Morning Session Capacity-based Power Control Şennur Ulukuş Department of Electrical and Computer Engineering University of Maryland
So Far, We Learned... Power control with SIR-based QoS guarantees Suitable for delay-intolerant services, e.g., voice Satisfy all SIR constraints with minimum transmit power Leading to energy-efficient communications Power control for capacity Suitable for delay-tolerant services, e.g., data Maximize rate with a given average transmit power Equivalently, support a given rate with minimum power Leading to energy-efficient communications
Capacity-Based Power Control Fading: random fluctuations in channel gains. Perfect CSI at both the transmitter and the receiver Maximize ergodic capacity subject to average power constraints Main operational difference: QoS based power control: compensate for channel fading Capacity-based power control: exploit the channel fading
Channel Fading r = hx + n 2 Change in Channel State as a Function of Time 6 Distribution of the States Assumed by the Channel.8.6 5 Fading Level.4.2.8.6 Density of Fading Levels (States) 4 3 2.4.2..2.3.4.5.6.7.8.9 Time.2.4.6.8.2.4.6.8 2 Fading Levels (States)
Channel Fading r = hx + n 4 Change in Channel State as a Function of Time 2 Distribution of the States Assumed By the Channel 2 Fading Level 8 6 4 Density of Fading Levels (States) 8 6 4 2 2 2 3 4 5 6 7 8 9 Time 2 4 6 8 2 4 Fading Levels (States)
Single-User Fading Channel (Goldsmith-Varaiya 94) Channel capacity for single user C = log ( + SNR) 2 = ( 2 log + p ) σ 2 In the presence of fading, the capacity for a fixed channel state h, C(h) = ( 2 log + p(h)h ) σ 2 Ergodic (expected) capacity under an average power constraint [ ( max E h p(h) 2 log + p(h)h )] σ 2 s.t. E h [p(h)] P p(h), h
Optimal Power Allocation using Waterfilling The Lagrangian function ( E h [log + p(h)h )] λ(e σ 2 h [p(h)] P) + µ(h)f (h)dh Optimality conditions where f (h) is the PDF of h. h p (h)h + σ + µ(h) 2 f (h) = λ The complementary slackness conditions µ(h)p (h) = for all h. If p (h) >, we get Otherwise, p (h) =. p (h) = λ σ2 h
Optimal Power Allocation using Waterfilling Optimal power allocation: waterfilling of power over time p(h) = ( ) + λ σ2 h Illustration of Waterfilling Over Inverse of the Channel States 3 Single User Optimum Power Allocation vs Fading States 9 8 2.5 7 2 6 σ 2 /h 5 p(h).5 4 3 2.5..2.3.4.5.6.7.8.9 h..2.3.4.5.6.7.8.9 h
Optimal Power Allocation using Waterfilling Optimal power allocation: waterfilling of power over time p(h) = ( ) + λ σ2 h 3 Single User Optimum Power Allocation vs Fading States 2.5 2 p(h).5.5..2.3.4.5.6.7.8.9 h
Optimal Power Allocation using Waterfilling Optimal power allocation: waterfilling of power over time p(h) = ( ) + λ σ2 h 3 Single User Optimum Power Allocation vs Fading States 2.5 2 p(h).5.5..2.3.4.5.6.7.8.9 h
Optimal Power Allocation using Waterfilling Optimal power allocation: waterfilling of power over time p(h) = ( ) + λ σ2 h 3 Single User Optimum Power Allocation vs Fading States 2.5 2 p(h).5.5..2.3.4.5.6.7.8.9 h
Optimal Power Allocation using Waterfilling Optimal power allocation: waterfilling of power over time p(h) = ( ) + λ σ2 h 3 Single User Optimum Power Allocation vs Fading States 2.5 2 p(h).5.5..2.3.4.5.6.7.8.9 h
Differences Between QoS-Based and Capacity-Based Power Control Single-user system SIR-based y = hx + n p(h)h γ p(h) = γσ2 σ 2 h Channel inversion; more power if bad channel, less if good channel Compensate for channel fading via power control Capacity-based [ ( max E h 2 log + p(h)h )] σ 2 p(h) = ( ) + λ σ2 h Waterfilling; more power if good channel, less if bad channel Exploit variations, opportunistic transmission
Fading Gaussian Multiple Access Channel user x h y user 2 x 2 h 2 n Channel model y = h x + h 2x 2 + n Simultaneously achievable ergodic rates for both users (R, R 2) Channel state vector h = (h, h 2) Adapt powers as functions of h: p (h) and p 2(h)
R Fading Gaussian Multiple Access Channel (Tse-Hanly 98) Union of pentagons [ R < E 2 log [ R 2 < E 2 log [ R + R 2 < E 2 log ( + p(h)h σ 2 ( + p2(h)h2 ( + over all feasible power distributions σ 2 )] )] ( C ) ( C 2) )] p(h)h + p2(h)h2 σ 2 ( C s) E [p (h)] P, E [p 2(h)] P 2, p (h), p 2(h) R 2
Determining the Boundary of the Capacity Region R 2 (R *, R* 2 ) µ + µ 2 R 2 = C * R R Capacity region is a convex region. To determine the boundary, maximize µ R + µ 2R 2 for all µ, µ 2. Any (R, R2 ) on the boundary is a corner of one of the pentagons. If µ 2 > µ then the upper corner; if µ > µ 2 then the lower corner.
