RADIO SYSTEMS – ETIN15 Lecture no: 6 Demodulation, bit-error probability and diversity arrangements
Anders J Johansson, Department of Electrical and Information Technology [email protected]
5 April 2017 1 Contents
• Receiver noise calculations [Covered briefly in Chapter 3 of textbook!] • Optimal receiver and bit error probability – Principle of maximum-likelihood receiver – Error probabilities in non-fading channels – Error probabilities in fading channels • Diversity arrangements – The diversity principle – Types of diversity – Spatial (antenna) diversity performance
5 April 2017 2 RECEIVER NOISE
5 April 2017 3 Receiver noise Noise sources
The noise situation in a receiver depends on several noise sources
Noise picked up Wanted by the antenna signal
Output signal Analog Detector with requirement circuits on quality Thermal noise
5 April 2017 4 Receiver noise Equivalent noise source
To simplify the situation, we replace all noise sources with a single equivalent noise source.
Wanted How do we determine Noise free signal N from the other sources? N
Output signal C Analog Detector with requirement circuits on quality Noise free
Same “input quality”, signal-to-noise ratio, C/N in the whole chain.
5 April 2017 5 Receiver noise Examples • Thermal noise is caused by random movements of electrons in circuits. It is assumed to be Gaussian and the power is proportional to the temperature of the material, in Kelvin. • Atmospheric noise is caused by electrical activity in the atmosphere, e.g. lightning. This noise is impulsive in its nature and below 20 MHz it is a dominating. • Cosmic noise is caused by radiation from space and the sun is a major contributor. • Artificial (man made) noise can be very strong and, e.g., light switches and ignition systems can produce significant noise well above 100 MHz.
5 April 2017 6 Receiver noise Noise sources
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]: Ns often given in dB and called 2) Noise temperature [Kelvin]: Ts noise figure.
3) Noise factor [1]: Fs
The relation between the three is
Ns kT s kF s T0
-23 where k is Boltzmann’s constant (1.38x10 W/Hz) and T0 is the, so called, room temperature of 290 K (17O C).
5 April 2017 7 Receiver noise Noise sources, cont.
Antenna example
N a Model
Noise temperature Noise free of antenna 1600 K antenna
Power spectral density of antenna noise is −23 −20 N a=1.38×10 ×1600=2.21×10 W/Hz=−196.6 dB [W/Hz]
5 April 2017 8 Receiver noise System noise
The noise factor or noise figure of a system component (with input and output) is: N sys
System System Model component component Noise factor F Noise free Due to a definition of noise factor as the ratio of noise powers on the output versus on the input, when a resistor in room
temperature (T0=290 K) generates the input noise, the PSD of the equivalent noise source (placed at the input) becomes
Nsys k F 1 T0 W/Hz
Don’t use dB value! Equivalent noise temperature
5 April 2017 9 Receiver noise Several noise sources
A simple example
T a
System 1 System 2 F F 1 2 Na kT a
Noise N1 k F 1 1 T 0 free N N N N2 k F 2 1 T 0 a 1 2
System 1 System 2
Noise Noise free free
5 April 2017 10 Receiver noise Several noise sources, cont.
After extraction of the noise sources from each component, we need to move them to one point.
When doing this, we must compensate for amplification and attenuation!
Amplifier: N NG
G G
Attenuator: N N/L
1/L 1/L
5 April 2017 11 Receiver noise Pierce’s rule
A passive attenuator in room temperature, in this case a transmission line, has a noise factor equal to its attenuation.
N f
Lf Lf F = L Noise free f f
Nf k F f 1 T0 k L f 1 T 0
Remember to convert from dB!
5 April 2017 12 Receiver noise Remember ...
Antenna noise is usually given as a noise temperature!
Noise factors or noise figures of different system components are determined by their implementation.
When adding noise from several sources, remember to convert from the dB-scale noise figures that are usually given, before starting your calculations.
A passive attenuator in (room temperature), like a transmission line, has a noise figure/factor equal to its attenuation.
