Difference between revisions of "Aufgaben:Exercise 4.16: Comparison between Binary PSK and Binary FSK"

Latest revision as of 11:52, 12 April 2022

Bit error probability curves:
Binary PSK and binary FSK

The graph shows the bit error probability for binary FSK modulation $\rm (BFSK)$ in

coherent demodulation , as well as in
non-coherent demodulation

in comparison with binary phase modulation $\rm (BPSK)$.

Orthogonality is always assumed. For coherent demodulation, the modulation index can be a multiple of $h = 0.5$ , so that the middle plot can also apply to Minimum Shift Keying $\rm (MSK)$ . In contrast, for non-coherent demodulation of BFSK, the modulation index must be a multiple of $h = 1$ .

This system comparison is once again based on the AWGN channel , characterized by the relationship $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows

for Binary Phase Shift Keying $\rm (BPSK)$:

$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$

for Binary Frequency Shift Keying $\rm (BFSK)$ with coherent demodulation:

$$p_{\rm B} = {\rm Q}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/(2 N_0 )} \hspace{0.1cm}\right ),$$

for Binary Frequency Shift Keying $\rm (BFSK)$ with incoherent demodulation:

$$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$

It was shown in Exercise 4.8 , that for BPSK, the log ratio $10 · \lg \ E_{\rm B}/N_0$ must be at least $9.6 \ \rm dB$ so that the bit error probability does not exceed $p_{\rm B} = 10^{–5}$ .

Hints:

This exercise belongs to the chapter Nonlinear_Digital_Modulation.
However, reference is also made to the page Linear Digital Modulation.

Use the approximation $\lg(2) ≈ 0.3$.

Questions

$10 · \lg \ E_{\rm B}/N_0 \ = \ $

$\ \rm dB$

	FSK with modulation index $h = 0.7$,
	FSK with modulation index $h = 1$?

$10 · \lg \ E_{\rm B}/N_0 \ = \ $

$\ \rm dB$

$p_{\rm B} \ = \ $

$\ \cdot 10^{-4}$

Solution

(1) A comparison of the first two equations on the exercise page makes it clear that for MSK with coherent demodulation, the AWGN ratio $E_{\rm B}/N_0$ must be doubled to achieve the same error probability as for BPSK.

In other words: the coherent BFSK curve is $10 · \lg (2) ≈ 3 \ \rm dB$ to the right of the the BPSK curve.
Thus, to guarantee $p_{\rm B} \le 10^{–5}$ it must hold that:

$$10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}} /{N_{\rm 0}}= 9.6\,\,{\rm dB} + 3\,\,{\rm dB} = \underline{12.6\,\,{\rm dB}}\hspace{0.05cm}.$$

(2) Answer 2 is correct:

The equation given does not just hold for MSK $($this is a FSK with $h = 0.5)$, but also for every form of orthogonal FSK.
This is the case whenever the modulation index $h$ is an integer multiple of $0.5$ , such as when $h = 1$.
When $h = 0.7$ , there is no orthogonal FSK.

(3) From the inverse function of the equation given we get:

$$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm ln}(50000)\approx 10.82 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{E_{\rm B}} /{N_{\rm 0}}= 21.64 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}}/ {N_{\rm 0}}\approx \underline{13.4\,\,{\rm dB}}\hspace{0.05cm}.$$

(4) From $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$ it follows that:

$${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} ({E_{\rm B}} /{N_{\rm 0}})/2 \approx 8.4 \hspace{0.25cm} \Rightarrow \hspace{0.25cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4} \approx \underline{1.12 \cdot 10^{-4}}\hspace{0.05cm}.$$

This means that for the same $E_{\rm B}/N_0$, the error probability of incoherent demodulation is increased by a factor of $11$ compared to that of coherent demodulation (see answer to question 1) .

