View source for Digital Signal Transmission/Carrier Frequency Systems with Coherent Demodulation

 
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== Signal space representation of linear modulation ==
<br>
In the first three chapters of this&nbsp; [[Digital_Signal_Transmission|fourth main chapter:]] &nbsp; "Generalized Description of Digital Modulation Methods",&nbsp; the structure of the optimal receiver and the signal representation by means of basis functions were treated by the example of baseband transmission.

[[File:EN_Dig_T_4_4_S1_v2.png|right|frame|Equivalent low-pass model of carrier-modulated transmission methods|class=fit]]

With the same systematics and the same uniformity,&nbsp; band&ndash;pass systems will now also be considered which have already been described in earlier books or chapters,&nbsp; namely
*in the main chapter 4:&nbsp; "Digital Modulation Methods"&nbsp; of the book&nbsp; [[Modulationsverfahren|"Modulation Methods"]],<br>

*in the chapter&nbsp; [[Digital_Signal_Transmission/Lineare_digitale_Modulation_–_Kohärente_Demodulation|"Linear Digital Modulation - Coherent Demodulation"]]&nbsp; of the present book.<br><br>

In the following,&nbsp; we restrict ourselves to&nbsp; '''linear modulation methods'''&nbsp; and&nbsp; '''coherent demodulation'''.&nbsp; This means that&nbsp; "the receiver must know exactly the frequency and phase of the carrier signal added to the transmitter".&nbsp; 

In the following chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation|"Carrier Frequency Systems with Non-Coherent Demodulation"]]&nbsp; are discussed.

In the case of coherent demodulation,&nbsp; the entire transmission system can be described in the&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation#System_description_using_the_equivalent_low-pass_signal| "equivalent low-pass domain"]],&nbsp; and the relationship to baseband transmission is even more obvious than when band-pass signals are considered.

This results in the sketched model.&nbsp; Complex quantities are marked by a yellow filled double arrow.&nbsp; It should be noted with regard to this graph: <br> 
*From the incoming bit stream&nbsp; $\langle q_k \rangle \in \{\rm 0, \ L \}$,&nbsp; &nbsp; $b$&nbsp; data bits each are converted serially/parallel.&nbsp; These output bits result in the message&nbsp; $m \in \{m_0, \hspace{0.05cm}\text{...} \hspace{0.05cm}, m_{M-1} \}$,&nbsp; where&nbsp; $M = 2^b$&nbsp; indicates the level number.&nbsp; For the following,&nbsp; the message&nbsp; $m = m_i$&nbsp; is assumed.<br>

*In the &nbsp;'''signal space allocation''', &nbsp; a complex amplitude coefficient&nbsp; $a_i = a_{{\rm I}i} + {\rm j} \cdot a_{{\rm Q}i}$&nbsp; is assigned to each message&nbsp; $m_i$,&nbsp; whose real part will form the&nbsp; "in-phase component"&nbsp; and whose imaginary part will form the&nbsp; "quadrature component"&nbsp; of the later transmitted signal.<br>

*At the output of the blue marked block &nbsp; '''generation of the low-pass signal''' &nbsp; the (in general) complex-valued&nbsp; [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function|"equivalent low-pass signal"]]&nbsp; is present,&nbsp; where&nbsp; $g_s(t)$&nbsp; shall be limited for the time being to the range&nbsp; $ 0 \le t \le T$&nbsp; just like&nbsp; $s_{\rm TP}(t)$.&nbsp; The index&nbsp; $i$&nbsp; again provides an indication of the message&nbsp; $m_i$ sent:
::<math>s_{\rm TP}(t) \big {|}_{m \hspace{0.05cm}= \hspace{0.05cm} m_i} = a_i \cdot g_s(t) = a_{{\rm I}i} \cdot g_s(t) + {\rm j} \cdot a_{{\rm Q}i} \cdot g_s(t)</math>
*By energy normalization one gets from the basic transmission pulse&nbsp; $g_s(t)$&nbsp; to the basis function

:$$\varphi_1(t) = { g_s(t)}/{\sqrt{E_{gs}}} \hspace{0.4cm} {\rm with} \hspace{0.4cm} E_{gs} = 
 \int_{0}^{T} g_s(t)^2 \,{\rm d} t$$
:$$ \hspace{0.3cm}
\Rightarrow \hspace{0.3cm} s_{\rm TP}(t) \big {|}_{m\hspace{0.05cm} =\hspace{0.05cm} m_i} = s_{{\rm I}i} \cdot \varphi_1(t) + s_{{\rm Q}i}  \cdot {\rm j} \cdot \varphi_1(t)
 \hspace{0.05cm}.$$

*While the coefficients &nbsp; $a_{{\rm I}i}$ &nbsp; and &nbsp; $a_{{\rm Q}i}$ &nbsp; are dimensionless,&nbsp; the new coefficients &nbsp; $s_{{\rm I}i}$ &nbsp; and &nbsp; $s_{{\rm Q}i}$ &nbsp; have the unit&nbsp; "root of energy" &nbsp; &rArr; &nbsp; see page&nbsp; [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#Nomenclature_in_the_fourth_chapter|"Nomenclature in the fourth main chapter"]]:

:$$s_{{\rm I}i} = {\sqrt{E_{gs}}} \cdot a_{{\rm I}i}\hspace{0.05cm}, $$
:$$ s_{{\rm Q}i} = {\sqrt{E_{gs}}} \cdot a_{{\rm Q}i}\hspace{0.05cm}. $$

