Exercise 4.4: Pointer Diagram for DSB-AM

Spectrum of the analytical signal

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

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

The frequency conversion is done by means of "Double-Sideband Amplitude Modulation with Carrier".

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 $\rm (OSB)$ and the last term the lower sideband $\rm (USB)$.

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 exercise belongs to the chapter Analytical Signal and its Spectral Function.
The interactive applet Physical and Analytical Signal illustrates the topic covered here.
In this task we use the following nomenclature because of the German original:

The index $\rm N$ stands for "source signal" ⇒ (German: "Nachrichtenignal").
The index $\rm T$ stands for "carrier" ⇒ (German: "Trägersignal").
$\rm OSB$ denotes the "upper sideband" ⇒ (German: "oberes Seitenband").
$\rm USB$ denotes the "lower sideband" ⇒ (German: "unteres Seitenband").

Questions

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

$\text{V}$

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

$\text{V}$

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

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

$\text{V}$

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

$\text{V}$

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

$\text{V}$

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

$\text{V}$

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

$\text{V}$

$t_{\text{min}}\ = \ $

${\rm µ} \text{s}$

Solution

(1) By inverse Fourier transform of $S_+(f)$ taking into account the "Shifting Theorem":

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

Three different analytical signals

The expression describes the sum of three pointers rotating at different circular 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 real value $s_+(t = 0) \;\underline{= 1.8 \ \text{V}}$.

(2) The first statement is correct and results from the "Hilbert transform". On the other hand, the next two statements are'nt correct:

$s_+(t)$ is always a complex time function with exception of the limiting case $s(t) \equiv 0$.
However, every complex function also has purely real values at some points in time.
The "pointer group" 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 duration 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 $90^{\circ}$.
The blue pointer $\rm (OSB)$ rotates $20\%$ faster, the green one $\rm (USB)$ $20\%$ 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 OSB and USB are $108^{\circ}$ and $72^{\circ}$ respectively.
Since at this time the real parts of OSB and USB 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 rotation 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 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:

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