Achieving Arbitrary Rate Tuples on the Boundary For given µ i, maximize C µ µ R + µ 2R 2 s.t. E h [p i (h)] P i, R C. Wlog, µ 2 > µ. Given power policy, the optimum R is the upper corner. The coordinates of the upper corner are: and R 2 = C 2, R = C s C 2 C µ = µ (C s C 2) + µ 2C 2 = (µ 2 µ )C 2 + µ C s R 2 (R *, R* 2 ) µ + µ 2 R 2 = C * R R
Achieving Arbitrary Rate Tuples on the Boundary Therefore, the optimum power allocation policy is the solution of: ( ) ( )] max E h [(µ 2 µ ) log + p2(h)h2 p(h)h + p2(h)h2 + µ p(h) σ 2 log + σ 2 s.t. E h [p i (h)] P i, i =, 2 p i (h), h, i =, 2 Objective function is concave, and constraint set is convex in powers. R 2 (R *, R* 2 ) µ + µ 2 R 2 = C * R R
Optimality Conditions p (h) achieves the global maximum of C µ iff it satisfies the KKTs, µ h h p (h) + h 2p 2(h) + σ 2 λ, µ h 2 (µ2 µ)h2 + h p (h) + h 2p 2(h) + σ2 h 2p 2(h) + σ λ2, 2 h h with equality at h, if p (h) > and p 2(h) >, respectively. Solution based on utilities in Tse-Hanly 98. Solve KKTs iteratively: generalized iterative waterfilling (Kaya-Ulukus)
Generalized Iterative Waterfilling Given p 2(h), find p (h) and λ such that µ h h p (h) + h 2p 2(h) + σ 2 λ, h with equality at h, if p (h)>. Waterfilling for user given the power of user 2 contributing to noise p (h) = Given p (h), find p 2(h) and λ 2. ( ) + σ2 + h 2p 2(h) λ h µ h 2 (µ2 µ)h2 + h p (h) + h 2p 2(h) + σ2 h 2p 2(h) + σ λ2, 2 h with equality at h, if p 2(h) >. A second order equation to solve.
Optimal Power Allocation via Generalized Iterative Waterfilling KKT conditions for the K-user case k (µ i µ i ) h k i j= p j(h)h j + p k (h)h k + σ λ k, h, k =,, K 2 i= with equality at h, if p k (h) >. Generalized iterative waterfilling Given p j (h), j < k, find p k (h) and λ k such that k (µ i µ i )h k i i= j= p j(h)h j + p k (h)h k + σ λ k, h, k =,, K 2 with equality at h, if p k (h) >. One-user-at-a-time algorithm, converges to the optimum. A Gauss-Seidel type of iteration.
Special Case: Sum Capacity (Knopp-Humblet 95) Ergodic sum capacity (µ = µ 2 = ) ( E h [log + max p(h) KKT conditions )] p(h)h + p2(h)h2 σ 2 s.t. E h [p i (h)] P i, p i (h), i =, 2 h p (h)h + p 2(h)h 2 + σ 2 λ, h h 2 λ2, h, p (h)h + p 2(h)h 2 + σ2 with equality at h, if p (h) > and p 2(h) >, respectively. For both users to transmit simultaneously at channel state h = (h, h 2), h h 2 = λ λ 2 For continuous channel gains, this is a zero-probability event. Only the strongest (after some scaling) user transmits at any given time.