*The equations show that the system considered here is completely described in the equivalent low-pass&nbsp; $($German:&nbsp; "Tiefpass" &nbsp; &rArr; &nbsp; "TP"$)$&nbsp; domain by one real basis function&nbsp; $\varphi_1(t)$&nbsp; and one purely imaginary basis function&nbsp; $\psi_1(t) = {\rm j} \cdot \varphi_1(t)$&nbsp; each,&nbsp; or by a single complex basis function&nbsp; $\xi_1(t)$.&nbsp;<br>

*The gray shaded part shows the model for generating the band-pass signal&nbsp; $s_{\rm BP}(t)$,&nbsp; first the generation of the&nbsp;  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|"analytical signal"]]&nbsp; $s_{\rm +}(t) = s_{\rm TP}(t) \cdot {\rm e}^{{\rm j}2\pi \cdot f_{\rm T} \cdot T}$&nbsp; and then the real part formation.<br>

*The two basis functions of the band-pass signal&nbsp; $s_{\rm BP}(t)$&nbsp; result here as energy-normalized and time-limited to the range &nbsp; $0 \le t \le T$ &nbsp; cosine and minus-sine oscillations, respectively.<br><br>


== Coherent demodulation and optimal receiver ==
<br>
In the following,&nbsp; we always assume the equivalent low-pass signal unless explicitly stated otherwise.&nbsp; In particular,&nbsp; the signals&nbsp;
[[File:EN_Dig_T_4_4_S2.png|right|frame|AWGN channel model for complex signals '''KORREKTUR: optimal '''|class=fit]]
 
*$s(t) = s_{\rm TP}(t)$&nbsp; and&nbsp;
 
*$r(t) = r_{\rm TP}(t)$&nbsp; 


in the graph are&nbsp; "low-pass signals"&nbsp; and thus generally complex.&nbsp; The suffix&nbsp; "TP"&nbsp; is omitted in the remainder of this paper .<br>

To this figure is to be noted:
*The phase delay of the channel&nbsp; $($i.e. a phase function increasing linearly with frequency$)$&nbsp; is expressed in the low-pass range by the time-independent rotation factor &nbsp; ${\rm e}^{{\rm j}\hspace{0.05cm} \phi}$.&nbsp; <br>

*The signal&nbsp; $n\hspace{0.05cm}'(t)$&nbsp; describes a complex white Gaussian random process in the low-pass domain,&nbsp; as given in the section&nbsp; [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#N-dimensional_Gaussian_noise|"N-dimensional Gaussian noise"]].&nbsp; The apostrophe was added in order to be able to work with&nbsp; $n(t)$&nbsp; later in the overall system.<br>

*The receiver knows the channel phase &nbsp;  $\phi$ &nbsp; and corrects it by the conjugate-complex rotation factor&nbsp; ${\rm e}^{-{\rm j}\hspace{0.05cm}\phi}$.&nbsp; Thus,&nbsp; the received signal in the equivalent low-pass range is:

::<math>r(t) = s(t) + n\hspace{0.05cm}'(t) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\phi}= s(t) + n(t) \hspace{0.05cm}.</math>

*The phase rotation does not change the properties of the circular symmetric noise &nbsp; &rArr; &nbsp; $n(t) = n\hspace{0.05cm}'(t) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\phi}$&nbsp; has exactly the same statistical properties as&nbsp; $n\hspace{0.05cm}'(t)$.&nbsp; The left graphic in the figure above illustrates the facts just described. 
:#The right graph shows the overall system as used for the rest of the fourth main chapter.
:#The AWGN channel is followed by an optimal receiver according to the section&nbsp; [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#N-dimensional_Gaussian_noise|"N-dimensional Gaussian noise"]]. 


{{BlaueBox|TEXT=  
$\text{Definition:}$&nbsp; A &nbsp;'''symbol error'''&nbsp; occurs whenever &nbsp;$\hat{m}$&nbsp; does not match the sent message &nbsp;$m$:
::<math>m = m_i \hspace{0.2cm} \cap \hspace{0.2cm} \hat{m} \ne m_i  \hspace{0.05cm}.</math>}}

== On–off keying  (2–ASK) ==
<br>
The simplest digital modulation method is&nbsp; "On&ndash;off keying"&nbsp; $\rm (OOK)$,&nbsp; which has already been described in detail in the book&nbsp; [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying|"Modulation Methods"]]&nbsp; on the basis of its band-pass signals.&nbsp; There,&nbsp; this method was partly also called&nbsp; "Amplitude Shift Keying"&nbsp; $\rm (2&ndash;ASK)$.<br>

[[File:EN_Dig_T_4_4_S3.png|right|frame|Signal space constellations for on-off keying|class=fit]]

This method can be characterized as follows:
*OOK is a one-dimensional modulation method&nbsp; $(N = 1)$&nbsp; with&nbsp; $s_{{\rm I}i} = \{0, E^{1/2}\}$&nbsp; <u>and</u> &nbsp;$s_{{\rm Q}i} \equiv 0$&nbsp; or &nbsp;$s_{{\rm I}i} \equiv 0$&nbsp; <u>and</u> &nbsp;$s_{{\rm Q}i} = \{0, -E^{1/2}\}$.&nbsp; As an abbreviation,&nbsp; $E = E_{g_s}$. 

*The first combination describes a cosinusoidal carrier signal,&nbsp; the second combination a sinusoidal carrier.<br>

*Each bit is assigned to a binary symbol&nbsp; $(b = 1, \ M = 2)$; thus,&nbsp; no serial/parallel converter is needed. 

