Modulation Methods/Quadrature Amplitude Modulation - Revision history

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2026-03-16T13:47:09Z

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	*The example considered in the graph is for the   $\text{16-QAM}$  with  $b = M = 4$  and correspondingly  $M^2 =16$  signal space points.  For a   $\text{256-QAM}$,  $b = 8$  and  $M = 16$  would apply:   $2^b = M^2 = 256$.		*The example considered in the graph is for the   $\text{16-QAM}$  with  $b = M = 4$  and correspondingly  $M^2 =16$  signal space points.  For a   $\text{256-QAM}$,  $b = 8$  and  $M = 16$  would apply:   $2^b = M^2 = 256$.
	*Next,  the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are each applied to a [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain\|$\text{Dirac comb}$]]  as pulse weights.  Thus,  after pulse shaping with the basic transmission pulse  $g_s(t)$,  the following holds for both branches of the circuit diagram:		*Next,  the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are each applied to a [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain\|$\text{Dirac comb}$]]  as pulse weights.  Thus,  after pulse shaping with the basic transmission pulse  $g_s(t)$,  the following holds for both branches of the circuit diagram:
	:$$s_{\rm I}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm I\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm},\hspace{1cm}s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm		:$$s_{\rm I}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm I\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm},\hspace{1cm}s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm Q\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$
	Q\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$
	*Note that because of the redundancy–free conversion,  the symbol duration  $T$  of these signals is larger by a factor of  $b$  than the bit duration  $T_{\rm B}$  of the binary source signal.  In the illustrated   $\text{16-QAM}$  example,  $T = 4 · T_{\rm B}$  holds.		*Note that because of the redundancy–free conversion,  the symbol duration  $T$  of these signals is larger by a factor of  $b$  than the bit duration  $T_{\rm B}$  of the binary source signal.  In the illustrated   $\text{16-QAM}$  example,  $T = 4 · T_{\rm B}$  holds.
	*The QAM transmitted signal  $s(t)$  is then the sum of the two signals multiplied by cosine and minus-sine, respectively:		*The QAM transmitted signal  $s(t)$  is then the sum of the two signals multiplied by cosine and minus-sine, respectively:
	:$$s_{\rm cos}(t) = s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t), \hspace{1cm} s_{\rm -sin}(t) = -s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t)$$		:$$s_{\rm cos}(t) = s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t), \hspace{1cm} s_{\rm -sin}(t) = -s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t)$$
	:$$\Rightarrow \hspace{0.3cm}s(t) = s_{\rm cos}(t)+ s_{\rm -sin}(t) =		:$$\Rightarrow \hspace{0.3cm}s(t) = s_{\rm cos}(t)+ s_{\rm -sin}(t) =s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t)\hspace{0.05cm}.$$
	s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t)
	\hspace{0.05cm}.$$

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
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	Please note the following:		Please note the following:
	*All double arrows in the baseband model denote complex quantities.  The operations associated with them should also be understood as complex.  For example,  the complex amplitude coefficient  $a_ν$  combines one inphase and one quadrature coefficient:		*All double arrows in the baseband model denote complex quantities.  The operations associated with them should also be understood as complex.  For example,  the complex amplitude coefficient  $a_ν$  combines one inphase and one quadrature coefficient:
	:$$a_\nu = a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm		:$$a_\nu = a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} \hspace{0.05cm}.$$
	Q\hspace{0.03cm}\it \nu} \hspace{0.05cm}.$$
	*The equivalent low–pass representation of the actual, physical and thus per se real transmitted QAM signal  $s(t)$  is always complex and with the partial signals  $s_{\rm I}(t)$  and  $s_{\rm Q}(t)$ it holds for the equivalent low-pass signal  $($German:  "äquivalentes Tiefpass–Signal"   ⇒   subscript:  "TP"$)$:		*The equivalent low–pass representation of the actual, physical and thus per se real transmitted QAM signal  $s(t)$  is always complex and with the partial signals  $s_{\rm I}(t)$  and  $s_{\rm Q}(t)$ it holds for the equivalent low-pass signal  $($German:  "äquivalentes Tiefpass–Signal"   ⇒   subscript:  "TP"$)$:
	:$$s_{\rm TP}(t) = s_{\rm I}(t) + {\rm j} \cdot s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty} a_\nu \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$		:$$s_{\rm TP}(t) = s_{\rm I}(t) + {\rm j} \cdot s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty} a_\nu \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$
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	Averaging over the $16$ squared distances to the origin yields:		Averaging over the $16$ squared distances to the origin yields:
	:$${\rm E}\big[\hspace{0.05cm} \vert a_{\nu}\vert^2 \hspace{0.05cm}\big] \hspace{0.18cm} = \hspace{0.18cm} \frac{4}{16} \cdot \left [1^2 + 1^2\right ]+ \frac{4}{16} \cdot \left[(1/3 )^2 +(1/3)^2 \right ]		:$${\rm E}\big[\hspace{0.05cm} \vert a_{\nu}\vert^2 \hspace{0.05cm}\big] \hspace{0.18cm} = \hspace{0.18cm} \frac{4}{16} \cdot \left [1^2 + 1^2\right ]+ \frac{4}{16} \cdot \left[(1/3 )^2 +(1/3)^2 \right ]+ \frac{8}{16} \cdot \left [1^2 + (1/3)^2\right ] $$
	+ \frac{8}{16} \cdot \left [1^2 + (1/3)^2\right ] $$
	:$$\Rightarrow \hspace{0.3cm}{\rm E}\big[\hspace{0.05cm} \vert a_{\nu}\vert^2 \hspace{0.05cm}\big] = {10}/{9}\approx 1.11 \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}{\rm E}\big[\hspace{0.05cm} \vert a_{\nu}\vert^2 \hspace{0.05cm}\big] = {10}/{9}\approx 1.11 \hspace{0.05cm}.$$

