Aufgaben:Exercise 3.3Z: Characteristics Determination

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Spectrum of the analytical signal

Let us consider the phase modulation of the harmonic oscillation

$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$

which, given a normalized carrier amplitude  $(A_{\rm T} = 1)$ , leads to the following transmitted signal:

$$ s(t) = \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \big]\hspace{0.05cm}.$$

The spectrum of the corresponding analytical signal $s_{\rm TP}(t)$  is generally:

$$S_{\rm TP}(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}(\phi_{\rm N}\hspace{0.05cm}+\hspace{0.05cm} 90^\circ) }\cdot \hspace{0.05cm} \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}$$

Here,  $η = K_{\rm PM} · A_{\rm N}$  is called the modulation index.

In the graph, the real and imaginary parts of the spectrum  $S_+(f)$  of the analytical signal  $s_+(t)$ are shown separately. This should be used to determine the characteristics  $f_{\rm T}$,  $f_{\rm N}$,  $ϕ_{\rm N}$  and  $η$ .

Hints:

• This exercise belongs to the chapter  Phase Modulation.
• Particular reference is made to the page  Equivalent low-pass signal in phase modulation.

• For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:

$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm J}_{2} (\eta)= {2}/{\eta} \cdot {\rm J}_{1} (\eta) - {\rm J}_{0} (\eta) \hspace{0.05cm}.$$

Questions

1 What are the frequencies  $f_{\rm T}$  and  $f_{\rm N}$?

$f_{\rm T} \ = \ $ $\ \rm kHz$

$f_{\rm N} \ = \ $ $\ \rm kHz$

2 Calculate the magnitude and the phase of  $S_{\rm TP}(f = 3 \ \rm kHz)$.

$|S_{\rm TP}(f = 3 \ \rm kHz)| \ = \ $

${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $ $\ \rm Grad$

3 Calculate the magnitude and the phase of  $S_{\rm TP}(f = 6 \ \rm kHz)$.

$|S_{\rm TP}(f = 6 \ \rm kHz)| \ = \ $

${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $ $\ \rm Grad$

4 What is the phase of the source signal  $q(t)$?

$ϕ_{\rm N} \ = \ $ $\ \rm Grad$

5 What is the magnitude of the modulation index  $η$ ?

$η \ = \ $

Solution

Solution

(1)  Regarding   $|S_+(f)|$  there is a symmetry with respect to the carrier frequency  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  The distance between the spectral lines is  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.

(2)  Considering  $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$ , it holds that:

$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$

$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$

(3)  Analogously to in question  (2) , at frequency   $f = 6 \ \rm kHz$ we obtain:

$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$

$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$

(4)  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   (2), the phase is:

$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$

• Checking this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  (3)  yields the same value:

$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$

(5)  The given equation can be rewritten as follows:

$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$

• With  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  and  ${\rm J}_2(η) = 0.232$  we thus get:

$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$