View source for Aufgaben:Exercise 2.1: DSB-AM with Cosine? Or with Sine?


{{quiz-Header|Buchseite=Modulation_Methods/Double-Sideband_Amplitude_Modulation
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[[File:P_ID985__Mod_A_2_1.png|right|frame|Spectrum of the analytical signal]]
Let us consider the amplitude modulation of the source signal &nbsp;$q(t)$&nbsp; with the carrier signal &nbsp;$z(t)$.&nbsp;These signals are given by:
:$$q(t)  =  A_{\rm N} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N})\hspace{0.05cm},$$
:$$z(t)  =  \hspace{0.15cm}1 \hspace{0.13cm} \cdot \hspace{0.1cm}\cos(2 \pi f_{\rm T} t + \phi_{\rm T})\hspace{0.05cm}.$$
The carrier frequency is known to be&nbsp;$f_{\rm T} = 40\text{ kHz}$.&nbsp; The other system parameters &nbsp;$A_{\rm N}$, &nbsp;$f_{\rm N}$, &nbsp;$ϕ_{\rm N}$&nbsp; and &nbsp;$ϕ_{\rm T}$&nbsp; are to be determined in this exercise.&nbsp; 

*The subscript&nbsp; "N"&nbsp; refers to the message signal&nbsp; (German:&nbsp; "Nachrichtensignal")&nbsp; $q(t)$&nbsp; 
*and&nbsp; "T" to the carrier&nbsp; (German:&nbsp; "Trägersignal")&nbsp; $z(t)$. 


The spectrum &nbsp;$S_+(f)$&nbsp; of the analytical signal &nbsp;$s_+(t)$&nbsp; at the modulator output is also given (see graph):
:$$S_+(f) = {\rm j}\cdot 2\,{\rm V} \cdot \delta ( f - f_{30} )+ {\rm j}\cdot 2\,{\rm V} \cdot \delta ( f - f_{50} )\hspace{0.05cm}.$$
Here,&nbsp; the abbreviations&nbsp; $f_{30} = 30\text{ kHz}$&nbsp; and&nbsp; $f_{50} = 50\text{ kHz}$&nbsp; are used. 

As a reminder:&nbsp; The spectrum &nbsp;$S_+(f)$&nbsp; is obtained from &nbsp;$S(f)$ by
*truncating the components at negative frequencies and
*doubling positive frequencies.





Hints: 
*This exercise belongs to the chapter &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
*Particular reference will be made to the pages &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_frequency_domain|Description in the frequency domain]]&nbsp; and  &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_time_domain|Description in the time domain]].
*The following trigonometric identities are given:
:$$\cos(\alpha)\cdot \cos(\beta)  =  {1}/{2} \cdot \big[ \cos(\alpha-\beta) + \cos(\alpha+\beta)\big ] \hspace{0.05cm}, \hspace{0.5cm} \cos(90^{\circ}- \hspace{0.05cm} \alpha)  =  \sin(\alpha) \hspace{0.05cm}, \hspace{0.5cm}  \cos(90^{\circ}+ \hspace{0.05cm} \alpha)  =  -\sin(\alpha) \hspace{0.05cm}.$$
 


===Questions===

<quiz display=simple>
{Find the spectrum &nbsp;$S(f)$.&nbsp; Which of the following statements are correct?
|type="[]"}
+ $S(f)$&nbsp; consists of four Dirac delta functions.
- All Dirac weights have the same magnitude&nbsp; $2\text{ V}$.
+ All Dirac weights are imaginary.

 
{What is the modulated signal &nbsp;$s(t)$?&nbsp; Which statement is true?
|type="()"}
+ It is DSB-AM without carrier &nbsp; &rArr; &nbsp; "DSB-AM with carrier suppression".
- It is DSB-AM with carrier.

{State the message signal frequency $f_{\rm N}$.
|type="{}"}
$f_{\rm N} \ = \ $ { 10 3% } $\ \text{kHz}$

{Determine the phases of the two signals.
|type="{}"}
$ϕ_{\rm N} \ = \ $ { 0. } $\ \text{degrees}$
$ϕ_{\rm T} \ = \ $ { 90 }  $\ \text{degrees}$

{What is the amplitude of the message signal?
|type="{}"}
$A_{\rm N} \ = \ $ { 4 3% } $\ \text{V}$


