ARC Future Fellow at The University of Melbourne Sessional Lecturer at Monash University
August 15, 2022
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 1 / 24
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ECE5884 Wireless Communications Week 4: Capacity of Wireless Channels
Course outline
This week: Ref. Ch. 4 of [Goldsmith, 2005]
● Week 1: Overview of Wireless Communications
● Week 2: Wireless Channel (Path Loss and Shadowing)
● Week 3: Wireless Channel Models
● Week 4: Capacity of Wireless Channels
● Week 5: Digital Modulation and Detection
● Week 6: Performance Analysis
● Week 7: Equalization
● Week 8: Multicarrier Modulation (OFDM)
● Week 9: Diversity Techniques
● Week 10: Multiple-Antenna Systems (MIMO Communications) ● Week 11: Multiuser Systems
● Week 12: Guest Lecture (Emerging 5G/6G Technologies)
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 2 / 24
Base sta on
Home internet plans
Uplink 17 50 Mbps
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 3 / 24
Capacity in AWGN
1 For a discrete-time AWGN channel, h = 1, input/out relationship at time i:
y[i] = x[i] + n[i] (1)
where x[i] is the channel input, y[i] is the corresponding channel
output, and n[i] is a white Gaussian noise random process.
C=Blog2(1+γ)bits/sec where SNR=γ= Pr (2)
3 Shannon proved that capacity is the maximum error-free data rate a channel can support.
4 So this theoretical limit may not be achievable.
5 Only dependent on channel characteristic, but not dependent on
design techniques.
6 C [bits/sec]; Pr received signal power [W]; N0 Noise power spectral
density [W/Hz]; B channel bandwidth [Hz].
7 Any code with rate R > C has a probability of error bounded away
from zero.
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 4 / 24
Capacity at asymptotic regimes
1 Large bandwidth regime:
lim Blog2(1+ Pr )=log2(e)Pr ; as limxlog2(1+a)=alog2(e) (3)
B→∞ BN0 N0x→∞ x
● There is not sufficient power to spread over the large amount of bandwidth available.
● Capacity no longer depends on the channel bandwidth.
2 Low power regime:
Blog2(1+ Pr )≈log2(e)Pr;asloge(1+x)≈xforsmallx (4)
● Capacity no longer depends on the channel bandwidth.
3 How capacity scales with bandwidth (B to kB for k > 1) at high power:
kBlog(1+Pr ) kBlog(Pr )
lim 2 kBN0 ≈ lim 2 kBN0 =k
Pr→∞ Blog (1+ Pr ) Pr→∞ Blog (Pr ) 2 BN0 2BN0
4 C is also defined as channel’s maximum mutual information (Not discussing here!) – Information theoretic perspective …
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022
Capacity of flat fading channels
Figure 1: Flat fading channel and system model.
1 Time-varying channel gain: ∣h∣ = √g and ∣h∣2 = g.
2 The channel gain g[i], called as channel state information or channel
side information (CSI), can change at each time i, e.g., an i.i.d. process.
3 Block fading channel: g[i] is constant over some blocklength T, after
which time g[i] changes to a new independent value based on the
distribution fg(t).
4 The instantaneous received SNR:
P ̄ g [ i ] N0B
; Expected value:γ ̄ =
P ̄ g ̄ N0B
; P ̄ is the average Tx power (6)
5 The CSI g[i] changes during the transmission of the codeword.
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 6 / 24
CSI Knowledge
The capacity depends on what is known about g[i] at the transmitter (Tx) and receiver (Rx) [Goldsmith, 1997, Goldsmith, 1999, Caire, 1999].
