SDH and PDH Networks: Characteristics and Differences

Posted on Dec 16, 2024 in Computers

1. Indicate the Correct Statement Regarding PDH Networks

  • PDH networks are limited to carrying voice signals.
  • In PDH networks, multiplexing is performed bitwise.
  • PDH networks have different hierarchies in Bolivia, the USA, and Japan.
  • A PDH network is based on synchronous multiplexers.

2. Which of the Following Policies Defines the Signal Format?

None of the above. (The correct answer would likely involve a standard like G.703, not G.708)

3. List Five Disadvantages of SDH

(Note: The original text lists advantages, not disadvantages. Here are some actual disadvantages)

  1. Complexity: SDH networks are more complex to design and manage than PDH networks.
  2. Bandwidth Efficiency: SDH can be less bandwidth-efficient for lower-rate signals due to fixed container sizes.
  3. Equipment Cost: SDH equipment can be more expensive than comparable PDH equipment.
  4. Synchronization Requirements: SDH requires precise network-wide synchronization.
  5. Legacy Compatibility: While SDH can accommodate PDH signals, integrating them can add complexity.

4. What are the Main Differences Between PDH and SDH?

  • Multiplexing: SDH multiplexing is more direct and efficient. Pointers allow for easy location of tributary signals.
  • Clock Synchronization: In PDH, the ends of a link operate with independent clocks. SDH uses a common, high-precision clock signal.
  • Frame Structure: SDH frames can accommodate plesiochronous loads, resulting in more flexible networks.

5. What is the Transmission Rate in Mbps of the Listed Synchronous Transport Modules?

  • STM-1: 155.52 Mbps
  • STM-4: 622.08 Mbps
  • STM-8: 1244.16 Mbps
  • STM-12: 1866.24 Mbps
  • STM-16: 2488.32 Mbps
  • STM-64: 9953.28 Mbps
  • STM-256: 39813.12 Mbps

6. What Does TLE Stand for in the Context of this Equipment?

TLE: Terminal Line Equipment

7. Diagram the Path from an STM-1 to a C-11

C-11 => VC-11 => TU-11 => TUG-2 => VC-3 => AU-3 => AUG => STM-1

8. Diagram the Structure of an STM-1 Frame

An STM-1 frame consists of a 9×270 matrix:

  • 9 rows x 270 columns
  • First 9 columns: Section Overhead (SOH) and Administrative Unit Pointer (AU PTR)
    • Rows 1-3: Regenerator Section Overhead (RSOH)
    • Row 4: AU PTR
    • Rows 5-9: Multiplexer Section Overhead (MSOH)
  • Remaining 261 columns: Payload (VC)

9. Show that the Actual Capacity of an STM-1 is 150.336 Mbps

Total STM-1 frame size: 270 bytes * 9 rows = 2430 bytes = 19440 bits

Overhead size: 9 bytes * 9 rows = 81 bytes = 648 bits

Useful information (payload): 19440 bits – 648 bits = 18792 bits

Repetition frequency: 8 kHz (8000 frames per second)

Effective capacity: 18792 bits * 8000 1/s = 150.336 Mbps

10. What is the Input Data Rate of a C-4?

C-4 Input Data Rate: 140 Mbps

11. Can the RSOH Bytes Monitor and Control Regenerating Pathways?

Yes

12. Was SDH the First Effort to Standardize Digital Hierarchy Worldwide?

True

13. What is the Size of a TUG-3 in Columns?

TUG-3 Size: 86 Columns

14. What Types of Pointers Exist in SDH?

Existing Pointers:

  • H1 and H2: Indicate the start of the VC (Virtual Container).
  • H3: Used for negative justification (frequency adjustment).

15. Calculate the Following Capacities

  • CAU-4: Approximately 150.920 Mbps
  • CVC-4: 150.336 Mbps
  • CC-4: 149.760 Mbps

16. List Three Disadvantages of an Asynchronous Network

  • The receiver does not know exactly when a message will arrive.
  • Lower transmission efficiency due to start and stop bits.
  • Errors can cause loss of synchronization and data.

17. What Container Size Fits into a T1?

A T1 signal fits into a C-11 container.

18. What is the Function of POH?

POH (Path Overhead): Transports information about the path and the status of the virtual container. It allows monitoring of the entire route.

19. What is the Purpose of the A1 and A2 Bytes?

A1 and A2 Bytes: Used for frame alignment (framing bytes). They identify the start of an STM-N frame.

20. Where is the J0 Byte Located?

J0 Byte: Located in the Regenerator Section Overhead (RSOH). It is used to transmit the section’s access point identifier.

21. Where is the J1 Byte Located?

J1 Byte: Located in the Path Overhead (POH). It is the path trace byte, used to verify the path connection.

22. What is the General Purpose of the B Bytes?

B Bytes: Used for error monitoring in the STM-N signal within a section.

23. How to Proceed with Positive, Negative, and Zero Bit Rate Justification

  • Equal to transmission capacity: No justification (0).
  • Greater than transmission capacity: Negative justification (-).
  • Less than transmission capacity: Positive justification (+).

24. What are the Codes and Interfaces that may Apply to STM-x Signals?

CMI (Coded Mark Inversion) is a possible code.

25. Explain the Principle of Light Confinement in an Optical Fiber

Light is confined within the core of an optical fiber due to total internal reflection. This occurs when the core has a higher refractive index than the cladding, causing light to reflect back into the core instead of refracting out.

26. What are the Exclusive Characteristics of Optical Fibers?

  • Numerical Aperture
  • Bandwidth
  • Modal Dispersion
  • Attenuation

27. What is the Purpose of Doping in Optical Fibers, and What are the Most Frequently Used Dopants?

Doping: Alters the refractive index of the core and cladding. This is essential for creating the conditions for total internal reflection. Common dopants include Germanium (Ge) and Phosphorus (P).

28. Which Characteristics are Responsible for Attenuation in Optical Fibers?

  • Absorption
  • Dispersion
  • Curvature (bending losses)

29. Are Multimode Fibers Most Commonly Used in Optical Transmission Systems in Buildings?

True