Aufgaben:Exercise 4.5: Locality Curve for DSB-AM - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:57:12Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:57, 16 March 2026
Line 18:		Line 18:
	For  $\phi(t)$,  the range  $–\pi < \phi(t) \leq +\pi$  is permissible and the generally valid equation applies:		For  $\phi(t)$,  the range  $–\pi < \phi(t) \leq +\pi$  is permissible and the generally valid equation applies:

	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\big[s_{\~~rmTP~~}(t)\big]}{{\rm Re}\big[s_{\rm TP}(t)\big]}.$$		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\big[s_{\rm TP}(t)\big]}{{\rm Re}\big[s_{\rm TP}(t)\big]}.$$


Line 74:		Line 74:


	'''(2)'''  The above equation can be transformed according to  [[Signal_Representation/~~Calculating_With_Complex_Numbers~~#~~Representation_by_Amplidute_and_Phase~~\|Euler's theorem]]  with  $T_0 = 1/f_{\rm N} = 100 \ {\rm µ} \text{s}$  as follows:		'''(2)'''  The above equation can be transformed according to  [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase\|Euler's theorem]]  with  $T_0 = 1/f_{\rm N} = 100 \ {\rm µ} \text{s}$  as follows:

	:$$\frac{s_{\rm TP}(t)}{{\rm 1 \hspace{0.05cm} V}}\hspace{-0.05cm} =\hspace{-0.05cm}1\hspace{-0.05cm} - \hspace{-0.05cm}{\~~rmj~~}\cdot \cos({ \omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm} \sin({\omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm}{\rm j}\cdot \cos({\omega_{\rm 10}\hspace{0.05cm} t })\hspace{-0.05cm} + \hspace{-0.05cm} \sin({ \omega_{\rm10}\hspace{0.05cm} t }) = 1+2 \cdot \sin(2 \pi{t}/{T_0}) .$$		:$$\frac{s_{\rm TP}(t)}{{\rm 1 \hspace{0.05cm} V}}\hspace{-0.05cm} =\hspace{-0.05cm}1\hspace{-0.05cm} - \hspace{-0.05cm}{\rm j}\cdot \cos({ \omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm} \sin({\omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm}{\rm j}\cdot \cos({\omega_{\rm 10}\hspace{0.05cm} t })\hspace{-0.05cm} + \hspace{-0.05cm} \sin({ \omega_{\rm10}\hspace{0.05cm} t }) = 1+2 \cdot \sin(2 \pi{t}/{T_0}) .$$

	*This shows that  $s_{\rm TP}(t)$  is real for all times  $t$.		*This shows that  $s_{\rm TP}(t)$  is real for all times  $t$.
Line 100:		Line 100:
	'''(4)'''  In general, the phase function is:		'''(4)'''  In general, the phase function is:

	:$$\phi(t)= {\rm arc} \left[s_{\rm TP}(t)\right]= {\rm arctan}\hspace{0.1cm}\frac{{\rm Im}\left[s_{\rm TP}(t)\right]}{{\~~rmRe~~}\left[s_{\rm TP}(t)\right]}$$		:$$\phi(t)= {\rm arc} \left[s_{\rm TP}(t)\right]= {\rm arctan}\hspace{0.1cm}\frac{{\rm Im}\left[s_{\rm TP}(t)\right]}{{\rm Re}\left[s_{\rm TP}(t)\right]}$$

	Due to the fact that here  ${\rm Im}[s_{\rm TP}(t)] = 0$  for all times, one obtains:		Due to the fact that here  ${\rm Im}[s_{\rm TP}(t)] = 0$  for all times, one obtains:
Line 125:		Line 125:
	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]
	[[de:Aufgaben:4.~~5_Ortskurve_bei_ZSB~~-AM]]		[[de:Aufgaben:Aufgabe 4.5: Ortskurve bei ZSB-AM]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:39Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 76:		Line 76:
	'''(2)'''  The above equation can be transformed according to  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase\|Euler's theorem]]  with  $T_0 = 1/f_{\rm N} = 100 \ {\rm µ} \text{s}$  as follows:		'''(2)'''  The above equation can be transformed according to  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase\|Euler's theorem]]  with  $T_0 = 1/f_{\rm N} = 100 \ {\rm µ} \text{s}$  as follows:

