# Exercise 2.1: DSB-AM with Cosine? Or with Sine?

Spectrum of the analytical signal

Let us consider the amplitude modulation of the source signal  $q(t)$  with the carrier signal  $z(t)$. 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 $f_{\rm T} = 40\text{ kHz}$.  The other system parameters  $A_{\rm N}$,  $f_{\rm N}$,  $ϕ_{\rm N}$  and  $ϕ_{\rm T}$  are to be determined in this exercise.

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

The spectrum  $S_+(f)$  of the analytical signal  $s_+(t)$  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,  the abbreviations  $f_{30} = 30\text{ kHz}$  and  $f_{50} = 50\text{ kHz}$  are used.

As a reminder:  The spectrum  $S_+(f)$  is obtained from  $S(f)$ by

• truncating the components at negative frequencies and
• doubling positive frequencies.

Hints:

$$\cos(\alpha)\cdot \cos(\beta) = {1}/{2} \cdot \big[ \cos(\alpha-\beta) + \cos(\alpha+\beta)\big ] \hspace{0.05cm}, \hspace{0.5cm} \cos(90^{\circ}- \hspace{0.05cm} \alpha) = \sin(\alpha) \hspace{0.05cm}, \hspace{0.5cm} \cos(90^{\circ}+ \hspace{0.05cm} \alpha) = -\sin(\alpha) \hspace{0.05cm}.$$

### Questions

1

Find the spectrum  $S(f)$.  Which of the following statements are correct?

 $S(f)$  consists of four Dirac delta functions. All Dirac weights have the same magnitude  $2\text{ V}$. All Dirac weights are imaginary.

2

What is the modulated signal  $s(t)$?  Which statement is true?

 It is DSB-AM without carrier   ⇒   "DSB-AM with carrier suppression". It is DSB-AM with carrier.

3

State the message signal frequency $f_{\rm N}$.

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

4

Determine the phases of the two signals.

 $ϕ_{\rm N} \ = \$ $\ \text{degrees}$ $ϕ_{\rm T} \ = \$ $\ \text{degrees}$

5

What is the amplitude of the message signal?

 $A_{\rm N} \ = \$ $\ \text{V}$

### Solution

#### Solution

(1)  Answers 1 and 3  are correct:

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

(2)  The inverse Fourier transform of   $S(f)$  with  $ω_{30} = 2π · f_{30}$  and  $ω_{50} = 2πf_{50}$  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 $f_{\rm T} = 40\text{ kHz}$,  so the  first statement  is true.

(3)  For DSB–AM without carrier,   $s(t)$  includes only the two frequencies $f_{\rm T} – f_{\rm N}$  and  $f_{\rm T} + f_{\rm N}$.

• Hence,  with $f_{\rm T} = 40\text{ kHz}$  for the message frequency,  it follows that   $f_{\rm N} \hspace{0.05cm}\underline {= 10\ \rm kHz}.$

(4)  For DSB–AM without carrier,  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  (2)  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   $ϕ_{\rm N} \hspace{0.05cm}\underline {= 0}$ .
• Additionally,  from the last given trigonometric relation it follows that  $ϕ_{\rm T} \hspace{0.05cm}\underline {= 90^\circ} = π/2$.

(5)  Comparing the results from subtasks  (2)  and  (4)  leads to  $A_{\rm N} \hspace{0.05cm}\underline {= 4 \ \rm V}$.  Thus,  the equations of the signals involved in the modulation are:

$$q(t ) = 4\,{\rm V} \cdot \cos (2 \pi \cdot 10\,{\rm kHz} \cdot t) \hspace{0.05cm},$$
$$z(t) = 1 \cdot \cos (2 \pi \cdot 40\,{\rm kHz} \cdot t + 90^{\circ}) = -\sin (2 \pi \cdot 40\,{\rm kHz} \cdot t )\hspace{0.05cm}.$$