Final answer:
To find the minimum number of PN bits for an FHSS system with a total bandwidth of 400 MHz and a channel bandwidth of 200 kHz, divide the total bandwidth by the channel bandwidth to get 2,000 channels.
Step-by-step explanation:
The student asked about an FHSS (Frequency-Hopping Spread Spectrum) system that uses a total bandwidth of 400 MHz and an individual channel bandwidth of 200 kHz. To determine the minimum number of Pseudo-Random Number (PN) bits required for each frequency hop, we need to calculate how many separate channels can fit into the total available bandwidth. Using arithmetic, divide the total bandwidth by the channel bandwidth:
Wₛ / Wₒ = 400 MHz / 200 kHz = 400,000 kHz / 200 kHz = 2,000 channels
Since the number of channels represents the number of different frequencies the system can hop to, we can represent this as a binary number. To find the minimum number of bits needed to represent all possible channels, we calculate the binary logarithm (log base 2) of the number of channels:
log₂(2,000) ≈ 11 bits
We round up because we can't have a fraction of a bit. Therefore, the minimum number of PN bits required for each frequency hop is 11 bits. This allows the FHSS system to uniquely address each of the 2,000 channels.