# Difference between revisions of "Aufgaben:Exercise 3.6: PM or FM? Or AM?"

Two different locus curves for angle modulation

To analyze a modulator, apply the signal

$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N})$$

to its input, where the signal amplitude is always  $A_{\rm N} = 2\ \rm V$ .

• Using the signal frequency  $f_{\rm N} = f_1 = 5 \ \rm kHz$  the locus  $\rm O_1$  is determined.
• If one uses the message frequency  $f_{\rm N} = f_2$, then the  $\rm O_2$  locus is established.

In your solutions, note that for angle modulation – the umbrella term for phase and frequency modulation – the following relationship holds between the modulation index  $η$  and the modulator constant  $K_{\rm WM}$:

$$\eta = \left\{ \begin{array}{c} K_{\rm WM} \cdot A_{\rm N} \\ {K_{\rm WM} \cdot A_{\rm N}}/({2 \pi \cdot f_{\rm N})} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{bei}} \\ {\rm{bei}} \\ \end{array}\begin{array}{*{20}c} {\rm PM} \hspace{0.05cm}, \\ {\rm FM}. \hspace{0.05cm} \\ \end{array}$$

Hints:

### Questions

1

Which modulator do we have here?

 AM modulator. PM modulator. FM modulator.

2

What is the modulation index when the message frequency is  $f_{\rm N} = f_1 = 5 \ \rm kHz$?

 $η_1 \ = \$

3

What is the value of the modulator constant?   Hint:   the "unit" stands for $\rm V^{-1}$ (for PM) or $\rm (Vs)^{-1}$ (for FM).

 $K_{\rm WM} \ = \$ $\ \cdot 10^4$ "unit"

4

What is the angle  $ϕ_0$  (with respect to the real axis) of the locus curve  $\rm O_1$  when  $ϕ_{\rm N} = 30^\circ$  at time  $t = 0$ ?

 $ϕ_0 \ = \$ $\ \rm degrees$

5

What message frequency  $f_{\rm N} = f_2$  was used to determine the locus  $\rm O_2$ ?

 $f_2 \ = \$ $\ \rm kHz$

### Solution

#### Solution

• Since the locus curve describes an arc, we have an angle modulator (PM or FM) with modulation index  $η$.
• But since  $η$  here clearly depends on the message frequency  $f_{\rm N}$ , we can rule out phase modulation.

(2)  The modulation index can be read off the graph. It is   $η_1 = 75^\circ/180^\circ · π\hspace{0.15cm}\underline { ≈ 1.3}$.

(3)  For frequency modulation, it holds that:

$$K_{\rm WM} = K_{\rm FM} = \frac{ 2 \pi \cdot f_{\rm N} \cdot \eta}{A_{\rm N}} = \frac{ 2 \pi \cdot 5 \cdot 10^3 \,\,{\rm Hz}\cdot 1.3}{2\,{\rm V}} \hspace{0.15cm}\underline {\approx 2.04 \cdot 10^4 \hspace{0.1cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$

(4)  The frequency modulator can be realized as a phase modulator if the source signal is integrated beforehand.  This is:

$$q_{\rm I}(t) = \int q(t)\hspace{0.15cm}{\rm d}t = A_{\rm N} \cdot\int \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.15cm}{\rm d}t =\frac{A_{\rm N}}{\omega_{\rm N}} \cdot \sin(\omega_{\rm N} \cdot t + \phi_{\rm N}) = \frac{A_{\rm N}}{\omega_{\rm N}} \cdot \cos(\omega_{\rm N} \cdot t + \phi_{\rm N} - 90^\circ)\hspace{0.05cm}.$$
• Thus, for the equivalent low-pass signal with  $ϕ_{\rm N} = 30^\circ$:
$$s_{\rm TP}(t) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\omega_{\rm N} \hspace{0.03cm}\cdot \hspace{0.03cm}t \hspace{0.03cm} - \hspace{0.03cm}60^\circ)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{\rm TP}(t = 0) = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta \hspace{0.03cm}\cdot \hspace{0.05cm}\cos(\hspace{0.03cm}60^\circ)} = {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\eta /2}\hspace{0.05cm}.$$
• Thus, the zero phase angle is equal to  $η/2$  corresponding to   $ϕ_0\hspace{0.15cm}\underline {\approx 37.5^\circ}$.

(5)  From the definition of modulation index in frequency modulation, it follows that:

$$\eta_1 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 1}}\hspace{0.05cm},\hspace{0.3cm} \eta_2 = \frac{K_{\rm WM} \cdot A_{\rm N}}{2 \pi \cdot f_{\rm 2}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{\eta_1}{\eta_2} = \frac{f_2}{f_1}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} f_2 = \frac{\eta_1}{\eta_2} \cdot f_1 = \frac{75^\circ}{125^\circ} \cdot 5\,{\rm kHz} \hspace{0.15cm}\underline {= 3\,{\rm kHz}}\hspace{0.05cm}.$$