Aufgaben:Exercise 1.10: BPSK Baseband Model - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]
	[[de:Aufgaben:1.~~10_BPSK–Basisbandmodell~~]]		[[de:Aufgaben:Aufgabe 1.10: BPSK–Basisbandmodell]]

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	'''(3)'''  $H_{\rm MKD}(f)$  is additively composed of a rectangle and a triangle,  each with width  $\Delta f_{\rm K}$  and height  $0.5$. It follows:		'''(3)'''  $H_{\rm MKD}(f)$  is additively composed of a rectangle and a triangle,  each with width  $\Delta f_{\rm K}$  and height  $0.5$. It follows:
	:$$h_{\rm MKD}(t) = \frac{\Delta f_{\rm K}}{2} \cdot {\rm sinc} ( \Delta f_{\rm K} \cdot t)+ \frac{\Delta f_{\rm K}}{4} \cdot {\rm sinc}^2 ( \frac{\Delta f_{\rm K}}{2} \cdot t)$$		:$$h_{\rm MKD}(t) = \frac{\Delta f_{\rm K}}{2} \cdot {\rm sinc} ( \Delta f_{\rm K} \cdot t)+ \frac{\Delta f_{\rm K}}{4} \cdot {\rm sinc}^2 ( \frac{\Delta f_{\rm K}}{2} \cdot t)$$
	:$$ \Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0) = \frac{\Delta f_{\rm K}}{2} + \frac{\Delta f_{\rm K}}{4} = 0.75 \cdot \Delta f_{\rm K}\hspace{0.3cm}		:$$ \Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0) = \frac{\Delta f_{\rm K}}{2} + \frac{\Delta f_{\rm K}}{4} = 0.75 \cdot \Delta f_{\rm K}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0)/{\Delta f_{\rm K}} \hspace{0.1cm}\underline {= 0.75} .$$
	\Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0)/{\Delta f_{\rm K}} \hspace{0.1cm}\underline {= 0.75} .$$


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	*The first sinc–function does have equidistant zero crossings at the distance  $1/\Delta f_{\rm K}$.		*The first sinc–function does have equidistant zero crossings at the distance  $1/\Delta f_{\rm K}$.
	*But the equidistant zero crossings of the whole time function  $h_{\rm MKD}(t)$  are determined by the second term:		*But the equidistant zero crossings of the whole time function  $h_{\rm MKD}(t)$  are determined by the second term:
	:$$h_{\rm MKD}(t = \frac{1}{\Delta f_{\rm K}}) = \ \frac{\Delta f_{\rm K}}{2} \cdot {\rm sinc} (1 )+		:$$h_{\rm MKD}(t = \frac{1}{\Delta f_{\rm K}}) = \ \frac{\Delta f_{\rm K}}{2} \cdot {\rm sinc} (1 )+\frac{\Delta f_{\rm K}}{4} \cdot {\rm sinc}^2 (0.5) = \frac{\Deltaf_{\rm K}}{4},$$
	\frac{\Delta f_{\rm K}}{4} \cdot {\rm sinc}^2 (0.5) = \frac{\~~Delta~~		:$$h_{\rm MKD}(t = \frac{2}{\Delta f_{\rm K}}) = \ \frac{\Deltaf_{\rm K}}{2} \cdot {\rm sinc} (2 )+ \frac{\Delta f_{\rm K}}{4}\cdot {\rm sinc}^2 (1) = 0.$$
	f_{\rm K}}{4},$$
	:$$h_{\rm MKD}(t = \frac{2}{\Delta f_{\rm K}}) = \ \frac{\~~Delta~~
	f_{\rm K}}{2} \cdot {\rm sinc} (2 )+ \frac{\Delta f_{\rm K}}{4}
	\cdot {\rm sinc}^2 (1) = 0.$$

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	[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]
			[[de:Aufgaben:1.10_BPSK–Basisbandmodell]]

Hwang at 15:15, 10 May 2022

2022-05-10T15:15:46Z

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	:$$s(t) \ = \ z(t) \cdot q(t),$$		:$$s(t) \ = \ z(t) \cdot q(t),$$
	:$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$		:$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$
	The designations chosen here are based on the  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Common_block_diagram_for_ASK_and_BPSK\|block diagram]]  in the theory section.		The designations chosen here are based on the  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Common_block_diagram_for_ASK_and_BPSK\|"block diagram"]]  in the theory section.

