Exercise 2.13: Quadrature Amplitude Modulation

$\rm QAM$ model under consideration

The quadrature amplitude modulation $\rm (QAM)$ explained by the diagram allows the transmission of two source signals $q_1(t)$ and $q_2(t)$ over the same channel, under certain boundary conditions, which are to be determined in this task.

In this exercise, with $A_1 = A_2 = 2\ \rm V$, let:

$$q_1(t) = A_1 \cdot \cos(2 \pi \cdot f_{\rm 1} \cdot t),$$

$$q_2(t) = A_2 \cdot \sin(2 \pi \cdot f_{\rm 2} \cdot t)\hspace{0.05cm}.$$

For $ω_{\rm T} = 2π · 25\ \rm kHz$, the four carrier signals shown in the diagram are:

$$z_1(t) = \cos(\omega_{\rm T} \cdot t),$$

$$ z_2(t) = \sin(\omega_{\rm T} \cdot t),$$

$$ z_{1,\hspace{0.05cm}{\rm E}}(t) = 2 \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}),$$

$$ z_{2,\hspace{0.05cm}{\rm E}}(t) = 2 \cdot \sin(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T})\hspace{0.05cm}.$$

Both lowpass filters $\rm TP_1$ and $\rm TP_2$ with input signals $b_1(t)$ and $b_2(t)$ , respectively, remove all frequency components $|f| > f_{\rm T}$.

Hints:

This exercise belongs to the chapter Further AM Variants.
Particular reference is made to the page Quadrature Amplitude Modulation (QAM).

It is worth noting that the carrier signals $z_2(t)$ and $z_{2,\hspace{0.05cm}{\rm E}}(t)$ are applied with positive signs here.
Often – as in the theory section – these carrier signals are given as "minus-sine".
The following trigonometric transformations are given:

$$ \cos(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$

$$ \sin(\alpha) \cdot \sin(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)- \cos(\alpha + \beta) \big],$$

$$ \sin(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \sin(\alpha - \beta)+ \sin(\alpha + \beta) \big] \hspace{0.05cm}.$$

Questions

Solution

(1) With the given trigonometric transformations we get:

$$s(t) = A_1 \cdot \cos(\omega_{\rm 1} \cdot t)\cdot \cos(\omega_{\rm T} \cdot t) + A_2 \cdot \sin(\omega_{\rm 2} \cdot t)\cdot \sin(\omega_{\rm T} \cdot t) $$

$$\Rightarrow \hspace{0.3cm}s(t) = \frac{A_1}{2}\cdot \cos((\omega_{\rm T} - \omega_{\rm 1})\cdot t) + \frac{A_1}{2}\cdot \cos((\omega_{\rm T} + \omega_{\rm 1})\cdot t) + \frac{A_2}{2}\cdot \cos((\omega_{\rm T} - \omega_{\rm 2})\cdot t) - \frac{A_2}{2}\cdot \cos((\omega_{\rm T} + \omega_{\rm 2})\cdot t)\hspace{0.05cm}.$$

The second answer is correct.

(2) With $A_1 = A_2 = 2 \ \rm V$ and $f_1 = f_2 = 5\ \rm kHz$, the first and third cosine oscillations constructively overlap and the other two cancel completely.

Thus, the following simple result is obtained:

$$ s(t) = 2\,{\rm V} \cdot \cos(2 \pi \cdot 20\,{\rm kHz} \cdot t) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} s(t = 50\,{\rm µ s}) \hspace{0.15cm}\underline {= 2\,{\rm V}} \hspace{0.05cm}.$$

(3) The first answer is correct:

For phase-synchronous demodulation $(Δϕ_T = 0)$ , the signals before the low-pass filters according to subtask (2) are obtained as:

$$b_1(t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \cos(\omega_{\rm 45} \cdot t),$$

$$ b_2(t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \sin(\omega_{\rm 45} \cdot t)\hspace{0.05cm}.$$

Thus, after eliminating the respective $45\ \rm kHz$ components, we get $v_1(t) = q_1(t)$ and $v_2(t) = q_2(t)$.

(4) Analogously to subtask (3) it now holds that:

$$ b_1(t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})= 2\,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )},$$

$$b_2(t)= 2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})= 2\,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )}\hspace{0.05cm}.$$

The sink signals $v_1(t)$ and $v_2(t)$ in this constellation exhibit delays and thus phase distortions compared with $q_1(t)$ and $q_2(t)$ .
These belong to the class of linear distortions ⇒ Answer 2.

(5) In general, it holds for the source signal that:

$$r(t) = s(t) = q_1(t) \cdot \cos(\omega_{\rm T} \cdot t) + q_2(t) \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

Multiplication by the receiver-side carrier signals $z_{1,\hspace{0.05cm}{\rm E}}(t)$ and $z_{2,\hspace{0.05cm}{\rm E}}(t)$ and band-limiting leads to the signals

$$v_1(t) = \cos(\Delta \phi_{\rm T}) \cdot q_1(t) - \sin(\Delta \phi_{\rm T}) \cdot q_2(t),$$

$$ v_2(t) = \sin(\Delta \phi_{\rm T}) \cdot q_1(t) + \cos(\Delta \phi_{\rm T}) \cdot q_2(t) \hspace{0.05cm}.$$

From this it can be seen:

With a phase offset of $Δϕ_{\rm T} = 30^\circ$ , the sink signal $v_1(t)$ includes not only the signal $q_1(t)$ attenuated by about $\cos(30^\circ) = 0.866$ , but also the frequency $f_2$ contained in $q_2(t)$ .
This is weighted by the factor $\sin(30^\circ) = 0.5$ .
Thus, nonlinear distortions are present ⇒ Answer 3.

	$s(t)$ is composed of two cosine and two sine oscillations.
	$s(t)$ is composed of four cosine oscillations.
	$s(t)$ is composed of four sine oscillations.

	$v_1(t) = q_1(t)$ and $v_2(t) = q_2(t)$ both hold.
	Linear distortions occur.
	Nonlinear distortions occur.