Closed-Form Solution Single-user waterfilling over favorable channel states { ( ) + p k (h) = λ k σ2 h k, if hk /λ k > h j /λ j, j k, otherwise Power Distribution of User Power Distribution of User 2 3 3 2.5 2.5 2 2 p.5 p 2.5.5.8.6.4 h.2.2.4 h 2.6.8.5.8.6 h.4.2.2.4.6 h 2.8
Issue of Simultaneous Transmissions Sum capacity achieving pentagon: a rectangle (orthogonal transmissions) R 2 Transmit Regions for Identical Sequences (R *, R* 2 ).9.8.7 User.6 R + R 2 = C * h.5.4.3.2 User 2. None This result is specific to sum capacity and scalar channel. R..2.3.4.5.6.7.8.9 h 2
Differences Between QoS-Based and Capacity-Based Power Control in MAC SIR-based p ( h γ γ p p p 2h 2 + σ2 2h 2 + σ 2) I (p) h p ( 2h 2 γ2 γ2 p2 p p h + σ2 h + σ 2) I 2(p) h 2 p > I (p) p 2 > I 2 (p) p γ 2σ 2 h 2 p γ 2σ 2 h 2 γ σ 2 h γ σ 2 h Both users transmit simultaneously More power if bad channels, less if good channels.
Differences Between QoS-Based and Capacity-Based Power Control in MAC Sum-capacity-based Power Distribution of User Power Distribution of User 2 p 3 2.5 2.5.5 p 2 3 2.5 2.5.8.6.4 h.2.2.4.6 h 2.8.5.8.6 h.4.2.2.4.6 h 2.8 User with the stronger channel transmits Stronger user transmits with more power at better channels Multi-user diversity, multi-user opportunistic transmission
Single-User MIMO Channel (Telatar 99) x y x 2 y 2 x t Tx H y r Rx Channel gain matrix H: r t matrix Transmitted signal t-dim. vector x and received vector r-dim. vector y y = Hx + n n is i.i.d. zero-mean Gaussian noise vector with equal variance H is deterministic or random and known perfectly at the receiver Average power constraint E[x T x] P
Single-User MIMO Channel (Telatar 99) Use singular value decomposition to express H as H = UDV T U and V are unitary matrices, D is a diagonal matrix of singular values Diagonal entries of D are square roots of the eigenvalues of HH T Columns of U are the normalized eigenvectors of HH T Columns of V are the normalized eigenvectors of H T H
Single-User MIMO Channel (Telatar 99) Obtain an equivalent channel y = UDV T x + n Let x = V T x, ỹ = U T y and ñ = U T n ỹ = D x + ñ ñ is also i.i.d. zero-mean Gaussian Equivalent channel: min{r, t} parallel channels with (squared) gains d i ỹ i = d i x i + ñ i N(, ) x d ỹ N(, ) x 2 d2 ỹ 2 N(, ) x i di ỹ i
Single-User MIMO Channel (Telatar 99) Independent signalling is optimal over parallel channels Let P i E[ x 2 i ], power over the ith parallel channel. MIMO capacity min{r,t} max 2 log ( + d ip i ) i= min{r,t} s.t. i= P i P Optimal power allocation: waterfilling ( P i = λ ) + d i
MIMO Multiple Access Channel H Tx Tx 2 H 2 Rx Tx k H k The received vector at the receiver, y = H x + H 2x 2 + n H and H 2 are r t matrices Additive noise n is i.i.d. Gaussian with covariance I
Capacity Region of MIMO MAC Channel (Yu-Rhee-Boyd-Cioffi ) Q = E[x x T ] and Q 2 = E[x 2x T 2 ] are covariances Transmit power constraints tr(q ) P, tr(q 2) P 2 For fixed Q, Q 2, define B(Q,Q 2) R HQ 2 log H T + I R 2 H2Q 2 log 2H T 2 + I R + R 2 HQ 2 log H T + H 2Q 2H T 2 + I The capacity region is C = tr(q i ) P i B(Q,Q 2)
Boundary of the Capacity Region of MIMO MAC (Yu-Rhee-Boyd-Cioffi ) Solve the following wlog for µ µ 2 max µ log H Q H T + H 2Q 2H T 2 + I + (µ 2 µ ) log H 2Q 2H T 2 + I Q,Q 2 s.t. tr (Q ) P, tr (Q 2) P 2 R 2 A B C slope = µ µ D R
Sum Capacity of the MIMO MAC Channel (Yu-Rhee-Boyd-Cioffi ) Maximize sum rate max Q,Q 2 log H Q H T + H 2Q 2H T 2 + I s.t. tr (Q ) P, tr (Q 2) P 2 Sum rate optimal allocation Q and Q 2 Necessary condition: Q is the single user waterfilling over the colored noise H 2Q 2 H T 2 + I. Q 2 is the single user waterfilling over the colored noise H Q H T + I. Multi-dimensional water-filling:
Iterative Waterfilling (Yu-Rhee-Boyd-Cioffi ) Perform single-user optimizations: Iterative waterfilling log H Q H T + H 2Q 2H T 2 + I }{{} effective colored noise Given Q 2, user waterfills over the effective colored noise and updates Q Q = arg max log Q HQHT + H 2Q 2H T 2 + I Given Q, user 2 waterfills over the effective colored noise and updates Q 2 Q 2 = arg max log H 2Q 2H T 2 + H Q H T + I Q 2
Iterative Waterfilling (Yu-Rhee-Boyd-Cioffi ) An illustration of the trajectory followed during the iterative waterfilling R 2 A B C D R
Conclusions So Far Power adaptation for maximizing the capacity in single-user and MAC fading channels, and MIMO and MIMO MAC channels Common tool: waterfilling Fading channel is equivalent to parallel channels over channel states. MIMO channel is equivalent to min{t, r} parallel channels. Fading scalar and MIMO multiple access channels Sum-rate optimal operating points are reached by iterative waterfilling. Any arbitrary point is reached by generalized iterative waterfilling.