	:$$\frac{s_{\rm TP}(t)}{{\rm 1 \hspace{0.05cm} V}}\hspace{-0.05cm} =\hspace{-0.05cm}1\hspace{-0.05cm} - \hspace{-0.05cm}{\rm		:$$\frac{s_{\rm TP}(t)}{{\rm 1 \hspace{0.05cm} V}}\hspace{-0.05cm} =\hspace{-0.05cm}1\hspace{-0.05cm} - \hspace{-0.05cm}{\rmj}\cdot \cos({ \omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm} \sin({\omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm}{\rm j}\cdot \cos({\omega_{\rm 10}\hspace{0.05cm} t })\hspace{-0.05cm} + \hspace{-0.05cm} \sin({ \omega_{\rm10}\hspace{0.05cm} t }) = 1+2 \cdot \sin(2 \pi{t}/{T_0}) .$$
	j}\cdot \cos({ \omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm} \sin({
	\omega_{\rm 10}\hspace{0.05cm} t }) \hspace{-0.05cm}+\hspace{-0.05cm}{\rm j}\cdot \cos({
	\omega_{\rm 10}\hspace{0.05cm} t })\hspace{-0.05cm} + \hspace{-0.05cm} \sin({ \omega_{\rm
	10}\hspace{0.05cm} t }) = 1+2 \cdot \sin(2 \pi
	{t}/{T_0}) .$$

	*This shows that  $s_{\rm TP}(t)$  is real for all times  $t$.		*This shows that  $s_{\rm TP}(t)$  is real for all times  $t$.

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:31:11Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:31, 16 March 2026
Line 18:		Line 18:
	For  $\phi(t)$,  the range  $–\pi < \phi(t) \leq +\pi$  is permissible and the generally valid equation applies:		For  $\phi(t)$,  the range  $–\pi < \phi(t) \leq +\pi$  is permissible and the generally valid equation applies:

	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\big[s_{\rm		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\big[s_{\rmTP}(t)\big]}{{\rm Re}\big[s_{\rm TP}(t)\big]}.$$
	TP}(t)\big]}{{\rm Re}\big[s_{\rm TP}(t)\big]}.$$


Line 67:		Line 66:
	*The equation for  $s_{\rm TP}(t)$  is with  $\omega_{10} = 2 \pi \cdot 10 \ \text{kHz}$:		*The equation for  $s_{\rm TP}(t)$  is with  $\omega_{10} = 2 \pi \cdot 10 \ \text{kHz}$:

	:$$s_{\rm TP}(t) = {\rm 1 \hspace{0.05cm} V} - {\rm j}\cdot {\rm 1		:$$s_{\rm TP}(t) = {\rm 1 \hspace{0.05cm} V} - {\rm j}\cdot {\rm 1\hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\omega_{\rm 10} \hspace{0.05cm} t }+{\rm j}\cdot {\rm 1\hspace{0.05cm} V} \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\omega_{\rm 10} \hspace{0.05cm} t }$$
	\hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm}
	\omega_{\rm 10} \hspace{0.05cm} t }+{\rm j}\cdot {\rm 1
	\hspace{0.05cm} V} \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}
	\omega_{\rm 10} \hspace{0.05cm} t }$$

	:$$\Rightarrow \hspace{0.3cm} s_{\rm TP}(t = 0) = {\rm 1 \hspace{0.05cm} V} - {\rm j}\cdot {\rm 1		:$$\Rightarrow \hspace{0.3cm} s_{\rm TP}(t = 0) = {\rm 1 \hspace{0.05cm} V} - {\rm j}\cdot {\rm 1\hspace{0.05cm} V} +{\rm j}\cdot {\rm 1 \hspace{0.05cm} V}= {\rm 1\hspace{0.05cm} V}.$$
	\hspace{0.05cm} V} +{\rm j}\cdot {\rm 1 \hspace{0.05cm} V}= {\rm 1
	\hspace{0.05cm} V}.$$

	:$$\Rightarrow \hspace{0.3cm} {\rm Re}[s_{\rm TP}(t = 0) ] \hspace{0.15 cm}\underline{= {+\rm 1 \hspace{0.05cm} V}}, \hspace{0.2cm}{\rm Im}[s_{\rm TP}(t = 0) ] \hspace{0.15 cm}\underline{= 0}		:$$\Rightarrow \hspace{0.3cm} {\rm Re}[s_{\rm TP}(t = 0) ] \hspace{0.15 cm}\underline{= {+\rm 1 \hspace{0.05cm} V}}, \hspace{0.2cm}{\rm Im}[s_{\rm TP}(t = 0) ] \hspace{0.15 cm}\underline{= 0}.$$
	.$$


