EE359 Discussion Session 7Adaptive Modulation, MIMO systems
November 15, 2017
EE359 Discussion 7 November 15, 2017 1 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 2 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 3 / 29
Adapt to what?
Idea
Use the estimate of the instantaneous channel state fed back to thetransmitter
Adapt what ?
Power
Modulation (rate)
Coding
QoS (quality of service or BER Pb)
. . .
Problems
Estimate of the channel may be:
Inaccurate (due to receiver measurement error)
Delayed (due to non zero delay of the feedback link)
EE359 Discussion 7 November 15, 2017 4 / 29
Adapt to what?
Idea
Use the estimate of the instantaneous channel state fed back to thetransmitter
Adapt what ?
Power
Modulation
Coding
QoS (quality of service or BER Pb)
. . .
Problems
Estimate of the channel may be:
Inaccurate (due to receiver measurement error)
Delayed (due to non zero delay of the feedback link)
EE359 Discussion 7 November 15, 2017 4 / 29
Adaptive schemes used in many high performance systems
Lots of tunable parameters
Adaptive modulation: 802.11ac (latest commercial Wifi standard)uses BPSK, QPSK, 16 QAM, 64 QAM, 256 QAM
Adaptive power: Used more in multiuser systems (e.g. power controlin CDMA systems) (we are discussing point to point channels here)
Adaptive rate coding: Also used for QoS requirements, e.g. 802.11achas coding rates of 1/2, 2/3, 3/4, 5/6. LTE has 29 distinctmodulation and coding schemes!
Adaptive spatial streams: Different configurations for up to 8× 8MIMO in 802.11ac. 4× 4 for older 802.11 standards and LTE. LTEdoes not achieve more than 3 spatial streams
EE359 Discussion 7 November 15, 2017 5 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 6 / 29
Spectral efficiency and target BER
Spectral efficiency
Simply defined as the number of bits sent on average per unit bandwidth(i.e. Ei[log2(Mi)] bits where an Mi-ary constellation is sent at time i)
Target BER
We can use a Chernoff bound to approximate the BER for MQAMconstellations:
Pb ≤ 0.2e−1.5γM−1
EE359 Discussion 7 November 15, 2017 7 / 29
Spectral efficiency and target BER
Spectral efficiency
Simply defined as the number of bits sent on average per unit bandwidth(i.e. Ei[log2(Mi)] bits where an Mi-ary constellation is sent at time i)
Target BER
We can use a Chernoff bound to approximate the BER for MQAMconstellations:
Pb≈0.2e−1.5γM−1
EE359 Discussion 7 November 15, 2017 7 / 29
An equivalent way of looking at Pb ≈ 0.2e−1.5γM−1
Given SNR γ, the maximum constellation size “supported” for BERPb is
M ≤ 1− 1.5
ln(5Pb)γ
Questions
What happens when K > 1?
Spectral efficiency greater than capacity
Is a bigger K better from a performance standpoint ?
Bigger K means more spectral efficiency for a given power
What does K depend on/how do we improve K ?
Coding, BER, blocklength/latency/delay
Does smaller BER mean smaller or larger K ?
Smaller K
EE359 Discussion 7 November 15, 2017 8 / 29
An equivalent way of looking at Pb ≈ 0.2e−1.5γM−1
Given SNR γ, the maximum constellation size “supported” for BERPb is
M ≤ 1 +Kγ
Questions
What happens when K > 1?
Spectral efficiency greater than capacity
Is a bigger K better from a performance standpoint ?
Bigger K means more spectral efficiency for a given power
What does K depend on/how do we improve K ?
Coding, BER, blocklength/latency/delay
Does smaller BER mean smaller or larger K ?
