1. 1 st semester 1436/1437 1

2.  When a signal is transmitted over a communication channel, it is subjected to different types of impairments because of imperfect characteristics of the channel.  As a consequence, the received and the transmitted signals are not the same.  These impairments introduce random modifications in analog signals leading to distortion. On the other hand, in case of digital signals, the impairments lead to error in the bit values. 2

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4.  Irrespective of whether a medium is guided or unguided, the strength of a signal falls off with distance.  When a signal travels through a medium it loses energy overcoming the resistance of the medium  Attenuation means loss of energy -> weaker signal  The attenuation leads to several problems: 4

5. 1. - To be able to detect correctly the signal, the signal strength should be sufficiently high. - If the strength of the signal is very low, the signal cannot be detected and interpreted properly at the receiving end. - An amplifier can be used to compensate the attenuation of the transmission line. - So, attenuation decides how far a signal can be sent without amplification through a particular medium. 5

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7. 2. Attenuation Distortion - Attenuation of all frequency components is not same. - Some frequencies are passed without attenuation, some are weakened and some are blocked. 7

8. - As an example, after sending a square wave through a medium, the output is no longer a square wave because of more attenuation of the high-frequency components in the medium. 8

9.  Means that the signal changes its form or shape  A composite signal made of different frequencies components  Each signal component has its own propagation speed through a medium and, therefore, its own delay in arriving at the final destination.  Differences in delay may create a difference in phase if the delay is not exactly the same as the period duration.  In other words, signal components at the receiver have phases different from what they had at the sender.  The shape of the composite signal is therefore not the same. 9

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11.  As signal is transmitted through a channel, undesired signal in the form of noise gets mixed up with the signal, along with the distortion introduced by the transmission media.  Noise is any unwanted energy tending to interfere with the signal to be transmitted. 11

12. The noise either be:  External Noise. This is noise originating from outside the communication system  Internal Noise: This is noise originating from within the communication system. 12

13.  Thermal Noise: This noise is due to the random and rapid movement of electrons in any resistive component. Electrons “bump” with each other.  Impulse noise is irregular pulses or noise spikes of short duration 13

14.  Cross talk is a result of bunching several conductors together in a single cable. Signal carrying wires generate electromagnetic radiation, which is induced on other conductors because of close proximity of the conductors. 14

15.  In the study of noise, it is not important to know the absolute value of noise.  Even if the power of the noise is very small, it may have a significant effect if the power of the signal is also small.  What is important is a comparison between noise and the signal.  The signal-to-noise ratio (SNR) is the ratio of signal power to noise power. 15

16. SNR = P s / P n 16

17.  Ideally, SNR = ∞ (when P n = 0). In practice, SNR should be high as possible.  A high SNR ratio means a good-quality signal.  A low SNR ratio means a low-quality signal.  The SNR is normally expressed in decibels, that is: SNR = 10 log 10 (P s / P n ) dB 17

18. 3.18 Figure 3.30 Two cases of SNR: a high SNR and a low SNR

19.  The power of a signal is 10 mW and the power of the noise is 1 μ W; what are the values of SNR and SNR dB SNR = 10 × 10 -3 / 10 -6 = 10,000 SNR dB = 10 log 10 (10 × 10 -3 / 10 -6 ) = 10 log 10 (10,000) = 40 dB 19

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21.  The maximum rate at which data can be correctly communicated over a channel in presence of noise and distortion is known as its channel capacity.  The capacity of an analog channel is its bandwidth  The capacity of a digital channel is the number of digital values the channel can convey in one second. It is usually measured in bits per second (bps) 21

22.  The bandwidth of a channel is the difference between the lowest and highest frequency an analog channel can convey to a receiver.  A channel with a wide bandwidth is called broadband channel  A channel with a narrow bandwidth is called baseband channel. 22

23.  Digital signals consist of a large number of frequency components. If digital signals are transmitted over a channel with a limited bandwidth, only those components that are within the bandwidth of the transmission medium are received.  The faster the data rate of a digital signal, the higher the bandwidth will be required since the frequency components will be spaced farther apart.  Therefore, a limited bandwidth will also limit the data rate that can be used for transmission 23

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25.  For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate of a transmission medium as a function of its bandwidth C = 2 x B x log 2 m bits/sec, C is known as the channel capacity, B is the bandwidth of the channel and m is the number of signal levels used. 25

26.  Consider a noiseless channel with a bandwidth of 3kHz transmitting a signal with two signal levels. What is the Nyquist bit rate? C=2 x 3000 x log 2 2 =6000 bps  Consider the same noiseless channel transmitting a signal with four signal levels. What is the Nyquist bit rate? C=2 x 3000 x log 2 4 = 12,000 bps 26

27.  In reality, we cannot have a noiseless channel; the channel is always noisy. In 1944, Claude Shannon introduced a formula, called the Shannon capacity, to determine the theoretical highest data rate for a noisy channel C = BW x log 2 (1 +SNR) 27

28.  Consider an extremely noisy channel in which the value of the SNR is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as C=B log 2 (1 + SNR) =B log 2 (1 + 0) =B log 2 1 =B x 0 = 0  This means that the capacity of this channel is zero regardless of the bandwidth.  In other words, we cannot receive any data through this channel. 28

29.  A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communications. The SNR is usually 3162. For this channel the capacity is calculated as C =B log2 (1 + SNR) =3000 log 2 (1 + 3162) = 3000 log 2 3163 = 3000 x 11.62 = 34,860 bps  This means that the highest bit rate for a telephone line is 34.860 kbps.  If we want to send data faster than this, we can either increase the bandwidth of the line or improve the SNR 29