Search any question & find its solution
Question:
Answered & Verified by Expert
An audio signal is modulated by a carrier wave of $20 \mathrm{MHz}$ such that the bandwidth required for modulation is $3 \mathrm{kHz}$. Could this wave be demodulated by a diode detector which has the values of $R$ and $C$ as
(i) $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=0.01 \mu \mathrm{F}$.
(ii) $\mathrm{R}=10 \mathrm{k} \Omega, \mathrm{C}=0.01 \mu \mathrm{F}$.
(iii) $\mathrm{R}=10 \mathrm{k} \Omega, \mathrm{C}=0.1 \mu \mathrm{F}$.
(i) $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=0.01 \mu \mathrm{F}$.
(ii) $\mathrm{R}=10 \mathrm{k} \Omega, \mathrm{C}=0.01 \mu \mathrm{F}$.
(iii) $\mathrm{R}=10 \mathrm{k} \Omega, \mathrm{C}=0.1 \mu \mathrm{F}$.
Solution:
2223 Upvotes
Verified Answer
As given, carrier wave frequency
$$
\begin{aligned}
&\mathrm{f}_{\mathrm{c}}=20 \mathrm{MHz}=20 \times 10^6 \mathrm{~Hz} \\
&\omega=2 \pi \mathrm{f}_{\mathrm{c}}=2 \pi \times 20 \times 10^6=40 \pi \times 10^6 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
Bandwidth required for modulation is $=2 \mathrm{f}_{\mathrm{m}}$
$$
\begin{aligned}
&\text { (given) } 2 \mathrm{f}_{\mathrm{m}}=3 \mathrm{kHz}=3 \times 10^3 \mathrm{~Hz} \\
&\Rightarrow \quad \mathrm{f}_{\mathrm{m}}=\frac{3 \times 10^3}{2}=1.5 \times 10^3 \mathrm{~Hz}
\end{aligned}
$$
Demodulation by a diode is possible if the condition $\frac{1}{\mathrm{f}_{\mathrm{c}}}< < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is satisfied
Thus, $\frac{1}{f_c}=\frac{1}{20 \times 10^6}=\left(5 \times 10^{-8}\right) \mathrm{s}$ and $\frac{1}{\mathrm{f}_{\mathrm{m}}}=\frac{1}{1.5 \times 10^3 \mathrm{~Hz}}=0.7 \times 10^{-3} \mathrm{~s}$
Now, gain through all the options of $\mathrm{R}$ and $\mathrm{C}$ one by one, we get
(i) $\mathrm{RC}=1 \mathrm{k} \Omega \times 0.01 \mu \mathrm{F}=10^3 \Omega \times\left(0.01 \times 10^{-6} \mathrm{~F}\right)=10^{-5} \mathrm{~s}$.
So from condition $\frac{1}{\mathrm{f}_{\mathrm{c}}} \ll < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is satisfied
because, $5 \times 10^{-8}< < 10^{-5} < 0.7 \times 10^{-3}$ is true.
Hence the wave can be demodulated.
(ii) $\mathrm{RC}=10 \mathrm{k} \Omega \times 0.01 \mu \mathrm{F}=10^4 \Omega \times 10^{-8} \mathrm{~F}=10^{-4} \mathrm{~S}$
Here condition $\frac{1}{\mathrm{f}_{\mathrm{c}}}< < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is satisfied because, $5 \times 10^{-8}< < 10^{-4} < 0.7 \times 10^{-3}$ is true.
So, the wave can be demodulated.
(iii) $\mathrm{RC}=10 \mathrm{k} \Omega \times 1 \mu \mu \mathrm{F}=10^4 \Omega \times 10^{-12} \mathrm{~F}=10^{-8} \mathrm{~s}$.
Here, condition $\frac{1}{\mathrm{f}_{\mathrm{c}}} < \mathrm{RC}$, Condition, $\frac{1}{\mathrm{f}_{\mathrm{c}}}< < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is, not satisfied because $5 \times 10^{-8}< < $ $10^{-8} < 0.7 \times 10^{-3}$ is not true.
So, the modulated carrier wave cannot be demodulated.
$$
\begin{aligned}
&\mathrm{f}_{\mathrm{c}}=20 \mathrm{MHz}=20 \times 10^6 \mathrm{~Hz} \\
&\omega=2 \pi \mathrm{f}_{\mathrm{c}}=2 \pi \times 20 \times 10^6=40 \pi \times 10^6 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
Bandwidth required for modulation is $=2 \mathrm{f}_{\mathrm{m}}$
$$
\begin{aligned}
&\text { (given) } 2 \mathrm{f}_{\mathrm{m}}=3 \mathrm{kHz}=3 \times 10^3 \mathrm{~Hz} \\
&\Rightarrow \quad \mathrm{f}_{\mathrm{m}}=\frac{3 \times 10^3}{2}=1.5 \times 10^3 \mathrm{~Hz}
\end{aligned}
$$
Demodulation by a diode is possible if the condition $\frac{1}{\mathrm{f}_{\mathrm{c}}}< < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is satisfied
Thus, $\frac{1}{f_c}=\frac{1}{20 \times 10^6}=\left(5 \times 10^{-8}\right) \mathrm{s}$ and $\frac{1}{\mathrm{f}_{\mathrm{m}}}=\frac{1}{1.5 \times 10^3 \mathrm{~Hz}}=0.7 \times 10^{-3} \mathrm{~s}$
Now, gain through all the options of $\mathrm{R}$ and $\mathrm{C}$ one by one, we get
(i) $\mathrm{RC}=1 \mathrm{k} \Omega \times 0.01 \mu \mathrm{F}=10^3 \Omega \times\left(0.01 \times 10^{-6} \mathrm{~F}\right)=10^{-5} \mathrm{~s}$.
So from condition $\frac{1}{\mathrm{f}_{\mathrm{c}}} \ll < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is satisfied
because, $5 \times 10^{-8}< < 10^{-5} < 0.7 \times 10^{-3}$ is true.
Hence the wave can be demodulated.
(ii) $\mathrm{RC}=10 \mathrm{k} \Omega \times 0.01 \mu \mathrm{F}=10^4 \Omega \times 10^{-8} \mathrm{~F}=10^{-4} \mathrm{~S}$
Here condition $\frac{1}{\mathrm{f}_{\mathrm{c}}}< < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is satisfied because, $5 \times 10^{-8}< < 10^{-4} < 0.7 \times 10^{-3}$ is true.
So, the wave can be demodulated.
(iii) $\mathrm{RC}=10 \mathrm{k} \Omega \times 1 \mu \mu \mathrm{F}=10^4 \Omega \times 10^{-12} \mathrm{~F}=10^{-8} \mathrm{~s}$.
Here, condition $\frac{1}{\mathrm{f}_{\mathrm{c}}} < \mathrm{RC}$, Condition, $\frac{1}{\mathrm{f}_{\mathrm{c}}}< < \mathrm{RC} < \frac{1}{\mathrm{f}_{\mathrm{m}}}$ is, not satisfied because $5 \times 10^{-8}< < $ $10^{-8} < 0.7 \times 10^{-3}$ is not true.
So, the modulated carrier wave cannot be demodulated.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.