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Suppose that we modulate the periodic message signal m(t) given in the figure by PM modulation with kp = 1000m. Assume that the bandwidth of the message signal is its fifth harmonic frequency. What is the bandwidth of the modulated signal? 4 his 6 9 5 -1 m(t) -2 1 3 t (ms)

Suppose that we modulate the periodic message signal m(t) given in the figure by PM modulation with kp = 1000m. Assume that the bandwidth of the message signal is its fifth harmonic frequency. What is the bandwidth of the modulated signal? 4 his 6 9 5 -1 m(t) -2 1 3 t (ms)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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PM BANDWIDTH( NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE).

**Phase Modulation and Bandwidth Calculation** **Problem:** Suppose that we modulate the periodic message signal \( m(t) \) given in the figure by PM modulation, with \( k_p = 1000\pi \). Assume that the bandwidth of the message signal is its fifth harmonic frequency. What is the bandwidth of the modulated signal? **Figure Description:** The figure shows a periodic message signal \( m(t) \) with time \( t \) in milliseconds (ms) on the horizontal axis and amplitude on the vertical axis. The signal has a repeating pattern with the following important points: - The signal starts at -2 amplitude at \( t=0 \) ms. - It rises linearly to -1 at \( t=1 \) ms, then increases to 4 amplitude by \( t=2 \) ms. - It remains constant at 4 amplitude from \( t=2 \) ms to \( t=3 \) ms. - The signal then decreases linearly to 0 amplitude at \( t=5 \) ms. - It further decreases to -2 amplitude by \( t=6 \) ms. - The signal remains constant at -2 amplitude from \( t=6 \) ms to \( t=7 \) ms. - It rises linearly to -1 at \( t=9 \) ms and continues the same pattern periodically. **Choices:** - 8 kHz - 4 kHz - 6 kHz **Graph Description:** The graph depicts \( m(t) \) as a periodic sequence of triangular and rectangular waveforms. It repeats every 6 ms, indicating a fundamental frequency (\( f \)) of \(\frac{1}{6} \)kHz = \( \frac{1}{0.006} \approx 166.67 \) Hz. Since the bandwidth of the message signal is its fifth harmonic frequency, the signal bandwidth in Hz is \( 5 \times 166.67 \approx 833.34 \) Hz. Given the phase modulation index \( k_p = 1000\pi \), the bandwidth of the modulated signal (\( W \)) is determined as: \[ W = 2 \times (5 \text{th harmonic bandwidth}) \] \[ W = 2 \times 833.34 \approx 1666.68 \] \[ W \approx 1.67 \text
Transcribed Image Text:**Phase Modulation and Bandwidth Calculation** **Problem:** Suppose that we modulate the periodic message signal \( m(t) \) given in the figure by PM modulation, with \( k_p = 1000\pi \). Assume that the bandwidth of the message signal is its fifth harmonic frequency. What is the bandwidth of the modulated signal? **Figure Description:** The figure shows a periodic message signal \( m(t) \) with time \( t \) in milliseconds (ms) on the horizontal axis and amplitude on the vertical axis. The signal has a repeating pattern with the following important points: - The signal starts at -2 amplitude at \( t=0 \) ms. - It rises linearly to -1 at \( t=1 \) ms, then increases to 4 amplitude by \( t=2 \) ms. - It remains constant at 4 amplitude from \( t=2 \) ms to \( t=3 \) ms. - The signal then decreases linearly to 0 amplitude at \( t=5 \) ms. - It further decreases to -2 amplitude by \( t=6 \) ms. - The signal remains constant at -2 amplitude from \( t=6 \) ms to \( t=7 \) ms. - It rises linearly to -1 at \( t=9 \) ms and continues the same pattern periodically. **Choices:** - 8 kHz - 4 kHz - 6 kHz **Graph Description:** The graph depicts \( m(t) \) as a periodic sequence of triangular and rectangular waveforms. It repeats every 6 ms, indicating a fundamental frequency (\( f \)) of \(\frac{1}{6} \)kHz = \( \frac{1}{0.006} \approx 166.67 \) Hz. Since the bandwidth of the message signal is its fifth harmonic frequency, the signal bandwidth in Hz is \( 5 \times 166.67 \approx 833.34 \) Hz. Given the phase modulation index \( k_p = 1000\pi \), the bandwidth of the modulated signal (\( W \)) is determined as: \[ W = 2 \times (5 \text{th harmonic bandwidth}) \] \[ W = 2 \times 833.34 \approx 1666.68 \] \[ W \approx 1.67 \text
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