Question:
Let a frequency modulated (FM) signal
\[
x(t) = A \cos\left(\omega_c t + k_f \int_{-\infty}^{t} m(\lambda) \, d\lambda \right),
\]
where \(m(t)\) is a message signal of bandwidth \(W\). It is passed through a non-linear system with output
\[
y(t) = 2x(t) + 5(x(t))^2.
\]
Let \(B_T\) denote the FM bandwidth. The minimum value of \(\omega_c\) required to recover \(x(t)\) from \(y(t)\) is _____________
Let a frequency modulated (FM) signal
\[
x(t) = A \cos\left(\omega_c t + k_f \int_{-\infty}^{t} m(\lambda) \, d\lambda \right),
\]
where \(m(t)\) is a message signal of bandwidth \(W\). It is passed through a non-linear system with output
\[
y(t) = 2x(t) + 5(x(t))^2.
\]
Let \(B_T\) denote the FM bandwidth. The minimum value of \(\omega_c\) required to recover \(x(t)\) from \(y(t)\) is _____________
Show Hint
After squaring an FM, a component appears at $2\omega_c$ with {double} FM bandwidth. Ensure band separation: upper edge at $\omega_c$ must lie below lower edge at $2\omega_c$.
Updated On: Feb 12, 2026
- $B_T+W$
- $\dfrac{3}{2}B_T$
- $2B_T+W$
- $\dfrac{5}{2}B_T$
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Expand the nonlinearity
The given signal passes through a nonlinear device, producing \[ y(t) = 2A\cos\phi(t) + \frac{5A^{2}}{2}\big[1 + \cos(2\phi(t))\big], \] where \(\phi(t) = \omega_c t + k_f \int m(\lambda)\, d\lambda.\)
Step 2: Identify the spectral components
The output has three distinct parts:
(i) A DC component: \(\tfrac{5A^{2}}{2}\), centered at 0 Hz.
(ii) An FM term: \(2A \cos\phi(t)\), centered at \(\omega_c\), with the same bandwidth as the original FM signal, namely \(B_T\).
(iii) A doubled FM term: \(\tfrac{5A^{2}}{2}\cos(2\phi(t))\), centered at \(2\omega_c\).
Step 3: Bandwidth of the doubled FM
For the doubled term, the instantaneous frequency is \[ \frac{d}{dt}(2\phi(t)) = 2\omega_c + 2k_f m(t). \] Thus, both the carrier frequency and the frequency deviation are doubled. This means the FM bandwidth doubles as well: the bandwidth of this component is \(2B_T\).
Step 4: Separation condition
To recover the original FM term, we must isolate the band around \(\omega_c\) and reject the band around \(2\omega_c\). These two spectral regions should not overlap. Hence the highest edge of the band around \(\omega_c\) must not exceed the lowest edge of the band around \(2\omega_c\): \[ \omega_c + \frac{B_T}{2} \;\le\; 2\omega_c - \frac{2B_T}{2}. \] Simplify: \[ \omega_c + \frac{B_T}{2} \le 2\omega_c - B_T \;\;\Rightarrow\;\; \omega_c \ge \frac{3}{2} B_T. \]
Step 5: Check the weaker condition
There is also the trivial requirement \(\omega_c > \tfrac{B_T}{2}\) to avoid overlap with DC. But since \(\tfrac{3}{2}B_T > \tfrac{1}{2}B_T\), the stricter requirement dominates.
Final Answer:
\[ \boxed{\omega_c^{\min} = \tfrac{3}{2}B_T} \]
The given signal passes through a nonlinear device, producing \[ y(t) = 2A\cos\phi(t) + \frac{5A^{2}}{2}\big[1 + \cos(2\phi(t))\big], \] where \(\phi(t) = \omega_c t + k_f \int m(\lambda)\, d\lambda.\)
Step 2: Identify the spectral components
The output has three distinct parts:
(i) A DC component: \(\tfrac{5A^{2}}{2}\), centered at 0 Hz.
(ii) An FM term: \(2A \cos\phi(t)\), centered at \(\omega_c\), with the same bandwidth as the original FM signal, namely \(B_T\).
(iii) A doubled FM term: \(\tfrac{5A^{2}}{2}\cos(2\phi(t))\), centered at \(2\omega_c\).
Step 3: Bandwidth of the doubled FM
For the doubled term, the instantaneous frequency is \[ \frac{d}{dt}(2\phi(t)) = 2\omega_c + 2k_f m(t). \] Thus, both the carrier frequency and the frequency deviation are doubled. This means the FM bandwidth doubles as well: the bandwidth of this component is \(2B_T\).
Step 4: Separation condition
To recover the original FM term, we must isolate the band around \(\omega_c\) and reject the band around \(2\omega_c\). These two spectral regions should not overlap. Hence the highest edge of the band around \(\omega_c\) must not exceed the lowest edge of the band around \(2\omega_c\): \[ \omega_c + \frac{B_T}{2} \;\le\; 2\omega_c - \frac{2B_T}{2}. \] Simplify: \[ \omega_c + \frac{B_T}{2} \le 2\omega_c - B_T \;\;\Rightarrow\;\; \omega_c \ge \frac{3}{2} B_T. \]
Step 5: Check the weaker condition
There is also the trivial requirement \(\omega_c > \tfrac{B_T}{2}\) to avoid overlap with DC. But since \(\tfrac{3}{2}B_T > \tfrac{1}{2}B_T\), the stricter requirement dominates.
Final Answer:
\[ \boxed{\omega_c^{\min} = \tfrac{3}{2}B_T} \]
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