{2 \cdot t/T + 1} - \frac {1}{ 2
	\cdot t/T - 1}\right] $$
	:$$\Rightarrow \hspace{0.3cm} \frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}\bigg ] = \ \frac {1}{2} \cdot {\rm cos}(\pi\cdot t/T )\cdot \frac{1- 2 \cdot t/T + 1+ 2 \cdot t/T}{(1+ 2\cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \cdot t/T)}{1 - (2\cdot t/T)^2}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} \frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}\bigg ] = \ \frac {1}{2} \cdot {\rm cos}(\pi\cdot t/T )\cdot \frac{1- 2 \cdot t/T + 1+ 2 \cdot t/T}{(1+ 2\cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \cdot t/T)}{1 - (2\cdot t/T)^2}\hspace{0.05cm}.$$
	*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:		*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:

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	:$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) = {\rm const.}$$		:$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) = {\rm const.}$$
	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:		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:
	:$$g( t )= g_0 \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \~~cdot~~		:$$g( t )= g_0 \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdott/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.5cm} \text{with}\hspace{0.5cm} {\rm sinc}(x)=\sin(\pi x)/(\pi x)\hspace{0.05cm}.$$
	t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.5cm} \text{with}\hspace{0.5cm} {\rm sinc}(x)=\sin(\pi x)/(\pi x)\hspace{0.05cm}.$$
	The questions for this exercise refer to the following properties:		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 spectral function  $G(f)$  is a special case of the cosine rolloff spectrum,  which is point symmetric about the Nyquist frequency  $f_{\rm Nyq}$.
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	* For  $\| f \| < f_{1}$,   $G(f) = g_{0} \cdot T = \rm const.$,  while the spectrum for  $\| f \| > f_{2}$  has no components.		* 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:		*The relation between the Nyquist frequency and the corner frequencies is:
	:$$f_{\rm Nyq}= \frac{f_1 +f_2 }		:$$f_{\rm Nyq}= \frac{f_1 +f_2 }{2 }\hspace{0.05cm}.$$
	{2 }\hspace{0.05cm}.$$
	*The edge steepness is characterized by the so-called rolloff factor:		*The edge steepness is characterized by the so-called rolloff factor:
	:$$r = \frac{f_2 -f_1 }		:$$r = \frac{f_2 -f_1 }{f_2 +f_1 }\hspace{0.2cm}(0 \le r \le 1) \hspace{0.05cm}.$$
	{f_2 +f_1 }\hspace{0.2cm}(0 \le r \le 1) \hspace{0.05cm}.$$


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	'''(2)'''  From the given equations we obtain:		'''(2)'''  From the given equations we obtain:
	:$$f_{\rm Nyq} = \ \frac{f_1 +f_2 }		:$$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}.$$
	{2 }\hspace{0.1cm}\underline { = 1\,{\rm MHz}}\hspace{0.05cm},\hspace{0.5cm} r = \ \frac{f_2 -f_1 } {f_2 +f_1 }\hspace{0.1cm}\underline { = 1 }\hspace{0.05cm}.$$