</quiz>

===Solution===
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'''(1)'''&nbsp; <u>Answers 1 and 3</u>&nbsp; are correct:
*At positive frequencies,&nbsp; $S_+(f)$&nbsp; is obtained from&nbsp; $S(f)$&nbsp; by doubling. 
*It follows that the impulse weights of&nbsp; $S(f)$&nbsp; are each only &nbsp; ${\rm j} · 1 \text{ V}$. 
*Because of the Assignment Theorem,&nbsp; $S(f)$&nbsp; must be an odd function. 
*Therefore,&nbsp; $S(f)$&nbsp; has two more Dirac delta functions at $f = -f_{30}$&nbsp; and $f = -f_{50}$,&nbsp; each with weight&nbsp; $-{\rm j} · 1 \text{ V}$:
:$$S(f) = 1\,{\rm V} \cdot \big[ {\rm j}\cdot \delta ( f - f_{30} )-{\rm j} \cdot \delta ( f + f_{30} )+ {\rm j} \cdot \delta ( f - f_{50} )-{\rm j} \cdot \delta ( f + f_{50} )\big] \hspace{0.05cm}.$$



'''(2)'''&nbsp; The inverse Fourier transform of &nbsp; $S(f)$&nbsp; with&nbsp; $ω_{30} = 2π · f_{30}$&nbsp; and&nbsp; $ω_{50} = 2πf_{50}$&nbsp;  leads to the following signal:
:$$ s(t) = -2\,{\rm V} \cdot \sin(\omega_{\rm 30} t )-2\,{\rm V} \cdot \sin(\omega_{\rm 50} t )\hspace{0.05cm}.$$
*This does not contain any component at the carrier frequency $f_{\rm T} = 40\text{ kHz}$,&nbsp; so the&nbsp; <u>first statement</u>&nbsp;  is true.



'''(3)'''&nbsp; For DSB–AM without carrier, &nbsp; $s(t)$&nbsp; includes only the two frequencies  $f_{\rm T} – f_{\rm N}$&nbsp; and&nbsp; $f_{\rm T} + f_{\rm N}$. 
*Hence,&nbsp; with $f_{\rm T} = 40\text{ kHz}$&nbsp; for the message frequency,&nbsp; it follows that &nbsp; $f_{\rm N} \hspace{0.05cm}\underline {= 10\ \rm  kHz}.$



'''(4)'''&nbsp; For DSB–AM without carrier,&nbsp; it holds that:
:$$s(t)  =  q(t) \cdot z(t) = A_{\rm N} \cdot \cos(\omega_{\rm N} t + \phi_{\rm N})\cdot \cos(\omega_{\rm T} t + \phi_{\rm T})$$
:$$\Rightarrow \hspace{0.5cm} s(t) =  \frac{A_{\rm N}}{2} \cdot \left[ \cos\left((\omega_{\rm T} +\omega_{\rm N})\cdot t + \phi_{\rm T}+ \phi_{\rm N}\right) + \cos\left((\omega_{\rm T} -\omega_{\rm N})\cdot t + \phi_{\rm T}- \phi_{\rm N}\right) \right] \hspace{0.05cm}.$$
*A comparison with the result from subtask&nbsp; '''(2)'''&nbsp; shows that:
:$$\cos(\omega_{\rm 30} \cdot t + \phi_{\rm T}- \phi_{\rm N})  =  -\sin(\omega_{\rm 30} \cdot t )\hspace{0.05cm},$$ 
:$$\cos(\omega_{\rm 50} \cdot t + \phi_{\rm T}+ \phi_{\rm N})  =  -\sin(\omega_{\rm 50} \cdot t ) \hspace{0.05cm}.$$
*Both equations can only be satisfied simultaneously with phase &nbsp; $ϕ_{\rm N} \hspace{0.05cm}\underline {= 0}$&nbsp;. 
*Additionally,&nbsp; from the last given trigonometric relation it follows that&nbsp; $ϕ_{\rm T} \hspace{0.05cm}\underline {= 90^\circ} = π/2$.



'''(5)'''&nbsp; Comparing the results from subtasks&nbsp; '''(2)'''&nbsp; and&nbsp; '''(4)'''&nbsp; leads to&nbsp; $A_{\rm N} \hspace{0.05cm}\underline {= 4 \ \rm V}$.&nbsp; Thus,&nbsp; the equations of the signals involved in the modulation are:
:$$q(t )  =  4\,{\rm V} \cdot \cos (2 \pi \cdot 10\,{\rm kHz} \cdot t) \hspace{0.05cm},$$
:$$z(t)  =  1 \cdot \cos (2 \pi \cdot 40\,{\rm kHz} \cdot t + 90^{\circ}) = -\sin (2 \pi \cdot 40\,{\rm kHz} \cdot t )\hspace{0.05cm}.$$


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[[Category:Modulation Methods: Exercises|^2.1 Double Sideband Amplitude Modulation^]]