Aufgaben:Exercise 4.14: Phase Progression of the MSK

• Return to book

Source signal and low-pass signals
in both branches of the MSK

One possible implementation of  Minimum Shift Keying  $\rm (MSK)$  is offered by   $\rm Offset–QPSK$, as shown in the block diagram  in the theory section.

• For this, a recoding of the source symbols  $q_k ∈ \{+1, –1\}$  into the similarly binary amplitude coefficients  $a_k ∈ \{+1, –1\}$  must first be undertaken.
• This recoding is discussed in detail in  Exercise 4.14Z .

The graph shows the two equivalent low-pass signals  $s_{\rm I}(t)$  and  $s_{\rm Q}(t)$  in the two branches below, which are obtained for the inphase and quadrature branches after recoding  $a_k = (-1)^{k+1} \cdot a_{k-1} \cdot q_k $ from the source signal  $q(t)$  sketched above.  Considered here is the MSK fundamental pulse,

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

This is normalized to  $1$ , as are the signals  $s_{\rm I}(t)$  and  $s_{\rm Q}(t)$ .

In keeping with the chapter  Equivalent Low-Pass Signal and its Spectral Function in the book "Signal Representation", the equivalent low-pass signal is:

$$ s_{\rm TP}(t) = s_{\rm I}(t) + {\rm j} \cdot s_{\rm Q}(t) = |s_{\rm TP}(t)| \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}\phi(t)}\hspace{0.05cm},$$

• with magnitude

$$|s_{\rm TP}(t)| = \sqrt{s_{\rm I}^2(t) + s_{\rm Q}^2(t)} $$

• and phase

$$ \phi(t) = {\rm arc} \hspace{0.15cm}s_{\rm TP}(t) = {\rm arctan}\hspace{0.1cm} \frac{s_{\rm Q}(t)}{s_{\rm I}(t)} \hspace{0.05cm}.$$

The physical MSK transmitted signal is then given by

$$ s(t) = |s_{\rm TP}(t)| \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi(t)) \hspace{0.05cm}.$$

Hints:

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

• Assume  $ϕ(t = 0) = ϕ_0 = 0$ .

Questions

1

Which statements are true for the envelope curve  $|s_{\rm TP}(t)|$  of MSK?

	The envelope curve is a cosine oscillation.
	The envelope curve is constant.
	The envelope curve is independent of the transmitted sequence.

2

Let  $T = 1 \ \rm µs$.  Calculate the phase response in the interval  $0 ≤ t ≤ T$.
What are the phase values for  $t = T/2$  and  $t = T$?

$ϕ(t = T/2)\ = \ $ $\ \rm degrees$

$ϕ(t = T) \hspace{0.63cm} = \ $ $\ \rm degrees$

3

Determine the phase values at  $t = 2T$,  $t = 3T$  and  $t = 4T$.

$ϕ(t = 2T) \ = \ $ $\ \rm degrees$

$ϕ(t = 3T) \ = \ $ $\ \rm degrees$

$ϕ(t = 4T) \ = \ $ $\ \rm degrees$

4

Sketch and interpret the phase response  $ϕ(t)$  in the range from  $0$  to  $8T$.
What are the phase values at the following times?

$ϕ(t = 5T) \ = \ $ $\ \rm degrees$

$ϕ(t = 6T) \ = \ $ $\ \rm degrees$

$ϕ(t = 7T) \ = \ $ $\ \rm degrees$

$ϕ(t = 8T) \ = \ $ $\ \rm degrees$

Solution

Solution

(1)  Answers 2 and 3 are correct:

• For example, in the range   $0 ≤ t ≤ T$, considering that   $a_0^2 = a_1^2 = 1$ :

$$ |s_{\rm TP}(t)| = \sqrt{a_0^2 \cdot \cos^2 (\frac{\pi \cdot t}{2 \cdot T}) + a_1^2 \cdot \sin^2 (\frac{\pi \cdot t}{2 \cdot T})} = 1 \hspace{0.05cm}.$$

• Thus, statement 2 is correct, while statement 1 is false.
• This result holds for any pair of values  $a_0 ∈ \{+1, \ –1\}$  and  $a_1 ∈ \{+1, \ –1\}$.
• From this, it can be further concluded that the envelope is independent of the transmitted sequence.

