What is the bandwidth occupying 98% of power, as per Carson's rule for the following FM signal 10 sin\(\left(2 \pi * 10^{6} t+4 \cos \left(2 \pi * 10^{4} t\right)\right)\)?

Explanation:

 According to Carson's rule, the bandwidth of an FM signal that occupies 98% of its power can be expressed as:

Bandwidth (B) ≈ 2 × (Δf + fm)

Where:

• Δf: The frequency deviation, which represents the maximum shift in the carrier frequency due to the modulation signal.
• fm: The maximum frequency of the modulating signal.

In this case, the FM signal is given as:

10 sin(2π × 10⁶t + 4 cos(2π × 10⁴t))

From the given FM signal, we can extract the following parameters:

• Carrier Frequency (fc): 10⁶ Hz (1 MHz)
• Frequency of the Modulating Signal (fm): 10⁴ Hz (10 kHz)
• Frequency Deviation (Δf): The frequency deviation is determined by the modulation index multiplied by the frequency of the modulating signal. In this case, the modulation index is 4, so:

Δf = Modulation Index × fm

Δf = 4 × 10⁴

Δf = 40 kHz

Now, applying Carson's rule to calculate the bandwidth:

Bandwidth (B) ≈ 2 × (Δf + fm)

B ≈ 2 × (40 kHz + 10 kHz)

B ≈ 2 × 50 kHz

B ≈ 100 kHz