Within the synchronous TDM hierarchy, multiple T1 carriers are multiplexed into

Question

Within the synchronous TDM hierarchy, multiple T1 carriers are multiplexed into higher order carriers. For example, four T1 channels are multiplexed onto one T2 channel running at 6.312 Mbps, seven T2 channels are multiplexed onto one T3 channel running at 44.736 Mbps, and six T3 channels are multiplexed onto one T4 channel running at 274.176 Mbps.

a.) How many T1 user slots are contained in a T4 frame?

b.) What is the time duration and total data rate of a T4 frame?

If I'm not mistaken part a is just 4 * 7 * 6 which is 168 T1 slots in a T4 frame but how do you calculate part b? Any help would be great!

Explanation / Answer

Answer:
a)
Since, four channels of T1 are multiplexed into one T2, seven T2 channels are multiplexed into one T3, six T3 channels frames are multiplexed into one T4.
Therefore, the total number of slots/channels is 4*7*6 = 168 channels or slots.

b)
Since, by the time T3 is multiplexed into T4, the total data rate will be the data rate of T3 multiplexed with T4.
Therefore, the data rate of T4 frame is 274.176 Mbps itself.
The time duration is, the minimum bandwidth obtained at T4 is 274.176 Mbps/2 = 137.088MHz.
If the bandwidth obtained in MHz, then it can be considered as the frequency of the cannel at T4.
The relation between the frequency and time are reciprocal to each other.
Therefore, the total time is given as,
t = 1/f
=1/137.088MHz
   =0.007294*10^6seconds
   =7.2*10-3*10^6 sec
   =7.2*10^9sec  
   =7.2nano-sec
Therefore, the total time taken is 7.2ns and the total data rate at T4 frame is 274.17Mbps.

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