	The influence of a channel frequency response  $H_{\rm K}(f)$  can be taken into account in a simple way if it is described together with modulator and demodulator by a common baseband frequency response:		The influence of a channel frequency response  $H_{\rm K}(f)$  can be taken into account in a simple way if it is described together with modulator and demodulator by a common baseband frequency response:
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	*Thus the modulator and demodulator are virtually shortened against each other,  and		*Thus the modulator and demodulator are virtually shortened against each other,  and

	*the ~~bandpass~~ channel  $H_{\rm K}(f)$  is transformed into the low-pass range.		*the band-pass channel  $H_{\rm K}(f)$  is transformed into the low-pass range.

Guenter at 12:57, 7 May 2022

2022-05-07T12:57:34Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Statements 2, 3 and 4</u> are correct:		'''(1)'''  <u>Statements 2, 3 and 4</u>  are correct:
	*$H_{\rm K,TP}(f)$ results from $H_{\rm K}(f)$ by cutting off the negative frequency components and shifting $f_{\rm T}$ to the left.		*$H_{\rm K,TP}(f)$  results from  $H_{\rm K}(f)$  by cutting off the negative frequency components and shifting  $f_{\rm T}$  to the left.
	* For frequency responses – in contrast to spectra – the doubling of the components at positive frequencies is omitted. Therefore:
			* For frequency responses  – in contrast to spectra –  the doubling of the components at positive frequencies is omitted.  Therefore:
	:$$H_{\rm K,\hspace{0.04cm} TP}(f= 0) = H_{\rm K}(f= f_{\rm T})=1.$$		:$$H_{\rm K,\hspace{0.04cm} TP}(f= 0) = H_{\rm K}(f= f_{\rm T})=1.$$
	*Because of the real and asymmetrical spectral functions $H_{\rm K,\hspace{0.04cm}TP}(f)$ the corresponding time function (Fourier ~~inverse~~ transform) $h_{\rm K,\hspace{0.04cm}TP}(t)$ is complex according to the ~~allocation theorem~~.		*Because of the real and asymmetrical spectral functions  $H_{\rm K,\hspace{0.04cm}TP}(f),$  the corresponding time function  (inverse Fourier transform)  $h_{\rm K,\hspace{0.04cm}TP}(t)$  is complex according to the  "Allocation Theorem".


			[[File:P_ID1684__Dig_A_4_3_a.png\|right\|frame\|Low-pass functions for  $H_{\rm K}(f)$]]

	~~[[File~~:~~P_ID1684__Dig_A_4_3_a.png\|center\|frame\|Lowpass functions for~~ $H_{\rm K}(f)$]]		'''(2)'''  Here only the  <u>third proposed solution</u>  is correct:
			*The spectral function  $H_{\rm MKD}(f)$  always has an even real part and no imaginary part.  Consequently  $h_{\rm MKD}(t)$  is always real.

	~~'''(2)'''~~  ~~Here only the <u>third proposed solution</u> is correct:~~		*If  $H_{\rm K}(f)$  had additionally an imaginary part odd by $f= f_{\rm T}$,  $H_{\rm MKD}(f)$  would have an imaginary part odd by $f = 0$.  Thus  $h_{\rm MKD}(t)$  would still be a real function.
	*The spectral function $H_{\rm MKD}(f)$ always has an even real part and no imaginary part. Consequently $h_{\rm MKD}(t)$ is always real.
	*If $H_{\rm K}(f)$ had additionally an imaginary part odd by $f_{\rm T}$, $H_{\rm MKD}(f)$ would have an imaginary part odd by $f = 0$. Thus $h_{\rm MKD}(t)$ would still be a real function.


	The diagram illustrates the differences between $H_{\rm K,\hspace{0.04cm}TP}(f)$ and $H_{\rm MKD}(f)$. The parts of $H_{\rm MKD}(f)$ in the range around $\pm 2f_{\rm T}$ need not be considered further.		The diagram illustrates the differences between  $H_{\rm K,\hspace{0.04cm}TP}(f)$  and  $H_{\rm MKD}(f)$.  The parts of  $H_{\rm MKD}(f)$  in the range around  $\pm 2f_{\rm T}$  need not be considered further.