*For equally probable symbols,&nbsp; which is assumed for what follows,&nbsp; both the&nbsp; "mean energy per symbol"&nbsp; $(E_{\rm S})$&nbsp; and the&nbsp; "mean energy per bit"&nbsp; $(E_{\rm B})$&nbsp; are equal to&nbsp; $E/2$.<br>

*The optimal OOK receiver virtually projects the complex&ndash;valued received signal&nbsp; $r(t)$&nbsp; onto the basis function&nbsp; $\varphi_1(t)$,&nbsp; if one starts from the left sketch&nbsp; (cosine carrier).<br>

*Because of&nbsp; $N = 1$,&nbsp; the noise can be one-dimensional with the variance&nbsp; $\sigma_n^2 = N_0/2$.&nbsp; 

*Using the statements in the section&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Error_probability_for_symbols_with_equal_probability|"Error probability for symbols with equal probability"]],&nbsp; we obtain for the (mean)&nbsp; "symbol error probability":

::<math>p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Q} \left ( \frac{d/2}{\sigma_n}\right ) =  {\rm Q} \left ( \sqrt{\frac{E}{2 N_0}}\right )
 =  {\rm Q} \left ( \sqrt{{E_{\rm S}}/{N_0}}\right ) \hspace{0.05cm}.</math>

*Since each bit is mapped to one symbol,&nbsp; the average bit error probability&nbsp; $p_{\rm B}$&nbsp; is exactly:

::<math>p_{\rm B}  
 =  {\rm Q} \left ( \sqrt{{E_{\rm S}}/{N_0}}\right ) =  {\rm Q} \left ( \sqrt{{E_{\rm B}}/{N_0}}\right ) \hspace{0.05cm}.</math>

== Binary phase shift keying  (BPSK) ==
<br>
The very often used method&nbsp; "Binary Phase Shift Keying"&nbsp; $\rm (BPSK)$,&nbsp; which was already described in detail in the chapter&nbsp; [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|"Linear Digital Modulation"]]&nbsp; of the book "Modulation Methods"&nbsp; using the band&ndash;pass signals&nbsp; $($typical: &nbsp; phase jumps$)$,&nbsp; differs from&nbsp; "On&ndash;off keying"&nbsp; by a constant envelope.<br>

For the signal space points,&nbsp; $\boldsymbol{s}_1 = -\boldsymbol{s}_0$ always holds.&nbsp; For example:
*with cosine carrier: &nbsp; $s_{{\rm I}i} = \{\pm E^{1/2}\}$&nbsp; and&nbsp; $s_{{\rm Q}i} \equiv 0$,&nbsp; <br>

*with sinusoidal carrier: &nbsp; $s_{{\rm I}i} \equiv 0$&nbsp; and&nbsp; $s_{{\rm Q}i} = \{\pm E^{1/2}\}$.

[[File:EN_Dig_T_4_4_S4_v2.png|right|frame|Signal space constellations of the BPSK|class=fit]]

The improvements compared to on&ndash;off keying can be seen from the equations given in the graphic&nbsp; $($in the field with green background$)$:
*For a given normalization energy&nbsp; $E$,&nbsp; the distance between&nbsp;  $\boldsymbol{s}_0$&nbsp; and&nbsp;  $\boldsymbol{s}_1$&nbsp; is twice as large as with OOK. 

* This gives the error probability (both related to symbols and bits):
::<math>p_{\rm S} = p_{\rm B} = {\rm Pr}({\cal{E}})  =  {\rm Q} \left ( \sqrt{{2 E}/{N_0}}\right )
 =  {\rm Q} \left ( \sqrt{{2  E_{\rm S}}/{N_0}}\right ) \hspace{0.05cm}.</math>

*Now&nbsp; $E_{\rm S} = E_{\rm B} = E$&nbsp; applies,&nbsp; which means that the average energies per symbol or per bit are now twice as large as with OOK.<br>

*Because of factor&nbsp; $2$&nbsp; in the square root in Q-function's argument,&nbsp; the BPSK error probability is noticeably lower than OOK with same &nbsp; $E_{\rm S}$&nbsp; and&nbsp; $N_0$.<br>

*In other words: &nbsp; With the same&nbsp; $N_0$,&nbsp; BPSK only requires half the symbol energy&nbsp;  $E_{\rm S}$&nbsp; in order to achieve the same error probability as OOK.&nbsp; The logarithmic gain is&nbsp; $3 \ \rm dB$.<br>

== M–level amplitude shift keying  (M–ASK) ==
<br>
In analogy to&nbsp; [[Digital_Signal_Transmission/Redundancy-Free_Coding#Error_probability_of_a_multilevel_system| "M&ndash;level baseband transmission"]],&nbsp; we now consider&nbsp; "M&ndash;level Amplitude Shift Keying"&nbsp;  $\text{(M&ndash;ASK)}$,&nbsp; whose low-pass signal space constellation for the parameters&nbsp; $b = 3$  &nbsp; &#8658; &nbsp; $M = 8$ &nbsp; &#8658; &nbsp; "$\text{8&ndash;ASK}$"&nbsp; looks as follows.<br>

The name &nbsp;"M&ndash;ASK"&nbsp; is not entirely accurate.&nbsp; Rather,&nbsp; it is a&nbsp; "combined ASK/PSK method",&nbsp; since e.g.
*the two innermost signal space points&nbsp; $(\pm 1)$&nbsp; do not differ in terms of amplitude&nbsp; ("envelope"), 
*but only in terms of phase&nbsp; $(0^\circ$ or $180^\circ)$.
[[File:EN_Dig_T_4_4_S5.png|right|frame|Signal room constellation of the 8-ASK|class=fit]] 


