Signal Transmission - Assignment Example

Signal Transmission Maximum data rate depends on the three main factors that are: The available Bandwidth The level of signals The quality of channel (considering noise level) As the bandwidth of the channel increases, the capacity of the channel to send data increases. It can be said that in order to achieve the maximum data rate, the bandwidth of the channel should be kept maximum. The levels of the signals also influence the data transmission rate through the channel. If the levels of signals are increased in a channel, the distortion within the channel increases due to the increased risk of frequent errors. The reliability of the system decreases as the probability of the error decreases. The bit rate is in direct relation to the number of signal level. With the increase in the signal levels, the bit rate of the system also increases. A symbol may have an “n” number of bits or sometimes a single bit. The error probability of the signal increases, as the number of levels of the signal increase. This is due to the decrease in the spacing between the signal levels. Nyquist’s Theorem A transmission system depends on the bit rate of the system. Nyquist’s theorem simplifies calculation of the bit rate from the signal levels and the bandwidth of the system. The theorem also evaluates the upper bound of the bit rate of the transmission system. Nyquist considers a noiseless channel for the calculation of the bit rate. If "C" is the capacity of the system orbit rate capacity of the system, "B" is the bandwidth of the channel and "2n” is the number of signal levels allowed by the channel, then C = 2 B log22n Shannon’s theorem also simplifies the mathematical expression for the bit rate capacity of the system. The theorem evaluates the upper bound bit rate capacity of the system through the bandwidth of the system and signal to noise ratio. If “B” is the allocated bandwidth of the channel and S/R (or SNR) is the signal to noise ration of the channel, then, C = B log2 (1 + S/R) Or C = B log2 (1 + SNR) The major difference between the Nyquist’s theorem and Shannon’s theorem is that the Nyquist’s theorem considers an ideal situation without the interference of noise; however, Shannon's theorem is based on the signal to noise ratio. “The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need”. (i) A telephone line has a bandwidth of 3000Hz. (300-3300Hz). The signal to noise ratio (SNR) is 35dB. What is the theoretical highest bit rate transmission over the line? Solution: Here we have: Bandwidth of telephone line = B = 3000 Hz Signal to noise ratio = SNR dB = 35 dB SNR dB = 10 log 10 SNR SNR = 10 SNR dB/ 10 SNR = 10 35/ 10 SNR = 10 3.5 SNR = 3162 Channel capacity or highest bit that channel can pass C = B log2 (1 + SNR) C = 3000 log2 (1 + 3162) C = 3000 log2 3163 C = 3000 X 11.62 C = 34860 bps or 34.86 Kbps (ii) A channel has a 1 MHz bandwidth. The SNR for this channel is 63dB. What is the appropriate bit rate and number of signal levels? Here we have: Bandwidth of the channel = B = 1MHz SNR = 63 dB, First, we apply Shannon's theorem to evaluate the Capacity of the channel (Bit rate). If SNR is high we use the formula as: In order to evaluate the number of signal levels, we use Nyquist’s theorem: C = 2 B log2L 21 Mbps = 2 (1 MHz) log2L Radio Frequency Carrier Waves for transmission Radio waves require carrier frequencies due to three basic reasons: Antenna length Operating range Wireless communication In order to transmit the wave efficiently, the length of the antenna should be more or equal to the wavelength. In order to transmit higher wavelength frequency, the length of the antenna should be larger that become unpractical. In order to limit the size of the antenna, carrier frequencies are used that have powerful signal frequencies. Carrier frequencies make it possible to send and receive the radio waves under a set operating range. It limits the distortion and interference with other frequencies. Carrier frequencies allow the transmission of the signal frequencies to be transmitted wirelessly. Radio waves require carrier frequencies for the transmission as carrier frequencies make it easier to transmit the information through space. Carrier frequencies work in a fixed band where the power transmitted through the carries is considerably higher as compared to that of the packed radio frequencies. Carrier frequencies are transmitted by using a similar physical transmission medium. Signal frequencies and carrier frequencies are modulated in a manner that the receiver can easily separate the two frequencies. Mainly two basic modulation techniques are used for the transmission of the carrier frequencies. Frequencies modulation and amplitude modulation are the basic types of modulation techniques used extensively. In frequency modulation, the signal frequencies are mixed with the frequency of the carrier frequency and in amplitude modulation techniques, the signal frequencies are mixed with respect to the amplitude of the carrier frequency. The mathematical expression for the frequency modulated wave: If V FM (t) is the frequency modulated wave, Vc is the carrier frequency, then Where, is the frequency component of the modulated and the carrier wave. And Amplitude modulated is then given by: b) An AM wave is represented by the expression : v = 5 (1 + 0.6 cos 6280 t) in 211 × 104 t volts (i) What are the minimum and maximum amplitudes of the AM wave? (ii) What frequency components are contained in the modulated wave and what is the amplitude of each component? The AM wave equation is given by: v = 5 (1 + 0.6 cos 6280 t) sin 211 × 104t volts ... (i) Compare it with standard AM wave eq., ... (ii) From eqs. (i) and (ii) , we get, EC = carrier amplitude = 5 V m = modulation factor = 0.6 f fs = signal frequency = ωs/2π = 6280/2π = 1 kHz fc = carrier frequency = ωc/2π = 211 × 104/2π = 336 kHz (i) Minimum amplitude of AM wave = EC − mEC = 5 − 0.6 × 5 = 2 V Maximum amplitude of AM wave = EC + mEC = 5 + 0.6 × 5 = 8 V ii) The AM wave will contain three frequencies fc − fs = 336 -1= 335 KHz, fc = 336 KHz fc + fs = 336 +1= 337 KHz, The amplitudes of the three components of AM wave are : m EC/ 2 = (0.6 X 5)/2 = 1.5 V, EC = 5V (i) Amplitude shift keying (ASK), Amplitude shift keying is a modulation technique that is used for the modulation of digital signals. A digital system works with only high and low pulses (Either 1 or zero). Amplitude shift keying modulates the digital signals in the manner that for the high pulse, it generates rising amplitude and for the low pulse, the wave remains the same as that of the digital pulse. The figure below depicts a digital and converted modulates signal by using amplitude shift keying (ii) Frequency shift keying (FSK), Frequency shift keying is a modulation technique used for modulating digital signals. The digital signals with high and low pulses are modulated in a different manner. To represent digital “1”, mark frequency is used and to represent a digital zero, space frequency is used. In such a manner, the digital signal is modulated. The image below depicts the modulation of the digital signal by using FSK. (iii) Phase shift keying (PSK), Phase shift keying is the modulation technique that is used to modulate the digital signals. In this technique, a number of bits are modulated by altering the phase of the carrier signal. In this way, each phase represents different bits. The figure below simplifies the description of PSK. Read More