1 Channel distribution information (CDI): The distribution of g[i] is known to the Tx and Rx.
2 Rx CSI: The value of g[i] is known to the receiver at time i, and both the TX and Rx know the distribution of g[i].
3 TxandRxCSI:Thevalueofg[i]isknowntotheTxandRxattimei, and both the Tx and Rx know the distribution of g[i].
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Channel estimation for CSI
Figure 2: GSM (2G) frame structure. With Least Squares (LS) estimator:
For a single bit training: y = h s + n ⇒ hˆ = y s
For m bits training: y = h s + n ⇒ hˆ = (sTs) sTy
where y = [y1,⋯,ym]T;s = [s1,⋯,sm]T;n = [n1,⋯,nm]T ECE5884 Wireless Communications @ Monash Uni. August 15, 2022
Channel estimation error/Imperfect CSI
Due to pilot contamination from neighboring cells, channel frequency offset, Doppler effect, time synchronization mismatch, etc.
h = hˆ + ε where ε is the channel estimation error (9) If h ∼ CN (0, σh2) and ε ∼ CN (0, σε2), then (Gaussian still Gaussian!!!)
hˆ ∼ C N ( 0 , σ 2 − σ 2 ) hε
SNR with channel estimation error:
y = hs + n
hˆ∗y = hˆ∗(hˆ + ε)s + hˆ∗n
+ hˆ∗n (13)
Received signal power
Est. error power+ Noise power
ECE5884 Wireless Communications @ Monash Uni.
∣hˆ∣2P ̄ (14) σε2P ̄ + N0
= ∣hˆ∣2s +
(hˆ∗s)ε
channel estimation error
August 15, 2022
CSI at receiver
● Shannon (ergodic) capacity: Shannon capacity for an AWGN channel with SNR γ, and then averaged over the distribution of γ.
Blog2(1+γ)fγ(γ)dγ (15)
= Eγ [B log2(1 + γ)] (16)
≤ B log2 (1 + Eγ [γ]) ; using Jensen’s inequality (17) = B log2 (1 + γ ̄) ; γ ̄ is the average SNR on the channel (18)
● CAWGN = B log2 (1 + γ ̄) is equivalent to the capacity of AWGN channel with the average SNR γ ̄.
● Ergodic capacity over Rayleigh fading channel: From Eq. (15),
1γB11 Blog2(1+γ) e−γ ̄ dγ=− eγ ̄ Ei(− )
CRayleigh =∫ 0
γ ̄ log(2) γ ̄ where Ei (x ) is the Exponential Integral, Ei (x ) = − ∞ e−t dt .
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022
Numerical example
Figure 3: Capacity (bits/sec/Hz) vs average SNR (dB) over AWGN and Rayleigh
channels for B = 1 Hz.
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Capacity outage probability
● The capacity outage probability is the probability that the (instantaneous) capacity C falls below a certain predetermined capacity threshold Cth
Similarly, you can evaluate the capacity outage probabilities for Rician and Nakagami-m fading channels!
Cout =Pr[C
∞ P(γ)fγ(γ)dγ=P ̄ ∫0
∫ Blog2 (1+ 0
)fγ(γ)dγ (27)
ECE5884 Wireless Communications @ Monash Uni.
August 15, 2022
Optimization problem: power allocation
Blog2 (1+ ̄ )fγ(γ)dγ
subject to ∫ 0
P(γ)fγ(γ)dγ ≤ P ̄ P(γ) > 0
Optimal power allocation
1 Lagrangian
J (P(γ),λ) = ∫ B log2 (1 + ̄ )fγ(γ)dγ − λ(∫
∞ P(γ)γ ∞ 0P0
P(γ)fγ(γ)dγ − P ̄) 2 Differentiate the Lagrangian and set the derivative equal to zero:
∂J(P(γ),λ) ∞ B γ ̄ =∫ P
fγ(γ)dγ−λ∫ fγ(γ)dγ=0 (29) 0
∂P(γ) 0 B γ ̄
ln(2) (1 + P(γ)γ ) P ̄
−λ=0⇒(1+P(γ)γ)= B γ (30) P ̄ ln(2) λP ̄
P ln(2) (1 + P(γ)γ )
3 Solution:
where γ0 is a cutoff/threshold value.
ECE5884 Wireless Communications @ Monash Uni.
