Aufgaben:Exercise 4.14Z: Offset QPSK vs. MSK

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Koeffizientenzuordnung bei O-QPSK und MSK

One possible implementation fordie  $\rm MSK$  is offered by  "Offset–QPSK"  $\rm (O–QPSK)$, as can be seen from the  block diagrams  in the theory section.

In "normal offset QPSK operation", two bits of the source symbol sequence $〈q_k〉$ are assigned to one bit 𝑎Iν $a_{{\rm I}ν}$  in the in-phase branch and one bit  $a_{{\rm Q}ν}$  in the quadrature branch, respectively.

The graph shows this serial-to-parallel conversion in the top three plots for the first four bits of the source signal  $q(t)$.  It should be noted:

• The Offset–QPSK plot is for for a rectangular-shaped fundamental pulse.  The coefficients  $a_{{\rm I}ν}$  and  $a_{{\rm Q}ν}$  can take the values  $±1$ .
• If the time index of the source symbols passes through the values  $k =1,$ ... $, 8$, then the time variable  $ν$  only takes on the values  $1,$ ... $, 4$  an.
• The sketch also takes the time offset for the quadrature branch into account.

For a  "MSK–implementation using Offset–QPSK"  a recoding is required.  Here, with  $q_k ∈ \{+1, –1\}$  and  $a_k ∈ \{+1, –1\}$, it holds that:

$$a_k = (-1)^{k+1} \cdot a_{k-1} \cdot q_k \hspace{0.05cm}.$$

For example, by assuming  $a_0 = +1$ one gets:

$$a_1 = a_0 \cdot q_1 = +1,\hspace{0.4cm}a_2 = -a_1 \cdot q_2 = +1,\hspace{0.4cm} a_3 = a_2 \cdot q_3 = -1,\hspace{0.4cm}a_4 = -a_3 \cdot q_4 = -1 \hspace{0.05cm}.$$

Additionally, one must take into account:

• The coefficients  $a_0 = +1$,  $a_2 = +1$,  $a_4 = -1$  and the coefficients  $a_6$  and  $a_8$  which are yet to be calculated, are assigned to the signal  $s_{\rm I}(t)$ .
• On the other hand, the coefficients  $a_1 = +1$  and  $a_3 = -1$  as well as all other coefficients with an odd index are applied to the signal  $s_{\rm Q}(t)$ .

Hints:

• This exercise belongs to the chapter  Nonlinear Digital Modulation.
• Particular reference is made to the section  Realizing MSK as Offset–QPSK.

• The associated phase function  $ϕ(t)$  is determined in  Exercise 4.14 , and is also based on the  (normalized)  MSK fundamental pulse:

$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} \cos (\pi/2 \cdot t/T ) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{otherwise}} \\ \end{array}\begin{array}{*{10}c} -T \le t \le +T \hspace{0.05cm}, \\ {\rm }. \\ \end{array}$$

Questions

1

What is the bit duration  $T_{\rm B}$  of the source signal?

$T_{\rm B} \ = \ $ $\ \rm µ s$

2

What is the symbol duration  $T$  of the offset QPSK?

$T \ = \ $ $\ \rm µ s$

3

Give the above amplitude coefficients of the offset QPSK.

$a_{\rm I3} \hspace{0.25cm} = \ $

$a_{\rm Q3} \ = \ $

$a_{\rm I4} \hspace{0.25cm} = \ $

$a_{\rm Q4} \ = \ $

4

What is the symbol duration  $T$  of the  MSK?

$T \ = \ $ $\ \rm µ s$

5

Give the above amplitude coefficients of the MSK.

$a_5 \ = \ $

$a_6 \ = \ $

$a_7 \ = \ $

$a_8 \ = \ $

Solution

Solution

(1)  It can be seen from the upper plot that   $T_{\rm B} \hspace{0.15cm}\underline{ = 1 \ \rm µ s}$ .

(2)  For QPSK or offset QPSK , the symbol duration $T$  is twice the bit duration  $T_{\rm B}$ due to serial-to-parallel conversion:

$$T = 2 \cdot T_{\rm B} \hspace{0.15cm}\underline {= 2\,{\rm µ s}} \hspace{0.05cm}.$$

(3)  According to the allocation evident in the plot for the first bits:

$$a_{\rm I3} = q_5 \hspace{0.15cm}\underline {= +1},$$

$$a_{\rm Q3} = q_6 \hspace{0.15cm}\underline {= +1},$$

$$a_{\rm I4} = q_7 \hspace{0.15cm}\underline { = -1},$$

$$a_{\rm Q4} = q_8 \hspace{0.15cm}\underline {= +1} \hspace{0.05cm}.$$

(4)  In MSK, the symbol duration  $T$ is equal to the bit duration   $T_{\rm B}$:

$$T = T_{\rm B}\hspace{0.15cm}\underline { = 1\,{\rm µ s}} \hspace{0.05cm}.$$

(5)  According to the given recoding rule, when   $a_4 = –1$, we get:

$$q_5 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_5 = a_4 \cdot q_5 \hspace{0.15cm}\underline {= -1},$$

$$q_6 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_6 = -a_5 \cdot q_6 \hspace{0.15cm}\underline {= +1},$$

$$q_7 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_7 = a_6 \cdot q_7 \hspace{0.15cm}\underline {= -1},$$

$$q_8 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_8 = -a_7 \cdot q_8\hspace{0.15cm}\underline { = +1}\hspace{0.05cm}.$$