Exercise 4.Ten: QPSK Channel Capacity

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Capacity curves for BPSK and QPSK

Given are the AWGN channel capacity limit curves for the modulation methods

• Binary Phase Shift Keying  (BPSK),
• Quaternary Phase Shift Keying  (4–PSK or QPSK).

The channel capacities  $C_\text{BPSK}$  and  $C_\text{QPSK}$  simultaneously indicate the maximum code rate  $R_{\rm max}$ , with which the bit error probability  $p_\text{B} ≡ 0$  can be asymptotically achieved with BPSK (or QPSK) with suitable channel coding.

The upper diagram shows the dependence on the parameter  $10 \cdot \lg (E_{\rm B}/{N_0})$  in  $\rm dB$, where  $E_{\rm B}$  indicates the "energy per information bit".

• For large  $E_{\rm B}/{N_0}$ values, the BPSK curve provides the maximum code rate  $R ≈ 1$.
• From the QPSK curve, on the other hand,  $R ≈ 2$  can be read.

The capacitance curves for digital input (each with the unit "bit/symbol"),

• green curve   ⇒   $C_\text{BPSK} (E_{\rm B}/{N_0})$  and
• blue curve   ⇒   $C_\text{QPSK} (E_{\rm B}/{N_0})$

are to be related in subtask  (3)  to two Shannon limit curves, each valid for a Gaussian input distribution:

$$C_1( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2\cdot R \cdot E_{\rm B}}{N_0}) ,$$

$$C_2( E_{\rm B}/{N_0}) = {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { R \cdot E_{\rm B}}{N_0}) .$$

The two curves simultaneously indicate the maximum code rate  $R_{\rm max}$  with which error-free transmission is possible by long channel codes according to the  channel coding theorem .  Of course, different boundary conditions apply to  $C_1( E_{\rm B}/{N_0})$   or    $C_2( E_{\rm B}/{N_0})$ .  Which ones, you shall find out.

On the other hand, the abscissa in the lower diagram is   $10 \cdot \lg (E_{\rm S}/{N_0})$  with the "energy per symbol"  $(E_{\rm S})$.  Notice that the two limits are not changed from the upper plot::

$$C_{\rm BPSK}( E_{\rm S}/{N_0} \to \infty) = C_{\rm BPSK}( E_{\rm B}/{N_0} \to \infty) = 1 \ \rm bit/symbol,$$

$$C_{\rm QPSK}( E_{\rm S}/{N_0} \to \infty) = C_{\rm QPSK}( E_{\rm B}/{N_0} \to \infty) = 2 \ \rm bit/symbol.$$

Hints:

• The task belongs to the chapter  AWGN channel capacity with discrete value input.
• Reference is made in particular to the page  Maximum code rate for QAM structures.

Questions

1

Do QPSK and 4-QAM differ from an information theoretic point of view?

	Yes.
	No.

2

How can  $C_{\rm QPSK}( E_{\rm B}/{N_0})$  be constructed from  $C_{\rm BPSK}( E_{\rm B}/{N_0})$ ?

	By doubling:   $C_{\rm QPSK}( E_{\rm B}/{N_0}) = 2 \cdot C_{\rm BPSK}( E_{\rm B}/{N_0})$.
	Additionally by a shift to the right.
	Additionally by a shift to the left.
	$C_{\rm QPSK}( E_{\rm B}/{N_0})$  cannot be constructed from  $C_{\rm BPSK}( E_{\rm B}/{N_0})$  .

3

What is the relation to the Shannon boundary curves?

	$C_{\rm BPSK}( E_{\rm B}/{N_0}) \le C_{\rm 1}( E_{\rm B}/{N_0})$ holds.
	$C_{\rm BPSK}( E_{\rm B}/{N_0}) \le C_{\rm 2}( E_{\rm B}/{N_0})$ holds.
	$C_{\rm QPSK}( E_{\rm B}/{N_0}) \le C_{\rm 1}( E_{\rm B}/{N_0})$ holds.
	$C_{\rm QPSK}( E_{\rm B}/{N_0}) \le C_{\rm 2}( E_{\rm B}/{N_0})$ holds.

