1. Which drawbacks of FDM transmission networks have resulted in the creation of digital transmission networks?
2. The T-1 name means:
A. Multiplexing equipment developed by AT&T - correct
B. Rate level of 1.544 Mbps - correct
C. International standard for a communications link
D. Method of multiplexing 64 Kbps digital flows - correct
3. What functions are delegated to the least significant bit of each byte in the T-1 channel when transmitting voice?
This bit carries information that is used for establishing of telephone connection; the main part of the information is a dialed number.
Is it possible to separate the DS-0 channel
directly from the DS-3 channel in a
5. What methods are used in practice for solving the previous problem?
- a use of two T-3 multiplexors at every network node to provide full demultiplexing of DS-3 channel onto DS-1 channels;
- “back hauling”
6. What mechanisms are implemented in the E-1 channel to replace the bit robbing of the T-1 channel?
A use of two dedicated bytes of a E-1 frame header: 0 and 16.
7. Why do transmission networks ensure high quality of service for all kinds of traffic?
Because they are based on circuit-switching technique that dedicate a fixed bandwidth to ecery connection. Moreover these networks transmit data in digital form which provides a high level of transmission reliability.
8. What property of the PDH technology is reflected by the term "plesiochronous"?
The term reflects the fact that this technology is “almost” synchronous. The synchronization is provided by inserting a number of bits into slower data flow.
How does the SDH technology compensate for the
lack of synchronism in
By means of “floating” virtual containers
What is the maximum number of E-1 channels that
can be multiplexed by the
11. How many T-1 channels can be multiplexed by the STM-1 frame, provided that it already contains 15 E-1 channels?
12. Which layers of the SDH protocol stack are responsible for network reconfiguration in the case of equipment failure?
Line and path layers
13. What is the maximum rate of the data communication channel between SDH regenerators?
14. Why does the STM-1 frame use three pointers?
Because the STM-1 frame can carry up to three VC-3 virtual containers
15. What is the purpose of using an interleaving byte in PDH and SDH technologies?
To make the speed of tributary bits in a aggregate channel equal to their spped in tributary channel.
16. What is the difference between 1+1 and 1:1 protection methods:
A. The 1+1 method multiplexes two streams into one, and the 1:1 method does not.
B. In the 1+1 method, the redundant element carries out the same functions as the main one, and in the 1:1 method, the redundant element is not used until the main element fails. - correct
C. The 1+1 method is used for port protection, and the 1:1 method is used for protecting traffic paths.
17. Under what conditions is MS-SPRing more efficient than SNC-P?
When traffic is distributed between network multiplexors more or less evenly (in contrast to star topology when all traffic goes to one multiplexor).
18. What are the common features between FDM and DWDM transmission networks?
Both networks use frequency division multiplexing but in different frequency ranges (light has a frequency of electromagnetic waves in THz range)
19. What type of networks are DWDM networks — analog or digital?
20. What is the goal of using regenerators that transform an optical signal to an electric one in DWDM networks?
For a reduction of nonlinear effects.
21. What are the reasons for optical signal deterioration when passing many passive DWDM sections?
Chromatic dispersion, when waves of different length propagate with different speed.
22. What principles of switching of light signals are used in OXCs?
1. What will be the frequency of negative alignment of the pointer of VC-4 in the STM-1 frame if the relative difference between the clock frequencies of transmitting and those of receiving SDH multiplexers is 10−5?
A negative alignment occurs when difference in a number of bits received and transmitted gets equal to 24 (3 bytes). Having the basic frequency of 155 MHz and the relative difference of 10-5 we can figure out that every second will result in the difference of:
10-5 õ 155 õ 10+6 = 1550 bits.
Therefore, a negative alignment will happen with a frequency of
1550 / 24 = 64,58 Hz
2. The SDH network comprises four STM-4 multiplexers: A, B, C, and D. Fig. 11.22 shows the traffic distribution among multiplexers. All flows have a rate of STM-1. Multiplexers are joined into the STM-4 ring. Which protection method should be chosen to protect all connections?
Fig. 11.22. Traffic distribution
MS-SPRing as traffic is distributed between multiplexors