Smaller K
EE359 Discussion 7 November 15, 2017 8 / 29
An equivalent way of looking at Pb ≈ 0.2e−1.5γM−1
Given SNR γ, the maximum constellation size “supported” for BERPb is
M ≤ 1 +Kγ
Questions
What happens when K > 1?Spectral efficiency greater than capacity
Is a bigger K better from a performance standpoint ?Bigger K means more spectral efficiency for a given power
What does K depend on/how do we improve K ?Coding, BER, blocklength/latency/delay
Does smaller BER mean smaller or larger K ?Smaller K
EE359 Discussion 7 November 15, 2017 8 / 29
Variable rate variable power MQAM
Idea
Maximize spectral efficiency (S.E.) subject to average power constraints
Math
maximizeP (γ) E[log2(M)]s.t. E[P (γ)] = P
Solution
Optimal P (γ) given by
P (γ)/P =
{1/γ0 − 1/(γK) γ ≥ γ0/K0 γ < γ0/K
where γ0 satisfies E[(K/γ0 − 1/γ)1(γ > γ0/K)] = K. Optimal S.E. is
E[log2(Kγ/γ0)1(γ > γ0/K)]
EE359 Discussion 7 November 15, 2017 9 / 29
Variable rate variable power MQAM
Idea
Maximize spectral efficiency (S.E.) subject to average power constraints
Math
maximizeP (γ) E[log2(1 +KP (γ)/P )]s.t. E[P (γ)] = P
Solution
Optimal P (γ) given by
P (γ)/P =
{1/γ0 − 1/(γK) γ ≥ γ0/K0 γ < γ0/K
where γ0 satisfies E[(K/γ0 − 1/γ)1(γ > γ0/K)] = K. Optimal S.E. is
E[log2(Kγ/γ0)1(γ > γ0/K)]
EE359 Discussion 7 November 15, 2017 9 / 29
More details about variable rate variable power MQAM
Comparison of solution with Shannon capacity expression
Same waterfilling ideas work in this case
“Rate” M is now the size of the constellation
Capacity expression if K = 1; does it imply BER for capacity is nonzero?
K represents “power loss” due to BER requirement
Extension of BER constrained schemes
Channel inversion: S.E. is log2(1 +Kσ), where σ = 1/E[1/γ]
Truncated channel inversion: S.E. is (1− P (γ < γ0)) log2(1 +Kσ),where σ = 1/E[1/γ1(γ > γ0)].
But....
Rate cannot be continuous in practice
EE359 Discussion 7 November 15, 2017 10 / 29
More details about variable rate variable power MQAM
Comparison of solution with Shannon capacity expression
Same waterfilling ideas work in this case
“Rate” M is now the size of the constellation
Capacity expression if K = 1; does it imply BER for capacity is nonzero? No!
K represents “power loss” due to BER requirement
Extension of BER constrained schemes
Channel inversion: S.E. is log2(1 +Kσ), where σ = 1/E[1/γ]
Truncated channel inversion: S.E. is (1− P (γ < γ0)) log2(1 +Kσ),where σ = 1/E[1/γ1(γ > γ0)].
But....
Rate cannot be continuous in practice
EE359 Discussion 7 November 15, 2017 10 / 29
More details about variable rate variable power MQAM
Comparison of solution with Shannon capacity expression
Same waterfilling ideas work in this case
“Rate” M is now the size of the constellation
Capacity expression if K = 1; does it imply BER for capacity is nonzero?
K represents “power loss” due to BER requirement
Extension of BER constrained schemes
Channel inversion: S.E. is log2(1 +Kσ), where σ = 1/E[1/γ]
Truncated channel inversion: S.E. is (1− P (γ < γ0)) log2(1 +Kσ),where σ = 1/E[1/γ1(γ > γ0)].
But....
