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:

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

$f_0 \ = \ $

$\ \cdot 1/T$

$R(f=0) \ = \ $

$\ \cdot g_0 \cdot T$

$G(f=0) \ = \ $

$\ \cdot g_0 \cdot T$

$f_1 \ = \ $

$\ \cdot 1/T$

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:

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