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	*the average energy per symbol:		*the average energy per symbol:
	:$$E_{\rm S} = 1/2 \cdot {\rm E}\big[\hspace{0.05cm}\vert a_{\nu} \vert ^2 \big]\hspace{0.05cm}\~~cdot~~		:$$E_{\rm S} = 1/2 \cdot {\rm E}\big[\hspace{0.05cm}\vert a_{\nu} \vert ^2 \big]\hspace{0.05cm}\cdotg_0^2 \cdot T = {5}/{9}\cdot g_0^2 \cdot T \hspace{0.05cm}\approx 0.555 \cdot g_0^2 \cdot T \hspace{0.05cm},$$
	~~g_0~~^2 \cdot T = {5}/{9}\cdot g_0^2 \cdot T \hspace{0.05cm}\approx 0.555 \cdot g_0^2 \cdot T \hspace{0.05cm},$$
	*the average energy per bit:		*the average energy per bit:
	:$$E_{\rm B} ={E_{\rm S} }/{4}= {5}/{36}\cdot g_0^2 \cdot T \approx 0.139 \cdot g_0^2 \cdot T \hspace{0.05cm}.$$		:$$E_{\rm B} ={E_{\rm S} }/{4}= {5}/{36}\cdot g_0^2 \cdot T \approx 0.139 \cdot g_0^2 \cdot T \hspace{0.05cm}.$$
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	Thus,  with the equations		Thus,  with the equations
	:$$p_{\rm B} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm}		:$$p_{\rm B} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm}\sigma_d^2 = {N_0}(2 \cdot T_{\rm B} )$$
	\sigma_d^2 = {N_0}(2 \cdot T_{\rm B} )$$
	one obtains the exact same result for the  »'''4-QAM error probability'''«  as for BPSK:		one obtains the exact same result for the  »'''4-QAM error probability'''«  as for BPSK:
	:$$p_{\rm B, \hspace{0.1cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$		:$$p_{\rm B, \hspace{0.1cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$
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	<br>		<br>
	Taking the equations for the  "$\text{4–QAM}$"   $($or  "$π/4\text{–QPSK}$"$)$  as our starting point,  we arrive at the   "$\text{Offset–4–QAM}$",  which we simplify by calling   "$\text{Offset–QPSK}$".  Here the transmitted signal is given by:		Taking the equations for the  "$\text{4–QAM}$"   $($or  "$π/4\text{–QPSK}$"$)$  as our starting point,  we arrive at the   "$\text{Offset–4–QAM}$",  which we simplify by calling   "$\text{Offset–QPSK}$".  Here the transmitted signal is given by:
	:$$s(t) =s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t) - s_{\rm Q}(t)		:$$s(t) =s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t) - s_{\rm Q}(t)\cdot \sin(2 \pi f_{\rm T} t),$$
	\cdot \sin(2 \pi f_{\rm T} t),$$
	:$$\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{\rm I}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm I\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm},\hspace{0.5cm} s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm Q\hspace{0.03cm}\it \nu} \cdot g_s (t -{T}/{2} - \nu \cdot T)\hspace{0.05cm}.$$		:$$\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{\rm I}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm I\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm},\hspace{0.5cm} s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm Q\hspace{0.03cm}\it \nu} \cdot g_s (t -{T}/{2} - \nu \cdot T)\hspace{0.05cm}.$$

Maintenance script: Add German interlanguage link

2026-03-16T12:28:17Z

Add German interlanguage link

← Older revision		Revision as of 14:28, 16 March 2026
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	{{Display}}		{{Display}}
			[[de:Modulationsverfahren/Quadratur–Amplitudenmodulation]]

