07 - CAPACITY OF WIRELESS CHANNELS v2 (3)

CAPACITY OF WIRELESS

CHANNELS

PREPARED BY S. SERFATY

BASED ON WIRELESS COMMUNICATIONS, ANDREA G OLDSMITH, STANFORD UNI VERSITY

CAPACITY OF WIRELESS CHANNELS 1

Outline

Shannon Theorem

◦Capacity in AWGN

Capacity of Flat-Fading Channels

◦Channel and System Model

◦Known Channel Side Information at Receiver

◦Known Channel Side Information at Transmitter and Receiver

◦Capacity Comparison

Capacity of Frequency-Selective Fading Channels

◦Time-Invariant Channels

◦Time-Varying Channels


Introduction

Up to now, we have seen what a mobile fading channel can do to

signals (narrow band fading, wideband fading) and how to combat

these adversities (diversity, multicarrier).

Let’s now ask an essential question:

◦How much data can a channel carry in one second or what is the

ultimate data rate that can be supported by a channel?

◦And… we will ask this question particularly for the fading channel

◦This question belongs to a discipline called “Information Theory”


Introduction

The question was answered by Claude E. Shannon in his

pioneering work in the late 1940s, the paper (book) “A

Mathematical Theory of Communication” where he

introduces the concepts of information, entropy, mutual

information, and channel capacity.

PHOTOGRAPH BY ALFRED EISENSTAEDT

/ THE LIFE PICTURE COLLECTION / GETTY


Introduction

The question was answered by Claude E. Shannon in his

pioneering work in the late 1940s, the paper (book) “A

Mathematical Theory of Communication” where he

introduces the concepts of information, entropy, mutual

information, and channel capacity.

Shannon’s theory was developed for the AWGN static

channel, leading to what is known as the Shannon's

capacity theorem (ultimate rate a channel can support)

PHOTOGRAPH BY ALFRED EISENSTAEDT

/ THE LIFE PICTURE COLLECTION / GETTY


Shannon’s theorem in words

Any given AWGN type communications system characterized by a

bandwidth B and a received SNR of γ has a maximum rate of

information C (expressed in bits/sec) known as the channel capacity.

C is the maximum theoretical number of bits that can be transmitted

through the channel in one second, with vanishing probability of

error.

◦Said differently, if we transmit in that channel at an information rate R

(bits/sec) which is less than C (bits/sec), then, by using intelligent coding

techniques, data transmission in the presence of additive white Gaussian noise

can have arbitrarily small error probabilities.

◦Moreover…


Shannon’s theorem in words

Moreover, Shannon shows that if the rate at which we transmit R > C, then

there is no possibility of achieving a vanishing probability of error.

Shannon demonstrates in his paper that this capacity C exists. He just forgot to

tell us how to achieve it... As a fact, he didn’t know…

Normally, this “intelligent coding techniques” will require long blocks of data,

which entails long delays.

Lately, new error codes have been re-invented which achieve rates close to

Shannon capacity with acceptable complexity

◦ LDPC (Low Density Parity Codes) invented by R. G. Gallager in 1962 and rediscovered in the

late 90’s

◦ Polar codes, fountain codes, etc. have also emanated since then (ask Benny ☺)


Shannon’s AWGN capacity theorem in math’s

C = B log 2 1 + γ


Shannon’s AWGN capacity theorem in math’s

C = B log 2 1 + γ

C is the maximum rate one can achieve in this AWGN channel in bits/second and is

called Shannon’s capacity of the channel,

B is the bandwidth of the channel in Hertz,

γ is the linear Signal to Noise Ratio (SNR), that can be expressed as γ = P/(N 0 B),

where P is the received power and N 0 is related to the PSD of the AWGN (N 0 /2)

C, the maximum rate, is limited by the bandwidth, the RX power and the noise level.

Shannon’s capacity is the ultimate rate (in bits/s) that can be achieved on a channel

with bandwidth B and SNR γ given any combination of coding scheme, transmission

and/or decoding scheme.

Side note: Normalized capacity = C N = C/B is measured in bits/s/Hz


Shannon’s theorem intuition

C = B log 2 1 + γ

Bandwidth limits how fast the information symbols can be sent over the given channel

The SNR limits how much information we can squeeze in each transmitted symbol.

◦ Increasing SNR makes the transmitted symbols more robust against noise.

To achieve a given information rate in a channel, the signal-to-noise ratio and the

allocated bandwidth must be traded against each other.

