Difference between revisions of "Aufgaben:Exercise 3.1: Phase Modulation Locus Curve"
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*After a quarter of the period T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s , ϕ(t) = 0 and s_{\rm TP}(t) = 1 \, \rm V. | *After a quarter of the period T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s , ϕ(t) = 0 and s_{\rm TP}(t) = 1 \, \rm V. | ||
*At time t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s} , ϕ(t_1) = -π and s_{\rm TP}(t_1) = -1 \, \rm V. | *At time t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s} , ϕ(t_1) = -π and s_{\rm TP}(t_1) = -1 \, \rm V. |
Revision as of 19:48, 10 March 2022
The locus curve is generally understood as the plot of the equivalent low-pass signal s_{\rm TP}(t) in the complex plane.
- The graph shows locus curves at the output of two modulators \rm M_1 and \rm M_2.
- The real and imaginary parts are each normalized to 1 \ \rm V in this graph.
Let the source signal be the same for both modulators:
q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}
{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.
One of the two modulators implements phase modulation, which is characterized by the following equations:
- s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},
- s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm},
- \phi(t) = K_{\rm PM} \cdot q(t)\hspace{0.05cm}.
The maximum value ϕ(t) is called the modulation index η. Often η is also called phase deviation in the literature.
Hints:
- This exercise belongs to the chapter Phase Modulation.
- Particular reference is made to the page Equivalent low-pass signal in phase modulation.
Questions
Solution
- If one moves in the mathematically positive direction on the circle, it is specifically an USB–AM, otherwise it is a LSB–AM.
- The phase function ϕ(t) as the angle of a point s_{\rm TP}(t) on the circle (arc) with respect to the coordinate origin can take values between ±π/2 and does not show a cosine progression.
- The envelope a(t) = |s_{\rm TP}(t)| is also not cosine.
- If an envelope demodulator were used for \rm M_1 at the receiver, nonlinear distortions would occur, in contrast to DSB–AM, which has a horizontal straight line for a locus curve.
(2) Here, we observe phase modulation ⇒ Answer 3:
- The envelope a(t) = A_{\rm T} is constant,
- while the phase ϕ(t) is cosinusoidal according to the source signal q(t) .
(3) In the case of phase modulation:
- s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.
- From the graph, we can read the carrier amplitude A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V} as the radius of the circle.
(4) The source signal q(t) is at its maximum at time t = 0 and therefore so is the phase function:
- \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.
- This gives the modulator constant:
K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.
(5) One moves clockwise along the circular arc.
- After a quarter of the period T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s , ϕ(t) = 0 and s_{\rm TP}(t) = 1 \, \rm V.
- At time t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s} , ϕ(t_1) = -π and s_{\rm TP}(t_1) = -1 \, \rm V.
- Afterwards, move counterclockwise along the arc.