Summary

This document provides an overview of digital and analog signals, along with the bandwidth of periodic and non-periodic signals. Mathematical problems are solved demonstrating concepts such as attenuation and calculation of power.

Full Transcript

Data Communication Data and Signals Signals and Communication A single-frequency sine wave is not useful in data communications We need to send a composite signal, a signal made of many simple sine waves. According to Fourier analysis, any composite signal is a combination of s...

Data Communication Data and Signals Signals and Communication A single-frequency sine wave is not useful in data communications We need to send a composite signal, a signal made of many simple sine waves. According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases. Composite Signals and Periodicity If the composite signal is periodic, the decomposition gives a series of signals with discrete frequencies. If the composite signal is non-periodic, the decomposition gives a combination of sine waves with continuous frequencies. Example Figure 3.9 A composite periodic signal Example Figure 3.10 Decomposition of a composite periodic signal in the time and frequency domains Bandwidth and Signal Frequency The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. Figure 3.12 The bandwidth of periodic and non-periodic composite signals Mathematical problems P1: If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Solution Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz. Figure 3.13 The bandwidth for Example P1 Mathematical problems P2: A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude. Solution Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then The spectrum contains all integer frequencies. We show this by a series of spikes Figure 3.13 The bandwidth for Example P2 Mathematical problems P3: A non-periodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal. Solution The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure 3.15 shows the frequency domain and the bandwidth. Figure 3.13 The bandwidth for Example P3 Digital Signals In addition to being represented by an analog signal, information can also be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage. A digital signal can have more than two levels. In this case, we can send more than 1 bit for each level. Figure 3.16 Two digital signals: one with two signal levels and the other with four signal levels Mathematical Problems P4: A digital signal has eight levels. How many bits are needed per level? Each signal level is represented by 3 bits. Mathematical Problems P4: A digital signal has nine levels. How many bits are needed per level? We calculate the number of bits by using the formula. Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level. TRANSMISSION IMPAIRMENT Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment. This means that the signal at the beginning of the medium is not the same as the signal at the end of the medium. What is sent is not what is received. Three causes of impairment are attenuation, distortion, and noise. Figure 3.25 Causes of impairment Attenuation Means loss of energy -> weaker signal When a signal travels through a medium it loses energy overcoming the resistance of the medium Amplifiers are used to compensate for this loss of energy by amplifying the signal. 3.20 Figure 3.26 Attenuation Measurement of Attenuation To show the loss or gain of energy the unit “decibel” is used. dB = 10log10P2/P1 P1 - input signal power P2 - output signal power 3.22 Example Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as A loss of 3 dB (–3 dB) is equivalent to losing one-half the power. Example A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1. In this case, the amplification (gain of power) can be calculated as Example One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure 3.27 a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as Example One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure 3.27 a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as Figure 3.27 Decibels for Example 3.28 3.27 Example Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm , where Pm is the power in milliwatts. Calculate the power of a signal with dBm = −30. Solution We can calculate the power in the signal as Example The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km? Solution The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as Distortion Means that the signal changes its form or shape Distortion occurs in composite signals Each frequency component has its own propagation speed traveling through a medium. The different components therefore arrive with different delays at the receiver. That means that the signals have different phases at the receiver than they did at the source. Figure 3.28 Distortion Noise There are different types of noise – Thermal - random noise of electrons in the wire creates an extra signal – Induced - from motors and appliances, devices act are transmitter antenna and medium as receiving antenna. – Crosstalk - same as above but between two wires. – Impulse - Spikes that result from power lines, lightning, etc. 3.33 Figure 3.29 Noise Signal to Noise Ratio (SNR) To measure the quality of a system the SNR is often used. It indicates the strength of the signal wrt the noise power in the system. It is the ratio between two powers. It is usually given in dB and referred to as SNRdB. Mathematical problem The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRdB ? Solution The values of SNR and SNRdB can be calculated as follows: Ideal Condition We can never achieve this ratio in real life; it is an ideal condition. Figure 3.30 Two cases of SNR: a high SNR and a low SNR

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