	'''(3)'''  $H_{\rm MKD}(f)$ is additively composed of a rectangle and a triangle, each with width $\Delta f_{\rm K}$ and height $0.5$. It follows:		'''(3)'''  $H_{\rm MKD}(f)$  is additively composed of a rectangle and a triangle,  each with width  $\Delta f_{\rm K}$  and height  $0.5$. It follows:
	:$$h_{\rm MKD}(t) = \frac{\Delta f_{\rm K}}{2} \cdot {\rm si} (~~\pi \cdot~~ \Delta f_{\rm K} \cdot t)+ \frac{\Delta f_{\rm K}}{4} \cdot {\rm si}^2 (~~\pi \cdot~~ \frac{\Delta f_{\rm K}}{2} \cdot t)$$		:$$h_{\rm MKD}(t) = \frac{\Delta f_{\rm K}}{2} \cdot {\rm sinc} ( \Delta f_{\rm K} \cdot t)+ \frac{\Delta f_{\rm K}}{4} \cdot {\rm sinc}^2 ( \frac{\Delta f_{\rm K}}{2} \cdot t)$$
	:$$ \Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0) = \frac{\Delta f_{\rm K}}{2} + \frac{\Delta f_{\rm K}}{4} = 0.75 \cdot \Delta f_{\rm K}\hspace{0.3cm}		:$$ \Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0) = \frac{\Delta f_{\rm K}}{2} + \frac{\Delta f_{\rm K}}{4} = 0.75 \cdot \Delta f_{\rm K}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0)/{\Delta f_{\rm K}} \hspace{0.1cm}\underline {= 0.75} .$$		\Rightarrow \hspace{0.3cm}h_{\rm MKD}(t = 0)/{\Delta f_{\rm K}} \hspace{0.1cm}\underline {= 0.75} .$$


	'''(4)'''  The <u>second proposed solution</u> is correct:		'''(4)'''  The  <u>second proposed solution</u>  is correct:
	*The first si function does have equidistant zero crossings at the distance $1/\Delta f_{\rm K}$.		*The first sinc–function does have equidistant zero crossings at the distance  $1/\Delta f_{\rm K}$.
	*But the equidistant zero crossings of the whole time function $h_{\rm MKD}$ are determined by the second term:		*But the equidistant zero crossings of the whole time function  $h_{\rm MKD}(t)$  are determined by the second term:
	:$$h_{\rm MKD}(t = \frac{1}{\Delta f_{\rm K}}) = \ \frac{\Delta f_{\rm K}}{2} \cdot {\rm si} (~~\pi~~ )+		:$$h_{\rm MKD}(t = \frac{1}{\Delta f_{\rm K}}) = \ \frac{\Delta f_{\rm K}}{2} \cdot {\rm sinc} (1 )+
	\frac{\Delta f_{\rm K}}{4} \cdot {\rm si}^2 (~~\pi/2~~) = \frac{\Delta		\frac{\Delta f_{\rm K}}{4} \cdot {\rm sinc}^2 (0.5) = \frac{\Delta
	f_{\rm K}}{4},$$		f_{\rm K}}{4},$$
	:$$h_{\rm MKD}(t = \frac{2}{\Delta f_{\rm K}}) = \ \frac{\Delta		:$$h_{\rm MKD}(t = \frac{2}{\Delta f_{\rm K}}) = \ \frac{\Delta
	f_{\rm K}}{2} \cdot {\rm si} (2~~\pi~~ )+ \frac{\Delta f_{\rm K}}{4}		f_{\rm K}}{2} \cdot {\rm sinc} (2 )+ \frac{\Delta f_{\rm K}}{4}
	\cdot {\rm si}^2 (~~\pi~~) = 0.$$		\cdot {\rm sinc}^2 (1) = 0.$$

Guenter at 12:40, 7 May 2022

2022-05-07T12:40:43Z

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	[[File:P_ID1683__Dig_A_4_3.png\|right\|frame\|Unbalanced channel frequency response]]		[[File:P_ID1683__Dig_A_4_3.png\|right\|frame\|Unbalanced channel frequency response]]
	In this exercise, we consider a BPSK system with coherent demodulation, i.e.		In this exercise,  we consider a BPSK system with coherent demodulation,  i.e.
	:$$s(t) \ = \ z(t) \cdot q(t),$$		:$$s(t) \ = \ z(t) \cdot q(t),$$
	:$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$		:$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$
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	:$$H_{\rm MKD}(f) = {1}/{2} \cdot \big [ H_{\rm K}(f-f_{\rm T}) + H_{\rm K}(f+f_{\rm T})\big ] .$$		:$$H_{\rm MKD}(f) = {1}/{2} \cdot \big [ H_{\rm K}(f-f_{\rm T}) + H_{\rm K}(f+f_{\rm T})\big ] .$$

	*Thus the modulator and demodulator are virtually shortened against each other, ~~and~~		*Thus the modulator and demodulator are virtually shortened against each other,  and
	*the bandpass channel  ~~$H_{\rm K}(f)$  is transformed into the lowpass range.~~

			*the bandpass channel  $H_{\rm K}(f)$  is transformed into the low-pass range.

	The resulting transmission function  $H_{\rm MKD}(f)$  should not be confused with the lowpass transmission function  $H_{\rm K, \, TP}(f)$  as described in the chapter  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function\|Equivalent Low-Pass Signal and its Spectral Function]]  of the book "Signal Representation", which results from  $H_{\rm K}(f)$  by truncating the components at negative frequencies as well as a frequency shift by  $f_{\rm T}$  to the left.