Line 94:		Line 86:
	*We obtain for the numerical values we are looking for:		*We obtain for the numerical values we are looking for:

	:$$s_{\rm TP}(t = {\rm 10 \hspace{0.1cm} {\rm µ} s}) = {\rm 1		:$$s_{\rm TP}(t = {\rm 10 \hspace{0.1cm} {\rm µ} s}) = {\rm 1\hspace{0.05cm} V} \cdot \left[1+2 \cdot\sin(36^\circ)\right]\hspace{0.15 cm}\underline{={{\rm +2.176 \hspace{0.05cm} V}}},$$
	\hspace{0.05cm} V} \cdot \left[1+2 \cdot
	\sin(36^\circ)\right]\hspace{0.15 cm}\underline{={{\rm +2.176 \hspace{0.05cm} V}}},$$

	:$$s_{\rm TP}(t = {\rm 25 \hspace{0.1cm} {\rm µ} s}) = {\rm 1		:$$s_{\rm TP}(t = {\rm 25 \hspace{0.1cm} {\rm µ} s}) = {\rm 1\hspace{0.05cm} V} \cdot \left[1+2 \cdot\sin(90^\circ)\right]\hspace{0.15 cm}\underline{={{\rm +3 \hspace{0.05cm} V}}},$$
	\hspace{0.05cm} V} \cdot \left[1+2 \cdot
	\sin(90^\circ)\right]\hspace{0.15 cm}\underline{={{\rm +3 \hspace{0.05cm} V}}},$$

	:$$s_{\rm TP}(t = {\rm 75 \hspace{0.1cm} {\rm µ} s}) = {\rm 1		:$$s_{\rm TP}(t = {\rm 75 \hspace{0.1cm} {\rm µ} s}) = {\rm 1\hspace{0.05cm} V} \cdot \left[1+2 \cdot \sin(270^\circ)\right]\hspace{0.15 cm}\underline{=-{{\rm 1 \hspace{0.05cm} V}}},$$
	\hspace{0.05cm} V} \cdot \left[1+2 \cdot \sin(270^\circ)\right]\hspace{0.15 cm}\underline{=
	-{{\rm 1 \hspace{0.05cm} V}}},$$

	:$$s_{\rm TP}(t = {\rm 100 \hspace{0.1cm}{\rm µ} s}) = s_{\rm TP}(t =		:$$s_{\rm TP}(t = {\rm 100 \hspace{0.1cm}{\rm µ} s}) = s_{\rm TP}(t =0) \hspace{0.15 cm}\underline{={{\rm +1 \hspace{0.05cm} V}}}.$$
	0) \hspace{0.15 cm}\underline{={{\rm +1 \hspace{0.05cm} V}}}.$$



	'''(3)'''  By definition,  $a(t) = \|s_{\rm TP}(t)\|$. This gives the following numerical values:		'''(3)'''  By definition,  $a(t) = \|s_{\rm TP}(t)\|$. This gives the following numerical values:
	:$$a(t = {\rm 25 \hspace{0.1cm} {\rm µ} s}) = s_{\rm TP}(t = {\rm 25		:$$a(t = {\rm 25 \hspace{0.1cm} {\rm µ} s}) = s_{\rm TP}(t = {\rm 25\hspace{0.05cm}{\rm µ} s}) \hspace{0.15 cm}\underline{= {\rm +3 \hspace{0.05cm} V}} ,\hspace{4.15 cm}$$
	\hspace{0.05cm}{\rm µ} s}) \hspace{0.15 cm}\underline{= {\rm +3 \hspace{0.05cm} V}} ,
	\hspace{4.15 cm}$$

	:$$a(t = {\rm 75 \hspace{0.1cm} {\rm µ} s}) = \|s_{\rm TP}(t = {\rm 75		:$$a(t = {\rm 75 \hspace{0.1cm} {\rm µ} s}) = \|s_{\rm TP}(t = {\rm 75\hspace{0.05cm} {\rm µ} s})\| \hspace{0.15 cm}\underline{= {\rm +1 \hspace{0.05cm} V}} .$$
	\hspace{0.05cm} {\rm µ} s})\| \hspace{0.15 cm}\underline{= {\rm +1 \hspace{0.05cm} V}} .$$