It should also be noted:
*The&nbsp;  "average energy per symbol"&nbsp; can be calculated as follows for this one-dimensional modulation method using symmetry:
::<math>E_{\rm S} = \frac{2}{M} \cdot \sum_{k = 1}^{M/2} (2k -1)^2 \cdot E =  \frac{M^2 -1}{3} \cdot E \hspace{0.05cm}.</math>

*Since each of the&nbsp; $M$&nbsp; symbols represents&nbsp; $b = \log_2  (M)$&nbsp; bits,&nbsp; the&nbsp; "average energy per bit"&nbsp; is:
:$$E_{\rm B} = \frac{E_{\rm S}}{b} =   \frac{E_{\rm S}}{{\rm log_2}\, (M)} =\frac{M^2 -1}{3 \cdot {\rm log_2}\, (M)} \cdot E$$ 
:$$\Rightarrow\hspace{0.3cm}M= 8\hspace{-0.1cm}: \hspace{0.2cm} E_{\rm S}/E = 21
\hspace{0.05cm}, \hspace{0.2cm}E_{\rm B}/E = 7\hspace{0.05cm}.$$

*The probability that one of the two outer symbols is falsified due to AWGN noise is therefore the same:
::<math>{\rm Pr}({\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} \text{outer symbol)} =  {\rm Q} \left ( \sqrt{{2 E}/{N_0}}\right )\hspace{0.05cm}.</math>

*The falsification probability of the&nbsp; $M-2$&nbsp; inner symbols is twice as large, since other decision regions border on both the right and the left.&nbsp; By averaging one obtains for the&nbsp; "symbol error probability":
::<math>p_{\rm S} = {\rm Pr}({\cal{E}}) = \frac{1}{M} \cdot \left [ 2 \cdot  1 \cdot {\rm Q} \left ( \sqrt{{2 E}/{N_0}}\right ) +
 (M-2) \cdot  2 \cdot {\rm Q} \left ( \sqrt{{2 E}/{N_0}}\right ) \right ] </math>
::<math>\Rightarrow \hspace{0.3cm} p_{\rm S} = \frac{2 \cdot (M-1)}{M} \cdot {\rm Q} \left ( \sqrt{{2 E}/{N_0}}\right ) =\frac{2 \cdot (M-1)}{M} \cdot {\rm Q} \left ( \sqrt{\frac{6 \cdot E_{\rm S}}{(M^2-1) \cdot N_0}}\right )
 \hspace{0.05cm}.</math>

*When using the&nbsp; [[Digital_Signal_Transmission/Redundancy-Free_Coding#Symbol_and_bit_error_probability|"Gray code"]]&nbsp; $($neighboring symbols each differ by one bit$)$,&nbsp; the&nbsp; "bit error probability"&nbsp; $p_{\rm B}$ is approximately factor&nbsp; $b = \log_2 \ (M)$&nbsp; smaller than the&nbsp; $p_{\rm S}$:

::<math>p_{\rm B} \approx \frac{p_{\rm S}}{b} =  \frac{2 \cdot (M-1)}{M \cdot {\rm log_2}\, (M)} \cdot {\rm Q} \left ( \sqrt{{6 \cdot {\rm log_2}\, (M)}/({M^2-1 }) \cdot { E_{\rm B}}/{ N_0}}\right )
 \hspace{0.05cm}.</math>

== Quadrature amplitude modulation (M-QAM) ==
<br>
[[Modulation_Methods/Quadrature_Amplitude_Modulation#General_description_and_signal_space_allocation|"Quadrature amplitude modulation"]]&nbsp;  $\text{(M&ndash;QAM)}$ results from a&nbsp; M&ndash;ASK each for the&nbsp; "in-phase component"&nbsp; and the "quadrature component"&nbsp; &#8658; &nbsp; $M^2$&nbsp; signal space points.<br>
[[File:P ID2057 Dig T 4 4 S6 version1.png|right|frame|Signal space constellation of 16-QAM]]
*Each symbol now represents&nbsp; $b = \log_2  (M)$&nbsp; binary characters&nbsp; (bits).
 
*The graphic shows the special case&nbsp; $M = 16$ &nbsp; &#8658; &nbsp;  $b = 4$. 

*The bit assignment for&nbsp; [[Digital_Signal_Transmission/Redundancy-Free_Coding#Symbol_and_bit_error_probability|"Gray coding"]]&nbsp; is shown in red (neighboring symbols each differ by one bit).<br>


The&nbsp; "average energy per symbol" &nbsp; $(E_{\rm S})$ &nbsp; or the&nbsp; "average energy per bit" &nbsp; $(E_{\rm B})$  &nbsp;can be easily derived from the result for the&nbsp; "M&ndash;ASK" &nbsp; $($note the difference in the equation between an energy&nbsp; "$E$"&nbsp; and the expected value&nbsp; "$\rm E[\text{...}]$"$)$:
::<math>E_{\rm S} = {\rm E} \left [ |s_{i}|^2 \right ] = {\rm E} \left [ |s_{{\rm I}i}|^2  \right ] + {\rm E} \left [ |s_{{\rm Q}i}|^2 \right ] = 2 \cdot {\rm E} \left [ |s_{{\rm I}i}|^2  \right ]</math>
::<math>\Rightarrow \hspace{0.3cm} E_{\rm S} = 2 \cdot \frac{M_{\rm I}^2-1}{3} \cdot E = \frac{2}{3} \cdot (M-1) \cdot E\hspace{0.01cm},\hspace{0.3cm}E_{\rm B} =\frac{2 \cdot (M-1)}{3 \cdot {\rm log_2}\, (M)} \cdot E \hspace{0.01cm}.</math>

