CS6956: Wireless and Mobile Networks Lecture Notes: 2/4/2015

Transcription

1 CS6956: Wireless and Mobile Networks Lecture Notes: 2/4/2015 [Most of the material for this lecture has been taken from the Wireless Communications & Networks book by Stallings (2 nd edition).] Effective Area The effective area of an antenna is the area of the incoming wavefront which is capture by the receiving antenna. The effective area varies with direction but when the direction is not mentioned, assume the one with the maximum. G = (4 π A eff )/λ 2 Since, λ = c/f ; (c is the speed of light) G = (4 π f 2 A eff )/c 2 Signal Impairments in Wireless transmission 1. attenuation (requires medium) 2. free space loss (no medium) 3. noise 4. multipath effects Attenuation: Strength of signal falls off with distance over any transmission medium. Introduces 3 factors: 1. A received signal must have sufficient signal strength so that the electronic circuit in the receiver can detect and interpret the signal. 2. The signal must maintain a sufficiently high level than noise to be received without error. 3. The attenuation is higher at higher frequencies, causing distortion. Can equalize attenuation across a band of frequencies (amplifiers that amplify high frequencies more than lower frequencies). Factor 1 and 2 can be handled by having a repeater or amplifier to strengthen the signals. Factor 3 can be handled through an equalizer. Free Space Loss (no medium): The energy keeps spreading in space and hence the signal strength decreases at every unit of distance. Even if no other sources of attenuation and impairments, a transmitted signal attenuates over distance because the signal is spread over a large area. For Isotropic Antenna: P t /P r =(4πd) 2 /λ 2 = (4πfd) 2 /c 2 ; Loss: L db = 10 log 10 (P t /P r ) = - 20 log λ + 20 log d db = 20 log f + 20 log d db For other antennas:

2 P t /P r =(4πd) 2 /(G r G t λ 2 ) G r = gain for receiving antenna G t = gain for transmitting antenna Substituting A eff r and A eff t P t /P r =(c 2 d 2 ) /(f 2 A eff t A eff r ) L db = - 20 log f + 20 log d 10 log (A eff t A eff r ) db. In the isotropic antenna case, the loss increases with frequency. However, this loss can be compensated with the use of directional antennas. In fact, as seen by the equation for loss for directional antennas, the loss actually decreases with increase in frequency. Higher the frequency, higher is the data rate. However, free space loss is huge at higher frequencies. E.g., the signals at 60 GHz have a massive data rate but also suffer from loss. This can be compensated with the use of directional antennas. [An interesting problem space.] Example: Determine the isotropic free space loss at 4 GHz for the path to a satellite from earth (at a distance 35,863 km). At 4 GHz, the wavelength is (3X10^8)/(4X10^6) = 0.075m. Then, L (in db) = - 20log(0.075) + 20log(35,863 X 10^3) = db Now consider the antenna gain of both the satellite and the ground- based station to be 44 db and 48 db, respectively. What is the free space loss now? L (in db) = 10 log 10 (P t /P r ) = 10 log 10 (4πd) 2 /(G r G t λ 2 ) = 10 log 10 (4πd) 2 /λ 2 10 log 10 G r - 10 log 10 G t = = db The above computations show that with the use of directional antennas the free space loss is reduced significantly. Now assume that the ground station transmits with a power of 250 W. What is the power received at the satellite antenna in dbw? 10 log 10 (P t /P r ) = db 10 log 10 P t 10 log 10 P r = 103.6dB. Given that 10 log 10 P t = 24 dbw, 10 log 10 P r = dbw. Noise: Unwanted signals inserted somewhere between transmission and reception. Major limiting factor in communication system performance. 4 categories: 1. Thermal

3 2. Intermodulation: When transmitting signals we modulate the carrier frequency with cosine components. The result will be signals with cosine terms with sum or difference of frequencies. These signals may interfere with genuine signals (with sum or differences of frequencies) causing intermodulation noise. 3. Cross talk: Unwanted signals picked up by an antenna 4. Impulse: non continuous, consists of irregular pulses or noise spikes of short duration and of relatively high amplitude, minor annoyance for analog data could result in lots of errors in digital data. Thermal noise: Due to thermal agitation, present in all electronic devices and transmission media, uniformly distributed across the spectrum, independent of frequency but depends on temperature. Noise power density, N 0 = K T (unit is Watts/Hz) K = Boltzmann s constant = 1.38 * J/K T = temperature in Kelvin. N = N 0 B, where B is the bandwidth N = KTB N dbw =10 log K + 10 log T + 10 log B = dbw + 10 log T + 10 log B E b = Energy per bit; E b /N 0 = (S T b )/N 0 [since T b =1/R is the duration of transmission of 1 bit, R is the data rate] =S/(RN 0 ) = S/(KTR) (E b /N 0 ) db = S dbw 10 log R 10 log K 10 log T 10 log K = dbw. E b /N 0 is like SNR but more convenient for digital data rates and error rates. How E b /N 0 (EBNO) is used in wireless communications? In the figure below, y- axis is Probability of Bit Error and the x- axis is E b /N 0 in db. Lower E b /N 0 results in higher the bit error rate. As R increases, the transmitted signal power relative to noise, must increase to maintain the required E b /N 0. Higher data rate means that the same level of noise can destroy multiple bits. Thus, for constant signal and noise strength, an increase in data rate also increases the error rate.

4 Probability of bit error (BER) Better performance Worse performance (E b /N 0 ) (db) Figure 5.9 General Shape of BER vs E b /N 0 Curves E b /N 0 = S/RN 0 = (S/N)*(B/R) [since N=N 0 B] From Shannon Channel Capacity Equation, C = B log 2 (1+(S/N)) S/N = 2 C/B 1 E b /N 0 = (B/R)(2 C/B 1)(B/R) When R = C, E b /N 0 = (B/C)(2 C/B 1)= (1/(C/B))*)(2 C/B 1) The above equation determines the E b /N 0 required for the data rate to be equal to the channel capacity. C/B (in bps/hz) is the achievable spectral efficiency or bandwidth efficiency. Multipath The multiple paths (multipath) between the transmitter and the receiver are caused by the reflections, diffractions, and scattering of the radio waves interacting with the physical environment. Each path has a different length, so a wave propagating along that path takes a different amount of time to arrive at

5 the receiver. Each path has attenuation caused by path losses and interactions with objects in the environment, so each wave undergoes a different attenuation and phase shift. Given the heterogeneous and changing nature of most of the real- world environments (other than free space), we expect the physical behavior of the wireless links to be different at different locations and at different times.