Aufgaben:Exercise 3.3: Sum of two Oscillations

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Two different Bessel spectra

The equivalent low-pass signal with phase modulation, when normalized to the carrier amplitude $(A_{\rm T} = 1)$ is: 

$$ s_{\rm TP}(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}K_{\rm PM}\hspace{0.05cm}\cdot \hspace{0.05cm}q(t) }\hspace{0.05cm},$$

The modulator constant is assumed to be  $K_{\rm PM} = \rm 1/V$  throughout the task.

The upper graph shows the corresponding spectral function  $B_1(f)$, when the source signal is

$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$

The weights of the Bessel-Dirac lines when  $η_1 = 0.9$  are obtained as follows:

$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm} {\rm J}_1 (0.9) = 0.406 \approx 0.4,$$

$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm} {\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$

Use the approximations given in the graph to simplify the calculations.

The Bessel function  $B_2(f)$  is obtained for the source signal

$$q_2(t) = 0.65\,{\rm V} \cdot \cos(2 \pi \cdot 3\,{\rm kHz} \cdot t)$$

The numerical values of the Dirac lines are obtained from the following:

$${\rm J}_0 (0.65) = 0.897 \approx 0.9,\hspace{0.3cm}{\rm J}_1 (0.65) = 0.308 \approx 0.3, \hspace{0.3cm}{\rm J}_2 (0.65) = 0.051 \approx 0\hspace{0.05cm}.$$

From the graph, it can be seen that due to the cosine source signal  $q_2(t)$  and the cosine carrier signal $z(t)$ , the spectral lines at $±3 \ \rm kHz$  are both positive and imaginary.

In the context of this task, we will now investigate the case where the source signal

$$q(t) = q_1(t) + q_2(t)$$

is applied to the input of the phase modulator.

• It is worth mentioning that  $|q(t)| < q_{\rm max} = 1.45 \ \rm V$ .
• This maximum value is slightly smaller than the sum  $A_1 + A_2$ of the individual amplitudes when a sine and a cosine function with the given amplitudes are added up.

In the following questionnaire,

• $S_{\rm TP}(f)$  denotes the spectral function of the equivalent low-pass signal,
• $S_+(f)$  denotes the spectral functions of the analytic signal,

in both cases assuming that  $q(t) = q_1(t) + q_2(t)$  holds and that the carrier frequency is $f_{\rm T} = 100 \ \rm kHz$ .

Hints:

• This exercise belongs to the chapter  Phase Modulation.
• Particular reference is made to the page  Equivalent low-pass signal in phase modulation.
• The values of the Bessel functions can be found in formula collections in table form.
• You can also use the interactive applet   Bessel functions of the first kind  to solve this task.

Questions

1

Let  $q(t) = q_1(t)+q_2(t)$.  Which geometric figure describes the given locus curve $s_{\rm TP}(t)$?

	The locus curve is an ellipse.
	The locus curve is a circle.
	The locus curve is approximately a semi-circle.
	The locus curve is an arc, with an approximate opening angle of  $90^\circ$.

2

Calculate the spectral function $S_{\rm TP}(f)$.  Between what frequencies  $f_{\rm min}$  and  $f_{\rm max}$  do the spectral lines lie?

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

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

3

Calculate the weight of the Dirac function at  $f = 0$.

${\rm Re}\big[S_{\rm TP}(f = 0)\big] \ = \ $

${\rm Im}\big[S_{\rm TP}(f = 0)\big] \ = \ $

4

Calculate the weight of the Dirac function at  $f = 1\ \rm kHz$.

${\rm Re}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $

${\rm Im}\big[S_{\rm TP}(f = 1 \ \rm kHz)\big] \ = \ $

5

Calculate the weight of the  $S_+(f)$–Dirac function at  $f = 98 \ \rm kHz$.

${\rm Re}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $

${\rm Im}\big[S_{\rm +}(f = 98 \ \rm kHz)\big] \ = \ $

Solution

Solution

(1)  The third answer is correct:

• In angle modulation, the complex pointer   $s_{\rm TP}(t)$  always moves on a circular arc with the following opening angle:

$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$

• Using the (admittedly very rough) approximation  $166^\circ \approx 180^\circ$  we indeed get a semicircle.

(2)  In general,   $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.

• Since  $B_1(f)$  is limited to the frequencies   $|f| ≤ 2 \ \rm kHz$  and  $B_2(f)$  is limited to the range   $±3 \ \rm kHz$ , the convolution product is limited to  $|f| ≤ 5 \ \rm kHz$ .
• It follows that:

$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$

$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$

(3)  The convolution product for frequency   $f = 0$  results from multiplying  $B_1(f)$  with  $B_2(f)$  and summing.

• Only for   $f = 0$  are both  $B_1(f)$  and  $B_2(f)$  non-zero.
• Thus, we get:

$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$

(4)  Now, before multiplication and summation there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$ .  This gives:

$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz}) + B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0) = 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$

$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$

(5)  The Dirac line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is

$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) + B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$

$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$