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Demodulation, Bit-Error Probability and Diversity Arrangements

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Demodulation, Bit-Error Probability and Diversity Arrangements RADIO SYSTEMS – ETI 051 Contents Lecture no: • Receiver noise calculations [Covered briefly in Chapter 3 of textbook!] 6 • Optimal receiver and bit error probability – Principle of maximum-likelihood receiver Demodulation, – Error probabilities in non-fading channels bit-error probability – Error probabilities in fading channels • Diversity arrangements and – The diversity principle – Types of diversity diversity arrangements – Spatial (antenna) diversity performance Ove Edfors, Department of Electrical and Information Technology [email protected] 2010-04-22 Ove Edfors - ETI 051 1 2010-04-22 Ove Edfors - ETI 051 2 Receiver noise Noise sources The noise situation in a receiver depends on several noise sources Noise picked up Wanted by the antenna signal RECEIVER NOISE Output signal Analog Detector with requirement circuits on quality Thermal noise 2010-04-22 Ove Edfors - ETI 051 3 2010-04-22 Ove Edfors - ETI 051 4 Receiver noise Receiver noise Equivalent noise source Examples To simplify the situation, we replace all noise sources • Thermal noise is caused by random movements of with a single equivalent noise source. electrons in circuits. It is assumed to be Gaussian and the power is proportional to the temperature of the material, in Kelvin. Wanted How do we Noise free • Atmospheric noise is caused by electrical activity in signal determine N from the other sources? the atmosphere, e.g. lightning. This noise is impulsive in N its nature and below 20 MHz it is a dominating. Output signal • Cosmic noise is caused by radiation from space and the C Analog Detector with requirement circuits sun is a major contributor. on quality Noise free • Artificial (man made) noise can be very strong and, e.g., light switches and ignition systems can produce significant noise well above 100 MHz. Same “input quality”, signal-to-noise ratio, C/N in the whole chain. 2010-04-22 Ove Edfors - ETI 051 5 2010-04-22 Ove Edfors - ETI 051 6 Receiver noise Receiver noise Noise sources Noise sources, cont. Antenna example The power spectral density of a noise source is usually given in one of the following three ways: This one is 1) Directly [W/Hz]: N Ns often given in a dB and called Model 2) Noise temperature [Kelvin]: Ts noise figure. Noise temperature Noise free 3) Noise factor [1]: Fs of antenna 1600 K antenna The relation between the three is Power spectral density of antenna noise is = = −23 −20 Ns kT s kF s T0 N a=1.38×10 ×1600=2.21×10 W/Hz=−196.6 dB [W/Hz] -23 where k is Boltzmann’s constant (1.38x10 W/Hz) and T0 is the, and its noise factor/noise figure is so called, room temperature of 290 K (17O C). F a=1600/290=5.52=7.42 dB 2010-04-22 Ove Edfors - ETI 051 7 2010-04-22 Ove Edfors - ETI 051 8 Receiver noise Receiver noise System noise Several noise sources A simple example The noise factor or noise figure of a system component (with input and output) is defined in a different way than for noise sources: T a N sys System 1 System 2 System System Model component component F F 1 2 = Noise factor F Noise free Na kT a Due to a definition of noise factor (in this case) as the ratio of noise N= k( F −1) T powers on the output versus on the input, when a resistor in room Noise 1 1 0 free temperature (T0=290 K) generates the input noise, the PSD of the =( − ) N N N N2 k F 21 T 0 equivalent noise source (placed at the input) becomes a 1 2 =( − ) Nsys k F1 T0 W/Hz System 1 System 2 Don’t use dB value! Noise Noise Equivalent noise temperature free free 2010-04-22 Ove Edfors - ETI 051 9 2010-04-22 Ove Edfors - ETI 051 10 Receiver noise Receiver noise Several noise sources, cont. Pierce’s rule After extraction of the noise sources from each component, we need to A passive attenuator in room temperature, in this case a feeder, move them to one point. has a noise figure equal to its attenuation. When doing this, we must compensate for amplification and attenuation! N f Amplifier: N NG Lf Lf F = L Noise free G G f f Attenuator: =( −) =( − ) N N/L Nf k F f1 T0 k L f 1 T 0 1/L 1/L Remember to convert from dB! 2010-04-22 Ove Edfors - ETI 051 11 2010-04-22 Ove Edfors - ETI 051 12 Receiver