# Exercise 4.15Z: MSK Basic Pulse and MSK Spectrum

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:

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