Tutorial_DIN_Cruickshank_2010 - DIN 2010 Tutorial(A1a In...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
DIN 2010 Tutorial (A1a) In the IEEE 802.15 standard (Bluetooth), describe the following two types of physical links between primary and secondary stations: (i) Synchronous Connection Oriented (SCO); (ii) Asynchronous Connection Link (ACL). Answer: Synchronous Connection-oriented (SCO) link: Used when latency is more important than error free data delivery. Here the physical link is created between the primary and a secondary by reserving time slots at regular intervals. Data rate of 64Kbps for each link. Asynchronous Connectionless Link (ACL): Used when error free is more important than latency in data delivery. Here retransmission of corrupted frames is allowed. Secondary return ACL frames only if previous slot is addressed to it. Maximum data rate is 721 Kbps (A1b) If an Ethernet destination address is 47:20:1B:2E:08:EE, (i) What type of address is this (unicast, multicast or broadcast)? (ii) Can the above Ethernet address be used as a source address? Explain. Answer (i) The first byte in binary is 0100011 1 . The least significant bit is 1. This means that the pattern defines an Ethernet multicast address . (ii) An Ethernet multicast address can be a destination address, but not a source address. If used as source address, the receiver will assume that there is an error, and discards the packet. (A1c ) The data rate for standard Ethernet is 10 Mbps, with a minimum frame size of 64 bytes. Assuming the LAN propagation speed is 2 * 10 8 m/s: (i) Calculate the frame transmission time; (ii) Calculate the theoretical maximum length of the LAN; (iii) Explain why the practical maximum length of an Ethernet LAN is less than this theoretical value.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Answer (i) The smallest frame is 64 bytes or 512 bits. With a data rate of 10 Mbps and Frame transmission time, T fr = (512 bits) / (10 Mbps) = 51.2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

Tutorial_DIN_Cruickshank_2010 - DIN 2010 Tutorial(A1a In...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online