Guenter at 13:50, 13 January 2023

2023-01-13T13:50:11Z

← Older revision		Revision as of 15:50, 13 January 2023
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	:$$s(t) = {\rm Re}[s_{\rm +}(t)] = {\rm Re}[s_{\rm TP}(t) \cdot{\rm e}^{{\rm j}2\pi f_{\rm T} t}] $$		:$$s(t) = {\rm Re}[s_{\rm +}(t)] = {\rm Re}[s_{\rm TP}(t) \cdot{\rm e}^{{\rm j}2\pi f_{\rm T} t}] $$
	:$$\Rightarrow \hspace{0.3cm} s(t) = {\rm Re} \left[\left ( \sum (a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} ) \cdot g_s (t - \nu \cdot T)\right )\left ( \cos(2 \pi f_{\rm T} t) + {\rm j} \cdot \sin(2 \pi f_{\rm T} t) \right )\right]= s_{\rm I}(t) \cdot \cos(2\pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t) \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} s(t) = {\rm Re} \left[\left ( \sum (a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} ) \cdot g_s (t - \nu \cdot T)\right )\left ( \cos(2 \pi f_{\rm T} t) + {\rm j} \cdot \sin(2 \pi f_{\rm T} t) \right )\right]= s_{\rm I}(t) \cdot \cos(2\pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t) \hspace{0.05cm}.$$
	*The influence of the ~~bandpass '''KORREKTUR:~~ band-pass~~'''~~  $H_{\rm BP}(f)$,  which in practice often has to be considered at the output of the QAM modulator,  can be assigned to the pulse shape filter  $g_s(t)$.  If the passband of the ~~bandpass '''KORREKTUR:~~ band-pass~~'''~~ filter is symmetric about $f_{\rm T}$,  its low-pass equivalent  (in the time domain)   $h_{\rm BP\ →\ TP}(t)$   is purely real and one can replace  $g_s(t)$  with  $g_s(t) \star h_{\rm BP\ →\ TP}(t)$  in the model.		*The influence of the band-pass  $H_{\rm BP}(f)$,  which in practice often has to be considered at the output of the QAM modulator,  can be assigned to the pulse shape filter  $g_s(t)$.  If the passband of the band-pass filter is symmetric about $f_{\rm T}$,  its low-pass equivalent  (in the time domain)   $h_{\rm BP\ →\ TP}(t)$   is purely real and one can replace  $g_s(t)$  with  $g_s(t) \star h_{\rm BP\ →\ TP}(t)$  in the model.

	==Power and energy of QAM signals==		==Power and energy of QAM signals==
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	*$\rm 4–QAM$  (left diagram)  can also be understood as a ~~four-stage '''KORREKTUR:~~ four-level~~'''~~ phase modulation  $($positions  $±45^\circ$,  $±135^\circ)$.  Compared to QPSK,  there is a rotation of  $±45^\circ$  $(π/4)$   ⇒   4–QAM is also called  ${\rm π/4}\text{–QPSK}$.		*$\rm 4–QAM$  (left diagram)  can also be understood as a four-level phase modulation  $($positions  $±45^\circ$,  $±135^\circ)$.  Compared to QPSK,  there is a rotation of  $±45^\circ$  $(π/4)$   ⇒   4–QAM is also called  ${\rm π/4}\text{–QPSK}$.

Hwang at 12:40, 11 January 2023

2023-01-11T12:40:40Z

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	[[Aufgaben:Aufgabe_4.11Z:_Fehlerwahrscheinlichkeit_bei_QAM\|Exercise 4.11Z: Error Probability with QAM]]		[[Aufgaben:Aufgabe_4.11Z:_Fehlerwahrscheinlichkeit_bei_QAM\|Exercise 4.11Z: Error Probability with QAM]]

	[[Aufgaben:~~Aufgabe_4~~.12:~~_Wurzel–Nyquist–Systeme~~\|Exercise 4.12: Root Nyquist Systems]]		[[Aufgaben:Exercise_4.12:_Root-Nyquist_Systems\|Exercise 4.12: Root-Nyquist Systems]]

	[[Aufgaben:Exercise_4.12Z:_4-QAM_Systems_again\|Exercise 4.12Z: 4-QAM Systems again]]		[[Aufgaben:Exercise_4.12Z:_4-QAM_Systems_again\|Exercise 4.12Z: 4-QAM Systems again]]

Hwang at 16:54, 9 January 2023

2023-01-09T16:54:41Z

← Older revision		Revision as of 18:54, 9 January 2023
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	:$$s(t) = {\rm Re}[s_{\rm +}(t)] = {\rm Re}[s_{\rm TP}(t) \cdot{\rm e}^{{\rm j}2\pi f_{\rm T} t}] $$		:$$s(t) = {\rm Re}[s_{\rm +}(t)] = {\rm Re}[s_{\rm TP}(t) \cdot{\rm e}^{{\rm j}2\pi f_{\rm T} t}] $$
	:$$\Rightarrow \hspace{0.3cm} s(t) = {\rm Re} \left[\left ( \sum (a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} ) \cdot g_s (t - \nu \cdot T)\right )\left ( \cos(2 \pi f_{\rm T} t) + {\rm j} \cdot \sin(2 \pi f_{\rm T} t) \right )\right]= s_{\rm I}(t) \cdot \cos(2\pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t) \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} s(t) = {\rm Re} \left[\left ( \sum (a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} ) \cdot g_s (t - \nu \cdot T)\right )\left ( \cos(2 \pi f_{\rm T} t) + {\rm j} \cdot \sin(2 \pi f_{\rm T} t) \right )\right]= s_{\rm I}(t) \cdot \cos(2\pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t) \hspace{0.05cm}.$$
	*The influence of the bandpass  $H_{\rm BP}(f)$,  which in practice often has to be considered at the output of the QAM modulator,  can be assigned to the pulse shape filter  $g_s(t)$.  If the passband of the bandpass filter is symmetric about $f_{\rm T}$,  its low-pass equivalent  (in the time domain)   $h_{\rm BP\ →\ TP}(t)$   is purely real and one can replace  $g_s(t)$  with  $g_s(t) \star h_{\rm BP\ →\ TP}(t)$  in the model.		*The influence of the bandpass '''KORREKTUR: band-pass'''  $H_{\rm BP}(f)$,  which in practice often has to be considered at the output of the QAM modulator,  can be assigned to the pulse shape filter  $g_s(t)$.  If the passband of the bandpass '''KORREKTUR: band-pass''' filter is symmetric about $f_{\rm T}$,  its low-pass equivalent  (in the time domain)   $h_{\rm BP\ →\ TP}(t)$   is purely real and one can replace  $g_s(t)$  with  $g_s(t) \star h_{\rm BP\ →\ TP}(t)$  in the model.