◦ One can either increase the bandwidth for a given SNR or increase the SNR for a given bandwidth

For the negligible noise case, the signal to noise ratio tends to infinity and so an infinite

information rate is possible with a very small bandwidth (if no noise, use M-QAM with

M=2 1000… )


On the Shannon’s Theorem

The Shannon’s theorem gives us the upper limit of the rate that can be

achieved in the AWGN channel.

As said, Shannon does not tell us how to achieve it.

Yet, the application of the Shannon theorem to different cases can

give us a hint (as we will see in several cases) on what is the best way

to approach the problem itself.


Our

cases

In this course,

we deal with

fading channels

and not AWGN

channels

We will analyze what happens

to this Capacity when the

channel is a fading channel

and not an AWGN channel

We will start with flat fading

channels and then look at

frequency selective channels.


Flat-Fading Channels: system model

SYSTEM MODEL

Avge. TX Power തP

For our analysis of capacity in fading channels, we will use a “discrete time

model” of the channel, which is a “sampled” version of the continuous channel.



SYSTEM MODEL

Avge. TX Power തP

An input message w (vector) needs to be sent from the transmitter to the

receiver.

The message is encoded into the codeword x (vector), which is transmitted over

the fading (time-varying) channel as x[i] at time i.



SYSTEM MODEL

Avge. TX Power തP

Flat fading channel

g i Rayleigh distribution

g i exponential distribution

The channel is a flat fading channel (Rayleigh fading) channel.

g[i] is the power gain of the channel (square of the channel gain) at every

discrete time i, which is random.

Remember that for, this channel, g[i] has an exponential distribution and

a Rayleigh distribution.

g i



SYSTEM MODEL

Avge. TX Power തP

Flat fading channel

g i Rayleigh distribution

g i exponential distribution

As said, the probability density function (pdf) of the channel power gain g[i] is

exponential with expected value g. ҧ

Let തP denotes the average transmit signal power, N 0 /2 the power spectral

density of the additive noise n[i], and B denotes the received signal bandwidth.

The instantaneous received SNR at any instant i is then γ[i]= തPg[i]/(N 0 B),

where 0 ≤ γ i < ∞ (linear, not dB)

The expected value of γ[i] over all times, denoted γ, ҧ is γ ҧ = തP g/(N ҧ 0 B)


Knowledge of the Channel State Information (CSI)

We will differentiate between two cases depending on what is known

about g[i] , called the Channel State Information (CSI), by the

transmitter and by the receiver.

Case 1: Receiver CSI (CSIR): The value of g[i] is known only at the

receiver at all time i, yet both the transmitter and receiver know the

distribution of g[i]

Case 2: Transmitter (CSIT) and Receiver CSI (CSIR) (also called CSIT/R):

The value of g[i] is known at the transmitter and the receiver at time

i, and both the transmitter and receiver know the distribution of g[i] .


CSIR case (only receiver knows the CSI)

In CSIR, two channel capacity definitions are relevant: ergodic

capacity and capacity with outage.

Ergodic capacity is the average capacity (over the distribution of the

SNR, i.e., over all possible values of the SNR)


Ergodic capacity

If the instantaneous SNR γ[i] of the channel varies, so does the instantaneous

capacity C γ[i] = B log 2 1 + γ[i]

Ergodic capacity is the capacity averaged over the distribution of γ

C = E γ C γ = E γ B log 2 1 + γ = න B log 2 1 + γ p γ dγ

0

◦ Note: This is very similar to what we did to compute the average probability of error in a

Rayleigh fading channel.

Note that only the receiver knows instantaneous SNR γ[i] (CSIR case)

The transmitter doesn’t know the instantaneous SNR, so it cannot take any

measure to counteract bad channel situations. The transmitter will transmit at a

constant rate (in bits/sec) and expect that the receiver does its best.

∞


Ergodic capacity

Ergodic capacity is the capacity averaged over the distribution of γ

Jensen’s inequality

E B log 2 1 + γ

C = E γ C γ = E B log 2 1 + γ = න

Capacity of a fading channel with CSIR only is always less than the Shannon

capacity of an AWGN channel with same average SNR.

Note on Jensen’s inequality: for any concave function (log is concave) g, E X [g(X)] ≤ g(E X [X]).

