Aufgaben:Exercise 4.15Z: MSK Basic Pulse and MSK Spectrum

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MSK basic pulse and its spectrum

The fundamental pulse that is always required to  realize MSK as Offset–QPSK   has the form shown in the graph above:

$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} | t | \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

The spectral function  $G(f)$ is drawn below, that is, the  Fourier transform  of  $g(t)$.

The corresponding equation is to be determined in this task, by considering:

$$g(t) = c(t) \cdot r(t)\hspace{0.05cm}.$$

The following abbreviations are used here:

• $c(t)$  is a cosine oscillation with amplitude  $1$  and  frequency  $f_0$ (yet to be determined).
• $r(t)$  is a square wave function with amplitude $g_0$  and duration $2T$.

Hints:

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

• The result obtained here is also used in  Exercise 4.15 .

Questions

1

How should one choose the frequency $f_0$  of the cosine oscillation  $c(t)$  so that  $g(t) = c(t) · r(t)$ ?

$f_0 \ = \ $ $\ \cdot 1/T$

2

What is the spectrum  $R(f)$  of the rectangular function  $r(t)$?  What spectral value occurs when $f = 0$ ?

$R(f=0) \ = \ $ $\ \cdot g_0 \cdot T$

3

Calculate the spectrun  $G(f)$  of the MSK pulse  $g(t)$, particularly the spectral value at  $f = 0$.

$G(f=0) \ = \ $ $\ \cdot g_0 \cdot T$

4

Summarize the result of question   (3)  in one term. At what frequency  $f_1$  does  $G(f)$  have its first zero?

$f_1 \ = \ $ $\ \cdot 1/T$

Solution

Solution

(1)  The period of the cosine signal must be  $T_0 = 4T$ .  Thus, the frequency is $f_0 = 1/T_0\hspace{0.15cm}\underline {= 0.25} · 1/T$.

(2)  The spectral function of a rectangular pulse of height  $g_0$  and duration $ 2T$  is:

$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm} \Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$

(3)  When  $g(t) = c(t) · r(t)$ , it follows from the convolution theorem that:   $ G(f) = C(f) \star R(f)\hspace{0.05cm}.$

• The spectral function  $C(f)$  consists of two Dirac functions at  $± f_0$, each with weight  $1/2$.  From this follows:

$$ G(f) = 2 \cdot g_0 \cdot T \cdot \big [ 1/2 \cdot \delta (f - f_0 ) + 1/2 \cdot \delta (f + f_0 )\big ] \star {\rm si} ( 2 \pi f T )= g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi T \cdot (f - f_0 ) ) + {\rm si} ( 2 \pi T \cdot (f + f_0 ) ) \big ] \hspace{0.05cm}.$$

• Using the result  $f_0 = 1/(4T)$  from question  (1) , it further holds that:

$$G(f) = g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi f T - \pi / 2 ) + {\rm si} ( 2 \pi f T + \pi / 2) \big ]$$

$$\Rightarrow \hspace{0.3cm} G(f = 0) = g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \big [ {\rm si} ( - \pi/2 ) + {\rm si} ( +\pi/2 ) \big ] = 2 \cdot g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} {\rm si} ( \pi/2 ) = 2 \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \frac {{\rm sin}({\pi}/{2}) } { {\pi}/{2} } ={4}/{\pi} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{0.15cm}\underline {\approx 1.273} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T .$$

(4)  By writing out the  $\rm si$–function, with   $\sin (α ± π/2) = ± \cos(α)$, one gets:

$$G(f) = g_0 \cdot T \cdot \left [ \frac{{\rm sin} ( 2 \pi f T - \pi / 2 )}{2 \pi f T - \pi / 2 } + \frac{{\rm sin} ( 2 \pi f T + \pi / 2 )}{2 \pi f T + \pi / 2 } \right ]= g_0 \cdot T \cdot \frac {2}{\pi}\cdot\left [ \frac{-{\rm cos} ( 2 \pi f T )}{4 f T - 1 } + \frac{{\rm cos} ( 2 \pi f T )}{4 f T + 1 } \right ]$$

$$\Rightarrow \hspace{0.3cm} G(f) = g_0 \cdot T \cdot \frac {2}{\pi}\cdot \frac{(1+4 f T ) \cdot {\rm cos} ( 2 \pi f T )+ (1-4 f T ) \cdot {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 } = \frac {4}{\pi}\cdot g_0 \cdot T \cdot \frac{ {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 }\hspace{0.05cm}.$$

• The zeroes of  $G(f)$  are exclusively determined by the cosine function in the numerator, and are found at the frequencies  $f · T = 0.25,\ 0.75,\ 1.25,$  ...
• However, the first zero at  $f · T = 0.25$  is cancelled out by the simultaneously occuring zero in the denominator.  Therefore:

$$f_1 \hspace{0.15cm}\underline {= 0.75} \cdot 1/T \hspace{0.05cm}.$$