Aufgaben:Exercise 4.2Z: Eight-level Phase Shift Keying

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Signal space points at 8-PSK

The  $M = 8$  possible transmitted signals at  "8–PSK"  are with   $i = 0, \ \text{...} \ , 7$   in the range  $0 ≤ t < T$:

$$s_i(t)= A \cdot \cos(2\pi f_{\rm T}t + i \cdot {\pi}/{4}) \hspace{0.05cm}.$$

Outside the symbol duration  $T$,  the signals  $s_i(t)$  are all zero.

In   "Exercise 4.2"  it was shown that this signal set is given by the basis functions

$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},$$

$$\varphi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}$$

which can be represented as follows  $(i = 0, \ \text{...} \ , 7)$:

$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}.$$

The equivalent low-pass representation of the signals  $s_i(t)$  is according to the section  "System description using the equivalent low-pass signal"  of the book  "Modulation methods":

$$s_{{\rm TP}i}(t)= a_{i} \cdot g_s(t) \hspace{0.05cm}, \hspace{0.2cm}a_{i} = a_{{\rm I}i} + {\rm j} \cdot a_{{\rm Q}i} \hspace{0.05cm}, \hspace{0.2cm}i = 0,\text{...} \hspace{0.1cm} , 7 \hspace{0.05cm},$$

where  $a_i$  are dimensionless complex coefficients and the energy of the basic transmission pulse  $g_s(t)$  is  $E_{\it g_s}$  in the low-pass region.  In the case shown here,   $g_s(t)$  describes a rectangular pulse, s but any other energy-limited pulse can be used for  $g_s(t)$. 

The graph shows the  8–PSK  signal space representation

1. for the band-pass signal  $s_{ \it i}(t)$  (top),  and
2. for the equivalent low-pass signal  $s_{\rm TP \it i}(t)$  (bottom):

It can be seen from this that the two representations differ only in the basis functions used,  with  $\varphi_1(t)$  representing different functions in the upper and lower graphs. In the low-pass representation,  $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$  holds.

Notes:

• The exercise belongs to the chapter  "Signals, Basis Functions and Vector Spaces".

• For abbreviation,  use the energy  $E = 1/2 \cdot A^2 \cdot T$.

• In contrast to the theory section and  "Exercise 4.2",  here the indexing variable  $i$  can take the values  $0, \ \text{...} \, ,M-1$. 

Questions

1

What are the coefficients of the signal  $s_0(t)$?

$s_{\rm 01} \ = \ $ $\ \cdot \sqrt{E}$

$s_{\rm 02} \ = \ $ $\ \cdot \sqrt{E}$

2

What are the coefficients of the signal  $s_2(t)$?

$s_{\rm 21} \ = \ $ $\ \cdot \sqrt{E}$

$s_{\rm 22} \ = \ $ $\ \cdot \sqrt{E}$

3

What are the coefficients of the signal  $s_5(t)$?

$s_{\rm 51} \ = \ $ $\ \cdot \sqrt{E}$

$s_{\rm 52} \ = \ $ $\ \cdot \sqrt{E}$

4

By which basis functions can the low–pass signals  $s_{\rm TP \it i}(t)$  be represented?  By

	one complex basis function  $\xi_1(t)$,
	two complex basis functions  $\xi_1(t)$  and  $\xi_2(t)$,
	two real functions  $\varphi_1(t)$  and  $\psi_1(t)$.

5

What are the real basis functions in the present case?

	$\varphi_1(t) = g_s(t)$,
	$\varphi_1(t) = g_s(t)/\sqrt{E_{\rm g_s}}$,
	$\psi_1(t) = \varphi_1(t)$,
	$\psi_1(t) = {\rm j} \cdot \varphi_1(t)$.

6

Let  $s_{\rm TP0}(t) = \sqrt{E}$. Which is true:

	The energy  $E$  refers to the low–pass signal.
	The energy  $E$  refers to the band–pass signal.