0

∞

B log 2 1 + γ p γ dγ

ҧ

∞

= න B log 2 1 + γ p γ dγ ≤ B log 2 1 + E γ = B log 2 1 + γ

0

E B log 2 1 + γ ≤ B log 2 1 + γҧ

קָ‏ עּור


Example

Consider a flat-fading channel with channel gain a i = g i which can take on three

possible values: a 1 = .05 with probability p 1 = .1, a 2 = .5 with probability p 2 = .5, and a 3

= 1 with probability p 3 = .4.

The transmit power is 10 mW, the noise spectral density is N 0 = 10 −9 W/Hz, and the

channel bandwidth is 30 KHz. The receiver has knowledge of the instantaneous value

of a i but the transmitter does not. Find the ergodic capacity of this channel and

compare with the capacity of an AWGN channel with the same average SNR.

Solution: P t = 10mW, N 0 = 10 −9 W/Hz, B = 30 KHz, three states: 1,2,3

The channel has 3 possible received SNRs,

◦ State 1 SNR: γ 1 = P t a 12 /(N 0 B) = 0.01 ×(0.05 2 )/(30,000∗10 −9 ) = 0.8333 = −.79 dB,

◦ State 2 SNR: γ 2 = P t a 22 /(N 0 B) = 0.01 ×(0.5 2 )/(30000 ∗ 10 −9 ) = 83.333 = 19.2 dB,

◦ State 3 SNR: γ 3 = P t a 32 /(N 0 B) = 0.01 × (1 2 )/(30000 ∗ 10 −9 ) = 333.33 = 25 dB.


Example – cont.

◦ State 1 SNR γ 1 = 0.8333, p 1 = 0.1

◦ State 2 SNR γ 2 = 83.333, p 2 = 0.5

◦ State 3 SNR γ 3 = 333.33, p 3 = 0.4

Ergodic capacity is the capacity

averaged over the distribution of γ

C ergodic = = 30,000‧[0.1‧ log 2 (1.8333) + 0.5‧log 2 (84.333)

What about AWGN?

+ 0.4‧log 2 (334.33)] = 199.26 Kbps.

Avge. SNR = γ ҧ = 0.1‧(0.8333) + 0.5‧(83.33) + 0.4‧(333.33) = 175.08 = 22.43 dB.

C AWGN = B log 2 (1 + 175.08) = 223.8 Kbps > C ergodic = 199.26 Kbps


CSIR case (only receiver knows the CSI)

In CSIR, two channel capacity definitions are relevant: ergodic

capacity and capacity with outage.

Ergodic capacity is the average capacity (over the distribution of the

SNR, i.e., over all possible values of the SNR)

In CSIR, the transmitter doesn’t know the channel, so there is nothing

it can do. The channel may turn so bad that information cannot be

decoded (outage).

Capacity with outage assumes that in some cases (outage), the

transmission cannot be decoded with negligible error probability and

asks what is the capacity over the rest of the cases (non-outage).


Capacity with Outage

As said, the transmitter doesn’t know what the instantaneous SNR γ[i]

is but it knows its distribution.

Let say that the transmitter assumes that the receiver can get a

minimum SNR γ min and transmits at a fixed data rate C = B log 2 (1+γ min ).

◦ If the received instantaneous SNR γ ≥ γ min , according to Shannon, we are

transmitting at a rate that is less than the instantaneous capacity, therefore

that data can be perfectly decoded.

◦ On the other hand, if due to the fading, the reveived instantaneous SNR

γ < γ min , again according to Shannon, the received data cannot be

correctly decoded → receiver declares an outage


Illustration of Capacity with Outage

SNR γ

C(γ min ) = B log 2 (1+γ min )

γ min

γ min

rate

C(γ min )=X bps

rate

C(γ min )=X bps

rate

0

rate

0

rate

C(γ min )=X bps

rate

C(γ min )=X bps

rate

0

rate

0

The transmitter transmits at a rate

of C(γ min ) say X bits/sec.

Shannon: If the SNR is above γ min

then the bits received at a rate

C(γ min ) = X bits/sec can be decoded,

otherwise, they cannot (we receive

0 bps).

time


Capacity with Outage

The probability of outage is thus p out = p(γ < γ min )

◦γ min is a design parameter based on the acceptable outage probability 1

The average rate of correctly received bits is called “Capacity with

outage” C o and can be computed as:

C o = (1 − p out )B log 2 (1 + γ min )

◦When in outage (with probability p out ), capacity is zero as information cannot

be decoded properly

1

Note: For Rayleigh fading channels, γ is exponentially distributed, so given γ min and ҧ γ we can calculate the probability of

outage. See the chapter on diversity.