5 April 2017 13 Receiver noise A final example
Let’s consider two (incomplete) receiver chains with equal gain from point A to B: T a
1 Lf
G1 G2 A B Would there be any reason to choose one F1 F2 over the other?
Ta Let’s calculate the equivalent noise 2 Lf at point A for both! G1 G2 A B
F1 F2
5 April 2017 14 Receiver noise A final example
1 2 Ta Ta
Lf Lf G1 G2 G1 G2 A B A B F1 F2 F1 F2 Equivalent noise sources at point A for the two cases would have the power spectral densities: NkTkF1 L 1 / GF 1 LGT / 1 0a 1 f 1 2 f 1 0
2 NkTkL1 F 1 LF 1 LGT / 0a f 1 f 2 f 1 0 Two of the noise contributions are equal and two are larger in (2), which makes (1) a better arrangement. This is why we want a low-noise amplifier (LNA) close to the antenna.
5 April 2017 15 Receiver noise Noise power
We have discussed noise in terms of power spectral density N0 [W/Hz].
For a certain receiver bandwidth B [Hz], we can calculate the equivalent noise power:
N =B×N 0 [W]
N ∣dB=B∣dBN 0∣dB [dBW]
This is the version we will use in our link budget.
5 April 2017 16 Receiver noise The link budget
”POWER” [dB]
PTX Lf, TX G a, TX Lp
Ga, RX Lf, RX C F [dB] is the noise figure of the equivalent C / N noise source at the reference point and N B [dBHz] the system bandwidth . B The receiver noise calculations N 0 F show up here. Noise reference level = kT0 = -204 dB[W/Hz]
Transmitter Receiver
In this version the reference point is here
5 April 2017 17 OPTIMAL RECEIVER AND BIT ERROR PROBABILITY
5 April 2017 18 Optimal receiver What do we mean by optimal?
Every receiver is optimal according to some criterion!
We would like to use optimal in the sense that we achieve a minimal probability of error.
In all calculations, we will assume that the noise is white and Gaussian – unless otherwise stated.
5 April 2017 19 Optimal receiver Transmitted and received signal
Transmitted signals Channel Received (noisy) signals
s1(t) r(t) 1: t n(t) t s(t) r(t)
s0(t) r(t)
0: t t
5 April 2017 20 Optimal receiver A first “intuitive” approach
“Look” at the received signal and compare it to the possible received noise free signals. Select the one with the best “fit”.
Assume that the following Comparing it to the two possible signal is received: noise free received signals:
r(t), s1(t)
r(t) 1: This seems to be t the best “fit”. We assume that “0” t was the
r(t), s2(t) transmitted bit.
0: t
5 April 2017 21 Optimal receiver Let’s make it more measurable
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:
s1(t) - r(t)
r(t), s1(t)
2 1: e1= ∣s1 t −r t ∣ dt t t ∫
s0(t) - r(t)
r(t), s0(t)
2 0: e 0=∫∣s0 t −r t ∣ dt t t
This residual energy is much smaller. We assume that “0” was transmitted.
5 April 2017 22 Optimal receiver The AWGN channel
The additive white Gaussian noise (AWGN) channel
n t s t r t s t n t
s t - transmitted signal In our digital transmission system, the transmitted - channel attenuation signal s(t) would be one of, n t - white Gaussian noise let’s say M, different alternatives s0(t), s1(t), ... , sM-1(t). r t - received signal
5 April 2017 23 Optimal receiver The AWGN channel, cont.
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) was transmitted over the AWGN channel (if they are equally probable).
For a received r(t), the residual energy ei for each possible transmitted
alternative si(t) is calculated as 2 * ei=∫∣r t− si t∣ dt=∫ r t− si t r t− si t dt
2 * * 2 2 =∫∣r t∣ dt−2 Re { ∫ r t si t dt }∣∣ ∫∣si t∣ dt
Same for all i Same for all i, if the transmitted The residual energy is minimized by signals are of maximizing this part of the expression. equal energy.
