STEM Problem Detail

Problem Statement

Prompt

GIVE A SUITABLE ANSWER FOR Example 3.43

Solution Rendering...

Solution: The output of the demodulator is given as
\[
v_o(t) = [s(t)+2]^2 = s^2(t) + 4 + 4s(t) \tag{i}
\]

Here \(s(t)\) is input to the modulator.

Given an AM modulated signal as under:
\[
s(t) = \left[1 + 0.2\cos\left\{2\pi\left(\frac{f_m}{3}\right)t\right\}\right]\cos(2\pi f_c t) \tag{ii}
\]

Substituting equation (ii) in equation (i), we get
\[
v_o(t) = s^2(t) + 4 + 4s(t)
\]

or
\[
v_o(t) =
\left[1 + 0.2\cos\left\{2\pi\left(\frac{f_m}{3}\right)t\right\}\right]^2
\cos^2(2\pi f_c t)
+ 4\left[1 + 0.2\cos\left\{2\pi\left(\frac{f_m}{3}\right)t\right\}\right]
\cos(2\pi f_c t)
\]

or
\[
v_o(t) =
\left[1 + 0.2\cos\left\{2\pi\left(\frac{f_m}{3}\right)t\right\}\right]^2
\cos^2(2\pi f_c t)
+ 4\left[1 + 0.2\cos\left\{2\pi\left(\frac{f_m}{3}\right)t\right\}\right]
\cos(2\pi f_c t) \tag{iii}
\]

Simplifying the last equation, we get
\[
\begin{aligned}
v_o(t) =\;& 4.51
+ 0.2\cos\left[2\pi\left(\frac{f_m}{3}\right)t\right]
+ 0.01\cos\left[2\pi\left(\frac{2f_m}{3}\right)t\right]
+ 1.02\cos(4\pi f_c t) \\
&+ 0.2\cos\left[2\pi\left(2f_c+\frac{f_m}{3}\right)t\right]
+ 0.2\cos\left[2\pi\left(2f_c-\frac{f_m}{3}\right)t\right] \\
&+ 0.01\cos\left[2\pi\left(2f_c+\frac{2f_m}{3}\right)t\right]
+ 0.01\cos\left[2\pi\left(2f_c-\frac{2f_m}{3}\right)t\right] \\
&+ 4\cos(2\pi f_c t)
+ 0.4\cos\left[2\pi\left(f_c+\frac{f_m}{3}\right)t\right]
+ 0.4\cos\left[2\pi\left(f_c-\frac{f_m}{3}\right)t\right]
\end{aligned} \tag{iv}
\]

Also, the frequency spectrum of the signal \(v_o(t)\) can be obtained by taking
Fourier transform of both sides of equation (iv), as under:
\[
\begin{aligned}
V_o(f) =\;& 4.51\,\delta(f)
+ 0.1\left[\delta\!\left(f-\frac{f_m}{3}\right)+\delta\!\left(f+\frac{f_m}{3}\right)\right] \\
&+ 0.005\left[\delta\!\left(f-\frac{2f_m}{3}\right)+\delta\!\left(f+\frac{2f_m}{3}\right)\right] \\
&+ 0.51\left[\delta(f-2f_c)+\delta(f+2f_c)\right] \\
&+ 0.01\left[\delta\!\left(f-2f_c-\frac{f_m}{3}\right)
+ \delta\!\left(f+2f_c+\frac{f_m}{3}\right)\right] \\
&+ 0.01\left[\delta\!\left(f-2f_c+\frac{f_m}{3}\right)
+ \delta\!\left(f+2f_c-\frac{f_m}{3}\right)\right] \\
&+ 0.005\left[\delta\!\left(f-2f_c-\frac{2f_m}{3}\right)
+ \delta\!\left(f+2f_c+\frac{2f_m}{3}\right)\right] \\
&+ 2\left[\delta(f-f_c)+\delta(f+f_c)\right] \\
&+ 0.2\left[\delta\!\left(f-f_c-\frac{f_m}{3}\right)
+ \delta\!\left(f+f_c+\frac{f_m}{3}\right)\right] \\
&+ 0.2\left[\delta\!\left(f-f_c+\frac{f_m}{3}\right)
+ \delta\!\left(f+f_c-\frac{f_m}{3}\right)\right]
\end{aligned}
\]

The spectrum of the low-pass filter (LPF) output has been shown in figure 3.62.

The low-pass filter with a cut-off frequency equal to \(f_m\) will only allow
the d.c. component and the components having frequencies
\(\frac{f_m}{3}\) and \(\frac{2f_m}{3}\) and reject all other frequency
components present in the output \(V_o(f)\) of the square-law modulator.

Hence, the spectrum of the filter output is given by
\[
S_{\text{LPF}}(f) =
4.51\,\delta(f)
+ 0.1\left[\delta\!\left(f-\frac{f_m}{3}\right)+\delta\!\left(f+\frac{f_m}{3}\right)\right]
+ 0.005\left[\delta\!\left(f-\frac{2f_m}{3}\right)+\delta\!\left(f+\frac{2f_m}{3}\right)\right]
\]

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Problem #88