Match communication services to their typical bandwidths: (A) AM broadcast, (B) Telephone (voice channel), (C) Wide-band FM, (D) Television (TV) — with: (1) 10 kHz, (2) 4 kHz, (3) 200 kHz, (4) 7 MHz.

Difficulty: Easy

Correct Answer: A-1, B-2, C-3, D-4

Explanation:


Introduction / Context:
Each communication service occupies a characteristic bandwidth dictated by modulation method, fidelity requirements, and regulatory standards. Matching these correctly is foundational for spectrum planning and receiver/transmitter design.


Given Data / Assumptions:

  • (A) AM broadcast
  • (B) Telephone (POTS) channel
  • (C) Wide-band FM (broadcast FM)
  • (D) TV channel
  • Candidate bandwidths: 10 kHz, 4 kHz, 200 kHz, 7 MHz


Concept / Approach:

AM broadcast stations transmit audio up to ~5 kHz, leading to ≈10 kHz occupied bandwidth (two sidebands). Voice-grade telephony is limited to ≈300–3400 Hz, i.e., about 4 kHz. Wide-band FM (commercial FM broadcast) allocates roughly 200 kHz channels (allowing for deviation and subcarriers). Analog TV channels historically occupy several MHz; a representative nominal figure here is ≈7 MHz.


Step-by-Step Solution:

A → (1): AM broadcast ≈ 10 kHz.B → (2): Telephone voice channel ≈ 4 kHz.C → (3): Broadcast FM ≈ 200 kHz spacing/occupancy.D → (4): TV channel on the order of MHz (≈ 7 MHz typical in many standards).


Verification / Alternative check:

Regulatory tables (e.g., ITU/legacy standards) list these canonical figures; Carson’s rule for FM supports the 200 kHz order for wide-band FM.


Why Other Options Are Wrong:

Swapping 10 kHz with 4 kHz misstates telephony limits; assigning kHz-scale bandwidth to TV ignores video RF spectrum needs.


Common Pitfalls:

Confusing narrowband FM (few kHz) with broadcast wide-FM; overlooking guard bands vs occupied bandwidth.


Final Answer:

A-1, B-2, C-3, D-4

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