Rate cannot be continuous in practice
EE359 Discussion 7 November 15, 2017 10 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 11 / 29
Continuous rate continuous power MQAM
Problem
maxP (γ) E[log2(M(γ))]s.t. E[P (γ)] = P , Pb(γ) ≤ Pb
The solution
Use waterfilling
M(γ) = γ/γ∗K
M1
M2
M3
γ1 γ2 γ3
Figure: M versus γ
EE359 Discussion 7 November 15, 2017 12 / 29
Discrete rate variable power MQAM
Problem
maxP (γ),M(γ) E[log2(M(γ))]s.t. E[P (γ)] = P ,M(γ) ∈M, Pb(γ) ≤ Pb
Ideas to approximate solution
Given available constellations M = {M1, . . . ,Mn}, choose aconstellation which meets the Pb target for a given channel state γ
Reduce transmit power to keep BER constant if channel is good
M(γ) = γ/γ∗K
M1
M2
M3
γ1 γ2 γ3
Figure: M versus γ
EE359 Discussion 7 November 15, 2017 13 / 29
Discrete rate variable power MQAM
Problem
maxP (γ),M(γ) E[log2(M(γ))]s.t. E[P (γ)] = P ,M(γ) ∈M, Pb(γ) ≤ Pb
Ideas to approximate solution
Given available constellations M = {M1, . . . ,Mn}, choose aconstellation which meets the Pb target for a given channel state γ
Reduce transmit power to keep BER constant if channel is good
M(γ) = γ/γ∗K
M1
M2
M3
γ1 γ2 γ3
Figure: M versus γ
EE359 Discussion 7 November 15, 2017 13 / 29
Power adaptation
Idea
Assuming that Pb ≈ 0.2e−1.5γM−1 , for constant M , Pb ↓ as γ ↑
To conserve power simply reduce power if channel state is good!
(M1 − 1)/K
(M2 − 1)/K
(M3 − 1)/K
γ1 γ2 γ3
Figure: γP (γ)/P versus γ
Resultant spectral efficiency (S.E.)
R/B =∑
i log2(Mi)p(γi ≤ γ < γi+1)
EE359 Discussion 7 November 15, 2017 14 / 29
Power adaptation
Idea
Assuming that Pb ≈ 0.2e−1.5γM−1 , for constant M , Pb ↓ as γ ↑
To conserve power simply reduce power if channel state is good!
(M1 − 1)/K
(M2 − 1)/K
(M3 − 1)/K
γ1 γ2 γ3
Figure: γP (γ)/P versus γ
Resultant spectral efficiency (S.E.)
R/B =∑
i log2(Mi)p(γi ≤ γ < γi+1)
EE359 Discussion 7 November 15, 2017 14 / 29
Power adaptation
Idea
Assuming that Pb ≈ 0.2e−1.5γM−1 , for constant M , Pb ↓ as γ ↑
To conserve power simply reduce power if channel state is good!
(M1 − 1)/K
(M2 − 1)/K
(M3 − 1)/K
γ1 γ2 γ3
Figure: P (γ)/P versus γ
Resultant spectral efficiency (S.E.)
R/B =∑
i log2(Mi)p(γi ≤ γ < γi+1)
EE359 Discussion 7 November 15, 2017 14 / 29
And so on . . .
What we have presented so far is achievable, but not necessarilyoptimal
Reduction in optimization complexity achieved by using insights fromwaterfilling solutions
Can also use insights from channel inversion/truncated channelinversion
Can extend ideas to discrete power, discrete rate
EE359 Discussion 7 November 15, 2017 15 / 29
Homework 6
Problem 11 Use the adaptive rate and power adaptation assuming continuousM(γ) for first two parts
2 Use the truncated channel inversion formulation for the last part(again using continuous modulation assumption)
Problem 21 Use the fact that the number of constellation points is finite;
maximize spectral efficiency over all possible γ0 (dictated by theavailable constellations).
2 For variable rate-variable power schemes, need to compute regionwhen constellation is transmitted - gives spectral efficiency.
EE359 Discussion 7 November 15, 2017 16 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 17 / 29
Using adaptive schemes
Idea
The channel cannot change too fast!
Valid when time that channel “stays” in a particular state is muchhigher than several symbol times
Calculation in terms of level crossing rates with Markov assumption(with jump only between adjacent regions)
R0 R1 . . .
Figure: Markov assumption (R0 = {γ : γ < γ0}, R1 = {γ : γ0 ≤ γ < γ1}, . . . )
EE359 Discussion 7 November 15, 2017 18 / 29
Errors in channel estimation or due to delay
Basic effect
True value γ, but our estimate is γ; what is the error ?
Some expressions
Can be characterized by p(γ, γ)
Approximation for Pb gives
Pb(γ, γ) ≈ 0.2[5Pb]γ/γ
(depends only on ε = γ/γ!)