	==Power and energy of QAM signals==		==Power and energy of QAM signals==

Hwang at 16:52, 9 January 2023

2023-01-09T16:52:26Z

← Older revision		Revision as of 18:52, 9 January 2023
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Conclusion:}$  These statements can be summarized as follows:		$\text{Conclusion:}$  These statements can be summarized as follows:
	*The two transmission branches   $\rm (I,\ Q)$  can be thought of as two completely separate  $M$-~~step~~ ASK systems <br>that do not interfere with each other as long as all components are optimally designed.		*The two transmission branches   $\rm (I,\ Q)$  can be thought of as two completely separate  $M$-level ASK systems <br>that do not interfere with each other as long as all components are optimally designed.
	*Quadrature amplitude modulation thus makes it  (ideally)  possible to double the data rate while maintaining the same quality. }}		*Quadrature amplitude modulation thus makes it  (ideally)  possible to double the data rate while maintaining the same quality. }}

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	*$\rm 4–QAM$  (left diagram)  can also be understood as a four-stage phase modulation  $($positions  $±45^\circ$,  $±135^\circ)$.  Compared to QPSK,  there is a rotation of  $±45^\circ$  $(π/4)$   ⇒   4–QAM is also called  ${\rm π/4}\text{–QPSK}$.		*$\rm 4–QAM$  (left diagram)  can also be understood as a four-stage '''KORREKTUR: four-level''' phase modulation  $($positions  $±45^\circ$,  $±135^\circ)$.  Compared to QPSK,  there is a rotation of  $±45^\circ$  $(π/4)$   ⇒   4–QAM is also called  ${\rm π/4}\text{–QPSK}$.

Hwang at 16:48, 9 January 2023

2023-01-09T16:48:57Z

← Older revision		Revision as of 18:48, 9 January 2023
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	This very general model can be described as follows:		This very general model can be described as follows:
	*The binary source symbol sequence $〈q_k〉$  with bit rate  $R_{\rm B}$ is applied to the input.  Thus,  the time interval between two symbols is  $T_{\rm B} = 1/R_{\rm B}$.		*The binary source symbol sequence $〈q_k〉$  with bit rate  $R_{\rm B}$ is applied to the input.  Thus,  the time interval between two symbols is  $T_{\rm B} = 1/R_{\rm B}$.
	*Two ~~multilevel~~ amplitude coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are derived from each  $b$  binary input symbols  $q_k$,  where  $\rm I$  stands for  "inphase component"  and  $\rm Q$  stands for  "quadrature component".		*Two multi-level amplitude coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are derived from each  $b$  binary input symbols  $q_k$,  where  $\rm I$  stands for  "inphase component"  and  $\rm Q$  stands for  "quadrature component".
	*If  $b$  is even and the signal space allocation is quadratic, then the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  can each take on one of the  $M = 2^{b/2}$  amplitude values with equal probability.  This is then referred to as  »'''quadrature amplitude modulation'''«  $\rm (QAM)$.		*If  $b$  is even and the signal space allocation is quadratic, then the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  can each take on one of the  $M = 2^{b/2}$  amplitude values with equal probability.  This is then referred to as  »'''quadrature amplitude modulation'''«  $\rm (QAM)$.
	*The example considered in the graph is for the   $\text{16-QAM}$  with  $b = M = 4$  and correspondingly  $M^2 =16$  signal space points.  For a   $\text{256-QAM}$,  $b = 8$  and  $M = 16$  would apply:   $2^b = M^2 = 256$.		*The example considered in the graph is for the   $\text{16-QAM}$  with  $b = M = 4$  and correspondingly  $M^2 =16$  signal space points.  For a   $\text{256-QAM}$,  $b = 8$  and  $M = 16$  would apply:   $2^b = M^2 = 256$.
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	#$\text{16-QAM}$  is thus nothing more than the parallel transmission of two digital signals with  $M = 4$  amplitude levels each.		#$\text{16-QAM}$  is thus nothing more than the parallel transmission of two digital signals with  $M = 4$  amplitude levels each.
	#For  $\text{64-QAM}$   ⇒   $M = 8$  would apply analogously.  For  $\text{256-QAM}$:  the "one-dimensional" number of levels is  $M = 16$.		#For  $\text{64-QAM}$   ⇒   $M = 8$  would apply analogously.  For  $\text{256-QAM}$:  the "one-dimensional" number of levels is  $M = 16$.
	#All the properties for redundancy-free ~~multilevel~~ signals mentioned in the later chapter  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"Redundancy-free Coding"]]  also apply here.		#All the properties for redundancy-free multi-level signals mentioned in the later chapter  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"Redundancy-free Coding"]]  also apply here.