In addition, &nbsp; the M&ndash;level quadrature amplitude modulation shows the following properties:
*The&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Union_Bound_-_Upper_bound_for_the_error_probability|"Union Bound"]]&nbsp; can be used as an upper bound for the symbol error probability,&nbsp; whereby it should be noted that an inner symbol can be falsified in four directions:
::<math>p_{\rm S} = {\rm Pr}({\cal{E}}) \le \left\{ \begin{array}{c}  4 \cdot p \\
 2 \cdot p  \end{array} \right.\quad
 \begin{array}{*{1}c} {\rm for}  \hspace{0.15cm} M \ge 16 \hspace{0.05cm},
\\  {\rm for}  \hspace{0.15cm} M = 4 \hspace{0.05cm},\\ \end{array}  \hspace{0.4cm} {\rm with} \hspace{0.4cm} p = {\rm Q} \left ( \sqrt{{2 E}/{N_0}}\right ) 
 \hspace{0.05cm}.</math>

*If one takes into account that only the&nbsp; $(b-2)^2$&nbsp; inner points are falsified in four directions,&nbsp; in contrast,&nbsp; the four vertices are falsified only in two and the remaining points in three directions&nbsp; (blue arrows in the graphic),&nbsp; therefore one obtains with&nbsp; $M = b^2$&nbsp; the better approximation

::<math>p_{\rm S} \approx   {1}/{M} \cdot \big [(b - 2)^2 \cdot 4p + 4 \cdot 2p + 4 \cdot (b - 2) \cdot 3p \big ] = {p}/{M} \cdot \big [ 4 \cdot M - 16 \cdot \sqrt{M} + 16 + 8  + 12 \cdot \sqrt{M} - 24\big ] </math>
::<math>\Rightarrow \hspace{0.3cm} p_{\rm S} \approx  {4 \cdot p}/{M} \cdot \big [ M -  \sqrt{M} \big ] = 4p \cdot \hspace{0.05cm} \big [ 1 -  {1}/{\sqrt{M}} \hspace{0.05cm}\big ] </math>
::<math>\Rightarrow\hspace{0.3cm} M = 16\hspace{-0.1cm}:  \hspace{0.1cm}
p_{\rm S} \approx  3 \cdot p = 3 \cdot {\rm Q} \big ( \sqrt{{2 E}/{N_0}}\big ) = 3 \cdot {\rm Q} \big ( \sqrt{{1/5 \cdot E_{\rm S}}/{ N_0}}\big ) \hspace{0.05cm}.</math>

{{BlaueBox|TEXT=  
$\text{Conclusion:}$&nbsp; With &nbsp;$M$&ndash;level QAM, &nbsp; $E_{\rm B} = E_{\rm S}/\log_2 \hspace{0.05cm} (M)$&nbsp; generally applies and with Gray coding,&nbsp; $p_{\rm B} = p_{\rm S}/\log_2 \hspace{0.05cm} (M)$ is also applicable.

*This gives the&nbsp; '''mean bit error probability''':
::<math>p_{\rm B} \approx \frac{4 \cdot (1 - 1/\sqrt{M})}{ {\rm log_2}\hspace{0.05cm} (M)} \cdot {\rm Q} \left ( \sqrt{ \frac{3 \cdot {\rm log_2}\, (M)}{M-1 } \cdot { E_{\rm B} }/{ N_0} }\right )
 \hspace{0.05cm}.</math>

*The approximation is exactly valid for&nbsp; $M \le 16$&nbsp; if&nbsp; &ndash; as assumed for the upper graphic &ndash;&nbsp; no&nbsp; "diagonal falsifications"&nbsp; occur.

*The special case&nbsp; "4&ndash;QAM"&nbsp; (without inner symbols)&nbsp; is dealt with in &nbsp;[[Aufgaben:Exercise_4.13:_Four-level_QAM|"Exercise 4.13"]].&nbsp; <br>}}

== Multi-level phase–shift keying  (M–PSK) ==
<br>
In the case of multi-level phase modulation,&nbsp; in which case the level number&nbsp; $M$&nbsp; is usually a power of two in practice,&nbsp; all signal space points are evenly distributed on a circle with radius&nbsp; $E^{1/2}$.&nbsp; This means that &nbsp;
[[File:P ID2064 Dig T 4 4 S7 version2.png|right|frame|Signal space constellation of the 8–PSK and 16–PSK|class=fit]]
 
#$E_{\rm S} = E$&nbsp; holds for the&nbsp; "'average symbol energy",&nbsp; and 
#$E_{\rm B} = E_{\rm S}/b = E/\hspace{-0.05cm}\log_2 \hspace{0.05cm} (M)$ for the&nbsp; "average energy per bit".&nbsp;<br>

*For the in-phase and quadrature components of the signal space points&nbsp; $\boldsymbol{s}_i$,&nbsp; the general rule is &nbsp;$(i = 0, \hspace{0.05cm}\text{...} \hspace{0.05cm}, \hspace{0.05cm}M-1)$:
:$$s_{{\rm I}i} = \cos \left ( { 2\pi i}/{ M} + \phi_{\rm off} \right ) \hspace{0.05cm},$$
:$$ s_{{\rm Q}i} = \sin \left ( { 2\pi i}/{ M} + \phi_{\rm off} \right )$$
:$$\Rightarrow \hspace{0.2cm} || \boldsymbol{ s}_i || = \sqrt{ s_{{\rm I}i}^2 +  s_{{\rm Q}i}^2} = 1 \hspace{0.05cm}.$$ 
*The phase offset is set to&nbsp; $\phi_{\rm off} = 0$&nbsp; in the graphic above. The 4&ndash;PSK with&nbsp; $\phi_{\rm off} = \pi/4 \ (45^\circ)$&nbsp; is identical to the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"4&ndash;QAM"]]. 