Question
What happens if γ > γ? (in terms of power and BER ?)
EE359 Discussion 7 November 15, 2017 19 / 29
Homework 6
Problem 3 hint
The BER under the approximation we saw depends only on the statisticsof ε (not the joint statistics of γ and γ)
Problem 4
For part a, modulation used when Pb(γ) ≤ Pfloor.
For part b, yt, ht, ht+1 are jointly gaussian. yt = xtht + nt. Can beused to find MMSE estimate ht = E[ht|yt, xt]For part c, ht+1 = aht +
√1− a2nt,2 for appropriate a (Jakes’
scattering). Can be used to find MMSE estimate.
Pb(|ht+1|2, |ht+1|2) is the probability of error of modulation scheme(chosen by estimate |ht+1|2) with SNR |ht+1|2. ht+1 is a function ofyt. Need to integrate to find mean over joint distribution of (yt, ht+1).
EE359 Discussion 7 November 15, 2017 20 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 21 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 22 / 29
Representing MIMO systems
Assumptions
Narrowband signals
Nt transmit antennas
Nr receive antennas
Noise n is zero mean with a covariance matrix of σ2INr
Idea
Represent gain from transmitter antenna j to receiver antenna i by hi,j
...
...
...
...
Tx antenna j Rx antenna ihi,j
EE359 Discussion 7 November 15, 2017 23 / 29
System model
Model y1...yNr
︸ ︷︷ ︸
y
=
h1,1 . . . h1,Nt...
. . ....
hNr,1 . . . hNr,Nt
︸ ︷︷ ︸
H
x1...xNt
︸ ︷︷ ︸
x
+
n1...nNr
︸ ︷︷ ︸
n
where x is what transmitter sends and y is what receiver sees
Transmit power constraint
E[xx∗] =∑Nt
i=1E[|xi|2] ≤ ρ
EE359 Discussion 7 November 15, 2017 24 / 29
Decomposition of H
Idea
Use the singular value decomposition (SVD) of channel matrix
H = UΣVH
Parallel channel decomposition
Transmitter sends x = Vx (transmit precoding)
Receiver obtains y = UHy (receiver shaping)
y = Σx
...
...
xi yi
σi
Figure: Equivalent parallel channels (no “crosstalk” or interchannel interference)
Question
How many such channels are there?
EE359 Discussion 7 November 15, 2017 25 / 29
Outline
1 Adaptive transmit schemesVariable rate variable powerDiscrete rate variable powerChannel estimation errors
2 MIMO systemsRepresentationCapacity
EE359 Discussion 7 November 15, 2017 26 / 29
Channel capacity with Tx and Rx CSI
Expression
Under system model, can be shown to be
C = maxRx:Tr(Rx)≤ρ
B log2∣∣I + HRxHH
∣∣Equivalent expression
By using parallel decomposition, we get
C = maxρ:∑i ρi≤ρ
B∑i
log2(1 + ρiσ2i )
Question
How do you acheive this?
Waterfill!
EE359 Discussion 7 November 15, 2017 27 / 29
Channel capacity with Tx and Rx CSI
Expression
Under system model, can be shown to be
C = maxRx:Tr(Rx)≤ρ
B log2∣∣I + HRxHH
∣∣Equivalent expression
By using parallel decomposition, we get
C = maxρ:∑i ρi≤ρ
B∑i
log2(1 + ρiσ2i )
Question
How do you acheive this? Waterfill!
EE359 Discussion 7 November 15, 2017 27 / 29
Channel capacity with Rx CSI only
Idea
Cannot use the channel realization at the transmitter; so spread energyequally at all the transmitters
Capacity expression
C = maxRx:Rx=ρ/NtINt
B log2∣∣I + HRxHH
∣∣Equivalent expression
C =∑i
B log2(1 + ρσ2i /Nt)
EE359 Discussion 7 November 15, 2017 28 / 29
Homework 6
Problem 6
For part a, use symmetry, Jensen’s inequality or AM-GM inequality
For part b, use Hadamard’s inequality
Problem 7
Use the law of large numbers and note that Mt is fixed
EE359 Discussion 7 November 15, 2017 29 / 29