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]

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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

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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.

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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

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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

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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!

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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

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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∣dBN 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.

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”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

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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.

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“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.

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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  rt− 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 tdt ∣∣ ∫∣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), ... , s (t). 0 1 M-1 if the transmitted r( t ) - received signal The residual energy is minimized by signals are of maximizing this part of the equal energy. expression.

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The central part of the comparison of different signal alternatives In antipodal signaling, the alternatives (for “0” and “1”) are is a correlation, that can be implemented as a correlator: ( ) = ϕ ( ) s0 t t r( t ) s( t) = −ϕ ( t ) ∫ 1 T s This means that we only need ONE correlation in the receiver * ( ) α * for simplicity: si t The real part of the output from or a matched filter either of these If the real part is sampled at t = Ts ( ) r t at T=T is ( ) s r t ∫T >0 decide “0” * ( − ) s si T s t * <0 decide “1” ϕ * ( t ) α α *

where Ts is the symbol time (duration).

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Optimal receiver Optimal receiver Orthogonal signals Interpretation in signal space

The correlations performed on the previous slides can be seen as In binary orthogonal signaling, with equal energy alternatives s (t) and s (t) 0 1 inner products between the received signal and a set of basis functions (for “0” and “1”) we require the property: for a signal space. * The resulting values are coordinates of the received signal in the 〈 so t , s1t〉=∫ s0t s1 t dt=0 signal space. The approach here is to use two correlators: Antipodal signals Orthogonal signals s( t ) Decision 1 ∫T s Compare real boundaries “1” * * part at t=T s( t ) α s r( t ) 0 and decide in favor of the “1” “0” “0” larger. ϕ ( ) ( ) ∫ t s0 t T s * ( ) α * s1 t * (Only one correlator is needed, if we correlate with (s0(t) - s1(t)) .)

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Assume a 2-dimensional signal space, here viewed as the complex plane What is the probability of deciding si if sj was transmitted? Im Im Noise-free We need the distance positions between the two symbols. s s i i d In this orthogonal case: Noise pdf. ji

E E 2 2 s s = + = dji E s E s2 E s sj This normalization of sj axes implies that the Re noise centered around Re The probability of the noise each alternative is pushing us across the boundary

complex Gaussian at distance dji / 2 is

E 2 2 E d / 2 E s N( 0,σ) + j N( 0, σ ) s ji s Pr  s j  si =Q =Q   N 0 /2   N 0  2  Fundamental question: What is the probability with variance σ = N0/2 that we end up on the wrong side of the decision in each direction. The book uses erfc() 1 E s = erfc boundary? instead of Q(). 2 2 N  0  2010-04-22 Ove Edfors - ETI 051 29 2010-04-22 Ove Edfors - ETI 051 30

Optimal receiver Optimal receiver The union bound Symbol- and bit-error rates Calculation of symbol error probability is simple for two signals! The calculations so far have discussed the probabilities of selecting When we have many signal alternatives, it may be impossible to the incorrect signal alternative (symbol), i.e. the symbol-error rate. calculate an exact symbol error rate. When each symbol carries K bits, we need 2K symbols. When s is the transmitted s 0 2 signal, an error occurs when Gray coding is used to assigning bits so that the nearest neighbors only s1 the received signal is outside this polygon. differ in one of the K bits. This minimizes the bit-error rate. s 0 s s 3 011 6 The UNION BOUND is the sum 010 001 of all pair-wise error probabilities, Gray-coded 8PSK and constitutes an upper bound 110 000 on the symbol error probability. s5 s4 111 100 s7 101 The higher the SNR, the better the approximation!

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0 10

2PAM/4QAM EXAMPLES: 2PAM 4QAM 8PSK 16QAM -1 10 8PSK

)

R 16QAM

E -2

B 10

(

e

t -3 a 10

r

Bits/symbol 1 2 3 4

r

o

r -4 Symbol energy Eb 2Eb 3Eb 4Eb r 10

e

-

t

i -5 2 E 2 E 2 Eb 3 E b, max B 10 BER Q b Q b ~ Q 0.87 ~ Q N N 3 N 2 2.25 N  0   0   0   0  -6 10 0 2 4 6 8 10 12 14 16 18 20 Gray coding is used when calculating these BER. ENb /0 [dB]

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Optimal receiver Optimal receiver

Where do we get Eb and N0? What about fading channels?

