Aufgaben:Exercise 4.9: Higher-Level Modulation

• Return to book

Some channel capacitance curves

The graph shows AWGN channel capacity curves over the  $10 \cdot \lg (E_{\rm S}/{N_0})$:

• $C_\text{Gauß}$:    Shannon's boundary curve,
• $C_\text{BPSK}$:    valid for Binary Phase Shift Keying.

The two other curves  $C_\text{rot}$  and  $C_\text{braun}$  should be analyzed and assigned to possible modulation schemes in subtasks  (3)  and  (4) .

Hints:

• The task belongs to the chapter  AWGN channel capacitance with discrete value input.
• Reference is made in particular to the page  The channel capacity $C$ as a function of $E_{\rm S}/{N_0}$.
• Since the results are to be given in "bit", "log"  ⇒  "log₂" is used in the equations.
• The modulation methods mentioned in the questions are described in terms of their signal space constellation:

Proposed signal space constellations

Notes on nomenclature::

• In the literature,"BPSK" is sometimes also referred to as "2–ASK"

$$x ∈ X = \{+1,\ -1\}.$$

• In contrast, in our learning tutorial $\rm LNTwww$ we understand as "ASK" the unipolar case

$$x ∈ X = \{0,\ 1 \}.$$

• Therefore, according to our nomenclature:

$$C_\text{AK} < C_\text{BPSK}$$

This fact is irrelevant for the solution of the present problem.

Questions

1

What equation underlies the Shannon boundary curve  $C_{\rm Gauß}$  ?

	$C_{\rm Gauß} = C_1= {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + E_{\rm S}/{N_0})$ ,
	$C_{\rm Gauß} = C_2= {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + 2 \cdot E_{\rm S}/{N_0})$ ,
	$C_{\rm Gauß} = C_3= {\rm log}_2 \hspace{0.1cm} ( 1 + E_{\rm S}/{N_0})$ .

2

Which statements are true for the green curve  $C_{\rm BPSK}$ ?

	$C_{\rm BPSK}$  cannot be given in closed form.
	$C_{\rm BPSK}$  is greater than zero if  $E_{\rm S}/{N_0} > 0$  is assumed.
	For  $E_{\rm S}/{N_0} < \ln (2)$ , $C_{\rm BPSK} ≡ 0$.
	In the whole range  $C_{\rm BPSK} < C_{\rm Gauß} $ is valid.

3

Which statements are true for the red curve  $C_{\rm rot}$ ?

	For the associated random variable  $X$  gilt  $M_X = |X| = 2$.
	For the associated random variable  $X$  gilt  $M_X = |X| = 4$.
	$C_{\rm rot}$  is simultaneously the channel capacity of the 4–ASK.
	$C_{\rm rot}$  is simultaneously the channel capacity of the 4–QAM.
	For all  $E_{\rm S}/{N_0} > 0$   $C_{\rm rot}$  is between "grün" and "braun".

4

Which statements are true for the brown curve  $C_{\rm braun}$  ?  ($p_{\rm B}$:   bit error probability)

	For the associated random variable  $X$ ,  $M_X = |X| = 8$.
	$C_{\rm braun}$  is simultaneously the channel capacity of the 8–ASK.
	$C_{\rm braun}$  is simultaneously the channel capacity of the 8–PSK.
	$p_{\rm B} ≡ 0$  is possible with 8–ASK,  $R = 2.5$  and  $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ .
	$p_{\rm B} ≡ 0$  is possible with 8–ASK,  $R = 2$  and  $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ .

Solution

Solution

(1)  Proposition 2 is correct, as shown by the calculation for  $10 \cdot \lg (E_{\rm S}/{N_0}) = 15 \ \rm dB$   ⇒   $E_{\rm S}/{N_0} = 31.62$ zeigt:

$$C_2(15\hspace{0.1cm}{\rm dB}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + 2 \cdot 31.62 ) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 64.25 ) \approx 3\,{\rm bit/Kanalzugriff}\hspace{0.05cm}. $$

• The other two proposed solutions provide the following numerical values:

$$C_3(15\hspace{0.1cm}{\rm dB}) \ = \ {\rm log}_2 \hspace{0.1cm} ( 1 + 31.62 ) \approx 5.03\,{\rm bit/Kanalzugriff}\hspace{0.05cm},$$

$$ C_1(15\hspace{0.1cm}{\rm dB}) \ = \ C_3/2 \approx 2.51\,{\rm bit/Kanalzugriff}\hspace{0.05cm}.$$

• The proposed solution 3 corresponds to the case of two independent Gaussian channels with half transmit power per channel.

(2) Proposed solutions 1, 2 and 4 are correct:

• If one would replace  $E_{\rm S}$  by  $E_{\rm B}$ , then also the statement 3 would be correct.
• For  $E_{\rm B}/{N_0} < \ln (2)$   $C_{\rm Gauß} ≡ 0$  is valid and therefore also  $C_{\rm BPSK} ≡ 0$ .

(3)  Statements 2, 3 and 5 are correct::

• The red curve  $C_{\rm rot}$  is always above  $C_{\rm BPSK}$ , but below  $C_{\rm braun}$  and the Shannon boundary curve  $C_{\rm Gauß}$.
• The statements also hold if for certain  $E_{\rm S}/{N_0}$ values curves are indistinguishable within the character precision.
• From the limit  $C_{\rm rot}= 2 \ \rm bit/channel use$  for  $E_{\rm S}/{N_0} → ∞$ , the symbol range  $M_X = |X| = 4$.
• Thus, the red curve describes the 4–ASK.  $M_X = |X| = 2$  would apply to the BPSK.
• The 4–QAM leads exactly to the same final value "2 bit/channel use".  For small  $E_{\rm S}/{N_0}$ values, however, the channel capacity  $C_{\rm 4–QAM}$  is above the red curve, since  $C_{\rm rot}$  is bounded by the Gaussian boundary curve  $C_2$ , but $C_{\rm 4–QAM}$  is bounded by  $C_3$.

The designations  $C_2$  and  $C_3$  here refer to subtask  (1).

Channel capacity limits for
BPSK, 4–ASK and 8–ASK

(4)  Proposed solutions 1, 2 and 5 are correct:

• From the brown curve, one can see the correctness of the first two statements.
• The 8–PSK with I– and Q–components – i.e. with  $K = 2$  dimensions – lies slightly above the brown curve for small  $E_{\rm S}/{N_0}$ values   ⇒   the answer 3 is incorrect.

In the graph, the two 8–ASK–systems are also drawn as dots according to propositions 4 and 5.

• The purple dot is above the  $C_{\rm 8–ASK}$ curve   ⇒   $R = 2.5$ and $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ are not enough to decode the 8–ASK without errors   ⇒   $R > C$   ⇒   the channel coding theorem is not satisfied   ⇒   answer 4 is wrong.
• However, if we reduce the code rate to $R = 2 < C_{\rm 8–ASK}$ according to the yellow dot for the same $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$, the channel coding theorem is satisfied   ⇒   Answer 5 is correct.