Line 123:		Line 105:
	'''(4)'''  In general, the phase function is:		'''(4)'''  In general, the phase function is:

	:$$\phi(t)= {\rm arc} \left[s_{\rm TP}(t)\right]= {\rm arctan}		:$$\phi(t)= {\rm arc} \left[s_{\rm TP}(t)\right]= {\rm arctan}\hspace{0.1cm}\frac{{\rm Im}\left[s_{\rm TP}(t)\right]}{{\rmRe}\left[s_{\rm TP}(t)\right]}$$
	\hspace{0.1cm}\frac{{\rm Im}\left[s_{\rm TP}(t)\right]}{{\rm
	Re}\left[s_{\rm TP}(t)\right]}$$

	Due to the fact that here  ${\rm Im}[s_{\rm TP}(t)] = 0$  for all times, one obtains:		Due to the fact that here  ${\rm Im}[s_{\rm TP}(t)] = 0$  for all times, one obtains:
Line 137:		Line 117:
	*To calculate  $t_1$ , the result of subtask  '''(2)'''  can be used:		*To calculate  $t_1$ , the result of subtask  '''(2)'''  can be used:

	:$$\sin(2 \pi \cdot {t_1}/{T_0}) = -0.5 \hspace{0.3cm} \Rightarrow		:$$\sin(2 \pi \cdot {t_1}/{T_0}) = -0.5 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} 2 \pi \cdot {t_1}/{T_0} = 2 \pi \cdot{7}/{12}\hspace{0.3cm}{\text{(corresponds to}}\hspace{0.2cm}210^\circ)$$
	\hspace{0.3cm} 2 \pi \cdot {t_1}/{T_0} = 2 \pi \cdot
	{7}/{12}\hspace{0.3cm}{\text{(corresponds to}}\hspace{0.2cm}210^\circ
	)$$

	*From this one obtains  $t_1 = 7/12 · T_0 = 58.33 \ {\rm µ} \text{s}$.		*From this one obtains  $t_1 = 7/12 · T_0 = 58.33 \ {\rm µ} \text{s}$.

Maintenance script: Add German interlanguage link

2026-03-16T12:33:42Z

Add German interlanguage link

← Older revision		Revision as of 14:33, 16 March 2026
Line 153:		Line 153:
	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]
			[[de:Aufgaben:4.5_Ortskurve_bei_ZSB-AM]]

Hwang at 13:22, 18 January 2023

2023-01-18T13:22:34Z

← Older revision		Revision as of 15:22, 18 January 2023
Line 64:		Line 64:

	[[File:EN_Sig_A_4_5_a.png\|250px\|right\|frame\|Locality curve at time  $t = 0$]]		[[File:EN_Sig_A_4_5_a.png\|250px\|right\|frame\|Locality curve at time  $t = 0$]]
	'''(1)'''  If all Dirac lines are shifted to the left by  $f_{\rm T} = 50 \ \text{kHz}$ , they are located at  $-\hspace{-0.08cm}10 \ \text{kHz}$,  $0$  and  $+10 \ \text{kHz}$.		'''(1)'''  If all Dirac delta lines are shifted to the left by  $f_{\rm T} = 50 \ \text{kHz}$ , they are located at  $-\hspace{-0.08cm}10 \ \text{kHz}$,  $0$  and  $+10 \ \text{kHz}$.
	*The equation for  $s_{\rm TP}(t)$  is with  $\omega_{10} = 2 \pi \cdot 10 \ \text{kHz}$:		*The equation for  $s_{\rm TP}(t)$  is with  $\omega_{10} = 2 \pi \cdot 10 \ \text{kHz}$:

Guenter at 14:44, 11 May 2021

2021-05-11T14:44:09Z

← Older revision		Revision as of 16:44, 11 May 2021
Line 63:		Line 63:
	{{ML-Kopf}}		{{ML-Kopf}}