Solution

Solution

(1)  The signal  $s_0(t)$  is:

$$s_0(t)= A \cdot \cos(2\pi f_{\rm T}t ) = s_{01} \cdot \varphi_1(t) + s_{02} \cdot \varphi_2(t) \hspace{0.05cm}.$$

• Since this signal has no sinusoidal part,  $s_{\rm 02} \hspace{0.15cm}\underline {= 0}$.
• Further,  with the given abbreviation:

$$A = s_{01} \cdot \sqrt{{2}/{T}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} s_{01}=\sqrt{1/2 \cdot A^2 \cdot T} = \sqrt{E}\hspace{0.05cm} \hspace{0.15cm}\underline { = 1 \cdot E^{\hspace{0.05cm}0.5}}\hspace{0.05cm}.$$

(2)  The signal  $s_2(t)$  is with  $i = 2$  (note that the second basis function is minus–sine):

$$s_2(t)= A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{2})= - A \cdot \sin(2\pi f_{\rm T}t )\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{21}\hspace{0.05cm} \underline{= 0}\hspace{0.05cm}, \hspace{0.2cm} s_{22}= \sqrt{E} \hspace{0.05cm} \hspace{0.15cm}\underline {=1 \cdot E^{\hspace{0.05cm}0.5}}\hspace{0.05cm}.$$

(3)  According to the solutions to subtasks  (1)  and  (2),  the following is true:

$$s_{51}= s_{52}= - \sqrt{E/2} \hspace{0.05cm} \hspace{0.15cm}\underline { = -0.707 \cdot E^{\hspace{0.05cm}0.5}}$$

$$\Rightarrow \hspace{0.3cm} s_{5}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - {A}/{ \sqrt{2}} \cdot \cos(2\pi f_{\rm T}t ) - {A}/{ \sqrt{2}} \cdot \sin(2\pi f_{\rm T}t )=A \cdot \cos(2\pi f_{\rm T}t + \phi_5)\hspace{0.2cm}{\rm with}\hspace{0.2cm}\phi_5 = -0.75 \cdot \pi \hspace{0.2cm}{\rm and}\hspace{0.2cm}\phi_5 = 1.25 \cdot \pi \hspace{0.05cm}.$$

(4)  Solutions 1 and 3  are correct.  The following relation holds:  

$$\xi_1 (t) = \varphi_1 (t) + {\rm j} \cdot \psi_1 (t)\hspace{0.05cm}.$$

(5)  Correct are the  alternatives 2 and 3:

• The basis function must be energy-normalized.

• Like  $\psi_1(t)$,  the basis  $\varphi_1(t)$  is a real function,  not an imaginary one:

$$\varphi_1 (t) = \psi_1 (t) = \left\{ \begin{array}{c} 1/\sqrt{T} \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

(6)  From the low–pass signal  $s_{\rm TP0}(t)$,  one can also calculate the band-pass signal  $s_0(t)$.

• In the range  $0 ≤ t ≤ T$,  the result from  (5)  gives the same result as in subtask  (1):

$$s_0(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\left[s_{{\rm TP}0}(t) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \right] = {\rm Re}\left[\sqrt{E} /{\sqrt{T}} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \right]= \sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t ) \hspace{0.05cm},$$

• It follows:  The energy  $E$  refers to the band–pass signal even when considered in the equivalent low–pass region.

• Accordingly,  for the signal  $s_2(t)$  marked with blue dot in the region of interest:

$$s_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\big[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \big] = {\rm Re}\big[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t)- \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t) \big] = - \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t) \hspace{0.05cm}.$$

• Finally,  for the  (green)  signal  $s_5(t)$  in the range $0 ≤ t < T$  can be written:

$$s_5(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\big[\frac{-1 - {\rm j}}{\sqrt{2}} \cdot \sqrt{{E}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \big] = \text{...} = - \sqrt{\frac{E}{2T}} \cdot \cos(2\pi f_{\rm T}t)+ \sqrt{\frac{E}{2T}} \cdot \sin(2\pi f_{\rm T}t)=\sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t + 1.25 \cdot \pi) \hspace{0.05cm}.$$

• These results also agree with those of subtasks  (2)  resp.  (3).  Therefore,  solution 2  is correct.