*The distance between two adjacent points is the same in all cases:<br>
::<math>d_{\rm min}  =  d_{\rm 0, \hspace{0.05cm}1} =  d_{\rm 1, \hspace{0.05cm}2} = \hspace{0.05cm}\text{...} \hspace{0.05cm} =  d_{M-1,  \hspace{0.05cm}0} = 2 \cdot \sqrt{E} \cdot \sin (\pi/M)</math>
::<math>\Rightarrow\hspace{0.3cm} M = 4\hspace{-0.1cm}:\hspace{0.1cm}d_{\rm min}/E^{1/2}  =  \sqrt{2} \approx 1.414  \hspace{0.05cm}, \hspace{0.8cm} M = 8\hspace{-0.1cm}:\hspace{0.1cm}d_{\rm min}/E^{1/2}   \approx 0.765  \hspace{0.05cm},\hspace{0.8cm} M = 16\hspace{-0.1cm}:\hspace{0.1cm}d_{\rm min}/E^{1/2}   \approx 0.390  \hspace{0.05cm}.</math>

*The upper bound&nbsp; $p_{\rm UB}$&nbsp; for the AWGN symbol error probability&nbsp; $p_{\rm S}$&nbsp; after the&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Union_Bound_-_Upper_bound_for_the_error_probability|"Union Bound"]]&nbsp; yields:

::<math>p_{\rm UB} =  2 \cdot {\rm Q} \left ( \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )  \ge  p_{\rm S}  
 \hspace{0.05cm}.</math>

*One recognises:
#For&nbsp; $M = 2$&nbsp; $\rm (BPSK)$&nbsp; one obtains the estimate &nbsp; $p_{\rm S}  \le p_{\rm UB} =2 \cdot {\rm Q} \left ( \sqrt{ 2E_{\rm S}/{ N_0} }\right )$.&nbsp; A comparison with the equation &nbsp; $p_{\rm S}   ={\rm Q} \left ( \sqrt{ 2E_{\rm S}/{ N_0} }\right )$ &nbsp; given on the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Binary_phase_shift_keying_.28BPSK.29|"BPSK section"]]&nbsp; shows that in this special case the&nbsp; "Union Bound"&nbsp; returns double the value as the upper limit.
#The larger&nbsp; $M$&nbsp; is,&nbsp; the more precisely&nbsp; $p_{\rm UB}$&nbsp; approximates the exact symbol error probability&nbsp; $p_{\rm S}$.&nbsp; The interactive SWF applet&nbsp; [[Applets:MPSK_%26_Union-Bound(Applet)|"Multi-level PSK & Union Bound"]]&nbsp; also gives the more accurate error probability obtained through simulation.<br>


{{BlaueBox|TEXT=  
$\text{Conclusion:}$&nbsp; The limit for the&nbsp; '''M&ndash;PSK bit error probability'''&nbsp; is,&nbsp; assuming Gray code &nbsp;&#8658;&nbsp; red labeling:

::<math>p_{\rm B}  \le \frac{2}{ {\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{ {\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm B} } }/{ N_0} }\right )  
 \hspace{0.05cm}.</math>

*However,&nbsp; this limit only has to be applied for&nbsp; $M > 4$.&nbsp;
 
*For&nbsp; $M = 2$&nbsp; (BPSK)&nbsp;  and&nbsp; $M = 4$&nbsp; (identity between 4&ndash;PSK and 4&ndash;QAM),&nbsp; the bit error probability can be specified exactly:
:$$p_{\rm B}  = {\rm Q} \left (  \sqrt{ { {2E_{\rm B} } }/{ N_0} }\right )  
 \hspace{0.05cm}.$$}}

== Binary frequency shift keying (2–FSK) ==
<br>
This type of modulation with parameter&nbsp; $b = 1$ &nbsp; &#8658; &nbsp; $M = 2$ &nbsp; has already been described in detail in the section&nbsp; [[Modulation_Methods/Non-Linear_Digital_Modulation#FSK_.E2.80.93_Frequency_Shift_Keying|"FSK &ndash; Frequency Shift Keying"]]&nbsp; of the book&nbsp; "Modulation Methods"&nbsp; using the band-pass signals.

*The two possible signal forms are represented by two different frequencies in the range&nbsp; $0 \le t \le T$:&nbsp; <br>
::<math>s_{\rm BP0}(t) \hspace{-0.1cm}  =  \hspace{-0.1cm}  A \cdot \cos( 2\pi \cdot( f_{\rm T} + \Delta f_{\rm A})\cdot t)\hspace{0.05cm},</math>
::<math> s_{\rm BP1}(t) \hspace{-0.1cm}  =  \hspace{-0.1cm}  A \cdot \cos( 2\pi \cdot( f_{\rm T} - \Delta f_{\rm A})\cdot t)\hspace{0.05cm}.</math>

*$f_{\rm T}$&nbsp; designates the&nbsp; "carrier frequency"&nbsp; and&nbsp; $\Delta f_{\rm A}$&nbsp; the (one-sided)&nbsp; "frequency deviation".&nbsp; The average energy per symbol or per bit is the same in each case:

::<math>E_{\rm S} = E_{\rm B} = E = \frac{A^2 \cdot T}{2}
 \hspace{0.05cm}.</math>