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1746__Mod_A_4_15.png|right|frame|Bit error probability curves <br>of binary PSK and binary FSK <b>KORREKTUR</b>]]
+[[File:EN_Mod_A_4_15_neu.png|right|frame|Bit error probability curves: <br>Binary PSK and binary FSK]]
-<b> KORREKTUR: Titel "Erercise" </b>
 The graph shows the bit error probability for binary&nbsp;[[Modulation_Methods/Nonlinear_Digital_Modulation#FSK_.E2.80.93_Frequency_Shift_Keying| FSK modulation]]&nbsp; $\rm (BFSK)$&nbsp; in
 *[[Modulation_Methods/Nonlinear_Digital_Modulation#Coherent_demodulation_of_FSK|coherent demodulation]]&nbsp;, as well as in
@@ Line 63: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Ein Vergleich der beiden ersten Gleichungen auf der Angabenseite macht deutlich, dass bei der MSK mit kohärenter Demodulation das AWGN–Verhältnis&nbsp; $E_{\rm B}/N_0$&nbsp; verdoppelt werden muss, damit die gleiche Fehlerwahrscheinlichkeit wie bei BPSK erreicht wird.
+'''(1)'''&nbsp; A comparison of the first two equations on the exercise page makes it clear that for MSK with coherent demodulation, the AWGN ratio &nbsp; $E_{\rm B}/N_0$&nbsp; must be doubled to achieve the same error probability as for BPSK.
-*In anderen Worten:&nbsp; Die kohärente BFSK–Kurve liegt um&nbsp; $10 · \lg (2) ≈ 3 \ \rm dB$&nbsp; rechts von der BPSK–Kurve.
+*In other words:&nbsp; the coherent BFSK curve is&nbsp; $10 · \lg (2) ≈ 3 \ \rm dB$&nbsp; to the right of the the BPSK curve.
-*Um&nbsp; $p_{\rm B} \le 10^{–5}$&nbsp; zu garantieren, muss daher gelten:
+*Thus, to guarantee&nbsp; $p_{\rm B} \le 10^{–5}$&nbsp; it must hold that:
 :$$10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}} /{N_{\rm 0}}= 9.6\,\,{\rm dB} + 3\,\,{\rm dB} = \underline{12.6\,\,{\rm dB}}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u>:
+'''(2)'''&nbsp; <u>Answer 2</u> is correct:
-*Die angegebene Gleichung gilt nicht nur für die MSK&nbsp; $($diese ist eine FSK mit&nbsp; $h = 0.5)$, sondern für jede Form von orthogonaler FSK.
+*The equation given does not just hold for MSK&nbsp; $($this is a FSK with&nbsp; $h = 0.5)$, but also for every form of orthogonal  FSK.
-*Eine solche liegt vor, wenn der Modulationsindex&nbsp; $h$&nbsp; ein ganzzahliges Vielfaches von&nbsp; $0.5$&nbsp; ist, zum Beispiel für&nbsp; $h = 1$.
+*This is the case whenever the modulation index&nbsp; $h$&nbsp; is an integer multiple of $0.5$&nbsp;, such as when &nbsp; $h = 1$.
-*Mit&nbsp; $h = 0.7$&nbsp; liegt keine orthogonale FSK vor.&nbsp;
+*When &nbsp; $h = 0.7$&nbsp;, there is no orthogonal FSK.&nbsp;
-'''(3)'''&nbsp; Aus der Umkehrfunktion der angegebenen Gleichung erhält man:
+'''(3)'''&nbsp; From the inverse function of the equation given we get:
 :$$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm ln}(50000)\approx 10.82 \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}{E_{\rm B}} /{N_{\rm 0}}= 21.64 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}}/ {N_{\rm 0}}\approx \underline{13.4\,\,{\rm dB}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp;    Aus $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$ folgt:
+'''(4)'''&nbsp; From  $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$ it follows that:
 :$${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} ({E_{\rm B}} /{N_{\rm 0}})/2 \approx 8.4 \hspace{0.25cm} \Rightarrow \hspace{0.25cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4} \approx \underline{1.12 \cdot 10^{-4}}\hspace{0.05cm}.$$
-Das heißt: &nbsp; Bei gleichem $E_{\rm B}/N_0$ wird die Fehlerwahrscheinlichkeit bei inkohärenter Demodulation gegenüber kohärenter Demodulation (siehe Teilaufgabe 1) um etwa den Faktor 11 vergrößert.
+This means that for the same $E_{\rm B}/N_0$, the error probability of incoherent demodulation is increased by a factor of  $11$ compared to that of coherent demodulation (see answer to question '''1''') .
 {{ML-Fuß}}