Exercise 3.5Z: Phase Modulation of a Trapezoidal Signal

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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.




How should the modulator constant  $K_{\rm PM}$  be chosen so that  $ϕ_{\rm max} = 3 \ \rm rad$ ?

$K_{\rm PM} \ = \ $

$\ \rm V^{-1}$


What range of values does the instantaneous frequency  $f_{\rm A}(t)$  take on in the time interval $0 < t < T$ ?

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

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

$\ \rm kHz$


What range of values does the instantaneous frequency  $f_{\rm A}(t)$ take on in the time interval  $T < t < 3T$ ?

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

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

$\ \rm kHz$


What range of values does the instantaneous frequency $f_{\rm A}(t)$ take on in the time interval  $3T < t < 5T$ ?

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

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

$\ \rm kHz$


How must the modulator constant  $K_{\rm FM}$ be chosen so that the signal $q_2(t)$  results in the same RF signal $s(t)$  after frequency modulation?

$K_{\rm FM} \ = \ $

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


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