5 April 2017 24 Optimal receiver The AWGN channel, cont.
The central part of the comparison of different signal alternatives is a correlation, that can be implemented as a correlator: r t ∫ T s * * si t The real part of the output from or a matched filter either of these
is sampled at t = Ts r t * si T s t
*
where Ts is the symbol time (duration).
5 April 2017 25 Optimal receiver Antipodal signals
In antipodal signaling, the alternatives (for “0” and “1”) are
s0 t t
s1 t t This means that we only need ONE correlation in the receiver for simplicity:
If the real part r t at T=T is ∫ s T s >0 decide “0” * <0 decide “1” * t
5 April 2017 26 Optimal receiver Orthogonal signals
In binary orthogonal signaling, with equal energy alternatives s0(t) and s1(t) (for “0” and “1”) we require the property: * 〈 so t, s1t〉=∫ s0t s1t dt=0
The approach here is to use two correlators: ∫ T s Compare real * * part at t=T s s0 t r t and decide in favor of the ∫ larger. T s * * s1 t * (Only one correlator is needed, if we correlate with (s0(t) - s1(t)) .)
5 April 2017 27 Optimal receiver Interpretation in signal space
The correlations performed on the previous slides can be seen as inner products between the received signal and a set of basis functions for a signal space. The resulting values are coordinates of the received signal in the signal space.
Antipodal signals Orthogonal signals
s1 t Decision boundaries “1”
“1” “0” “0”
t s0 t
5 April 2017 28 Optimal receiver The noise contribution Assume a 2-dimensional signal space, here viewed as the complex plane Im Noise-free positions si Noise pdf.
Es
sj This normalization of axes implies that the Re noise centered around each alternative is complex Gaussian E s N 0,2 j N 0, 2 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. boundary?
5 April 2017 29 Optimal receiver Pair-wise symbol error probability
What is the probability of deciding si if sj was transmitted? Im We need the distance s between the two symbols. i d ji In this orthogonal case:
Es 2 2 dji E s E s 2 E s sj
Re The probability of the noise pushing us across the boundary
at distance dji / 2 is E d / 2 E s Pr s s =Q ji =Q s j i N /2 N 0 0
The book uses erfc() 1 E s = erfc instead of Q(). 2 2 N 0 5 April 2017 30 Optimal receiver The union bound Calculation of symbol error probability is simple for two signals!
When we have many signal alternatives, it may be impossible to calculate an exact symbol error rate. When s is the transmitted s 0 2 signal, an error occurs when s1 the received signal is outside this polygon. s 0 s s 3 6 The UNION BOUND is the sum of all pair-wise error probabilities, and constitutes an upper bound on the symbol error probability. s5 s4 s7 The higher the SNR, the better the approximation!
5 April 2017 31 Optimal receiver Symbol- and bit-error rates
The calculations so far have discussed the probabilities of selecting the incorrect signal alternative (symbol), i.e. the symbol-error rate.
When each symbol carries K bits, we need 2K symbols.
Gray coding is used to assigning bits so that the nearest neighbors only differ in one of the K bits. This minimizes the bit-error rate.
011 010 001 Gray-coded 8PSK 110 000
111 100 101
5 April 2017 32 Optimal receiver Bit-error rates (BER)
EXAMPLES: 2PAM 4QAM 8PSK 16QAM
Bits/symbol 1 2 3 4
Symbol energy Eb 2Eb 3Eb 4Eb
2 E 2 E 2 Eb 3 E b, max Q b Q b ~ Q 0.87 ~ Q BER N N 3 N 2 2.25 N 0 0 0 0
Gray coding is used when calculating these BER.
5 April 2017 33 Optimal receiver Bit-error rates (BER), cont.
0 10
-1 2PAM/4QAM 10 8PSK ) 16QAM R -2 E 10 B (
e
t -3
a 10 r
r o r
r -4 10 e - t i B -5 10
-6 10 0 2 4 6 8 10 12 14 16 18 20
ENb /0 [dB]
5 April 2017 34 Optimal receiver
Where do we get Eb and N0?
