The document discusses bandwidth and how it relates to transmitting signals through communication channels. It defines bandwidth as the range of frequencies a signal or channel can carry. For an analog signal, bandwidth is the maximum frequency minus the minimum frequency. The bandwidth required for a digital signal is 2 times the highest signaling rate. Time division multiplexing allows multiple calls to be transmitted simultaneously by dividing the channel into time slots.

1. Chapter 3 – Bandwidth How much information can you send down a wire or fibre..?

2. Bandwidth Bandwidth of analogue signal - The spectrum of an analogue signal contains a range of frequencies. The range, B = maximum frequency – minimum frequency , is the bandwidth of the analogue signal. In many cases, this is in effect equal to the maximum frequency (the signal can extend down to dc). e.g. A telephone carries a voice signal between the range of 300Hz to 4000Hz. What is it’s bandwidth? Bandwidth, B = Maximum Frequency – Minimum Frequency = 4000 – 300 = 3700Hz What affects how many of these signals can be carried by a channel?

3. Bandwidth Bandwidth of a channel - A signal has to be sent along a communication channel, for example wires. The bandwidth of a channel is the range of frequencies it can carry. Often, this is the same as the maximum frequency it can carry. Fundamentally, this is determined by the fastest response time of the channel to a change in input voltage. To transmit an analogue signal across a channel, the channel must have a bandwidth greater than or at least equal to the signal bandwidth. This is affected by noise. How is the bandwidth required for a digital signal calculated?

4. Bandwidth and Digital Signals What determines the Bandwidth Requirements of a digital signal? The digital signal requires the highest bandwidth when it is a series of alternating 1s and 0s i.e. 101010101010101010 This can be sent as a sine wave at the fundamental frequency of the signal Maximum Digital Signalling rate = 2B = 2/T

5. Time Division Multiplexing

6. Fitting the Calls In Time to send 1 call (8-bit) sample = 1ns = 1 x 10 -9 s Gap between samples = 100  s Call samples in 100  s = 100 x 10 -6 / 1 x 10 -9 s = 100 000 Bits in 100  s = 100000 x 8 = 800 000 So 100 000 calls can be sent down one fibre/channel at 8 bits resolution

7. Answers to Sheet 1: Signal Bandwidth = High – Lowest Frequency = 1000KHz – 500KHz = 500KHz (notice units) 2: Bandwidth required = 2 x B = 1000KHz Therefore in a 1GHz bandwidth channel 1 x 10 9 / 1 x 10 6 = 1000 signals 3: Maximum Signalling Rate = 2/T = 2/(1 x 10 -3 ) = 2 x 10 3 s -1 4: Calls can be passed more than 1 at a time using Time Division Multiplexing. Number of calls at once = Time Interval / Time to send 1 call sample =200  s / 2ns = 1 x 10 5 calls In bits (x 8) = 8 x 10 5 bits