Aufgaben:Exercise 2.3Z: DSB-AM due to Nonlinearity

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DSB–AM with nonlinearity

In this task, we consider the realization of a double-sideband amplitude modulation  $\text{DSB-AM}$  using the nonlinear characteristic curve:

$$y = g(x) = c_1 \cdot x + c_2 \cdot x^2+ c_3 \cdot x^3\hspace{0.05cm}$$

$$ \Rightarrow \hspace{0.5cm} c_1 = 2,\hspace{0.5cm}c_2 = 0.25/{\rm V},\hspace{0.5cm}c_3 = 0 \hspace{0.5cm}{\rm or}\hspace{0.5cm}c_3 = 0.01/{\rm V^2}\hspace{0.05cm}.$$

The sum of the carrier signal  $ z(t)$  and the source signal  $ q(t)$  is present at the input of this characteristic curve:

$$ x(t) = z(t) + q(t) = A_{\rm T} \cdot \cos(\omega_{\rm T} t)+ q(t),\hspace{0.2cm} A_{\rm T} = 4\,{\rm V}\hspace{0.05cm}.$$

• It is known that the source signal  $q(t)$  contains spectral components between  $1 \ \rm kHz$  and  $9 \ \rm kHz$   (up to and including these limits).
• From subtask   (5)  onwards, the following source signal should be assumed:

$$q(t) = A_{\rm 1} \cdot \cos(\omega_{\rm 1} t)+A_{\rm 9} \cdot \cos(\omega_{\rm 9} t) \hspace{0.05cm}.$$

• Let the angular frequencies be  $ω_1 = 2 π · 1 \ \rm kHz$  and  $ω_9 = 2 π · 9\ \rm kHz$.  The corresponding amplitudes are given as:  $A_1 = 1\ \rm V$  and  $A_9 = 2\ \rm V$.

The following abbreviations are used in the questions for this task:

$$ y(t) = y_1(t) + y_2(t)+y_3(t),$$

$$y_1(t) = c_1 \cdot [z(t) + q(t)],$$

$$ y_2(t) = c_2 \cdot[z(t) + q(t)]^2,$$

$$y_3(t) = c_3 \cdot [z(t) + q(t)]^3 \hspace{0.05cm}.$$

Thus,  the transmitted signals  $s(t)$  i.e.,  $s_1(t)$,  $s_2(t)$  and  $s_3(t)$,  result from band-limiting to the range from  $90 \ \rm kHz$  to  $110 \ \rm kHz$.

Hints:

• This exercise belongs to the chapter  Double-Sideband Amplitude Modulation.
• Particular reference is made to the page  Amplitude modulation with a quadratic characteristic curve.

• The following trigonometric transformations are given:

$$ \cos^2(\alpha) = {1}/{2} \cdot \left[ 1 + \cos(2\alpha)\right] \hspace{0.05cm}, \hspace{0.5cm} \cos^3(\alpha) = {1}/{4} \cdot \left[ 3 \cdot \cos(\alpha) + \cos(3\alpha)\right] \hspace{0.05cm}.$$

Questions

1

What is a reasonable choice for the carrier frequency?

$f_{\rm T} \ = \ $ $\ \text{kHz}$

2

What signal components does  $s_1(t)$  include?

	The term  $c_1 \cdot z(t)$,
	the term  $c_1 \cdot q(t)$.

3

What signal components does  $s_2(t)$  include?

	The term  $c_2 · z^2(t)$,
	the term  $c_2 · q^2(t)$,
	the term  $2c_2 · z(t) · q(t)$.

4

What signal components does  $s_3(t)$  (at least partially)  include?

	The term  $c_3 · z^3(t)$,
	the term  $3 · c_3 · z^2(t) · q(t)$,
	the term  $3 · c_3 · z(t) · q^2(t)$,
	the term  $c_3 · q^3(t)$.

5

Calculate  $s(t)$,  when  $c_3 = 0$  and the source signal  $q(t)$  is composed of two cosine oscillations.  What is the modulation depth  $m$?

$m \ = \ $

6

Now calculate the transmitted signal  $s(t)$  given  $c_3 = \rm 0.01/V^{2}$.  Which of the following statements are true?

	Because  $c_3 ≠ 0$  the spectral line changes at  $f_{\rm T}$.
	Because  $c_3 ≠ 0$  linear distortions arise,  which can be compensated for.
	Because  $c_3 ≠ 0$  nonlinear  (i.e. irreversible) distortions arise.

Solution

Solution

(1)  The carrier frequency is most sensibly equal to the center frequency of the bandpass:  $f_{\rm T}\hspace{0.15cm}\underline{ = 100\ \rm kHz}$.

• If  $f_{\rm T}$  does not deviate from this by more than  $±1 \ \rm kHz$,  this also results in a  "DSB-AM".

(2)  $s_1(t)$  includes only the carrier  $z(t)$   ⇒   Answer 1. The source signal  $q(t)$  is removed by the band-pass.

(3)  The quadratic term  $z^2(t)$  consists of a DC component   $($at  $f = 0)$  and a component at   $2f_{\rm T}$.