	[[File:EN_Sig_A_4_5_a.png\|250px\|right\|frame\|~~Locus~~ curve at time  $t = 0$]]		[[File:EN_Sig_A_4_5_a.png\|250px\|right\|frame\|Locality curve at time  $t = 0$]]
	'''(1)'''  If all ~~diraclines~~ are shifted to the left by  $f_{\rm C} = 50 \ \text{kHz}$ , they are located at  $-\hspace{-0.08cm}10 \ \text{kHz}$,  $0$  and  $+10 \ \text{kHz}$.		'''(1)'''  If all Dirac lines are shifted to the left by  $f_{\rm T} = 50 \ \text{kHz}$ , they are located at  $-\hspace{-0.08cm}10 \ \text{kHz}$,  $0$  and  $+10 \ \text{kHz}$.
	*The equation for  $s_{\rm TP}(t)$  is with  $\omega_{10} = 2 \pi \cdot 10 \ \text{kHz}$:		*The equation for  $s_{\rm TP}(t)$  is with  $\omega_{10} = 2 \pi \cdot 10 \ \text{kHz}$:

Line 91:		Line 91:
	{t}/{T_0}) .$$		{t}/{T_0}) .$$

	*This shows that  $s_{\rm TP}(t)$  is real for all times  $t$~~&nbsp~~ .		*This shows that  $s_{\rm TP}(t)$  is real for all times  $t$.
	*~~For~~ the numerical values we are looking for~~, we obtain~~:		*We obtain for the numerical values we are looking for:

	:$$s_{\rm TP}(t = {\rm 10 \hspace{0.1cm} {\rm µ} s}) = {\rm 1		:$$s_{\rm TP}(t = {\rm 10 \hspace{0.1cm} {\rm µ} s}) = {\rm 1
Line 127:		Line 127:
	Re}\left[s_{\rm TP}(t)\right]}$$		Re}\left[s_{\rm TP}(t)\right]}$$

	Due to the fact that here  ${\rm Im}[s_{\rm TP}(t)] = 0$  for all times, one obtains ~~the result from this~~:		Due to the fact that here  ${\rm Im}[s_{\rm TP}(t)] = 0$  for all times, one obtains:
	* If  ${\rm Re}[s_{\rm TP}(t)] > 0$  holds, the phase  $\phi(t) = 0$.		* If  ${\rm Re}[s_{\rm TP}(t)] > 0$  holds, the phase  $\phi(t) = 0$.
	* On the other hand, if the real part is negative:     $\phi(t) = \pi$.		* On the other hand, if the real part is negative:     $\phi(t) = \pi$.
Line 139:		Line 139:
	:$$\sin(2 \pi \cdot {t_1}/{T_0}) = -0.5 \hspace{0.3cm} \Rightarrow		:$$\sin(2 \pi \cdot {t_1}/{T_0}) = -0.5 \hspace{0.3cm} \Rightarrow
	\hspace{0.3cm} 2 \pi \cdot {t_1}/{T_0} = 2 \pi \cdot		\hspace{0.3cm} 2 \pi \cdot {t_1}/{T_0} = 2 \pi \cdot
	{7}/{12}\hspace{0.3cm}{\rm (corresponds to}\hspace{0.2cm}210^\circ		{7}/{12}\hspace{0.3cm}{\text{(corresponds to}}\hspace{0.2cm}210^\circ
	)$$		)$$

Guenter at 14:36, 11 May 2021

2021-05-11T14:36:53Z

← Older revision		Revision as of 16:36, 11 May 2021
Line 26:		Line 26:

	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/~~Equivalent_Low_Pass_Signal_and_its_Spectral_Function~~\|Equivalent Low-Pass Signal and its Spectral Function]].		*This exercise belongs to the chapter  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function\|Equivalent Low-Pass Signal and its Spectral Function]].

	*You can check your solution with the interactive applet  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal\|Physical Signal & Equivalent Low-Pass Signal]]    ⇒   "Locality Curve".		*You can check your solution with the interactive applet  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal\|Physical Signal & Equivalent Low-Pass Signal]]    ⇒   "Locality Curve".

Guenter at 14:35, 11 May 2021

2021-05-11T14:35:25Z

← Older revision		Revision as of 16:35, 11 May 2021
Line 5:		Line 5:
	[[File:P_ID751__Sig_A_4_5_neu.png\|250px\|right\|frame\|Spectrum of the analytical signal]]		[[File:P_ID751__Sig_A_4_5_neu.png\|250px\|right\|frame\|Spectrum of the analytical signal]]