	~~For frequency responses~~, in ~~contrast to spectral functions,~~ the ~~doubling~~ of the components at ~~positive~~ frequencies ~~must be omitted~~.		The resulting transmission function  $H_{\rm MKD}(f)$  should not be confused with the low-pass transmission function  $H_{\rm K, \, TP}(f)$  as described in the chapter  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function\|"Equivalent Low-Pass Signal and its Spectral Function"]]  of the book "Signal Representation",  which results from  $H_{\rm K}(f)$  by truncating the components at negative frequencies as well as a frequency shift by the carrier frequency $f_{\rm T}$  to the left.

			For frequency responses,  in contrast to spectral functions,  the doubling of the components at positive frequencies must be omitted.




			Notes:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation\|"Linear Digital Modulation - Coherent Demodulation"]].

	~~''Notes:''~~		*Reference is made in particular to the section  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Baseband_model_for_ASK_and_BPSK\|"Baseband model for ASK and BPSK"]].
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation\|Linear Digital Modulation - Coherent Demodulation]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Baseband_model_for_ASK_and_BPSK\|Baseband model for ASK and BPSK]].		*The subscript  "MKD"  stands for  "modulator – channel – demodulator"  German:  "Modulator – Kanal – Demodulator").


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	<quiz display=simple>		<quiz display=simple>

	{Which statements are valid for the equivalent ~~lowpass~~ function  $H_{\rm K, \, TP}(f)$ ?		{Which statements are valid for the equivalent low-pass function  $H_{\rm K, \, TP}(f)$ ?
	\|type="[]"}		\|type="[]"}
	-  $H_{\rm K, \, TP}(f=0)= 2$ ~~holds~~.		-  $H_{\rm K, \, TP}(f=0)= 2$.
	+  $H_{\rm K, \, TP}(f = \Delta f_{\rm K}/4) = 1$ ~~holds~~.		+  $H_{\rm K, \, TP}(f = \Delta f_{\rm K}/4) = 1$.
	+  $H_{\rm K, \, TP}(f = –\Delta f_{\rm K}/4) = 0.75$ ~~holds~~.		+  $H_{\rm K, \, TP}(f = -\Delta f_{\rm K}/4) = 0.75$.
	+ The corresponding time function  $h_{\rm K, \, TP}(t)$  is complex.		+ The corresponding time function  $h_{\rm K, \, TP}(t)$  is complex.

	{Which statements are valid for the frequency response  $H_{\rm MKD}(f)$ ?		{Which statements are valid for the frequency response  $H_{\rm MKD}(f)$ ?
	\|type="[]"}		\|type="[]"}
	-  $H_{\rm MKD}(f=0)= 2$ ~~holds~~.		-  $H_{\rm MKD}(f=0)= 2$.
	-  $H_{\rm MKD}(f = \Delta f_{\rm K}/4) = 1$ ~~holds~~.		-  $H_{\rm MKD}(f = \Delta f_{\rm K}/4) = 1$.
	+  $H_{\rm MKD}(f = –\Delta f_{\rm K}/4) = 0.75$ ~~holds~~.		+  $H_{\rm MKD}(f = -\Delta f_{\rm K}/4) = 0.75$.
	- The corresponding time function  $h_{\rm MKD}(t)$  is complex.		- The corresponding time function  $h_{\rm MKD}(t)$  is complex.

	{Calculate the time function  $h_{\rm MKD}(t)$. Specify the value at  $t = 0$.		{Calculate the time function  $h_{\rm MKD}(t)$.  Specify the value at  $t = 0$.
	\|type="{}"}		\|type="{}"}
	$ h_{\rm MKD}(t = 0)/\Delta f_{\rm K} \ = \ $ { 0.75 3% }		$ h_{\rm MKD}(t = 0)/\Delta f_{\rm K} \ = \ $ { 0.75 3% }

Hwang at 15:18, 29 March 2022

2022-03-29T15:18:38Z

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Guenter at 14:05, 19 March 2022

2022-03-19T14:05:41Z

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	[[Category:Digital Signal Transmission: Exercises\|^1.5 ~~Lineare digitale~~ Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.10: BPSK–Basisbandmodell to Aufgaben:Exercise 1.10: BPSK Baseband Model

2022-03-19T14:02:32Z

Guenter moved page Aufgaben:Aufgabe 1.10: BPSK–Basisbandmodell to Aufgaben:Exercise 1.10: BPSK Baseband Model

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	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^1.5 Lineare digitale Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Lineare digitale Modulation^]]