Chapter 14. Timing

Timing refers to how the receiving system knows that it received the start and end of a group of bits. Two major timing schemes are used: Asynchronous and Synchronous Transmission.

1. Asynchronous Transmission sends only 1 character at a time. A character can be a letter of the alphabet, a number or a control character. Preceding each character is a Start bit and ending each character is 1 or more Stop bits .

2. Synchronous Transmission sends packets of characters at a time. Each packet is preceded by a Start Frame. This Start Frame is used to tell the receiving station that a new packet of characters is arriving, and it synchronizes the receiving station's internal clock. The packets also have End Frames to indicate the end of the packet. The packet can contain up to 64,000 bits. Both Start and End Frames have a special bit sequence that the receiving station recognizes (to indicate the start and end of a packet). The Start and End frames can be only 2 bytes each. Packet

Conventional representation has asynchronous data flowing left to right and synchronous data flowing right to left.

Asynchronous vs. Synchronous Transmission

Asynchronous transmission is simple and inexpensive to implement. It is used mainly with Serial Ports and dialup connections. It requires start and stop bits for each character: this adds a high overhead to transmission. For example, for every byte of data add 1 Start Bit and 2 Stop Bits (i.e. 11 bits are required to send 8 bits!). Asynchronous is used in slow transfer rates (typically, up to a maximum of 56 kbps).

Synchronous transmission is more efficient. For example, only 4 bytes (3 Start Framing bytes and 1 Stop Framing byte) are required to transmit up to 64 kbits. Synchronous transmission is more difficult and expensive to implement than asynchronous. It is used with all higher-communication transfer rates (Ethernet, Token Ring, etc.). Synchronous is used in fast transfer rates (typically, from 56 kbps to 100 Mbps).

Historically, synchronous communications were operating over 2400/4800 baud modems and on point-to-point communications. Some of these modems,for example, are as follows: IBM2770/IBM2780/IBM3780 (historical information courtesy of Jacques Sincennes, University of Ottawa).

Example: Compare a 10K Byte data transmission using Asynchronous transmission vs. Synchronous Transmission. Determine the efficiency (10 kBytes = 80 kbits).

Asynchronous: Add 3 bits (1 Start and 2 Stop bits) for every byte transmitted.

80 kbits + 30 kbits = total of 110 kbits transmitted

Synchronous: Add 4 bytes (32 bits) for the complete 10K byte data packet.

80 kbits + 32 bits = total of 80.032 kbits transmitted

efficiency = data transmitted x 100 = 80 kbits x 100 = 99.9%

If this section was helpful, why not donate to further development?

Donate $1.25.

Donate $2.50.

Donate $5.00.