Aufgaben:Exercise 4.4: Pointer Diagram for DSB-AM

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Spectrum of the analytical signal

We assume a cosine-shaped source signal  $q(t)$  with

• the amplitude  $A_{\rm N} = 0.8 \ \text{V}$  and
• the frequency  $f_{\rm N}= 10 \ \text{kHz}$.

The frequency conversion is done by means of  Zweiseitenband–Amplitudenmodulation mit Träger.

The modulated signal  $s(t)$  is with the (normalised) carrier  $z(t) = \text{cos}(\omega_{\rm T} \cdot t)$  and the DC component  $q_0 = 1 \ \text{V}$:

$$\begin{align*} s(t) & = \left(q_0 + q(t)\right) \cdot z(t)= \left({\rm 1 \hspace{0.05cm} V} + {\rm 0.8 \hspace{0.05cm}V}\cdot {\cos} ( \omega_{\rm N}\cdot t)\right) \cdot {\cos} ( \omega_{\rm T}\cdot t) = \\ & = q_0 \cdot {\cos} ( \omega_{\rm T}\cdot t) + {A_{\rm N}}/{2} \cdot {\cos} ( (\omega_{\rm T}+ \omega_{\rm N}) \cdot t) + {A_{\rm N}}/{2} \cdot {\cos} ( (\omega_{\rm T}- \omega_{\rm N}) \cdot t).\end{align*}$$

The first term describes the carrier, the second term the so-called upper sideband (USB) and the last term the lower sideband (LSB).

The sketch shows the spectrum  $S_+(f)$  of the corresponding analytical signal for  $f_{\rm T} = 50 \ \text{kHz}$. You can see

• the carrier (red),
• the upper sideband (blue) and
• the lower sideband (grün).

In subtask  (5)  the magnitude of  $s_+(t)$  is asked for. This is the length of the resulting pointer.

Hints:

• This task belongs to the chapter  Analytical Signal and Its Spectral Function.

• You can check your solution with the interaction module  Physikalisches Signal & Analytisches Signal .

Questions

1

What is the analytical signal  $s_+(t)$. What is its magnitude at time  $t = 0$?

$\text{Re}[s_+(t=0)]\ = \ $  $\text{V}$

$\text{Im}[s_+(t=0)]\ = \ $  $\text{V}$

2

Which of the following statements is true?

	$s_+(t)$  results from  $s(t)$, if  $\cos(\text{...})$  is replaced  ${\rm e}^{{\rm j}(\text{...})}$ .
	If  $s(t)$  is an even time function,  $s_+(t)$  is purely real.
	At no time does the imaginary part of  $s_+(t)$ disappear.

3

What is the value of the analytical signal at time  $t = 5 \ {\rm µ}\text{s}$?

$\text{Re}[s_+(t=5 \ {\rm µ} \text{s})]\ = \ $  $\text{V}$

$\text{Im}[s_+(t=5 \ {\rm µ} \text{s})]\ = \ $  $\text{V}$

4

What is the value of  $s_+(t)$  at time  $t = 20 \ {\rm µ}\text{s}$?

$\text{Re}[s_+(t=20 \ {\rm µ} \text{s})]\ = \ $  $\text{V}$

$\text{Im}[s_+(t=20 \ {\rm µ} \text{s})]\ = \ $  $\text{V}$

5

What is the smallest possible pointer length? At what time   $t_{\text{min}}$  does this value occur for the first time?

$|s_+(t)|_{\text{min}}\ = \ $  $\text{V}$

$t_{\text{min}}\ = \ $  ${\rm µ} \text{s}$

Solution

Solution

(1)  By inverse Fourier transformation of  $S_+(f)$  considering the  Verschiebungssatzes  holds:

$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } + {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 40}\hspace{0.05cm} t }.$$

The expression describes the sum of three pointers rotating at different angular velocities.

• In the above equation, for example,  $\omega_{60} = 2\pi (f_{\rm T} + f_{\rm N}) = 2\pi \cdot 60 \ \text{kHz}$.
• At time  $t = 0$  all three pointers point in the direction of the real axis (see left graph).

One obtains the purely real value  $s_+(t = 0) \;\underline{= 1.8 \ \text{V}}$.

Three different analytical signals

(2)  The first statement is correct and results from the Hilbert transform. On the other hand, the next two statements are not correct:

• $s_+(t)$  is always a complex time function with the exception of the limiting case  $s(t) = 0$.
• However, every complex function also has purely real values at some points in time.
• The pointer composite always rotates in a mathematically positive direction.
• If the sum vector crosses the real axis, the imaginary part disappears at this point and  $s_+(t)$  is purely real.

(3)  The period of the carrier signal is  $T_0 = 1/f_T = 20 \ {\rm µ} \text{s}$.

• After  $t = 5 \ {\rm µ} \text{s}$  (see middle graph) the carrier has thus rotated by  $90^{\circ}$  gedreht.
• The blue pointer (USB) rotates  $20\%$  faster, the green one (LSB)  $20\%$  slower than the red rotary pointer (carrier signal):

$$s_{+}({\rm 5 \hspace{0.05cm} {\rm µ} s}) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}50 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005 } + {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}60 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005 }+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}40 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005 } = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 90^\circ }+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 108^\circ }+{\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 72^\circ }.$$

• Thus, the angles travelled in  $ 5 \ {\rm µ} \text{s}$  by USB and LSB are  $108^{\circ}$  and  $72^{\circ}$ respectively.
• Since at this time the real parts of USB and LSB compensate,  $s_+(t=5 \ {\rm µ} \text{s})$  is purely imaginary and we obtain:

$${\rm Im}\left[s_{+}(t = {\rm 5 \hspace{0.05cm} {\rm µ} s})\right] = {\rm 1 \hspace{0.05cm} V} + 2 \cdot {\rm 0.4 \hspace{0.05cm} V}\cdot \cos (18^\circ ) \hspace{0.15 cm}\underline{= {\rm 1.761 \hspace{0.05cm} V}}.$$

(4)  After one revolution of the red carrier, i.e. at time $t$ = $T_0 = 20 \ {\rm µ} \text{s}$, the blue pointer has already covered $72^{\circ}$ more and the green pointer correspondingly $72^{\circ}$ less. The sum of the three pointers is again purely real and results in accordance with the graph on the right:

$${\rm Re}\left[s_{+}({\rm 20 \hspace{0.05cm} {\rm µ} s})\right] = {\rm 1 \hspace{0.05cm} V} + 2 \cdot {\rm 0.4 \hspace{0.05cm} V}\cdot \cos (72^\circ ) \hspace{0.15 cm}\underline{= {\rm 1.236 \hspace{0.05cm} V}}.$$

(5)  The magnitude is minimum when the pointers of the two sidebands are offset from the carrier by  $180^{\circ}$ . It follows:

$$|s_{+}(t)|_{\rm min} = {\rm 1 \hspace{0.05cm} V} - 2 \cdot {\rm 0.4 \hspace{0.05cm} V} \hspace{0.15 cm}\underline{= {\rm 0.2 \hspace{0.05cm} V}}.$$

Within one period  $T_0$  of the carrier, a phase offset of  $\pm72^{\circ}$  occurs with respect to the pointers of the two sidebands. From this follows:

$$t_{\text{min}} = 180^{\circ}/72^{\circ} \cdot T_0 = 2.5 \cdot T_0 \;\underline{= 50 \ {\rm µ} \text{s}}.$$