	We consider a similar transmission scenario as in  [[Aufgaben:~~Aufgabe_4~~.4:~~_Zeigerdiagramm_bei_ZSB~~-AM\|Exrcise 4.4]]  (but not the same):		We consider a similar transmission scenario as in  [[Aufgaben:Exercise_4.4:_Pointer_Diagram_for_DSB-AM\|Exrcise 4.4]]  (but not the same):
	* A sinusoidal source signal with amplitude  $A_{\rm N} = 2 \ \text{V}$  and frequency  $f_{\rm N} = 10 \ \text{kHz}$,		* A sinusoidal source signal with amplitude  $A_{\rm N} = 2 \ \text{V}$  and frequency  $f_{\rm N} = 10 \ \text{kHz}$,
	*Double-Sideband Amplitude Modulation without carrier suppression with carrier frequency  $f_{\rm T} = 50 \ \text{kHz}$.		*Double-Sideband Amplitude Modulation without carrier suppression with carrier frequency  $f_{\rm T} = 50 \ \text{kHz}$.

Guenter at 14:34, 11 May 2021

2021-05-11T14:34:46Z

Guenter at 14:17, 11 May 2021

2021-05-11T14:17:47Z

← Older revision		Revision as of 16:17, 11 May 2021
Line 1:		Line 1:

	{{quiz-Header\|Buchseite=Signal_Representation/~~Equivalent_Low_Pass_Signal_and_its_Spectral_Function~~		{{quiz-Header\|Buchseite=Signal_Representation/Equivalent Low-Pass Signal and its Spectral Function
	}}		}}

← Older revision		Revision as of 16:34, 11 May 2021
Line 5:		Line 5:
	[[File:P_ID751__Sig_A_4_5_neu.png\|250px\|right\|frame\|Spectrum of the analytical signal]]		[[File:P_ID751__Sig_A_4_5_neu.png\|250px\|right\|frame\|Spectrum of the analytical signal]]

	We consider a similar transmission scenario as in  [[Aufgaben:Aufgabe_4.4:_Zeigerdiagramm_bei_ZSB-AM\|~~task~~ 4.4]]  (but not the same):		We consider a similar transmission scenario as in  [[Aufgaben:Aufgabe_4.4:_Zeigerdiagramm_bei_ZSB-AM\|Exrcise 4.4]]  (but not the same):
	* a sinusoidal ~~message~~ signal with amplitude  $A_{\rm N} = 2 \ \text{V}$  and ~~the~~ frequency  $f_{\rm N} = 10 \ \text{kHz}$,		* A sinusoidal source signal with amplitude  $A_{\rm N} = 2 \ \text{V}$  and frequency  $f_{\rm N} = 10 \ \text{kHz}$,
	*~~DSB~~-Amplitude Modulation without carrier suppression with carrier frequency  $f_{\rm T} = 50 \ \text{kHz}$.		*Double-Sideband Amplitude Modulation without carrier suppression with carrier frequency  $f_{\rm T} = 50 \ \text{kHz}$.


	Opposite you see the spectral function  $S_+(f)$  of the analytical signal  $s_+(t)$.		Opposite you see the spectral function  $S_+(f)$  of the analytical signal  $s_+(t)$.

	When solving, take into account that the equivalent low-pass signal is ~~also~~ in the form		When solving, take into account that the equivalent low-pass signal is in the form

	:$$s_{\rm TP}(t) = a(t) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \phi(t)} $$		:$$s_{\rm TP}(t) = a(t) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \phi(t)},\hspace{0.5cm} a(t) ≥ 0.$$

	~~where  $a(t) ≥ 0$  shall hold.~~ For  $\phi(t)$ , the range ~~of values~~  $–\pi < \phi(t) \leq +\pi$  is permissible and the generally valid equation applies:		For  $\phi(t)$,  the range  $–\pi < \phi(t) \leq +\pi$  is permissible and the generally valid equation applies:

	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\big[s_{\rm		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\big[s_{\rm
	TP}(t)\big]}{{\rm Re}\big[s_{\rm TP}(t)\big]}.$$		TP}(t)\big]}{{\rm Re}\big[s_{\rm TP}(t)\big]}.$$





Line 29:		Line 26:

	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/~~Equivalent_Low_Pass_Signal_and_Its_Spectral_Function~~\|Equivalent Low-Pass Signal and ~~Its~~ Spectral Function]].		*This exercise belongs to the chapter  [[Signal_Representation/Equivalent_Low_Pass_Signal_and_its_Spectral_Function\|Equivalent Low-Pass Signal and its Spectral Function]].

