Aufgaben:Exercise 2.13: Quadrature Amplitude Modulation

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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 exercise.

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

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

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

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   $\rm LP_1$  and  $\rm LP_2$  with input signals  $b_1(t)$  resp.  $b_2(t)$ , 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

1

Calculate the transmitted signal  $s(t)$  in the case that  $f_1 ≠ f_2$.  Which of the following statements apply?

	$s(t)$  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.

2

What is  $s(t)$  with  $f_1 = f_2 = 5 \ \rm kHz$?  What signal value arises for  $t = 50 \ \rm µ s$ ?

$s(t = 50 \ \rm µ s) \ = \ $ $\ \rm V$

3

Calculate the sink signals  $v_1(t)$  and  $v_2(t)$ for  $f_1 = f_2$  and  $Δϕ_{\rm T} = 0$  (no phase offset).  Which statements are true?

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

4

Calculate the sink signals  $v_1(t)$  and  $v_2(t)$  for  $f_1 = f_2$  and a phase offset  $Δϕ_{\rm T} = 30^\circ$.  Which statements are true?

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

5

Which of the following statements apply when  $f_1 ≠ f_2$  and  $Δϕ_{\rm T} ≠ 0$  (with an arbitrary phase offset)?

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

Solution

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 received signal:

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

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   $f_2$ is contained in   $q_2(t)$.
• This is weighted by the factor   $\sin(30^\circ) = 0.5$.
• Thus,  nonlinear distortions are present   ⇒   Answer 3.