Exercise 4.9: Higher-Level Modulation

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Some channel capacity curves

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

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

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

Hints:

• The task belongs to the chapter  AWGN channel capacity with discrete value input.
• Reference is made in particular to the page  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 
  (see lower graph).

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 we understand as  "ASK"  the unipolar case:

$$x ∈ X = \{0,\ 1 \}.$$

• Therefore, according to our nomenclature:

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

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

Questions

1 What equation underlies Shannon's boundary curve  $C_{\rm Gaussian}$?

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

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 Gaussian} $  is valid.

3 Which statements are true for the red curve  $(C_{\rm red})$?

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

4 Which statements are true for the brown curve  $(C_{\rm brown})$?
Note:  $p_{\rm B}$  denotes the bit error probability here.

	For the associated random variable  $X$   ⇒   $M_X = |X| = 8$.
	$C_{\rm brown}$  is simultaneously the channel capacity of the  "8–ASK".
	$C_{\rm brown}$  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",  $R = 2.0$  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  $10 \cdot \lg (E_{\rm S}/{N_0}) = 15 \ \rm dB$   ⇒   $E_{\rm S}/{N_0} = 31.62$:

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

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

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

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

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

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

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

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

• The red curve  $(C_{\rm red})$  is always above  $C_{\rm BPSK}$,  but below  $C_{\rm brown}$  and Shannon's boundary curve  $(C_{\rm Gaussian})$.
• The statements also hold if  (for certain  $E_{\rm S}/{N_0}$ values)  curves are indistinguishable within the drawing precision.
• From the limit  $C_{\rm red}= 2 \ \rm bit/use$  for  $E_{\rm S}/{N_0} → ∞$,  the symbol set size  $M_X = |X| = 4$.
• Thus, the red curve describes  "4–ASK".  $M_X = |X| = 2$  would apply to the  "BPSK".
• The  "4–QAM" leads exactly to the same final value  "2 bit/use".  For small  $E_{\rm S}/{N_0}$ values,  however,  the channel capacity  $C_{\rm 4–QAM}$  is above the red curve,  since  $C_{\rm red}$  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"nbsp; 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   ⇒   $R > C_{\rm 8–ASK}$   ⇒   channel coding theorem is not satisfied   ⇒   answer 4 is wrong.
• However, if we reduce the code rate to  $R = 2 < C_{\rm 8–ASK}$  for the same  $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$  according to the yellow dot,  the channel coding theorem is satisfied   ⇒   answer 5 is correct.