Exam RCDD-001 All QuestionsBrowse all questions from this exam
Question 15

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?

    Correct Answer: C

    A SONET OC-1 channel has a data rate of 51.84 Mbps and there are 48 OC-1 channels within an OC-48 channel. Therefore, the minimum data rate required for the OC-48 channel is calculated by multiplying the data rate of one OC-1 channel by 48. This gives us 51.84 Mbps * 48 = 2488.32 Mbps or approximately 2.5 Gbps. This means the minimum data rate required for the OC-48 channel is 2.5 Gbps.

Discussion
LiONz5Option: C

(51.84 Mbps per channel)*(48 channels) = 2488.32 Mbps or 2.5 Gbps

Davario

TDDM14 - Pag. 234 - Table 1.37

waldo31Option: C

C48= 2488.32 mbps or 2.5GBPS