Where do those magic numbers Eb and N0 come from? We have (or can calculate) BER expressions for non-fading AWGN channels.

The noise power spectral density N0 is calculated according to If the channel is Rayleigh-fading, then E /N will have an N =k T F ⇔ N =−204F b 0 0 0 0 0∣dB 0∣dB exponential distribution (N0 is assumed to be constant)

where F0 is the noise factor of the “equivalent” receiver noise source. -- E /N b b 0 1 −b / b The bit energy E can be calculated from the received pdf  b = e b  -- average E /N b b b 0 power C (at the same reference point as N0). Given a certain

data-rate db [bits per second], we have the relation The BER for the Rayleigh fading channel is obtained by averaging: E b=C /d b ⇔ E b∣dB=C∣dB−d b∣dB ∞

BERRayleigh  b = BERAWGN  b × pdf  b  d b THESE ARE THE EQUATIONS THAT RELATE DETECTOR ∫ 0 PERFORMANCE ANALYSIS TO LINK BUDGET CALCULATIONS!

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THIS IS A SERIOUS PROBLEM!

Bit error rate (4QAM) 0 10

-1 10 dB Rayleigh fading 10 10 x -2 10 DIVERSITY ARRANGEMENTS -3 10

-4 10

-5 10 No fading

-6 10 0 2 4 6 8 10 12 14 16 18 20

Eb/N0 [dB]

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Diversity arrangements Diversity arrangements Let’s have a look at fading again The diversity principle

Illustration of interference pattern from above Received power [log scale] The principle of diversity is to transmit the same information on M statistically independent channels.

By doing this, we increase the chance that the information will be received properly. Movement A B The example given on the previous slide is one such arrangement: antenna diversity. Position Transmitter A B Reflector

Having TWO separated antennas in this case may increase the probability of receiving a strong signal on at least one of them.

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Spatial (antenna) diversity

Bit error rate (4PSK) 0 We will focus on this 10 ... one today!

-1 10 dB Rayleigh fading TX Signal combiner 10 No diversity 10 x -2 10 Frequency diversity 10 dB -3 10 Rayleigh fading D D D M:th order diversity TX -4 Signal combiner 10 10M x

-5 Temporal diversity 10 No fading

-6 Inter- De-inter- 10 Coding De-coding 0 2 4 6 8 10 12 14 16 18 20 leaving leaving

Eb/N0 [dB] (We also have angular and polarization diversity)

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Spatial (antenna) diversity Spatial (antenna) diversity Fading correlation on antennas Selection diversity

RSSI = received signal strength indicator

With several Isotropic antennas, we uncorrelated want the fading on scattering. them to be as independent as possible.

An antenna spacing less than half a wavelength gives zero correlation.

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By measuring BER instead of RSSI, we have a better guarantee that we obtain a low BER. This is the optimal way (SNR sense) of combining antennas.

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Spatial (antenna) diversity Spatial (antenna) diversity Performance comparison

Cumulative distribution of SNR

MRC Comparison of SNR distribution RSSI selection for different number of antennas M and two different diversity techniques.

Simpler than MRC, but almost the same performance. [Fig. 13.10] These curves can be used to calculate fading margins.

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• Optimal (maximum likelihood) receiver in AWGN channels MRC • Interpretation of received signal as a point in a signal space Comparison of RSSI selection • Euclidean distances between symbols determine the 2ASK/2PSK BER probability of symbol error for different number • Bit error rate (BER) calculations for some signal of antennas M and constellations two different diversity techniques. • Union bound (better at high SNRs) can be to derive approximate BER expressions • Fading leads to serious BER problems • Diversity is used to combat fading • Focus on spatial (antenna) diversity • Performance comparisons for RSSI selection and maximum ratio combining diversity.

[Fig. 13.11]

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