*The FSK in the equivalent low-pass signal space is now to be considered here.&nbsp; Then:

::<math>s_{\rm TP0}(t) \hspace{-0.1cm}  =  \hspace{-0.1cm}  \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} 2\pi \hspace{0.03cm}\cdot \hspace{0.03cm} \Delta f_{\rm A} \hspace{0.03cm}\cdot t}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},</math>
::<math> s_{\rm TP1}(t) \hspace{-0.1cm}  =  \hspace{-0.1cm}  \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} 2\pi \hspace{0.03cm}\cdot \hspace{0.03cm} \Delta f_{\rm A} \hspace{0.03cm}\cdot t}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},</math>

:and for the&nbsp; "inner product"&nbsp; one obtains:

::<math>< \hspace{0.02cm} s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{0.02cm}> \hspace{0.1cm}  =  \hspace{-0.1cm}  
 \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t = A^2 \cdot \int_{0}^{T} {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} 4\pi \hspace{0.03cm}\cdot \hspace{0.03cm} \Delta f_{\rm A} \hspace{0.03cm}\cdot t} \,{\rm d} t =  \frac{A^2}{{\rm j} \cdot 4\pi \cdot  \Delta f_{\rm A}} \cdot \big [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} 4\pi \hspace{0.03cm}\cdot \hspace{0.03cm} \Delta f_{\rm A} \hspace{0.03cm}\cdot T} - 1 \big ] \hspace{0.05cm}.</math>

{{BlaueBox|TEXT=  
$\text{Definition:}$&nbsp; The&nbsp; '''modulation index'''&nbsp; $h = 2 \cdot \Delta f_{\rm A}\hspace{0.03cm}\cdot T$&nbsp; is the ratio 
*between the total (bilateral) frequency deviation&nbsp; $(2 \cdot \Delta f_{\rm A})$&nbsp; 

*and the symbol rate&nbsp; $(1/T)$.}} 


*The two signals are&nbsp; "orthogonal"&nbsp; if this inner product is equal to zero:

[[File:P ID2076 Dig T 4 4 S8 version2.png|right|frame|Signal space constellation of the FSK, if &nbsp;$h$&nbsp; is an integer|class=fit]]
:$$< \hspace{0.02cm} s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{0.02cm}> \hspace{0.1cm}  =    \frac{A^2\cdot T}{{\rm j} \cdot 2\pi \cdot  h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} 2h} - 1 \right ] = 0$$
:$$ \Rightarrow \hspace{0.3cm} h = 2 \cdot \Delta f_{\rm A} \cdot T = 1,\hspace{0.1cm} 2, \hspace{0.1cm}3,\ \text{ ... }\hspace{0.05cm}.$$

*If the modulation index&nbsp; $h$&nbsp; is assumed to be an integer, the low-pass signals can be written in the form

::<math>s_{\rm TP0}(t)  = \sqrt{E} \cdot \xi_1(t) \hspace{0.05cm},</math>
::<math>s_{\rm TP1}(t) = \sqrt{E} \cdot \xi_2(t)</math>

:with complex basis functions:

:$$\xi_1(t) = \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
:$$ \xi_2(t)= \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T \hspace{0.05cm}.$$

*The result is the signal space representation of the binary FSK outlined here.<br>


{{BlaueBox|TEXT=  
$\text{Conclusions:}$&nbsp; 
*With an integer modulation index&nbsp; $h$,&nbsp; the low-pass signals &nbsp; $s_{\rm TP0}(t)$&nbsp; and &nbsp; $s_{\rm TP1}(t)$&nbsp; of the binary FSK are orthogonal to one another.<br>

*This results in the symbol error probability&nbsp; (derivation in the graphic):

::<math>p_{\rm S} = {\rm Pr}({\cal{E} }) = {\rm Q} \left (  \sqrt{ { {E_{\rm S} } }/{ N_0} }\right ) 
 \hspace{0.05cm}.</math>

*The bit error probability has the same value: &nbsp; $p_{\rm B} = p_{\rm S}$.}}<br>

<u>Note:</u> 
#In contrast to the representation in&nbsp; [KöZ08]<ref>Kötter, R., Zeitler, G.: Nachrichtentechnik 2. Vorlesungsmanuskript, Lehrstuhl für Nachrichtentechnik, Technische Universität München, 2008.</ref>,&nbsp; the frequency deviation&nbsp; $\Delta f_{\rm A}$&nbsp; is defined here on one side. 
#Therefore,&nbsp; the equations sometimes differ by a factor of&nbsp; $2$.&nbsp; However,&nbsp; if you work with the modulation index&nbsp; $h$,&nbsp; there are no differences.<br>

== Minimum Shift Keying (MSK) ==
<br>
[[Modulation_Methods/Non-Linear_Digital_Modulation#MSK_.E2.80.93_Minimum_Shift_Keying| "Minimum Shift Keying"]]&nbsp; $\rm (MSK)$&nbsp; is a binary FSK system with the modulation index&nbsp; $h = 0.5$ &nbsp; &#8658; &nbsp; frequency deviation $\Delta f_{\rm A} = 1/(2T)$. The graphic shows an MSK signal for the carrier frequency&nbsp; $ f_{\rm T} = 4/T$:
[[File:P ID2072 Dig T 4 4 S9 version2.png|right|frame|Source signal and band-pass MSK signal|class=fit]]
 
*The two frequencies within the transmitted signal are&nbsp; $ f_{\rm 0} = f_{\rm T} + 1/(4T)$&nbsp; to represent the message&nbsp; $m_0$&nbsp; (yellow background)&nbsp; and&nbsp; $ f_{\rm 1} = f_{\rm T} -1/(4T)$ &nbsp; &#8658; &nbsp; message&nbsp; $m_1$&nbsp; (green background).
 
