Exercise 2.1Z: DSB-AM without/with Carrier

The signals involved in this AM

The red curve on the graph shows a section of the transmitted signal $s(t) = q(t) · z(t)$ of a double-sideband amplitude modulation (abbreviated as DSB-AM) without carrier. The duration of the time interval is $\rm 200 \ µ s$.

Additionally plotted in the graph are:

the source signal (as a blue dashed curve):

$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$

the carrier signal (as a grey dashed trace):

$$z(t) = 1 \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T}).$$

From subtask (4) onwards, the "DSB-AM with carrier" is considered. In that case, with $A_{\rm T} = 2\text{ V}$:

$$s(t) = \left(q(t) + A_{\rm T} \right) \cdot z(t) \hspace{0.05cm}.$$

Hints:

This exercise belongs to the chapter Double-Sideband Amplitude Modulation.
Particlar reference is made to the pages Description in the time domain and Double-Sideband Amplitude Modulation with carrier.

Questions

$\phi_{\rm N} \ = \ $

$\ \text{degrees}$

$\phi_{\rm T} \ = \ $

$\ \text{degrees}$

$f_{\rm N} \ = \ $

$\ \text{kHz}$

$f_{\rm T} \ = \ $

$\ \text{kHz}$

	All zero crossings of $z(t)$ are preserved in $s(t)$.
	There are additional zero crossings caused by $q(t)$.
	$s(t) = a(t) · \cos(ω_T · t)$ holds with $a(t) = \|q(t)\|$.

$f_1 \ = \ $

$\ \text{kHz}$

$f_2\ = \ $

$\ \text{kHz}$

$m \ = \ $

	$S(f)$ now includes Dirac delta functions at $±f_{\rm T}$.
	The weights of these Dirac delta lines are each $2\text{ V}$.
	$q(t)$ can be seen in the envelope of $s(t)$.
	Due to the additional carrier component, the power remains unchanged.

Solution

(1) Both signals are cosine ⇒ $ϕ_{\rm N} \hspace{0.15cm}\underline { = 0}$ and $ϕ_{\rm T} \hspace{0.15cm}\underline { = 0}$.

(2) From the graph, the period durations of $200$ μs and $20$ μs can be seen for $q(t)$ and $z(t)$, respectively.

This gives the frequencies as $f_{\rm N} \hspace{0.15cm}\underline { = 5}$ kHz and $f_{\rm T} \hspace{0.15cm}\underline { = 50}$ kHz.

(3) Answers 1 and 2 are correct:

The zero crossings of $z(t)$ at $±5$ μs, $±15$ μs, $±25$ μs, ... ... are also present in the signal $s(t)$ ⇒ Answer 1 is correct.
Other zero intersects of $s(t)$ – cause by $q(t)$ – are present at $±50$ μs, $±150$ μs, $±250$ μs, .... ⇒ Answer 2 is also correct.
In contrast, the third statement is not true. Instead, $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$

DSB–AM spectra $Z(f)$, $Q(f)$ and $S(f)$

For $q(t) > 0$ the phase function is $ϕ(t) = 0$ and $s(t)$ coincides with $z(t)$.
In contrast, for $q(t) < 0$: $ϕ(t) = π = 180^\circ$.
At the zero crossings of $q(t)$, the modulated signal $s(t)$ exhibits phase jumps.

(4) The spectrum $S(f)$ results from the convolution of the spectral functions $Z(f)$ and $Q(f)$, each consisting of only two Dirac delta functions. The graph displays the result.

The Dirac delta functions plotted in red apply only to the "DSB-AM with carrier" and refer to subtask (6).
Convolution of the two $Z(f)$ Dirac delta functions at $f_{\rm T} = 50\text{ kHz}$ with $Q(f)$ leads to the Dirac delta lines at $f_{\rm T} - f_{\rm N}$ and $f_{\rm T} + f_{\rm N}$, each with weight $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
Thus, the desired values are $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$ and $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
The Dirac function $0.5 · δ(f + f_{\rm T})$ with two markers leads to two more Dirac delta lines at $-f_1$ and $-f_2$.

(5) The modulation depth is calculated as:

$$ m = \frac{q_{\rm max}}{A_{\rm T}} = \frac{A_{\rm N}}{A_{\rm T}} \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$

(6) Answers 1 and 3 are correct:

According to the sketch, Dirac delta lines result at $±f_{\rm T}$, both with impulse weight $A_{\rm T}/2 = 1\text{ V}$.
At $m ≤ 1$, $q(t)$ is detectable in the envelope ⇒ envelope demodulation is applicable.
However, this simpler receiver variant must be accounted for with a much larger transmission power.
In this example $(m = 0.5)$ the addition of a carrier multiplies the transmission power by nine.