Normalized capacity (C/B)

Capacity with outage (cont.)

C o = (1 − p out )B log 2 (1 + γ min )

If we want the outage probability to be

very low, then we need to make sure that

p(γ < γ min ) is very low.

We need to chose γ min so that this

seldom happen.

If the transmitter choses a very low γ min ,

it will transmit at a rate C = B‧ log 2

( 1

+ γ min

) which will be a low rate and,

although (1 − pout) will be close to 1,

the capacity with outage will be very low.

Rayleigh fading channel (γ exponential) with

γ = 20 dB


Normalized capacity (C/B)

Capacity with outage (cont.)

C o = (1 − p out )B log 2 (1 + γ min )

The capacity increases dramatically as the

outage probability increases.

These high-capacity values for large outage

probabilities have higher probability of

incorrect data reception.

The average rate of correctly received can

be maximized by finding the γ min that

maximizes C o .

Rayleigh fading channel (γ exponential) with

γ = 20 dB

4%


Example

Assume the same channel as in the previous example, with a bandwidth of 30

KHz and three possible received SNRs: γ 1 = .8333 with p(γ 1 ) = .1, γ 2 = 83.33 with

p(γ 2 ) = .5, and γ 3 = 333.33 with p(γ 3 ) = .4.

Find the capacity versus outage for this channel, and find the average correctly

received rates for outage probabilities (a)p out < .1, (b) p out = .1 and (c) p out = .6.

Solution

(a) For p out < .1, this means that we never want to be in outage, and we must

decode correctly in all channel states, That is, no matter how low the SNR gets.

The minimum received SNR for p out in this range of values is that of the weakest

channel: γ min = γ 1 , and the corresponding rate is:

C = B log 2 (1 + γ min ) = 30000‧log 2 (1+0.833) = 26.23 Kbps.


Example – cont.

(b) For 0.1 ≤ p out < 0.6 we can decode incorrectly when the channel is in the

weakest state only. That is, we give up the capacity when the channel is at γ 1

Then, the transmitter choses to encode at γ min = γ 2 and the corresponding rate is

C = B log 2 (1 + γ min ) = 30000 log 2 (1+83.33) = 191.94 Kbps.

(c) For .6 ≤ p out < 1 we can decode incorrectly if the channel has a received SNR

γ 1 or γ 2 , giving up that capacity.

Then the transmitter choses to encode at γ min = γ 3 and the corresponding rate is

C = B log 2 (1 + γ min ) = 30000 log 2 (1+333.33) = 251.55 Kbps.

γ 1 = 0.8333, p 1 = 0.1

γ 2 = 83.333, p 2 = 0.5

γ 3 = 333.33, p 3 = 0.4

Note: Observe that we found C AWGN = 223.8 Kbps for an AWGN channel with same average

SNR, what happened? We are not done yet!


Example – cont.

For p out < .1 data transmitted at rate C = 26.23 Kbps are always correctly

received

For p out = .1 we transmit at rate C = 191.94 Kbps, but we can only correctly

decode these data when the channel SNR is γ 2 or γ 3 , so the rate correctly

received is (capacity with outage C o )

C o = (1 − .1)191.94 = 172.75 Kbps.

For p out = .6 we transmit at rates close to C = 251.55 Kbps but we can only

correctly decode these data when the channel SNR is γ 3 , so the rate correctly

received is (capacity with outage C o )

C o = (1− .6)251.55 = 125.78 Kbps.

In this case, optimum capacity is 172.75 Kbps


Channel Side Information at Transmitter and Receiver

When both sides have CSI, the transmitter can adapt transmission

strategy relative to this CSI

SYSTEM MODEL

The transmitter can choose at what rate to transmit and at what

power, depending on the “known state” of the channel


Channel Side Information at Transmitter and Receiver

SYSTEM MODEL

We assume an ideal system where the transmit power P(γ) is allowed

to vary with γ, subject to an average power constraint തP and the

capacity is defined as

constrained to

Instantaneous power

or

pdf of


Optimum capacity

This optimization is solved using Lagrange multipliers:

With the constraint that P(γ) ≥ 0, the optimal power adaptation is:

and setting

Interpretation:

for some “cutoff” value γ 0

When the condition is met (γ γ 0 ), transmit with power

1

γ 0

− 1 γ

If the instantaneous SNR γ is below this cutoff γ 0 then no data is transmitted

over that time interval, so the channel is only used at times where γ 0 ≤γ<∞

ത P


Optimum capacity (cont.)