Hwang at 16:37, 9 January 2023

2023-01-09T16:37:16Z

Hwang at 23:55, 12 November 2022

2022-11-12T23:55:38Z

← Older revision		Revision as of 01:55, 13 November 2022
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	[[File:P_ID1708__Mod_T_4_3_S2_Ganz_neu.png \|right\|frame\| Linear modulator  $(\rm I$ and  $\rm Q$ components$)$  in the equivalent low-pass range]]		[[File:P_ID1708__Mod_T_4_3_S2_Ganz_neu.png \|right\|frame\| Linear modulator  $(\rm I$ and  $\rm Q$ components$)$  in the equivalent low-pass range]]
	#The graph shows the simplified model in the baseband.  This is equivalent to the block diagram considered so far.		#The graph shows the simplified model in the baseband.  This is equivalent to the block diagram considered so far.
	#The serial-parallel conversion and the signal space allocation drawn in red in  [[Modulation_Methods/Quadrature_Amplitude_Modulation#General_description_and_signal_space_allocation\|the block diagram on the last ~~page~~]] are retained.  This block is no longer drawn here.		#The serial-parallel conversion and the signal space allocation drawn in red in  [[Modulation_Methods/Quadrature_Amplitude_Modulation#General_description_and_signal_space_allocation\|the block diagram in the last section]] are retained.  This block is no longer drawn here.
	#We also initially disregard the band-pass  $H_{\rm BP}(f)$,  which is often introduced for technical reasons.		#We also initially disregard the band-pass  $H_{\rm BP}(f)$,  which is often introduced for technical reasons.
	<br clear=all>		<br clear=all>
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	:$$s_{\rm TP}(t) = s_{\rm I}(t) + {\rm j} \cdot s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty} a_\nu \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$		:$$s_{\rm TP}(t) = s_{\rm I}(t) + {\rm j} \cdot s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty} a_\nu \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$
	*The analytical signal  $s_+(t)$  is obtained from the equivalent low-pass signal  $s_{\rm TP}(t)$  by multiplying by the complex exponential function.  The physical signal  $s(t)$  is then obtained as the real part of  $s_+(t)$.		*The analytical signal  $s_+(t)$  is obtained from the equivalent low-pass signal  $s_{\rm TP}(t)$  by multiplying by the complex exponential function.  The physical signal  $s(t)$  is then obtained as the real part of  $s_+(t)$.
	*In order for the signs in the block diagram on the previous ~~page~~ and the sketched baseband model here to match,  multiplication by the negative sine wave is required in the quadrature branch,  as shown in the following calculation:		*In order for the signs in the block diagram in the previous section and the sketched baseband model here to match,  multiplication by the negative sine wave is required in the quadrature branch,  as shown in the following calculation:
	:$$s(t) = {\rm Re}[s_{\rm +}(t)] = {\rm Re}[s_{\rm TP}(t) \cdot{\rm e}^{{\rm j}2\pi f_{\rm T} t}] $$		:$$s(t) = {\rm Re}[s_{\rm +}(t)] = {\rm Re}[s_{\rm TP}(t) \cdot{\rm e}^{{\rm j}2\pi f_{\rm T} t}] $$
	:$$\Rightarrow \hspace{0.3cm} s(t) = {\rm Re} \left[\left ( \sum (a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} ) \cdot g_s (t - \nu \cdot T)\right )\left ( \cos(2 \pi f_{\rm T} t) + {\rm j} \cdot \sin(2 \pi f_{\rm T} t) \right )\right]= s_{\rm I}(t) \cdot \cos(2\pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t) \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} s(t) = {\rm Re} \left[\left ( \sum (a_{\rm I\hspace{0.03cm}\it \nu} + {\rm j} \cdot a_{\rm Q\hspace{0.03cm}\it \nu} ) \cdot g_s (t - \nu \cdot T)\right )\left ( \cos(2 \pi f_{\rm T} t) + {\rm j} \cdot \sin(2 \pi f_{\rm T} t) \right )\right]= s_{\rm I}(t) \cdot \cos(2\pi f_{\rm T} t) - s_{\rm Q}(t) \cdot \sin(2 \pi f_{\rm T} t) \hspace{0.05cm}.$$

Guenter at 14:45, 16 April 2022

2022-04-16T14:45:32Z

← Older revision		Revision as of 16:45, 16 April 2022
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	The following graph shows the signal waveforms of  $\rm 4-QAM$,  where the coloring corresponds to the signal space allocation defined above.		The following graph shows the signal waveforms of  $\rm 4-QAM$,  where the coloring corresponds to the signal space allocation defined above.

	[[File:P_ID1711__Mod_T_4_3_S4_neu.png\|right\|frame\| Signal waveforms for  $\rm 4-QAM$]]		[[File:P_ID1711__Mod_T_4_3_S4_neu.png\|right\|frame\| Signal waveforms for  $\text{4-QAM}$]]