	*You can check your solution with the interactive applet  [[Applets:~~Physikalisches_Signal_~~%~~26_Äquivalentes_TP-Signal~~\|Physical ~~signal~~ & ~~equivalent LP signal~~]]    ⇒   ~~&bdquo;locus curve”~~.		*You can check your solution with the interactive applet  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal\|Physical Signal & Equivalent Low-Pass Signal]]    ⇒   "Locality Curve".


Line 37:		Line 34:

	<quiz display=simple>		<quiz display=simple>
	{Calculate the equivalent low-pass signal  $s_{\rm TP}(t)$  in the frequency and time domain. What is the value of  $s_{\rm TP}(t)$  at the start time  $t = 0$?		{Calculate the equivalent low-pass signal  $s_{\rm TP}(t)$  in the frequency and time domain.  What is the value of  $s_{\rm TP}(t)$  at the start time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$\text{Re}[s_{\text{TP}}(t=0)]\ = \ $ { 1 3% }  $\text{V}$		$\text{Re}[s_{\text{TP}}(t=0)]\ = \ $ { 1 3% }  $\text{V}$
	$\text{Im}[s_{\text{TP}}(t=0 )]\ = \ $ { 0. }  $\text{V}$		$\text{Im}[s_{\text{TP}}(t=0 )]\ = \ $ { 0. }  $\text{V}$

	{What are the values of  $s_{\rm TP}(t)$  at  $t = 10 \ {\rm µ} \text{s}= T_0/10$,     $t = 25 \ {\rm µ} \text{s}= T_0/4$,     $t = 75 \ {\rm µ} \text{s}= 3T_0/4$  and  $T_0 = 100 \ {\rm µ}s$? <br>Show that all values are purely real.		{What are the values of  $s_{\rm TP}(t)$  at  $t = 10 \ {\rm µ} \text{s}= T_0/10$,     $t = 25 \ {\rm µ} \text{s}= T_0/4$,     $t = 75 \ {\rm µ} \text{s}= 3T_0/4$  and  $T_0 = 100 \ {\rm µs}$? <br>Show that all values are purely real.
	\|type="{}"}		\|type="{}"}
	$\text{Re}[s_{\text{TP}}(t=10 \ {\rm µ} \text{s})]\ = \ $ { 2.176 3% }  $\text{V}$		$\text{Re}[s_{\text{TP}}(t=10 \ {\rm µ} \text{s})]\ = \ $ { 2.176 3% }  $\text{V}$
Line 49:		Line 46:
	$\text{Re}[s_{\text{TP}}(t=100 \ {\rm µ} \text{s})]\ = \ $ { 1 3% }  $\text{V}$		$\text{Re}[s_{\text{TP}}(t=100 \ {\rm µ} \text{s})]\ = \ $ { 1 3% }  $\text{V}$

	{What is the magnitude function  $a(t)$  ~~im Zeitbereich?~~ in the time domain? What are the values at times  $t = 25 \ {\rm µ} \text{s}$  and  $t = 75 \ {\rm µ} \text{s}$?		{What is the magnitude function  $a(t)$  in the time domain?  What are the values at times  $t = 25 \ {\rm µ} \text{s}$  and  $t = 75 \ {\rm µ} \text{s}$?
	\|type="{}"}		\|type="{}"}
	$a(t=25 \ {\rm µ} \text{s})\ = \ $ { 3 3% }  $\text{V}$		$a(t=25 \ {\rm µ} \text{s})\ = \ $ { 3 3% }  $\text{V}$
	$a(t=75 \ {\rm µ} \text{s})\ = \ $ { 1 3% }  $\text{V}$		$a(t=75 \ {\rm µ} \text{s})\ = \ $ { 1 3% }  $\text{V}$

	{Give the phase function  $\phi(t)$  in the time domain ~~in general~~. What values result at the times  $t = 25 \ {\rm µ} \text{s}$  and  $t = 75 \ {\rm µ} \text{s}$?		{Give the phase function  $\phi(t)$  in the time domain.  What values result at the times  $t = 25 \ {\rm µ} \text{s}$  and  $t = 75 \ {\rm µ} \text{s}$?
	\|type="{}"}		\|type="{}"}
	$\phi(t=25 \ {\rm µ} \text{s}) \ = \ $ { 0. }  $\text{Grad}$		$\phi(t=25 \ {\rm µ} \text{s}) \ = \ $ { 0. }  $\text{Grad}$