Exercise 3.8: Modulation Index and Bandwidth

Bessel function values

A harmonic oscillation of the form

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

is angle-modulated and then the one-sided magnitude spectrum $|S_+(f)|$ is obtained.

with a message frequency of $f_{\rm N} = 2 \ \rm kHz$ the following spectral lines can be seen with the following weights:

$$|S_{\rm +}(98\,{\rm kHz})| = |S_{\rm +}(102\,{\rm kHz})| = 1.560\,{\rm V}\hspace{0.05cm},$$ $$|S_{\rm +}(96\,{\rm kHz})| = |S_{\rm +}(104\,{\rm kHz})| = 1.293\,{\rm V}\hspace{0.05cm},$$

$$ |S_{\rm +}(94\,{\rm kHz})| = |S_{\rm +}(106\,{\rm kHz})| = 0.594\,{\rm V}\hspace{0.05cm}.$$

Further spectral lines follow each with frequency spacing $f_{\rm N} = 2 \ \rm kHz$, but are not given here and can be ignored.

If one increases the message frequency to $f_{\rm N} = 4 \ \rm kHz$, there occur dominant lines

$$|S_{\rm +}(100\,{\rm kHz})| = 2.013\,{\rm V}\hspace{0.05cm},$$

$$|S_{\rm +}(96\,{\rm kHz})|\hspace{0.2cm} = |S_{\rm +}(104\,{\rm kHz})| = 1.494\,{\rm V}\hspace{0.05cm},$$

$$ |S_{\rm +}(92\,{\rm kHz})|\hspace{0.2cm} = |S_{\rm +}(108\,{\rm kHz})| = 0.477\,{\rm V},$$

as well as further, negligible Dirac lines with spacing $f_{\rm N} = 4 \ \rm kHz$.

Hints:

This exercise belongs to the chapter Frequency Modulation.
Reference is also made to the chapter Phase Modulation and particularly to the section Signal characteristics with frequency modulation.

Questions

	Phase modulation.
	Frequency modulation.

$η_2 \ = \ $

$A_{\rm T} \ = \ $

$\ \rm V$

$B_2 \ = \ $

$\ \rm kHz$

$η_4\ = \ $

$B_4 \ = \ $

$\ \rm kHz$

Solution

(1) We are dealing with a frequency modulation⇒ Answer 2.

In phase modulation, the weights of the Dirac lines would not change when the frequency is doubled.

(2) The spectral function given suggests the carrier frequency $f_{\rm T} = 100 \ \rm kHz$ due to the symmetry properties.

Since at $f_{\rm N} = 2 \ \rm kHz$ the spectral line disappears at $f_{\rm T} = 100 \ \rm kHz$ , we can assume $η_2 \hspace{0.15cm}\underline { ≈ 2.4}$ .
A check of the other pulse weights confirms this result:

$$\frac { |S_{\rm +}(f =102\,{\rm kHz})|}{ |S_{\rm +}(f =104\,{\rm kHz})|} = 1.206,\hspace{0.2cm} \frac { {\rm J}_1(2.4)}{ {\rm J}_2(2.4)}= 1.206 \hspace{0.05cm}.$$

(3) The weights of the Dirac lines at $f_{\rm T} + n · f_{\rm N}$ are generally:

$$D_n = A_{\rm T} \cdot { {\rm J}_n(\eta)} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} D_1 = A_{\rm T} \cdot { {\rm J}_1(\eta)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} A_{\rm T} = D_1/{\rm J}_1(η) = 1.560\ \rm V/0.520\hspace{0.15cm}\underline { = 3 \ V}.$$

(4) Given the requirement $K < 1\%$ , one can use the following rule of thumb (Carson's rule):

$$B_{\rm 2} = 2 \cdot f_{\rm N} \cdot (\eta +2) \hspace{0.15cm}\underline {= 17.6\,{\rm kHz}}\hspace{0.05cm}.$$

Thus, the Fourier coefficients $D_{–4}$, ... , $D_4$ are available.

(5) For frequency modulation, the general rule is:

$$\eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.05cm}.$$

Thus, by doubling the message frequency $f_{\rm N}$, the modulation index is halved: $η_4 = η_2/2\hspace{0.15cm}\underline { = 1.2}$.

(6) Using the same calculation as in question (4) , the channel bandwidth necessary for $K < 1\%$ is obtained using

$$B_4 = 3.2 · 8\ \rm kHz \hspace{0.15cm}\underline {= 25.6 \ \rm kHz}.$$

Because the modulation index is only half as large, transmitting the Fourier coefficients $D_{–3}$, ... , $D_3$ is sufficient for limiting the distortion factor to $1\%$.