Aufgaben:Exercise 3.8: Modulation Index and Bandwidth

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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

1

Which modulation method is used here?

	Phase modulation.
	Frequency modulation.

2

What is the modulation index  $η_2$  at message frequency  $f_{\rm N} = 2 \ \rm kHz$?

$η_2 \ = \ $

3

What is the carrier amplitude?

$A_{\rm T} \ = \ $ $\ \rm V$

4

Specify the bandwidth  $B_2$ for  $f_{\rm N} = 2 \ \rm kHz$  if a distortion factor  $K < 1\%$  is desired.

$B_2 \ = \ $ $\ \rm kHz$

5

What is the modulation index $η_4$  at message frequency  $f_{\rm N} = 4 \ \rm kHz$?

$η_4\ = \ $

6

What channel bandwidth  $B_4$  is now required to ensure  $K < 1\%$ ?

$B_4 \ = \ $ $\ \rm kHz$

Solution

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 given spectral function 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  $1\%$.