Power Adaptation for Energy Minimal Transmission (Uysal-Biyikoglu, Prabhakar, El Gamal 2) Rate-power relation in the AWGN channel R = 2 log ( + P σ 2 ) Energy-per-bit (E pb ) in an AWGN channel E pb = σ2 (2 2R ) R E pb monotonically decreases as R E pb σ 2 2ln(2) q = R
Power Adaptation for Energy Minimal Transmission Serving bits with slower rates is more energy-efficient. q: Number of transmissions to send a bit Codebook of fixed but sufficiently large block length. Code rate is adapted by changing its average power. q R Given B bits at the transmitter, decreasing transmit power yields longer transmission durations smaller transmission energy This is true for any system with a concave rate-power relation. E pb σ 2 2ln(2) q = R
Energy Minimal Packet Scheduling b b 2 b 3 b 4 t t 2 T t 3 t 4 Packets arriving at different times, deadline constrained by T Scheduling packets: τ i is the time allocated for packet i b b 2 b 3 b 4 τ τ 2 τ 3 τ 4 T Deadline constraint τ i T i Energy of a schedule τ: ω(τ) Find the energy minimal schedule τ to send all packets by T Adapt the transmission times of the packets. Equivalently, adapt the rate and hence the power.
Offline Packet Scheduling Offline schedule: bit arrival times are known a priori Bits cannot be served before they arrive at the data buffer: causality A necessary condition for optimality in b i = b case: τi τi+ with τi = T The necessity is due to the convexity of ω(τ). Assume τ i < τ i+ for some i. This is a contradiction since σ i = σ i+ = τ i+τ i+ 2 and σ j = τ j elsewhere ω(σ) ω(σ) = ω(τ i ) + ω(τ i+ ) ω(σ i ) ω(σ i+ ) i = ω(τ i ) + ω(τ i+ ) 2ω( τ i + τ i+ ) < 2 Equate τ i subject to feasibility constraints.
Optimal Offline Packet Scheduling A schedule with idle intervals is suboptimal. b b 2 = Not optimal τ τ idle 2 idle T Equate τ i subject to feasibility constraints. Split the transmission times to the available deadline as much as possible. b = b 2 = B b b 2 = Less equalized τ τ 2 T b b 2 = More equalized τ τ 2 T
Optimal Offline Packet Scheduling The optimal policy has the structure: τ i = b i max k {,...,M k i } k i = k i + arg k j= d k i +j max k j= b j k {,...,M k i } k j= d k i +j k j= b j Optimal offline schedule is called lazy scheduling. The lazy schedule start slowly and work harder as the deadline approaches Key reason: convexity of energy per bit in transmission time.
A Calculus Approach for Energy Minimal Packet Scheduling (Zafer-Modiano 9) Uses a geometric framework Let h(t) be the last time a packet arrived before t h(t) i= B i is the cumulative data arrival curve A policy is feasible if its cumulative service curve lies below h(t) i= B i The optimal rate policy is the tightest string. B B B 2 B 3 T h(t) i= Bi r 4 r 3 r 2 r s s 2 s 3 T
Algorithm to Find the Optimal Policy Connect the points at data arrival times to form lines Select the line which is feasible and which has minimum slope Deciding r and r 2 Deciding r 3 h(t) i= Bi h(t) i= Bi selected line selected line T T s s2 s3 s s2 s3 h(t) i= Bi r 4 r 3 r 2 r T s s2 s3
Conclusions Energy per bit monotonically decreases as increases, i.e., R decreases. R The slower the transmission, the more energy-efficient it is. Scheduling packets that arrive at different times. Optimal offline schedule has a majorization structure. Serve bits with a rate as constant as possible subject to bit feasibility.
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