	One can see from these plots:		One can see from these plots:
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	==Offset–Quadrature amplitude modulation==		==Offset–Quadrature amplitude modulation==
	<br>		<br>
	Taking the equations for the  "$\text{4–QAM}$"   $($or  "$π/4\text{–QPSK}$"$)$  as our starting point,  we arrive at the   "$\text{Offset–4–QAM}$",  which we simplify by calling   "$\text{Offset–QPSK}$".  Here the ~~transmit signal '''KORREKTUR:~~ transmitted signal~~'''~~ is given by:		Taking the equations for the  "$\text{4–QAM}$"   $($or  "$π/4\text{–QPSK}$"$)$  as our starting point,  we arrive at the   "$\text{Offset–4–QAM}$",  which we simplify by calling   "$\text{Offset–QPSK}$".  Here the transmitted signal is given by:
	:$$s(t) =s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t) - s_{\rm Q}(t)		:$$s(t) =s_{\rm I}(t) \cdot \cos(2 \pi f_{\rm T} t) - s_{\rm Q}(t)
	\cdot \sin(2 \pi f_{\rm T} t),$$		\cdot \sin(2 \pi f_{\rm T} t),$$

← Older revision		Revision as of 14:40, 11 January 2023
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	[[Aufgaben:Aufgabe_4.11Z:_Fehlerwahrscheinlichkeit_bei_QAM\|Exercise 4.11Z: Error Probability with QAM]]		[[Aufgaben:Aufgabe_4.11Z:_Fehlerwahrscheinlichkeit_bei_QAM\|Exercise 4.11Z: Error Probability with QAM]]

	[[Aufgaben:~~Aufgabe_4~~.12:~~_Wurzel–Nyquist–Systeme~~\|Exercise 4.12: Root Nyquist Systems]]		[[Aufgaben:Exercise_4.12:_Root-Nyquist_Systems\|Exercise 4.12: Root-Nyquist Systems]]

	[[Aufgaben:Exercise_4.12Z:_4-QAM_Systems_again\|Exercise 4.12Z: 4-QAM Systems again]]		[[Aufgaben:Exercise_4.12Z:_4-QAM_Systems_again\|Exercise 4.12Z: 4-QAM Systems again]]

← Older revision		Revision as of 18:37, 9 January 2023
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	*The binary source symbol sequence $〈q_k〉$  with bit rate  $R_{\rm B}$ is applied to the input.  Thus,  the time interval between two symbols is  $T_{\rm B} = 1/R_{\rm B}$.		*The binary source symbol sequence $〈q_k〉$  with bit rate  $R_{\rm B}$ is applied to the input.  Thus,  the time interval between two symbols is  $T_{\rm B} = 1/R_{\rm B}$.
	*Two multilevel amplitude coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are derived from each  $b$  binary input symbols  $q_k$,  where  $\rm I$  stands for  "inphase component"  and  $\rm Q$  stands for  "quadrature component".		*Two multilevel amplitude coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are derived from each  $b$  binary input symbols  $q_k$,  where  $\rm I$  stands for  "inphase component"  and  $\rm Q$  stands for  "quadrature component".
	*If  $b$  is even and the signal space allocation is quadratic, then the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  can each take on one of the  $M = 2^{b/2}$  amplitude values with equal probability.  This is then referred to as  '''quadrature amplitude modulation'''  $\rm (QAM)$.		*If  $b$  is even and the signal space allocation is quadratic, then the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  can each take on one of the  $M = 2^{b/2}$  amplitude values with equal probability.  This is then referred to as  »'''quadrature amplitude modulation'''«  $\rm (QAM)$.
	*The example considered in the graph is for the   $\text{16-QAM}$  with  $b = M = 4$  and correspondingly  $M^2 =16$  signal space points.  For a   $\text{256-QAM}$,  $b = 8$  and  $M = 16$  would apply:   $2^b = M^2 = 256$.		*The example considered in the graph is for the   $\text{16-QAM}$  with  $b = M = 4$  and correspondingly  $M^2 =16$  signal space points.  For a   $\text{256-QAM}$,  $b = 8$  and  $M = 16$  would apply:   $2^b = M^2 = 256$.
	*Next,  the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are each applied to a [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain\|"Dirac comb"]]  as pulse weights.  Thus,  after pulse shaping with the basic transmission pulse  $g_s(t)$,  the following holds for both branches of the circuit diagram:		*Next,  the coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  are each applied to a [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain\|$\text{Dirac comb}$]]  as pulse weights.  Thus,  after pulse shaping with the basic transmission pulse  $g_s(t)$,  the following holds for both branches of the circuit diagram:
	:$$s_{\rm I}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm I\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm},\hspace{1cm}s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm		:$$s_{\rm I}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm I\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm},\hspace{1cm}s_{\rm Q}(t) = \sum_{\nu = - \infty}^{+\infty}a_{\rm
	Q\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$		Q\hspace{0.03cm}\it \nu} \cdot g_s (t - \nu \cdot T)\hspace{0.05cm}.$$
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	[[File:P_ID1708__Mod_T_4_3_S2_Ganz_neu.png \|right\|frame\| Linear modulator  $(\rm I$ and  $\rm Q$ components$)$  in the equivalent low-pass range]]		[[File:P_ID1708__Mod_T_4_3_S2_Ganz_neu.png \|right\|frame\| Linear modulator  $(\rm I$ and  $\rm Q$ components$)$  in the equivalent low-pass range]]
	#The graph shows the simplified model in the baseband.  This is equivalent to the block diagram considered so far.		#The graph shows the simplified model in the baseband.  This is equivalent to the block diagram considered so far.
	#The serial-parallel conversion and the signal space allocation drawn in red in  [[Modulation_Methods/Quadrature_Amplitude_Modulation#General_description_and_signal_space_allocation\|the block diagram in the last section]] are retained.  This block is no longer drawn here.		#The serial-parallel conversion and the signal space allocation drawn in red in  [[Modulation_Methods/Quadrature_Amplitude_Modulation#General_description_and_signal_space_allocation\|"the block diagram in the last section"]] are retained.  This block is no longer drawn here.
	#We also initially disregard the band-pass  $H_{\rm BP}(f)$,  which is often introduced for technical reasons.		#We also initially disregard the band-pass  $H_{\rm BP}(f)$,  which is often introduced for technical reasons.
	<br clear=all>		<br clear=all>
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	==Power and energy of QAM signals==		==Power and energy of QAM signals==
	<br>		<br>
	As shown in the chapter  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function\|Equivalent Low-Pass Signal and its Spectral Function]]  in the book  "Signal Representation",  the   '''power'''  of the transmitted QAM signal  $s(t)$  can also be calculated from the equivalent low-pass signal  $s_{\rm TP}(t)$,  which is always complex.  Thus,  it is equally valid to write:		As shown in the chapter  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function\|"Equivalent Low-Pass Signal and its Spectral Function"]]  in the book  "Signal Representation",  the   »'''power'''«  of the transmitted QAM signal  $s(t)$  can also be calculated from the equivalent low-pass signal  $s_{\rm TP}(t)$,  which is always complex.  Thus,  it is equally valid to write:
	:$$P = \lim_{T_{\rm M} \rightarrow \infty} \frac{\rm 1}{T_{\rm M}}\cdot \int_{-T_{\rm M}/2}^{+T_{\rm M}/2} s^2(t)\,{\rm d} t = {\rm 1}/{2} \cdot \lim_{T_{\rm M} \rightarrow \infty} \frac{\rm 1}{T_{\rm M}}\cdot \int_{-T_{\rm M}/2}^{+T_{\rm M}/2} \|s_{\rm TP}(t)\|^2\,{\rm d} t \hspace{0.05cm}.$$		:$$P = \lim_{T_{\rm M} \rightarrow \infty} \frac{\rm 1}{T_{\rm M}}\cdot \int_{-T_{\rm M}/2}^{+T_{\rm M}/2} s^2(t)\,{\rm d} t = {\rm 1}/{2} \cdot \lim_{T_{\rm M} \rightarrow \infty} \frac{\rm 1}{T_{\rm M}}\cdot \int_{-T_{\rm M}/2}^{+T_{\rm M}/2} \|s_{\rm TP}(t)\|^2\,{\rm d} t \hspace{0.05cm}.$$