Where do those magic numbers Eb and N0 come from?
The noise power spectral density N0 is calculated according to
N 0=k T 0 F 0 ⇔ N 0∣dB=−204F 0∣dB
where F0 is the noise factor of the “equivalent” receiver noise source.
The bit energy Eb can be calculated from the received
power C (at the same reference point as N0). Given a certain
data-rate db [bits per second], we have the relation
E b=C /d b ⇔ E b∣dB=C∣dB−d b∣dB
THESE ARE THE EQUATIONS THAT RELATE DETECTOR PERFORMANCE ANALYSIS TO LINK BUDGET CALCULATIONS!
5 April 2017 35 Optimal receiver What about fading channels?
We have (or can calculate) BER expressions for non-fading AWGN channels.
If the channel is Rayleigh-fading, then Eb/N0 will have an
exponential distribution (N0 is assumed to be constant)
-- E /N b b 0 1 −b / b pdf b = e -- average E /N b b b 0
The BER for the Rayleigh fading channel is obtained by averaging: ∞
BERRayleigh b =∫ BERAWGN b × pdf b d b 0
5 April 2017 36 Optimal receiver What about fading channels?
THIS IS A SERIOUS PROBLEM!
Bit error rate (4QAM) 0 10
-1 10 dB Rayleigh fading 10 10 x -2 10
-3 10
-4 10
-5 10 No fading
-6 10 0 2 4 6 8 10 12 14 16 18 20
Eb/N0 [dB]
5 April 2017 37 DIVERSITY ARRANGEMENTS
5 April 2017 38 Diversity arrangements Let’s have a look at fading again
Illustration of interference pattern from above Received power [log scale]
Movement A B
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.
5 April 2017 39 Diversity arrangements The diversity principle
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.
The example given on the previous slide is one such arrangement: antenna diversity.
5 April 2017 40 Diversity arrangements General improvement trend
Bit error rate (4PSK) 0 10
-1 10 dB Rayleigh fading 10 No diversity 10 x -2 10 10 dB -3 10 Rayleigh fading M:th order diversity -4 10 10M x
-5 10 No fading
-6 10 0 2 4 6 8 10 12 14 16 18 20
Eb/N0 [dB]
5 April 2017 41 Diversity arrangements Some techniques
Spatial (antenna) diversity
We will focus on this ... one today! TX Signal combiner
Frequency diversity
D D D
TX Signal combiner
Temporal diversity
Inter- De-inter- Coding leaving leaving De-coding
(We also have angular and polarization diversity)
5 April 2017 42 Spatial (antenna) diversity Fading correlation on antennas
With several Isotropic antennas, we uncorrelated want the fading on scattering. them to be as independent as possible.
E.g.: An antenna spacing of about 0.4 wavelength gives zero correlation.
5 April 2017 43 Spatial (antenna) diversity Selection diversity
RSSI = received signal strength indicator
5 April 2017 44 Spatial (antenna) diversity Selection diversity, cont.
By measuring BER instead of RSSI, we have a better guarantee that we obtain a low BER.
5 April 2017 45 Spatial (antenna) diversity Maximum ratio combining
This is the optimal way (SNR sense) of combining antennas.
5 April 2017 46 Spatial (antenna) diversity
Simpler than MRC, but almost the same performance.
5 April 2017 47 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.
[Fig. 13.10] These curves can be used to calculate fading margins.
5 April 2017 48 Spatial (antenna) diversity Performance comparison, cont.
MRC Comparison of RSSI selection 2ASK/2PSK BER for different number of antennas M and two different diversity techniques.
[Fig. 13.11]
5 April 2017 49 Summary
• Optimal (maximum likelihood) receiver in AWGN channels • Interpretation of received signal as a point in a signal space • Euclidean distances between symbols determine the probability of symbol error • Bit error rate (BER) calculations for some signal constellations • 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.
5 April 2017 50