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GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 64 of 65

Marks: +1, -0

A voice signal

m(t)

5\ \text{kHz}

15\ \text{kHz}

f(t)=A(1+m(t))\cos 2\pi f_{c} t

f_{c}

=600\ \text{kHz}

f(t)

f(t)

1.2

K

K

Solution:

m(t) has frequency range 5 kHz to 15 kHz

Now it is amplitude modulated

f(t)=A(1+m(t))\cos 2\pi f_{c} t

where

f_{c}=6000\ \text{kHz}

∴ AM signal will have highest frequency

=f_{c}+f_{m}(\max)

=600+15=615\ \text{kHz}

And AM signal will have lowest frequency

=f_{c}-f_{m}(\max)

=600-15=585\ \text{kHz}

It is a band pass signal so we use bandpass sampling

f_{s}=1.2\times \frac{2 fH}{k}

K=\frac{f_{H}}{f_{H}-f_{L}}

=\frac{615}{615-585}

K=20.5

We select

K=20

\therefore f_{s}=1.2\times \frac{2\times 615}{20}

\therefore f_{s}=73.8\ \text{kHz}

Now

L=256

And

2^{n}=L=256

\therefore n=8

Bitrate

=R_{b}=nf_{s}

\therefore R_{b}=8\times 73.8\times 103

\therefore R_{b}=0.59\ \text{Mbps}

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