Aufgaben:Exercise 1.08: Comparison of ASK and BPSK

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Bit error probabilities  (ASK and BPSK)

The bit error probabilities of  "Amplitude Shift Keying"  $\rm (ASK)$  and  "Binary Shift Keying"  $\rm (BPSK)$  modulation are often given by the following two equations:

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

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

These equations are evaluated in the attached table.  The following applies:

• $E_{\rm B}$  indicates the average energy per bit.
• $N_{0}$  is the noise power density.
• There is a fixed relationship between the error functions  ${\rm Q}(x)$  and  ${\rm erfc}(x)$. 

It should be noted that these equations do not apply in general,  but only under certain idealized conditions.  These conditions are to be worked out in this exercise.

Notes:

• The exercise belongs to the chapter  "Linear Digital Modulation - Coherent Demodulation".
• You can check the results with the HTML5/JavaScript applet  "Complementary Gaussian Error Functions". 

Questions

1

What is the relationship between  ${\rm Q}(x)$  and  ${\rm erfc}(x)$?

	${\rm Q}(x)= 2 \cdot{\rm erfc}(x)$,
	${\rm Q}(x)= 0.5 \cdot{\rm erfc}(x)/\sqrt{2})$,
	${\rm erfc}(x)= 0.5 \cdot{\rm Q}(x)/\sqrt{2})$.

2

When do the given equations for the error probability apply?

	They apply only to the AWGN channel.
	They apply only to the matched filter receiver  (or variants).
	The equations take into account intersymbol interfering.
	The equations apply only to rectangular signals.

3

What are the error probabilities for  $10 \cdot \lg \ E_{\rm B}/N_{0} = 12\, \rm dB$?

$ p_{\rm ASK} \ = \ $ $\ \cdot 10^{-4}$

$ p_{\rm BPSK} \ = \ $ $\ \cdot 10^{-8}$

4

What are the error probabilities for  $E_{\rm B}/N_{0} = 8$?

$ p_{\rm ASK} \ = \ $ $\ \cdot 10^{-2}$

$ p_{\rm BPSK} \ = \ $ $\ \cdot 10^{-4}$

5

The error probability should not exceed  $10^{-8}$.  What is the required  $10 \cdot \lg \ E_{\rm B}/N_{0}$  for ASK?

$(E_{\rm B}/N_{0})_{\rm min} \ = \ $ $\ \rm dB $

Solution

Solution

(1)  It is already obvious from the equations on the information page that solution 2 is correct. The defining equations are:

$$\rm Q ({\it x}) = \ \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u \hspace{0.05cm},$$

$$\rm erfc ({\it x}) = \ \frac{\rm 2}{\sqrt{\rm \pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u \hspace{0.05cm}.$$

By simple substitutions, the above relationship can be easily proved:

$${\rm Q} ( x) = 1/2 \cdot {\rm erfc} (x/\sqrt{2}) \hspace{0.05cm}.$$

(2)  The first two solutions are correct:

• The equations are valid only for the AWGN channel and for an optimal binary receiver, for example, according to the matched filter approach.
• Intersymbol interfering – caused by the channel or the receiver filter – is not covered by this.
• The exact transmission pulse shaping, on the other hand, does not matter as long as the receiver filter $H_{\rm E}(f)$ is matched to the transmission spectrum. Rather:
• Two different transmission pulse shapers $H_{\rm S}(f)$ lead to exactly the same error probability if they have the same energy per bit.

(3)  The results can be read directly from the table:

$$p_{\rm ASK} \hspace{0.1cm}\underline {= 0.343 \cdot 10^{-4}},\hspace{0.3cm}p_{\rm BPSK} \hspace{0.1cm}\underline {= 0.901 \cdot 10^{-8}}.$$

(4)  With  $E_{\rm B}/N_{0} = 8\ \Rightarrow \ 10 \cdot \lg \ E_{\rm B}/N_{0} \approx 9 \ \rm dB$,  the following error probabilities are obtained:

$$p_{\rm ASK} \hspace{0.1cm}\underline {= 0.241 \cdot 10^{-2}},\hspace{0.3cm}p_{\rm BPSK} \hspace{0.1cm}\underline {= 0.336 \cdot 10^{-4}}.$$

(5)  From question (3), it follows that for binary phase modulation,  $10 \cdot \lg \ E_{\rm B}/N_{0} \approx 12 \ \rm dB$  must be satisfied for  $p_{\rm BPSK} \approx 10^{-8}$  to be possible.

• However, the given equations also show that the ASK curve is  $3 \ \rm dB$ $($exactly $3.01 \ \rm dB)$  to the right of the BPSK curve.
• It follows that:

$$10 \cdot {\rm lg}\hspace{0.1cm}(E_{\rm B}/N_{\rm 0})_{\rm min}\hspace{0.1cm}\underline {\approx 15\,\,{\rm dB}} \hspace{0.05cm}.$$