(2)  With the given equation, it holds that:

$$\phi(t) = {\rm arctan}\hspace{0.1cm} \frac{s_{\rm Q}(t)}{s_{\rm I}(t)} = {\rm arctan}\hspace{0.1cm} \frac{a_1 \cdot \sin (\frac{\pi \cdot t}{2 \cdot T})}{a_0 \cdot \cos (\frac{\pi \cdot t}{2 \cdot T})}= {\rm arctan}\hspace{0.1cm}\left [ \frac{a_1}{a_0}\cdot \tan \hspace{0.1cm}(\frac{\pi \cdot t}{2 \cdot T})\right ] \hspace{0.05cm}.$$

• The quotient  $a_1/a_0$  is always  $+1$  or  $-1$.  Thus, this quotient is preferable and we get:

$$\phi(t) = \frac{a_1}{a_0}\cdot {\rm arctan}\hspace{0.1cm}\left [ \tan \hspace{0.1cm}(\frac{\pi \cdot t}{2 \cdot T})\right ]= \frac{a_1}{a_0}\cdot \frac{\pi \cdot t}{2 \cdot T} \hspace{0.05cm}.$$

• The initial phase  $ϕ_0 = 0$  can rule out ambiguities.  In particular, because  $a_0 = a_1 = +1$:

$$\phi(t = T/2 = 0.5\,{\rm µ s}) = {\pi}/{4}\hspace{0.15cm}\underline { = +45^\circ},\hspace{0.2cm}\phi(t = T= 1\,{\rm µ s}) = {\pi}/{2}\hspace{0.15cm}\underline {= +90^\circ} \hspace{0.05cm}.$$

(3)  The easiest way to solve this problem is to use the unit circle:

$$ {\rm Re} = s_{\rm I}(2T) = +1, \hspace{0.2cm} {\rm Im} = s_{\rm Q}(2T) = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi(t = 2T= 2\,{\rm µ s}) \hspace{0.15cm}\underline {= 0^\circ},$$

$$ {\rm Re} = s_{\rm I}(3T) = 0, \hspace{0.2cm} {\rm Im} = s_{\rm Q}(3T) = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi(t = 3T= 3\,{\rm µ s}) \hspace{0.15cm}\underline {= -90^\circ},$$

Source signal and phase response in MSK

$${\rm Re} = s_{\rm I}(4T) = -1, \hspace{0.2cm} {\rm Im} = s_{\rm Q}(4T) = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi(t = 4T= 4\,{\rm µ s})= \pm 180^\circ \hspace{0.05cm}.$$

• From the sketch below, we can see that   $\phi(t = 4T= 4\,{\rm µ s})\hspace{0.15cm}\underline { = - 180^\circ}\hspace{0.05cm}$  is correct.

(4)  The graph shows the MSK phase  $ϕ(t)$  together with the source signal  $q(t)$.  It can be seen that:

• At symbol  $a_\nu =+1$  the phase increases linearly by $90^\circ \ (π/2)$  within the symbol duration $T$ .
• At symbol  $a_\nu =-1$  the phase decreases linearly by $90^\circ \ (π/2)$  within the symbol duration $T$ .

• Thus, the remaining phase values are:

$$\phi(5T) \hspace{0.15cm}\underline { = -90^\circ},\hspace{0.2cm}\phi(t = 6T) \hspace{0.15cm}\underline {= 0^\circ} \hspace{0.05cm}.$$

$$ \phi(7T)\hspace{0.15cm}\underline { = -90^\circ},\hspace{0.2cm} \phi(t = 8T) \hspace{0.15cm}\underline {= 0^\circ} \hspace{0.05cm}.$$