An amplitude modulated wave is as shown in figure. Calculate
(i) the percentage modulation,
(ii) peak carrier voltage and
(iii) peak value of information voltage
From the diagram,
Maximum voltage $V_{\max }=\frac{100}{2}=50 \mathrm{~V}$
Minimum voltage $V_{\min }=\frac{20}{2}=10 \mathrm{~V}$
(i) Percentage modulation, $\propto=\frac{V_{\max }-V_{\min }}{V_{\max }+V_{\min }} \times 100=\frac{50-10}{50+10} \times 100$
$$=\frac{40}{60} \times 100=66.67 \%$$
(ii) Peak carrier voltage, $\quad V_c=\frac{V_{\max }+V_{\min }}{2}=\frac{50+10}{2}=30 \mathrm{~V}$
(iii) Peak value of information voltage,
$$V_m=\propto V_c=\frac{66.67}{100} \times 30=20 \mathrm{~V}$$
(i) Draw the plot of amplitude versus $\omega$ for an amplitude modulated were whose carrier wave $\left(\omega_c\right)$ is carrying two modulating signals, $\omega_1$ and $\omega_2 \left(\omega_2>\omega_1\right)$.
(ii) Is the plot symmetrical about $\omega_c$ ? Comment especially about plot in region $\omega<\omega_c$.
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
(i) The plot of amplitude versus $\omega$ can be shown in the figure below
(ii) From figure, we note that frequency spectrum is not symmetrical about $\omega_{\mathrm{c}}$. Crowding of spectrum is present for $\omega<\omega_c$.
(iii) If more waves are to be modulated then there will be more crowding in the modulating signal in the region $\omega<\omega_c$. That will result more chances of mixing of signals.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves $\omega_c$. This shows that large carrier frequency enables to carry more information (i.e., more $\omega_m$ ) and the same will in turn increase bandwidth.
An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz . Could this wave be demodulated by a diode detector which has the values of $R$ and $C$ as
(i) $R=1 \mathrm{k} \Omega, C=0.01 \times \mathrm{F}$.
(ii) $R=10 \mathrm{k} \Omega, C=0.01 \propto \mathrm{~F}$.
(iii) $R=10 \mathrm{k} \Omega, C=0.1 \times \mathrm{F}$.
$$\begin{aligned} \text { Given, carrier wave frequency } f_C & =20 \mathrm{MHz} \\ & =20 \times 10^6 \mathrm{~Hz} \end{aligned}$$
Bandwidth required for modulation is
$$\begin{aligned} 2 f_m & =3 \mathrm{kHz}=3 \times 10^3 \mathrm{~Hz} \\ f_m & =\frac{3 \times 10^3}{2}=1.5 \times 10^3 \mathrm{~Hz} \end{aligned}$$
Demodulation by a diode is possible if the condition $\frac{1}{f_c}<< R C<\frac{1}{f_m}$ is satisfied
$$\begin{aligned} \text{Thus,}\quad & \frac{1}{t_c}=\frac{1}{20 \times 10^6}=0.5 \times 10^{-7} \quad\text{.... (i)}\\ \text{and}\quad & \frac{1}{f_m}=\frac{1}{1.5 \times 10^3} \mathrm{~Hz}=0.7 \times 10^{-3} \mathrm{~s}\quad\text{.... (ii)} \end{aligned}$$
Now, gain through all the options of $R$ and $C$ one by one, we get
(i) $R C=1 \mathrm{k} \Omega \times 0.01 \propto \mathrm{~F}=10^3 \Omega \times\left(0.01 \times 10^{-6} \mathrm{~F}\right)=10^{-5} \mathrm{~S}$
Here, condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied.
Hence it can be demodulated.
(ii) $R C=10 \mathrm{k} \Omega \times 0.01 \propto \mathrm{~F}=10^4 \Omega \times 10^{-8} \mathrm{~F}=10^{-4} \mathrm{~s}$
Here condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied.
Hence, it can be demodulated.
(iii) $R C=10 \mathrm{k} \Omega \times 1 \propto \propto \mathrm{~F}=10^4 \Omega \times 10^{-12} \mathrm{~F}=10^{-8} \mathrm{~S}$
Here, condition $\frac{1}{f_c}>R C$, so this cannot be demodulated.