Spectrum planning (PM/FM style): Approximately how many phase-modulated (PM) radio channels can be accommodated in a total usable spectrum of 300 MHz, assuming a typical allocated channel spacing of about 200 kHz (as used for wideband FM/PM broadcasting) and allowing practical guard separation?

Introduction:
This problem tests spectrum-allocation reasoning for phase modulation (PM) channels that are similar in occupied bandwidth to wideband FM broadcasting. The key idea is to convert the total usable spectrum into an integer count of channels using typical per-channel spacing with practical guard bands already implicit in that spacing.

Given Data / Assumptions:

• Total available spectrum = 300 MHz.
• Typical wideband PM/FM channel spacing ≈ 200 kHz per channel (includes guard for adjacent-channel interference under standard allocations).
• Uniform channelization across the band.

Concept / Approach:

Channel count is total bandwidth divided by per-channel spacing. For broadcast-style PM (akin to wideband FM), practical planning commonly adopts 200 kHz spacing. This spacing folds in Carson’s-rule considerations (peak deviation and modulating frequency) and adjacent-channel protections.

Step-by-Step Solution:

Verification / Alternative check:

If one chose a slightly more conservative spacing (for example 250 kHz), the count becomes 1200; if the band plan is explicitly WBFM-like, 200 kHz is standard and yields 1500. Hence 1500 is the reasonable textbook answer.

Why Other Options Are Wrong:

• 10/100/500: Far too small; they imply per-channel spacings in multi-MHz or ~0.6 MHz ranges, inconsistent with WBFM/PM planning.
• 1200: Matches 250 kHz spacing, which is more conservative than the common 200 kHz broadcast plan.

Common Pitfalls:

Confusing Carson’s occupied bandwidth with regulatory channel spacing. The spacing already bakes in adjacent-channel protection and planning rules.

Final Answer:

1500