Let's break down the problem step by step: **PROBLEM:** **Show, using the table of Bessel functions, that the total power in an FM signal is equal to the power in the unmodulated carrier for modulation index \( m = 2 \). Compare with full-carrier AM and SSB AM.** --- ## 1. **FM Signal and Power Distribution** The FM signal is: \[ s(t) = A_c \cos\left(\omega_c t + m \sin \omega_m t\right) \] where: - \( A_c \) = carrier amplitude - \( \omega_c \) = carrier angular frequency - \( \omega_m \) = modulating frequency - \( m = \frac{\Delta f}{f_m} \) = modulation index (here, \( m = 2 \)) ### **Expanding the FM Signal (Fourier/Bessel Expansion)** \[ s(t) = A_c \left[ J_0(m) \cos \omega_c t + \sum_{n=1}^{\infty} J_n(m) \left( \cos(\omega_c + n\omega_m)t + \cos(\omega_c - n\omega_m)t \right) \right] \] where \( J_n(m) \) are Bessel functions of the first kind. --- ## 2. **Total Power in the FM Signal** Since the signal consists of multiple frequencies, the **total power** is the sum of the powers in each frequency component. - **Amplitude of each component:** \( A_c \cdot J_n(m) \) - **Power in each component:** \( \frac{[A_c \cdot J_n(m)]^2}{2} \) **Total FM power:** \[ P_{\text{FM}} = \frac{A_c^2}{2} \left[ J_0^2(m) + 2 \sum_{n=1}^\infty J_n^2(m) \right] \] But since \[ J_0^2(m) + 2 \sum_{n=1}^\infty J_n^2(m) = 1 \] (this is a property of Bessel functions: the sum of the squares for all orders equals 1) So, \[ P_{\text{FM}} = \frac{A_c^2}{2} \] which is **exactly equal to the power in the unmodulated carrier**. **This is true for any value of \( m \), including \( m = 2 \).** --- ## 3. **Check Using Bessel Table for \( m = 2 \)** Let's use Bessel values for \( m = 2 \): | \( n \) | \( J_n(2) \) | |---------|-----------------| | 0 | 0.2239 | | 1 | 0.5767 | | 2 | 0.3528 | | 3 | 0.1289 | | 4 | 0.0349 | Now, the sum: \[ \begin{align*} \text{Sum} &= J_0^2 + 2(J_1^2 + J_2^2 + J_3^2 + J_4^2 + \ldots) \\ &= (0.2239)^2 + 2\left[ (0.5767)^2 + (0.3528)^2 + (0.1289)^2 + (0.0349)^2 \right] \\ &= 0.0501 + 2 \left[ 0.3326 + 0.1245 + 0.0166 + 0.0012 \right] \\ &= 0.0501 + 2 \times 0.4749 \\ &= 0.0501 + 0.9498 \\ &= 0.9999 \approx 1 \end{align*} \] Thus, \( P_{\text{FM}} = \frac{A_c^2}{2} \), same as the unmodulated carrier. --- ## 4. **Comparison with AM** ### **Full-carrier AM (standard AM):** - Modulated signal: \( s(t) = A_c [1 + m_a \cos(\omega_m t)] \cos(\omega_c t) \) - **Total power:** \[ P_{\text{AM}} = \frac{A_c^2}{2} \left( 1 + \frac{m_a^2}{2} \right) \] - For 100% modulation (\( m_a = 1 \)): \[ P_{\text{AM}} = \frac{A_c^2}{2} \left( 1 + \frac{1}{2} \right) = \frac{A_c^2}{2} \cdot \frac{3}{2} = \frac{3 A_c^2}{4} \] - **AM power increases with modulation depth.** ### **SSB AM (Single Sideband):** - Only one sideband is transmitted, no carrier. - **Total power:** Only the message (sideband) component is sent. \[ P_{\text{SSB}} = \frac{A_c^2 m_a^2}{8} \] - **Much less power than full-carrier AM or FM.** --- ## 5. **Summary Table** | Modulation Type | Total Power (for \( A_c \) carrier amplitude) | |-----------------|------------------------------------------------------| | FM | \( \frac{A_c^2}{2} \) (independent of modulation) | | AM (full) | \( \frac{A_c^2}{2} \left( 1 + \frac{m_a^2}{2} \right) \)| | SSB AM | \( \frac{A_c^2 m_a^2}{8} \) | --- ## 6. **Final Answer and Explanation** ### **FM signal with \( m = 2 \):** Using the Bessel function expansion and table, the sum of the squares of all Bessel coefficients for \( m = 2 \) adds to 1, so the total power in the FM signal is: \[ P_{\text{FM}} = \frac{A_c^2}{2} \] which is **equal to the power in the unmodulated carrier**. **In contrast:** - **Full-carrier AM**: Power increases with the modulation index (\( m_a \)), maximum at \( \frac{3A_c^2}{4} \) for 100% modulation. - **SSB AM**: Power is much less, only the sideband is transmitted, so \( P = \frac{A_c^2 m_a^2}{8} \). --- ### **Conclusion:** **For FM, regardless of the modulation index (even for \( m = 2 \)), the total transmitted power remains equal to the unmodulated carrier power. For AM, the power increases with modulation, while for SSB AM, the transmitted power is much lower. This demonstrates the constant power property of FM.**