4

How can  $C_{\rm QPSK}( E_{\rm S}/{N_0})$  be constructed from  $C_{\rm BPSK}( E_{\rm S}/{N_0})$ ?

	By doubling:   $C_{\rm QPSK}( E_{\rm S}/{N_0}) = 2 \cdot C_{\rm BPSK}( E_{\rm S}/{N_0})$  and Additionally by a shift to the right.
	By doubling:   $C_{\rm QPSK}( E_{\rm S}/{N_0}) = 2 \cdot C_{\rm BPSK}( E_{\rm S}/{N_0})$  and Additionally by a shift to the left.
	$C_{\rm QPSK}( E_{\rm S}/{N_0})$  cannot be constructed from  $C_{\rm BPSK}( E_{\rm S}/{N_0})$ .

Solution

Solution

QPSK– und 4–QAM–Signalraumkonstellation

(1)  The diagram shows the signal space constellations for

• Quaternary Phase Shift Keying  (QPSK), and
• four-level quadrature amplitude modulation  (4–QAM).

The latter is also referred to as  π/4–QPSK .  Both are identical from an information-theoretic point of view   ⇒   answer NO.

(2)  Correct is the proposed solution 1:

• The 4–QAM can be viewed as two BPSK constellations in orthogonal planes, where the energy per information bit $(E_{\rm B})$  is the same in both cases.
• Since, according to subtask  (1)  the 4–QAM is identical to the QSPK, in fact:

$$C_{\rm QPSK}( E_{\rm B}/{N_0}) = 2 \cdot C_{\rm BPSK}( E_{\rm B}/{N_0}).$$

(3)  In the lower graph, the two Shannon boundary curves given are sketched together with  $C_{\rm BPSK}( E_{\rm B}/{N_0})$  and  $C_{\rm QPSK}( E_{\rm B}/{N_0})$ :

$$C_1( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) ,$$

$$C_2( E_{\rm B}/{N_0}) = {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { R \cdot E_{\rm B}}{N_0}) .$$

Four capacity curves with different statements

One can see from this sketch:   Proposed solutions 1, 2 and 4 are correct.

• The green–dashed curve  $C_1( E_{\rm B}/{N_0})$  is valid for the AWGN channel with Gaussian distributed input. 
• For code rate  $R =1$ ,  $10 \cdot \lg (E_{\rm B}/{N_0}) = 1.76\ \rm dB$  is required according to this curve.  
• For  $R =2$ , on the other hand  $10 \cdot \lg (E_{\rm B}/{N_0}) = 5.74\ \rm dB$ is required.
• The blue–dashed curve  $C_2( E_{\rm B}/{N_0})$  gives the Shannon limit for  $K=2$  parallel Gaussian channels.  Here one needs  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0\ \rm dB$  for  $R =1$  or  $10 \cdot \lg (E_{\rm B}/{N_0}) = 1.76\ \rm dB$  for  $R =2$.
• The one–dimensional BPSK is below  $C_1$  in the entire range and thus, of course, below  $C_2 > C_1$.
• As expected, the two–dimensional QPSK lies below the  $C_2$ limit curve relevant for it.  However, it is above $C_1$ in the lower range   $($up to almost  $\text{6 dB)}$ .

(4)  The  $C_{\rm QPSK}( E_{\rm B}/{N_0})$ curve can also be constructed from  $C_{\rm BPSK}( E_{\rm B}/{N_0})$, namely

• on the one hand by doubling:

$$C_{\rm BPSK}(10 \cdot {\rm lg} \hspace{0.1cm}E_{\rm S}/{N_0}) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 2 \cdot C_{\rm BPSK}(10 \cdot {\rm lg} \hspace{0.1cm}E_{\rm S}/{N_0}) ,$$

• as well as by a shift of  $3\ \rm dB$  to the right:

$$C_{\rm QPSK}(10 \cdot {\rm lg} \hspace{0.1cm}E_{\rm S}/{N_0}) = 2 \cdot C_{\rm BPSK}(10 \cdot {\rm lg} \hspace{0.1cm}E_{\rm S}/{N_0} - 3\,{\rm dB}) .$$

• The  proposed solution 1  is correct.  This takes into account that with QPSK the energy in one dimension is only  $E_{\rm S}/2$.