Question Solved1 Answer The message signal for a phase modulation (PM) is as given in Figure 1.a. Your carrier mark amplitude ?? = ? ? ? ? ? and frequency ?? = ?? kHz and frequency deviation ∆? = ? kHz. Since the PM signal formed is demodulated as in Figure 1.b; Signs formed at points c, d, e and f (The YGF in the figure suppresses only the DC component. The amplitude and frequency in the figures you will draw. Also include information.)  4ms1-1Şekil 1.aİntegratörPMdZarfYGFm(t)ModülatördtDetektörüdemodülatörŞekil 1.b

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The message signal for a phase modulation (PM) is as given in Figure 1.a. Your carrier mark amplitude ?? = ? ? ? ? ? and frequency ?? = ?? kHz and frequency deviation ∆? = ? kHz. Since the PM signal formed is demodulated as in Figure 1.b; Signs formed at points c, d, e and f (The YGF in the figure suppresses only the DC component. The amplitude and frequency in the figures you will draw. Also include information.)

 

Transcribed Image Text: 4ms1-1Şekil 1.a İntegratörPMdZarfYGFm(t)ModülatördtDetektörüdemodülatörŞekil 1.b
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Transcribed Image Text: 4ms1-1Şekil 1.a İntegratörPMdZarfYGFm(t)ModülatördtDetektörüdemodülatörŞekil 1.b
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Step 1Given messige signal is,From above fiqure,a) Given imput signal is m(t) [modulation imput]Step 2(b) Output of PM modulitor:-A_(c)cos(omega_(c)t+K_(p)m(t))whereA_(c)= carrier amplitude of the signal =1sqrt()D_(c)=2pif_(c)quad[f_(c)=" frequency "=100kHz]K_(p)=p^("hase ") sensitivity or modulation constantInstantareous frequency is given by,f_(i)=f_(c)+(K_(p))/(2pi)m^(˙)(t)where m^(˙)(t)=(d)/(dt)m(t)The output waveform is in the following,[ Differentiation of triangubr wave becones square wave]frequency deviation is a KHzSo, waveform at (b) will be of the following form,PM Modulates wave formStep 3(c) At (c) output will be,{:[(d)/(dt){A_(c)cos(omega_(c)t+K_(p)m(t)}:}],[=-A_(c)(omega_(c)t+K_(p)(m^(˙))(t))sin(omega_(c ... See the full answer