noise Receiver noise Remember ... A final example Let’s consider two (incomplete) receiver chains with equal gain from point A to B: Antenna noise is usually given as a noise temperature! T a Noise factors or noise figures of different system components 1 are determined by their implementation. Lf G 1 G 2 A B Would there be any When adding noise from several sources, remember to reason to choose one convert from the dB-scale noise figures that are usually given, F1 F2 over the other? before starting your calculations. Ta Let’s calculate the A passive attenuator in (room temperature), like a feeder, has 2 L equivalent noise a noise figure/factor equal to its attenuation. f G 1 G 2 at point A for both! A B F1 F2 2010-04-22 Ove Edfors - ETI 051 13 2010-04-22 Ove Edfors - ETI 051 14 Receiver noise Receiver noise A final example Noise power 1 2 We have discussed noise in terms of power spectral density N [W/Hz]. Ta 0 Ta Lf Lf G 1 G 2 G 1 G 2 For a certain receiver bandwidth B [Hz], we can calculate the equivalent A B A B noise power: F1 F2 F1 F2 Equivalent noise sources at point A for the two cases would have the power spectral densities: N =B×N 0 [W] = +( −) + − +( − ) 1 NkTkF0a( 11( L f 1) / GF 1 2 1 LGT f / 1) 0 N ∣dB=B∣dBN 0∣dB [dBW] 2 NkTkL= +( ( −1) +( F − 1) LF +( − 1) LGT / ) 0a f 1 f 2 f 1 0 This is the version Two of the noise contributions are equal and two are larger in (2), we will use in our which makes (1) a better arrangement. link budget. This is why we want a low-noise amplifier (LNA) close to the antenna. 2010-04-22 Ove Edfors - ETI 051 15 2010-04-22 Ove Edfors - ETI 051 16 Receiver noise The link budget ”POWER” [dB] PTX Lf, TX G a, TX Lp G a, RX Lf, RX C OPTIMAL RECEIVER F [dB] is the noise figure of the C / N equivalent noise source at the N reference point and The receiver AND B [dBHz] the system bandwidth . B N noise calculations 0 show up here. F BIT ERROR PROBABILITY Noise reference level = kT0 = -204 dB[W/Hz] Transmitter Receiver In this version the reference point is here 2010-04-22 Ove Edfors - ETI 051 17 2010-04-22 Ove Edfors - ETI 051 18 Optimal receiver Optimal receiver What do we mean by optimal? Transmitted and received signal Transmitted signals Channel Received (noisy) signals Every receiver is optimal according to some criterion! s1(t) r(t) 1: n(t) We would like to use optimal in the sense that we achieve a t t minimal probability of error. s(t) r(t) s0(t) r(t) 0: In all calculations, we will assume that the noise is white and t t Gaussian – unless otherwise stated. 2010-04-22 Ove Edfors - ETI 051 19 2010-04-22 Ove Edfors - ETI 051 20 Optimal receiver Optimal receiver A first “intuitive” approach Let’s make it more measurable “Look” at the received signal and compare it to the possible received noise free signals. Select the one with the best “fit”. To be able to better measure the “fit” we look at the energy of the residual (difference) between received and the possible noise free signals: Assume that the following Comparing it to the two possible signal is received: noise free received signals: s1(t) - r(t) r(t), s1(t) r(t), s1(t) 1: e = s t −r t 2 dt r(t) 1: This seems to t t 1 ∫∣ 1 ∣ t be the best “fit”. We s0(t) - r(t) t assume that r(t), s0(t) r(t), s2(t) “0” was the transmitted bit. 2 0: e 0=∫∣s0 t −r t ∣ dt 0: t t t This residual energy is much smaller. We assume that “0” was transmitted. 2010-04-22 Ove Edfors - ETI 051 21 2010-04-22 Ove Edfors - ETI 051 22 Optimal receiver Optimal receiver The AWGN channel The AWGN channel, cont. The additive white Gaussian noise (AWGN) channel It can be shown that finding the minimal residual energy (as we did before) is the optimal way of deciding which of s0(t), s1(t), ... , sM-1 (t) n( t ) was transmitted over the AWGN channel (if they are equally probable). s( t ) r( t ) =αs( t) + n( t ) For a received r(t), the residual energy ei for each possible transmitted alternative si(t) is calculated as α 2 * ei=∫∣r t− si t ∣ dt=∫ r t− si t rt− si t dt ( ) - transmitted signal In our digital transmission s t 2 2 system, the transmitted * * 2 α =∫∣r t ∣ dt−2 Re { ∫ r t si tdt ∣∣ ∫∣si t ∣ dt - channel attenuation signal s(t) would be one of, } let’s say M, different alternatives ( ) - white Gaussian noise Same for all i Same for all i, n t s (t), s (t), ..
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