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	*On the other hand,  the middle statement is false,  since $g(t = T/2) \neq 0$.		*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:		*The condition for the second Nyquist criterion is in the frequency domain:
	:$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} \frac {G \left ( f -		:$$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.}$$
	{k}/{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}=
	{\rm const.}$$
	*The condition is indeed fulfilled for the  cos$^{2}$–spectrum,  as can be shown after a longer calculation.		*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:		*We restrict ourselves here to the frequency range  $\| f · T \| \leq 1$  and set  $g_{0} \cdot T = 1$  for simplicity:
	:$$G_{\rm Per}(f) = \frac {\cos^2 \left [\pi/2 \cdot ( f_{\rm Nyq}		:$$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}.$$
	- f) \cdot T \right ]}{\cos \left [\pi \cdot ( f_{\rm Nyq} - f)
	\cdot T \right ]}+\frac {\cos^2 \left [\pi/2 \cdot ( f_{\rm Nyq}
	+ f) \cdot T \right ]}{\cos \left [\pi \cdot ( f_{\rm Nyq} + f)
	\cdot T \right ]}\hspace{0.05cm}.$$
	*Further holds:		*Further holds:
	:$$\frac {\cos^2 (x)}{\cos(2x)} = {1}/{2} \cdot \frac		:$$\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 ]$$
	{1+\cos(2x)}{\cos(2x)}= {1}/{2} \cdot \left [1+ \frac		:$$\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}.$$
	{1}{\cos(2x)}\right ]$$
	:$$\Rightarrow \hspace{0.3cm} G_{\rm Per}(f) = {1}/{2}
	\cdot \left [1+ \frac {1}{\cos \left [\pi \cdot ( f_{\rm Nyq}
	- f) \cdot T \right ]} +1- \frac {1}{\cos \left [\pi \cdot ( f_{\rm Nyq}
	+ f) \cdot T \right ]}\right ]\hspace{0.05cm}.$$
	* Because of  $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos		* Because of  $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos
	\left ( {\pi}/{2} \pm \pi f T \right) = \sin \left ( \pm		\left ( {\pi}/{2} \pm \pi f T \right) = \sin \left ( \pm
	\pi f T \right)\text{:}$		\pi f T \right)\text{:}$
	:$$\Rightarrow \hspace{0.3cm} G_{\rm Per}(f) = 2 - \frac {1}{\sin		:$$\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}.$$
	(\pi f T)} + \frac {1}{\sin (\pi f T)} = 2 = {\rm const}\hspace{0.05cm}.$$


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	*A second solution method leads to the expression:		*A second solution method leads to the expression:
	:$$\frac{g( t )}{g_0} = {\rm sinc}( \frac{t}{T}) \cdot		:$$\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}.$$
	\frac {\pi}{4} \cdot \big [ {\rm sinc}( t/T + 1/2) +
	{\rm sinc}(t/T - 1/2)\big] \hspace{0.05cm}.$$
	*The second bracket expression can be transformed as follows:		*The second bracket expression can be transformed as follows:
	:$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}		:$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}
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	\cdot t/T )\cdot \left [ \frac {1}{2 \cdot t/T + 1} - \frac {1}{ 2		\cdot t/T )\cdot \left [ \frac {1}{2 \cdot t/T + 1} - \frac {1}{ 2
	\cdot t/T - 1}\right] $$		\cdot t/T - 1}\right] $$
	:$$\Rightarrow \hspace{0.3cm} \frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}		:$$\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}.$$
	\bigg ] = \ \frac {1}{2} \cdot {\rm cos}(\pi
	\cdot t/T )\cdot \frac{1- 2 \cdot t/T + 1+ 2 \cdot t/T}{(1+ 2
	\cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \cdot t/T)}{1 - (2
	\cdot t/T)^2}\hspace{0.05cm}.$$
	*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:		*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:
	:$$\frac{g( t = T/2)}{g_0} = {\rm sinc}( 0.5) \cdot \frac		:$$\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}.$$
	{\pi}{4} \cdot \big [ {\rm sinc}(1 ) + {\rm sinc}(0)\big]= \frac
	{2}{\pi}\cdot \frac {\pi}{4} = 0.5 \hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]		[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]
			[[de:Aufgaben:1.5_Cosinus-Quadrat-Spektrum]]

Guenter at 12:24, 2 May 2022

2022-05-02T12:24:05Z

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	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:		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:
	:$$g( t )= g_0 \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot		:$$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}.$$		t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.5cm} \text{with}\hspace{0.5cm} {\rm sinc}(x)=\sin(\pi x)/(\pi x)\hspace{0.05cm}.$$
	The questions for this exercise refer to the following properties:		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 spectral function  $G(f)$  is a special case of the cosine rolloff spectrum,  which is point symmetric about the Nyquist frequency  $f_{\rm Nyq}$.