P(γ)=1−1 where γ0= B (31)
P ( γ ) ⎧⎪ ⎪ 1 − 1 ; γ ≥ γ 0 = ⎨⎩γ0 γ
P ̄ ⎪0; γ<γ0
August 15, 2022
Optimal power allocation: water-filling
P ( γ ) ⎧⎪ ⎪ 1 − 1 ; γ ≥ γ 0
̄ = ⎨⎪γ0 γ then
)fγ(γ)dγ (33)
P ⎪⎩0; γ<γ0
Figure 7: Water-filling technique [Goldsmith, 2005]. The better channel, the more power and the higher data rate!
The line 1/γ sketches out the bottom of a bowl, and power is poured into the bowl to a constant water level of 1/γ0.
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 18 / 24
Calculate γ0!
● γ0 is found from the power constraint.
● Since using the maximum available power will always be optimal, we
have eq. (25) as
∫0 P ̄ fγ(γ)dγ=1 (34)
∞(1 −1)f (γ)dγ=1 (35) ∫γ0 γ0 γ γ
The value for γ0 must be found numerical integration and iteration, because no closed-form solutions exist for typical continuous distributions fγ (γ) [Goldsmith, 1999].
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Numerical Example (from the text book) [Goldsmith, 2005]
Received SNR γi
γ1 = 0.8333 γ2 = 83.33 γ3 = 333.33
Probability p(γi )
0.1 0.5 0.4
Table 1: Parameters
When B = 30 kHz, find
1 the capacity for AWGN channel
2 the ergodic capacity assuming that only Rx has CSI
3 the ergodic capacity assuming that both Tx and Rx have CSI.
Average SNR γ ̄ = ∑3i =1 γi p(γi ) = 175.08
1 the capacity for AWGN channel: C = B log2(1 + γ ̄) = 223.80 kbits/sec
2 the ergodic capacity assuming that only Rx has CSI:
C = ∫ Blog2 (1+γ)fγ(γ)dγ = ∑Blog2 (1+γi)p(γi) = 199.26kbits/sec.
3 the ergodic capacity when both Tx and Rx have CSI (next pages).
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● Calculate γ0: Use average power constraint in eq. (35)
∞(1 −1)f (γ)dγ=1⇒∑(1 − 1)p(γ)=1
Again check for γi < γ0. Now all SNRs satisfy γi ≥ γ0 condition! So start power allocation
∫γ0 γ γ γ γ γ i 0 i0i
0.1(1 − 1 )+0.5(1 − 1 )+0.4(1 − 1 )=1 γ0 0.833 γ0 83.33 γ0 333.33
γ0 = 0.887116
● Neglect channel’s SNR with γi < γ0, i.e., no data is transmitted over the ith time interval. In this case we avoid γ1.
● Again calculate γ0 for the remaining channel’s SNRs: 0.5(1− 1 )+0.4(1− 1 )=1
γ0 83.33 γ0 333.33 γ0 = 0.893566
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● The ergodic capacity assuming that both Tx and Rx have instantaneous CSI: The capacity of the channel can be evaluated by using Eq. (33) as
C=∫ Blog2( )fγ(γ)dγ=∑Blog2( )p(γi) (36)
γ0 γ0 i=2 γ0 =30×103[0.5log ( 83.33 )+0.4log ( 333.33 )]
2 0.893566 2 0.893566 = 200.82 kbits/sec
● This rate (200.82 kbits/sec) is only slightly higher than for the case of receiver CSI only (199.26 kbits/sec), and it is still significantly below that of an AWGN channel with the same average SNR
(199.26 kbits/sec). However, this may not be the case always!
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References
A. Goldsmith, Wireless Communications, Cambridge University Press, USA, 2005. A. J. Goldsmith and P. P. Varaiya, “Capacity of fading channels with channel side
information,” IEEE Trans. Inform. Theory, pp. 1986–92, November 1997.
G. Caire and S. Shamai, “On the capacity of some channels with channel state
information,” IEEE Trans. Inform. Theory, pp. 2007–19, September 1999.
M.-S. Alouini and A. J. Goldsmith, “Capacity of Rayleigh fading channels under different adaptive transmission and diversity combining techniques,” IEEE Trans. Veh. Tech. pp. 1165–81, July 1999.
ECE5884 Wireless Communications @ Monash Uni. August 15, 2022 23 / 24
Thank You! See you again
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