*The graph also accounts for continuous phase adjustment at the transitions to further reduce the signal bandwidth.&nbsp; This is then referred to as&nbsp; [[Modulation_Methods/Non-Linear_Digital_Modulation#Binary_FSK_with_Continuous_Phase_Matching|"Continuous Phase Modulation"]]&nbsp; $\rm (CPM)$.<br>


Without this phase adjustment,&nbsp; the two band-pass waveforms are:

::<math>s_{\rm BP0}(t) = \sqrt{2E/T} \cdot \cos( 2\pi  f_0  t)\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},</math>
::<math> s_{\rm BP1}(t) = \sqrt{2E/T} \cdot \cos( 2\pi  f_1  t)\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm}.</math>

If you form the inner product of the band-pass signals,&nbsp; you get with the abbreviation&nbsp; $f_{\rm \Delta} = f_0 - f_1$&nbsp; and&nbsp; $f_{\rm \Sigma} = f_0 + f_1$:

::<math>< \hspace{0.02cm} s_{\rm BP0}(t) \hspace{0.2cm}  \cdot  \hspace{0.2cm} s_{\rm BP1}(t) \hspace{0.02cm}> \hspace{0.2cm} =  
  {2E}/{T} \cdot \int_{0}^{T} \cos( 2\pi  f_0  t) \cdot \cos( 2\pi  f_1  t)\,{\rm d} t =   {E}/{T} \cdot \int_{0}^{T} \cos( 2\pi f_{\rm \Delta} t) \,{\rm d} t +    {E}/{T} \cdot \int_{0}^{T} \cos( 2\pi f_{\rm \Sigma} t) \,{\rm d} t</math>
::<math> \Rightarrow \hspace{0.3cm}< \hspace{0.02cm} s_{\rm BP0}(t) \hspace{0.2cm}  \cdot  \hspace{0.2cm} s_{\rm BP1}(t) \hspace{0.02cm}> \hspace{0.2cm} =
   {E}/{T} \cdot \int_{0}^{T} \hspace{-0.1cm} \cos( \pi \cdot {t}/{T}) \,{\rm d} t + 
   {E}/{T} \cdot \int_{0}^{T} \hspace{-0.1cm}\cos( 2\pi \cdot 2 f_{\rm T}  \cdot t) \,{\rm d} t
   \hspace{0.05cm}.</math>

#The first integral is zero&nbsp; $($integral over&nbsp; "cosine"&nbsp; from&nbsp; $0$&nbsp; to &nbsp;$\pi)$. 
#For&nbsp; $f_{\rm T} \gg 1/T$,&nbsp; which can be assumed in practice,&nbsp; the second integral also vanishes. 
#This gives the inner product: &nbsp;  &nbsp; $< \hspace{0.02cm} s_{\rm BP0}(t)  \cdot  s_{\rm BP1}(t) \hspace{0.02cm}> \hspace{0.2cm}=   0 \hspace{0.05cm}.$


{{BlaueBox|TEXT=  
$\text{Conclusions:}$&nbsp;
 
'''(1)''' &nbsp; This shows that the two band-pass signals are orthogonal for the modulation index&nbsp;  $h = 0.5$&nbsp; $($i.e. &nbsp;$\rm MSK)$&nbsp; and all multiples thereof.

'''(2)''' &nbsp; With the new real basis functions

::<math>\varphi_1(t) = \sqrt{2/T} \cdot \cos( 2\pi  f_0  t)\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},</math>
::<math> \varphi_2(t) = \sqrt{2/T} \cdot \cos( 2\pi  f_1  t)\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T</math>

:one obtains exactly the same signal space constellation as for even-numbered &nbsp; $h = 1, 2, 3, \ \text{ ...}$.
'''(3)''' &nbsp; This results in the same&nbsp; $($symbol resp. bit$)$&nbsp; error probability:

::<math>p_{\rm S} = {\rm Pr}({\cal{E} }) = {\rm Q} \left (  \sqrt{ { {E_{\rm S} } }/{ N_0} }\right ) = p_{\rm B} 
 \hspace{0.05cm}.</math>}}

==Exercises for the chapter==
<br>
[[Aufgaben:Exercise_4.11:_On-Off_Keying_and_Binary_Phase_Shift_Keying|Exercise 4.11: On-Off Keying and Binary Phase Shift Keying]]

[[Aufgaben:Exercise_4.11Z:_OOK_and_BPSK_once_again|Exercise 4.11Z: OOK and BPSK once again]]

[[Aufgaben:Exercise_4.12:_Calculations_for_the_16-QAM|Exercise 4.12: Calculations for the 16-QAM]]

[[Aufgaben:Exercise_4.13:_Four-level_QAM|Exercise 4.13: Four-level QAM]]

[[Aufgaben:Exercise_4.14:_8-PSK_and_16-PSK|Exercise 4.14: 8-PSK and 16-PSK]]

[[Aufgaben:Exercise_4.14Z:_4-QAM_and_4-PSK|Exercise 4.14Z: 4-QAM and 4-PSK]]

[[Aufgaben:Exercise_4.15:_Optimal_Signal_Space_Allocation|Exercise 4.15: Optimal Signal Space Allocation]]

[[Aufgaben:Exercise_4.16:_Binary_Frequency_Shift_Keying|Exercise 4.16: Binary Frequency Shift Keying]]

==References==

<references/>

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