Guenter at 10:55, 2 May 2022

2022-05-02T10:55:10Z

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	*A second solution method leads to the expression:		*A second solution method leads to the expression:
	:$$\frac{g( t )}{g_0} = {\rm si}(~~\pi \cdot~~ \frac{t}{T}) \cdot		:$$\frac{g( t )}{g_0} = {\rm sinc}( \frac{t}{T}) \cdot
	\frac {\pi}{4} \cdot \big [ {\rm si}~~(\pi \cdot~~ (t/T + 1/2)) +		\frac {\pi}{4} \cdot \big [ {\rm sinc}( t/T + 1/2) +
	{\rm si}~~(\pi \cdot~~ (t/T - 1/2))\big] \hspace{0.05cm}.$$		{\rm sinc}(t/T - 1/2)\big] \hspace{0.05cm}.$$
	*The second bracket expression can be transformed as follows:		*The second bracket expression can be transformed as follows:
	:$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}		:$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}
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	\cdot t/T)^2}\hspace{0.05cm}.$$		\cdot t/T)^2}\hspace{0.05cm}.$$
	*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:		*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:
	:$$\frac{g( t = T/2)}{g_0} = {\rm si}( ~~\frac{\pi}{2}~~) \cdot \frac		:$$\frac{g( t = T/2)}{g_0} = {\rm sinc}( 0.5) \cdot \frac
	{\pi}{4} \cdot \~~left~~ [ {\rm si}(~~\pi~~ ) + {\rm si}(0)\~~right~~]= \frac		{\pi}{4} \cdot \big [ {\rm sinc}(1 ) + {\rm sinc}(0)\big]= \frac
	{2}{\pi}\cdot \frac {\pi}{4} = 0.5 \hspace{0.05cm}.$$		{2}{\pi}\cdot \frac {\pi}{4} = 0.5 \hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 10:51, 2 May 2022

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Guenter at 10:39, 2 May 2022

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	[[File:P_ID1282__Dig_A_1_5.png\|right\|frame\|Cosine-square Nyquist spectrum]]		[[File:P_ID1282__Dig_A_1_5.png\|right\|frame\|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:		The spectrum  $G(f)$  with  $\cos^{2}$–shaped course is considered according to the sketch.  This satisfies the first Nyquist criterion:
	:$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) = {\rm const.}$$		:$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) = {\rm const.}$$
	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:		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:
	:$$g( t )= g_0 \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot		:$$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}.$$		t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.05cm}.$$
	The questions for this exercise refer to the following properties:		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 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  $f_{2}$.		*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.		* 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:		*The relation between the Nyquist frequency and the corner frequencies is:
	:$$f_{\rm Nyq}= \frac{f_1 +f_2 }		:$$f_{\rm Nyq}= \frac{f_1 +f_2 }
	{2 }\hspace{0.05cm}.$$		{2 }\hspace{0.05cm}.$$
	*The ~~transition~~ steepness is characterized by the so-called rolloff factor:		*The edge steepness is characterized by the so-called rolloff factor:
	:$$r = \frac{f_2 -f_1 }		:$$r = \frac{f_2 -f_1 }
	{f_2 +f_1 }\hspace{0.2cm}(0 \le r \le 1) \hspace{0.05cm}.$$		{f_2 +f_1 }\hspace{0.2cm}(0 \le r \le 1) \hspace{0.05cm}.$$
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			Note:  The exercise belongs to the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems\|"Properties of Nyquist Systems"]].

	''Note:''
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems\|Properties of Nyquist Systems]].


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	{Which of the following statements is true?		{Which of the following statements is true?
	\|type="[]"}		\|type="[]"}
	+ $g(t)$  satisfies the first Nyquist criterion because of the  $\rm si$ term.		+ $g(t)$  satisfies the first Nyquist criterion because of the  $\rm sinc$–term.
	- $g(t)$  has further zero crossings at  $\pm 0.5T,  \pm 1.5T,  \pm 2.5 T, \text{...}$		- $g(t)$  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.		+ The  $\cos^{2}$ spectrum also satisfies the second Nyquist criterion.


