Exercise 3.5Z: Phase Modulation of a Trapezoidal Signal

Trapezoidal and rectangular signals

A phase modulator with input signal $q_1(t)$ a modulated signal $s(t)$ at the output are described as follows:

$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\psi(t) \big ]= A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q_1(t) \big ] \hspace{0.05cm}.$$

The carrier angular frequency is $ω_{\rm T} = 2π · 10^5 \cdot {1}/{\rm s}$.
The instantaneous angular frequency $ω_{\rm A}(t)$ is equal to the derivative of the angle function $ψ(t)$ with respect to time.
The instantaneous frequency is thus $f_{\rm A}(t) = ω_{\rm A}(t)/2π$.

The trapezoidal signal $q_1(t)$ is applied as a test signal, where the nominated time duration is $T = 10 \ \rm µ s$ .

The same modulated signal $s(t)$ would result from a frequency modulator with the angular function

$$\psi(t) = \omega_{\rm T} \cdot t + K_{\rm FM} \cdot \int q_2(t)\hspace{0.15cm}{\rm d}t$$

if the rectangular source signal $q_2(t)$ is applied according to the lower plot.

Hints:

This exercise belongs to the chapter Frequency Modulation.
Reference is also made to the chapter Phase Modulation.

Questions

$K_{\rm PM} \ = \ $

$\ \rm V^{-1}$

$f_\text{A, min} \ = \ $

$\ \rm kHz$

$f_\text{A, max} \hspace{0.06cm} = \ $

$\ \rm kHz$

$f_\text{A, min} \ = \ $

$\ \rm kHz$

$f_\text{A, max} \hspace{0.06cm} = \ $

$\ \rm kHz$

$f_\text{A, min} \ = \ $

$\ \rm kHz$

$f_\text{A, max} \hspace{0.06cm} = \ $

$\ \rm kHz$

$K_{\rm FM} \ = \ $

$\ \cdot 10^5 \ \rm V^{-1}s^{-1}$

Solution

(1) The phase function is calculated as $ϕ(t) = K_{\rm PM} · q_1(t)$. The phase deviation $ϕ_{\rm max}$ is equal to the phase resulting from the maximum value of the source signal:

$$ \phi_{\rm max} = K_{\rm PM} \cdot 2\,{\rm V} = 3\,{\rm rad}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm PM} \hspace{0.15cm}\underline {= 1.5\,{\rm V^{-1}}} \hspace{0.05cm}.$$

(2) In the range $0 < t < T$ , the angular function can be represented as follows:

$$ \psi(t) = \omega_{\rm T} \cdot t + K_{\rm PM} \cdot 2\,{\rm V} \cdot {t}/{T}\hspace{0.05cm}.$$

For the instantaneous angular frequency $ω_{\rm A}(t)$ or the instantaneous frequency $f_{\rm A}(t)$ , the following holds:

$$\omega_{\rm A}(t) = \frac{{\rm d}\hspace{0.03cm}\psi(t)}{{\rm d}t}= \omega_{\rm T} + K_{\rm PM} \cdot \frac{2\,{\rm V}}{10\,{\rm µ s}}\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T} + \frac{1.5\,{\rm V}^{-1}}{2 \pi} \cdot 2 \cdot 10^5 \rm {V}/{ s} = 100\,{\rm kHz}+ 47.7\,{\rm kHz}= 147.7\,{\rm kHz}\hspace{0.05cm}.$$

The instantaneous frequency is constant, so $f_\text{A, min} = f_\text{A, max}\hspace{0.15cm}\underline{ = 147.7 \ \rm kHz}$ holds.

(3) Due to the constant source signal, the derivative is zero throughout the time interval $T < t < 3T$ under consideration, so the instantaneous frequency is equal to the carrier frequency:

$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} \hspace{0.15cm}\underline {= 100\,{\rm kHz}}\hspace{0.05cm}.$$

(4) The linear decay of $q_1(t)$ in the time interval $3T < t < 5T$ with slope as calculated in (2) leads to the result:

$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} - 47.7\,{\rm kHz} \hspace{0.15cm}\underline {= 52.3\,{\rm kHz}}\hspace{0.05cm}.$$

(5) By differentiation, we arrive at the instantaneous angular frequency:

$$ \omega_{\rm A}(t) = \omega_{\rm T} + K_{\rm FM} \cdot q_2(t) \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T}+\frac{ K_{\rm FM}}{2 \pi} \cdot q_2(t)\hspace{0.05cm}.$$

Using the result from (2) , we get:

$$\frac{ K_{\rm FM}}{2 \pi} \cdot 2\,{\rm V} = \frac{ 3 \cdot 10^5}{2 \pi} \cdot {\rm s^{-1}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} \hspace{0.15cm}\underline {= 1.5 \cdot 10^5 \hspace{0.15cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$