A SONET OC-1 channel can carry 672 voice signals and has a data rate of 51.84 Mbps. A SONET OC-48 channel can carry 32,256 voice channels. What MINIMUM data rate is required for the OC-48 channel?

The correct answer is C. 2.5 Gbps. SONET OC (Optical Carrier) levels scale linearly: OC-N runs at N × 51.84 Mbps. For OC-48: 48 × 51.84 Mbps = 2,488.32 Mbps ≈ 2.489 Gbps. This is confirmed by the voice channel scaling: OC-1 carries 672 channels, and OC-48 carries 672 × 48 = 32,256 channels, which matches the…

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Question

Options

A155 Mbps
B622 Mbps
C2.5 Gbps
D5 Gbps
E10 Gbps

How the community answered

(42 responses)

A
2% (1)
B
12% (5)
C
79% (33)
D
2% (1)
E
5% (2)

Explanation

SONET OC (Optical Carrier) levels scale linearly: OC-N runs at N × 51.84 Mbps. For OC-48: 48 × 51.84 Mbps = 2,488.32 Mbps ≈ 2.489 Gbps. This is confirmed by the voice channel scaling: OC-1 carries 672 channels, and OC-48 carries 672 × 48 = 32,256 channels, which matches the given figure. The minimum data rate must be at least 2,488.32 Mbps, making 2.5 Gbps (the next standard rate above that value) the correct answer.

Topics

#SONET#OC-48#data rate calculation#voice channels

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