	In contrast,  the energy of the unbounded signals  $s(t)$  and   $s_{\rm TP}(t)$  is infinite.		In contrast,  the energy of the unbounded signals  $s(t)$  and   $s_{\rm TP}(t)$  is infinite.

	However,  if we restrict ourselves to a symbol duration $T$,  we obtain the  '''energy per symbol''':		However,  if we restrict ourselves to a symbol duration $T$,  we obtain the  »'''energy per symbol'''«:
	:$$E_{\rm S} = \frac{{\rm E}[\hspace{0.05cm}\|a_{\nu} \|^2\hspace{0.05cm}]}{2}\cdot \int_{-\infty}^{+\infty} \|g_s(t)\|^2\,{\rm d} t = \frac{{\rm E}[\|a_{\nu} \|^2]}{2}\cdot \int_{-\infty}^{+\infty} \|G_s(f)\|^2\,{\rm d} f \hspace{0.05cm}.$$		:$$E_{\rm S} = \frac{{\rm E}[\hspace{0.05cm}\|a_{\nu} \|^2\hspace{0.05cm}]}{2}\cdot \int_{-\infty}^{+\infty} \|g_s(t)\|^2\,{\rm d} t = \frac{{\rm E}[\|a_{\nu} \|^2]}{2}\cdot \int_{-\infty}^{+\infty} \|G_s(f)\|^2\,{\rm d} f \hspace{0.05cm}.$$

	On the other hand,   $E_{\rm B} = E_{\rm S}/b$   gives the  '''energy per bit'''  if according to the signal space assignment,  $b$  binary symbols are combined to the complex coefficient  $a_\nu$.		On the other hand,   $E_{\rm B} = E_{\rm S}/b$   gives the  »'''energy per bit'''«  if according to the signal space assignment,  $b$  binary symbols are combined to the complex coefficient  $a_\nu$.

	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
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	Using the additional equations		Using the additional equations
	:$$p_{\rm B} = {\rm Q}\left ( {s_0}/{\sigma_d } \right ), \hspace{0.2cm} E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/{T_{\rm B} }$$		:$$p_{\rm B} = {\rm Q}\left ( {s_0}/{\sigma_d } \right ), \hspace{0.2cm} E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/{T_{\rm B} }$$
	the  '''BPSK error probability''' is given by:		the  »'''BPSK error probability'''« is given by:
	$$p_{\rm B, \hspace{0.1cm}BPSK} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$		$$p_{\rm B, \hspace{0.1cm}BPSK} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$

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	:$$p_{\rm B} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm}		:$$p_{\rm B} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm}
	\sigma_d^2 = {N_0}(2 \cdot T_{\rm B} )$$		\sigma_d^2 = {N_0}(2 \cdot T_{\rm B} )$$
	one obtains the exact same result for the  '''4-QAM error probability'''  as for BPSK:		one obtains the exact same result for the  »'''4-QAM error probability'''«  as for BPSK:
	:$$p_{\rm B, \hspace{0.1cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$		:$$p_{\rm B, \hspace{0.1cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$