	{What is the (normalized) value of the pulse at time  $t = T/2$?		{What is the  (normalized)  value of the pulse at time  $t = T/2$?
	\|type="{}"}		\|type="{}"}
	$g(t = T/2)/g_{0} \ = \ $ { 0.5 3% }		$g(t = T/2)/g_{0} \ = \ $ { 0.5 3% }

Hwang at 13:34, 8 March 2022

2022-03-08T13:34:56Z

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Guenter at 15:30, 2 February 2022

2022-02-02T15:30:31Z

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	[[Category:Digital Signal Transmission: Exercises\|^1.3 ~~Eigenschaften von Nyquistsystemen~~^]]		[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]

← Older revision		Revision as of 12:51, 2 May 2022
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	+ $g(t)$  satisfies the first Nyquist criterion because of the  $\rm sinc$–term.		+ $g(t)$  satisfies the first Nyquist criterion because of the  $\rm sinc$–term.
	- $g(t)$  has further zero crossings at  $\pm 0.5T,  \pm 1.5T,  \pm 2.5 T, \text{...}$		- $g(t)$  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.		+ The  $\cos^{2}$–spectrum also satisfies the second Nyquist criterion.


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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(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, $f_{1} \underline {= 0}$.		'''(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, $f_{1} \underline {= 0}$.


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	'''(4)'''  <u>Statements 1 and 3</u> are correct:		'''(4)'''  <u>Statements 1 and 3</u>  are correct:
	*The first statement is correct:   The function $si(~~π ·~~ t/T)$ leads to zero crossings at $\nu T (\nu \neq 0)$.		*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.		*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 $g(t = T/2) \neq 0$.		*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:		*The condition for the second Nyquist criterion is in the frequency domain:
	:$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} \frac {G \left ( f -		:$$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)}=		{k}/{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}=
	{\rm const.}$$		{\rm const.}$$
	*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:		*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:
	:$$G_{\rm Per}(f) = \frac {\cos^2 \left [\pi/2 \cdot ( f_{\rm Nyq}		:$$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)		- f) \cdot T \right ]}{\cos \left [\pi \cdot ( f_{\rm Nyq} - f)
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	- f) \cdot T \right ]} +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}.$$		+ f) \cdot T \right ]}\right ]\hspace{0.05cm}.$$
	* Because of $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos		* Because of  $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos
	\left ( {\pi}/{2} \pm \pi f T \right) = \sin \left ( \pm		\left ( {\pi}/{2} \pm \pi f T \right) = \sin \left ( \pm
	\pi f T \right)\text{:}$		\pi f T \right)\text{:}$
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	'''(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.		'''(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:		*To do this,  form the derivatives of the numerator and denominator and insert the desired time  $t = T/2$  into the result:

	:$$\frac{g( t = T/2)}{g_0} = \ {{\rm si}(~~\pi \cdot~~ \frac{t}{T})		:$$\frac{g( t = T/2)}{g_0} = \ {{\rm sinc}( \frac{t}{T})
	\cdot \frac{{\rm d}/{\rm d}t \left [ \cos(\pi \cdot		\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]}}		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		\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		\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}.$$		\frac {\pi}{4}\hspace{0.1cm}\underline { = 0.5}\hspace{0.05cm}.$$
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	\cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \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}.$$		\cdot t/T)^2}\hspace{0.05cm}.$$
	*It follows that both expressions are actually equal. Thus, for time $t = T/2$, the following is still true:		*It follows that both expressions are actually equal.  Thus,  for time  $t = T/2$,  the following is still true:
	:$$\frac{g( t = T/2)}{g_0} = {\rm si}( \frac{\pi}{2}) \cdot \frac		:$$\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		{\pi}{4} \cdot \left [ {\rm si}(\pi ) + {\rm si}(0)\right]= \frac