• Additionally,  all spectral components of   $q^2(t)$  are outside the band-pass.
• Therefore the  last answer  is correct.

(4) Answers 1 and 3  are correct:

• The term  $\cos^3(ω_Tt)$  has its largest signal component at  $f = f_{\rm T}$.
• The third answer  $(3 · c_3 · z(t) · q^2(t))$  lies between  $100\ \rm kHz ± 18 \ \rm kHz $.
• Some parts – namely the frequency components between   $90\ \rm kHz $ and $110 \ \rm kHz$  – are not removed by the band-pass and are thus also included in  $s(t)$.

Generated DSB-AM spectrum

(5)  The transmitted signal consists of a total of five frequencies:

$$s(t) = c_1 \cdot A_{\rm T} \cdot \cos(\omega_{\rm T} t)+ c_2 \cdot A_{\rm T} \cdot A_{\rm 1} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 1})t) + c_2 \cdot A_{\rm T} \cdot A_{\rm 2} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 2})t) \hspace{0.05cm}.$$

• Note that the second term and the third term each contain two signal frequencies::
  • $\text{99 kHz}$  and  $\text{101 kHz}$ and
  • $\text{91 kHz}$  and  $\text{109 kHz}$, respectively.

• Given   $A_{\rm T} = 4 \ \rm V$,  $A_1 = 1 V$,  $A_9 = 2 \ \rm V$,  $c_1 = 1$  and  $c_2 = 1/A_{\rm T} = \rm 0.25/V$,  it holds that:

$$s(t) = 4\,{\rm V} \cdot \cos(\omega_{\rm T} t) + 1\,{\rm V} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 1})t) + 2\,{\rm V}\cdot \cos((\omega_{\rm T} \pm \omega_{\rm 2})t) \hspace{0.05cm}.$$

• From this it can be seen that for the modulation depth:

$$m =\frac{A_1 + A_9}{A_{\rm T}} = \rm \frac{1\ V + 2 \ V}{4 \ V} \hspace{0.15cm}\underline{=0.75}.$$

(6)  The graph above shows the spectrum   $S_+(f)$  – that is, only positive frequencies – where  $c_3 = 0$. 

• When  $c_3 ≠ 0$  the following additional spectral components arise:

$$c_3 \cdot z^3(t)= \frac{c_3 \cdot A_{\rm T}^3}{4} \cdot \left[ 3 \cdot \cos(\omega_{\rm T} t) + \cos(3\omega_{\rm T} t)\right] \hspace{0.05cm}.$$

• The first component falls in the pass-band of the band-pass.  As a result,  the Dirac weight at   $f_{\rm T} = 100\ \rm kHz$  is increased from the original  $8 \ \rm V$  to 

$$\text{8 V + 0.75 · 0.01/V}^2 · 4^3 \text{ V}^3 = 8.48 \ \rm V.$$

• Furthermore,  the third spectral component of subtask  (4)  provides an unwanted contribution to  $S_+(f)$.  In this case:

$$q^2(t) = \left[A_{\rm 1} \cdot \cos(\omega_{\rm 1} t)+A_{\rm 9} \cdot \cos(\omega_{\rm 9} t)\right]^2 = A_{\rm 1}^2 \cdot \cos^2(\omega_{\rm 1} t)+ A_{\rm 9}^2 \cdot \cos^2(\omega_{\rm 9}t) + 2 \cdot A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 1} t)\cdot \cos(\omega_{\rm 9} t)$$

$$ \Rightarrow \hspace{0.2cm} q^2(t) = \frac{A_{\rm 1}^2}{2} +\frac{A_{\rm 1}^2}{2} \cdot \cos(\omega_{\rm 2} t)+ \frac{A_{\rm 9}^2}{2} + \frac{A_{\rm 9}^2}{2} \cdot \cos(\omega_{\rm 18} t) + A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 8} t)+ A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 10} t).$$

• After multiplying by   $z(t)$,  all but the fourth of these contributions fall within the   $\text{90 kHz}$  to  $\text{110 kHz}$ range.  The weight at  $f_{\rm T} = 100\ \rm kHz$  is further increased by  $3 · c_3 · A_{\rm T} · 0.5 (A_1^2 + A_9^2) = 0.6\ \rm V$  and is thus   $9.08 \ \rm V$.

Further components are obtained at:

• $98 \ \rm kHz$  and  $102 \ \rm kHz$  with weights  $c_3 · A_{\rm T}/2 · A_1^2/2 = 0.03\ \rm V$,
• $92 \ \rm kHz$  and  $108 \ \rm kHz$  with weights  $3c_3 · A_{\rm T}/2 · A_1 · A_9 = 0.12\ \rm V$,
• $90 \ \rm kHz$  and  $110 \ \rm kHz$  with weights  $3c_3 · A_{\rm T}/2 · A_1 · A_9 = 0.12\ \rm V$.

The lower plot in the above graph shows the spectrum   $S_+(f)$  considering the cubic components.  It can be seen that new frequencies have appeared, indicating nonlinear distortions.  The correct solutions are therefore Answers 1 and 3.