Aufgaben:Exercise 2.2Z: Power Consideration

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Analytical signal - Line spectrum

Let us consider two harmonic oscillations

$$ s_1(t) = A_1 \cdot \cos(\omega_{\rm 1} \cdot t ) \hspace{0.05cm},$$

$$s_2(t) = A_2 \cdot \cos(\omega_{\rm 2} \cdot t + \phi) \hspace{0.05cm},$$

where  $f_2 ≥ f_1$  should hold for the frequencies.

• The graph on the right shows the spectrum of the analytical signal  $s_+(t)$, which is additively composed of the two components  $s_{1+}(t)$  and  $s_ {2+}(t)$ .

• Here,  the transmission power  $P_{\rm S}$  should be understood as the second order moment of the signal  $s(t)$,  averaged over the largest measurement period possible:

$$P_{\rm S} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {s^2(t) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$

• According to this definition:  If  $s(t)$ describes a voltage waveform,  $P_{\rm S}$  has unit   $\rm V^2$  and refers to resistance  $R = 1 \ \rm Ω$. 
• Dividing by $R$  gives the physical power in   $\rm W$.

Hints:

• This exercise belongs to the chapter  Double-Sideband Amplitude Modulation.
• Reference is also made to the chapter  Quality Criteria.
• Use the numerical values  $A_1 = 2\ \rm V$,  $A_2 = 1 \ \rm V$,  and  $R = 50 \ \rm Ω$.

Questions

1 Calculate the power of the cosine signal  $s_1(t)$.

$P_1 \ = \ $ $\ \rm V^ $

2 Let  $R = 50 \ \rm Ω$.  What is the physical power of the signal  $s_1(t)$?

$P_1 \ = \ $ $\ \text{mW}$

3 What is the power of the phase-shifted oscillation  $s_2(t)$?

$P_2 \ = \ $ $\ \rm V^ $

4 What is the power of the sum signal  $s(t)$  when  $f_2 ≠ f_1$?

$P_{\rm S} \ = \ $ $\ \rm V^ $

5 What power is obtained for $f_2 = f_1$  with  $ϕ = 0$,  $ϕ = 90^\circ$  and  $ϕ = 180^\circ$?

$ϕ = 0\text{:}\hspace{0.3cm} P_{\rm S} \ = \ $ $\ \rm V^ $

$ϕ = 90^\circ\text{:}\hspace{0.3cm} P_{\rm S} \ = \ $ $\ \rm V^ $

$ϕ = 180^\circ\text{:}\hspace{0.3cm} P_{\rm S} \ = \ $ $\ \rm V^ $

Solution

Solution

(1)  According to the equations specified on the exercise page:

$$P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {A_1^2 \cdot \cos^2(\omega_{\rm 1} t + \phi_1) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$

• For more general calculation,  we consider the phase  $ϕ_1$,  which is actually zero here.  Using the equation  $\cos^{2}(α) = 0.5 · (1 + \cos(2α))$,  we get:

$$ P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {\frac{A_1^2}{2}}\hspace{0.1cm}{\rm d}t + \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {\frac{A_1^2}{2}\cdot \cos(2\omega_{\rm 1} t + 2\phi_1)}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}.$$

• Regardless of the phase  $ϕ_1$,  the second term does not contribute to the division by  $T_{\rm M}$  and subsequent boundary transition due to integration over the cosine function.  Thus,  we get:

$$P_{\rm 1} = \frac{A_1^2}{2} = \frac{(2\,{\rm V})^2}{2} \hspace{0.15cm}\underline {= 2\,{\rm V}^2}\hspace{0.05cm}.$$

(2)  With  $R = 50\ \rm Ω$,  we get the  "unnormalized"  power:

$$P_{\rm 1} = \frac{2\,{\rm V}^2}{50\,{\rm \Omega}} \hspace{0.15cm}\underline {= 40\,{\rm mW}}\hspace{0.05cm}.$$

(3)  It has already been shown in the solution to subtask  (1)  that the phase has no influence on the power.  It follows that:

$$P_{\rm 2} = \frac{A_2^2}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm V}^2}\hspace{0.05cm}.$$

(4) To calculate this power,  we have to average over $s^{2}(t)$,  where:

$$s^2(t) = s_1^2(t) + s_2^2(t) + 2 \cdot s_1(t) \cdot s_2(t).$$

• Due to the division by the measurement duration  $T_{\rm M}$  and the required boundary transition,  the last term does not contribute regardless of the phase   $ϕ$ .  Thus:

$$P_{\rm S} = P_{\rm 1} + P_{\rm 2} \hspace{0.15cm}\underline {= 2.5\,{\rm V}^2}\hspace{0.05cm}.$$

(5)  When  $f_2 = f_1$,  the spectrum of the analytical signal is:

$$S_+(f) = (A_{\rm 1} + A_{\rm 2} \cdot {\rm e}^{{\rm j}\hspace{0.03cm} \cdot \hspace{0.03cm} \phi})\cdot \delta (f - f_1) \hspace{0.05cm}.$$

• This results in the signal:

$$s(t) = A_3 \cdot \cos(\omega_{\rm 1} t + \phi_3) \hspace{0.05cm},$$

whose phase  $ϕ_3$  does not matter for the power calculation. The amplitude of this signal is

$$A_3 = \sqrt{ \left(A_1 + A_2 \cdot \cos(\phi)\right)^2 + A_2^2 \cdot \sin^2(\phi)} =\sqrt{ A_1^2 + A_2^2 + 2 \cdot A_1 \cdot A_2 \cdot \cos(\phi)}\hspace{0.05cm}.$$

• For  $ϕ = 0$,  the sum of the amplitudes is scalar:

$$A_3 = \sqrt{ A_1^2 + A_2^2 + 2 \cdot A_1 \cdot A_2 } = A_1 + A_2 = 3\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} \hspace{0.15cm}\underline {= 4.5\,{\rm V}^2}\hspace{0.05cm}.$$

• On the other hand,  the amplitudes for  $ϕ = 90^\circ$  are added as vectors  ⇒   same result as in subtask  (4):

$$ A_3 = \sqrt{ A_1^2 + A_2^2 } = \sqrt{5}\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} = \frac{5\,{\rm V}^2}{2}\hspace{0.15cm}\underline {= 2.5\,{\rm V}^2}\hspace{0.05cm}.$$

• For  $ϕ = 180^\circ$,  the cosine oscillations overlap destructively:

$$A_3 = A_1 - A_2 = 1\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} \hspace{0.15cm}\underline {= 0.5\,{\rm V}^2}\hspace{0.05cm}.$$