GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 29 of 65

Marks: +1, -0

The baseband signal

{m}( {t})

shown in the figure is phase-modulated to generate the PM signal

ϕ( {t})=\cos (2 π {f}_{ {c}} {t}+ {k} {m}( {t}))

. The time

{t}

on the

{x}

-axis in the figure is in milliseconds. If the carrier frequency is

{f}_{ {c}}=50 {kHz}

and

{k}=10 {n}

, then the ratio of the minimum instantaneous frequency (in

{kHz}

to the maximum instantaneous frequency (in

{kHz}

) is ______________ (rounded off to 2 decimal places).

Solution:

x(t)=\cos (2 π f c t+k m(t))

∴ Q(t)=2 π f c t+k m(t)

And

{fi}=\;{1}/{2 π} \;{∂}/{∂ {t}}( {Q}( {t}))

=\;{1}/{2 π} \;{∂}/{∂ {t}}[2 π f {ct}+ {km}( {t})]

{fi}= {fc}+\;{ {k}}/{2 π} \;{∂}/{∂ {t}} {m}( {t})

∴ {fi}_{\max }= {fc}+\;{ {k}}/{2 π}[\;{∂}/{∂ {t}} {m}( {t})]_{\max }

∴ {fi}_{\max }=50 {kHz}+5 × \;{1-(-1)}/{(7-6) × 10^{-3}}

∴ {f}_{\max }=50 {kHz}+10 {kHz}

∴ {fi}_{\max }=60 {kHz}

And

f i_{\min }=f c+\;{k}/{2 π}[\;{∂}/{∂ t}(m(t))]_{\min }

50 {kHz}+5 × \;{-1-1}/{(9-7) × 10^{-3}}

=50 {kHz}-5 {kHz}

{fi}_{\min }=45 {kHz}

∴ \;{ {f}_{\min }}/{ {f}_{\max }}=\;{45}/{60}=0.75

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