Question

An angle-modulated signal with carrier frequency \(\omega_c = 2\pi \times 10^5\) is described by the equation. \[ \Phi_{em}(t)=10\cos(\omega_c t+5\sin 3000t+10\sin 2000t) \] The power of the modulated signal is :

• A. \(10\)
• B. \(100\)
• C. \(50\)
• D. \(25\)

Hint:

In FM and PM:
• Amplitude remains constant
• Total transmitted power remains constant \[ P=\frac{A_c^2}{2} \]

Answer 1

The Correct Option is C

Concept: In angle modulation:
• Amplitude remains constant.
• Only phase or frequency changes. The average power of angle-modulated signal is: :contentReference[oaicite:4]{index=4} for \(1\Omega\) load.

Step 1: Identify carrier amplitude. Given signal: \[ \Phi_{em}(t)=10\cos(\omega_c t+5\sin3000t+10\sin2000t) \] Carrier amplitude: \[ A_c=10 \]

Step 2: Use power formula. For angle modulation: \[ P=\frac{A_c^2}{2} \] Substitute: \[ P=\frac{10^2}{2} \] \[ P=\frac{100}{2} \] \[ P=50 \]

Step 3: Write final answer. Therefore power of the signal is: \[ \boxed{50} \] Hence correct option is: \[ \boxed{(C)} \]