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	The figures apply to all Nyquist systems		The figures apply to all Nyquist systems
	*such as the  "rectangular-rectangular"  configuration  (basic transmission pulse and receiver filter impulse response are both rectangular)		*such as the  "rectangular-rectangular"  configuration  (basic transmission pulse and receiver filter impulse response are both rectangular)
	*or a  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#Root_Nyquist_systems\|~~Root~~ cosine rolloff Nyquist system]].		*or a  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#Root_Nyquist_systems\|$\text{root cosine rolloff Nyquist system}$]].


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	==Nyquist and Root-Nyquist QAM systems==		==Nyquist and Root-Nyquist QAM systems==
	<br>		<br>
	So far in this chapter,  for reasons of easy representation,  the rectangular basic transmission pulse has always been assumed.  In practice,  however,  a  '''root-Nyquist characteristic'''  is usually used according to the [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#Root_Nyquist_systems\|description]]  in the book  "Digital Signal Transmission".  Briefly,  these systems can be characterized as follows:		So far in this chapter,  for reasons of easy representation,  the rectangular basic transmission pulse has always been assumed.  In practice,  however,  a  »'''root-Nyquist characteristic'''«  is usually used according to the [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#Root_Nyquist_systems\|$\text{description}$]]  in the book  "Digital Signal Transmission".  Briefly,  these systems can be characterized as follows:

	#The receiver  (German:  "Empfänger"   ⇒   subscript;  "E")  frequency response  $H_{\rm E}(f)$  is chosen here to be equal in shape to the (normalized) transmission pulse  (German:  "Sendeimpuls"   ⇒   subscript;  "S")  spectrum  $H_{\rm S}(f)$,  which leads to the smallest possible error probability under the constraints of power limitation.		#The receiver  (German:  "Empfänger"   ⇒   subscript;  "E")  frequency response  $H_{\rm E}(f)$  is chosen here to be equal in shape to the (normalized) transmission pulse  (German:  "Sendeimpuls"   ⇒   subscript;  "S")  spectrum  $H_{\rm S}(f)$,  which leads to the smallest possible error probability under the constraints of power limitation.
	#The overall frequency response  $H_{\rm Nyq}(f) = H_{\rm S}(f) · H_{\rm E}(f)$  satisfies the [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_time_domain\|first Nyquist criterion]],  so there is no intersymbol interference  $\text{(ISI)}$  at the receiver.  Thus,  $H_{\rm S}(f)$  and  $H_{\rm E}(f)$  each have a root-Nyquist characteristic.		#The overall frequency response  $H_{\rm Nyq}(f) = H_{\rm S}(f) · H_{\rm E}(f)$  satisfies the [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_time_domain\|$\text{first Nyquist criterion}$]],  so there is no intersymbol interference  $\text{(ISI)}$  at the receiver.  Thus,  $H_{\rm S}(f)$  and  $H_{\rm E}(f)$  each have a root-Nyquist characteristic.
	#For the frequency response  $H_{\rm Nyq}(f)$,  one often uses a  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Cosine.E2.80.93square_low-pass_filter\|cosine rolloff low-pass]]  $H_{\rm CRO}(f)$  with equivalent bandwidth  $Δf_{\rm CRO} = 1/T$  and freely chosen rolloff factor $ (0 ≤ r ≤ 1)$.  The interactive applet  [[Applets:Frequency_%26_Impulse_Responses\|"Frequency and Impulse Responses"]]  illustrates frequency response and impulse response of this low-pass filter.		#For the frequency response  $H_{\rm Nyq}(f)$,  one often uses a  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Cosine.E2.80.93square_low-pass_filter\|$\text{cosine rolloff low-pass}$]]  $H_{\rm CRO}(f)$  with equivalent bandwidth  $Δf_{\rm CRO} = 1/T$  and freely chosen rolloff factor $ (0 ≤ r ≤ 1)$.  The interactive applet  [[Applets:Frequency_%26_Impulse_Responses\|"Frequency and Impulse Responses"]]  illustrates frequency response and impulse response of this low-pass filter.
	#The advantage of these root-Nyquist systems is the much smaller bandwidth  $(1 + r)/T$  compared to the previously considered configuration with rectangular  $g_s(t)$  and rectangular  $h_{\rm E}(t)$,  whose spectrum is  (theoretically)  infinitely extended.		#The advantage of these root-Nyquist systems is the much smaller bandwidth  $(1 + r)/T$  compared to the previously considered configuration with rectangular  $g_s(t)$  and rectangular  $h_{\rm E}(t)$,  whose spectrum is  (theoretically)  infinitely extended.
	#In terms of error probability,  nothing changes compared to the  "rectangular-rectangular"  configuration because the basic detection pulse  $g_d(t)$  before the decision has equidistant zero crossings,  thus avoiding intersymbol interference.		#In terms of error probability,  nothing changes compared to the  "rectangular-rectangular"  configuration because the basic detection pulse  $g_d(t)$  before the decision has equidistant zero crossings,  thus avoiding intersymbol interference.