The optimal capacity becomes:

To obtain γ 0 , we solve the following equation which is a function of p(γ)

Normally, this value cannot be derived analytically and is found through

numerical integration


Interpretation: Water-filling principle

Since γ is time-varying, the maximizing power

adaptation policy is a “water-filling” formula in

time.

The figure shows how much power is allocated to

the channel for instantaneous SNR γ(i) = γ.

The water-filling terminology refers to the fact that

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/γ 0 .

“water”

The relative amount of power allocated for a given γ equals [1/γ 0 − 1/γ], i.e.,

the amount of water between the bottom of the bowl (1/γ) and the constant

water line (1/γ 0 ).

CAPACITY OF WIRELESS CHANNELS

36

Bad channel

Good channel

Water-filling intuition

The intuition behind water-filling is to take

advantage of good channel conditions:

1. When channel conditions are good (γ large, 1/γ small)

more power and a higher data rate is sent over the

channel.

2. As channel quality degrades (γ small, 1/γ large) less

power and rate are sent over the channel.

3. If the instantaneous channel SNR falls below the cutoff

value, the channel is not used.

cutoff

“water”


Example

Assume the same channel as in the previous example, with a bandwidth of 30

KHz and three possible received SNRs: γ 1 = .8333 with p(γ 1 ) = .1, γ 2 = 83.33 with

p(γ 2 ) = .5, and γ 3 = 333.33 with p(γ 3 ) = .4.

Find the capacity of this channel assuming both transmitter and receiver have

instantaneous CSI knowledge (CSIT/R).

Solution

We know the optimal power allocation is water-filling, and we need to find the

cutoff value γ 0 that satisfies

discrete version of integral


Example

discrete version of integral

This type of problems is normally solved through iterations.

We first assume that all channel states are used to obtain γ 0 and look at the

power that is assigned to the weakest (lowest SNR) channel.

P γ 1

ሜP = 1 γ 0

− 1 γ 1

If this power is negative (we really don’t know how to transmit negative power),

we conclude that during this weakest channel we don’t transmit and go back

and solve again for the remaining channel states and so on until all assigned

powers are positive.

It just sounds complicated. We will present a straightforward one slide iterative

procedure to solve it.


Water-filling general algorithm for discrete SNR channels

List the r linear SNR values from low to high: γ 1 < γ 2 < ⋯ < γ r

1. Set the iteration count k to 1 and compute Τ 1 γ 0 by solving the following equation:

1

γ 0

=

σr

i=k

1

p γ i

r

1 + σ i=k

2. Using the value of 1Τγ 0 obtained above, solve for the relative power, P γ i Τ

the i th SNR using the following equation:

p γ i

P γ i

ሜP = 1 γ 0

− 1 γ i

i = k, . . . , r − 1, r

γ i

ሜP, for

3. If the relative power allocated to the channel with the lowest SNR is negative (i.e.,

if P γ k Τ ሜP< 0), discard that channel by setting P γ k Τ ሜP= 0 and rerun the algorithm

with the iteration count k incremented by 1.

4. Repeat steps 1–3 until all channels have been allocated (positive) power

You can verify that this is the

same discrete equation of

last slide written differently


Water-filling algorithm application

γ 1 = 0.8333, p 1 = 0.1

γ 2 = 83.333, p 2 = 0.5

γ 3 = 333.33, p 3 = 0.4

List the r = 3 SNRs from low to high: γ 1 = 0.833 < γ 2 = 83.333 < γ 3 = 333.33


1 1

=

γ 0 σ3

1 + σ3 p γ i

i=1

p γ i=1 = 1 + 0.1

+ 0.5

+ 0.4

= 1.13 γ

i

γ i 0.833 83.333 333.33

0 = 0.89

2. Using the value of 1Τγ 0 obtained above, solve for the relative power, P γ i Τ ሜP, for

the i th SNR using the following equation:

P γ i

ሜP

= 1 γ 0

− 1 γ i

i = 1, . . . , 3



with the iteration count k incremented by 1. (k =2)

4. Repeat steps 1–3 until all channels have been allocated power

APPLICATIONS OF THE MIMO CAPACITY EQUATION 41

P γ 1

ሜP = 1 γ 0

− 1 γ 1

= 1.13 − 1

0.8333 = −0.070

P γ 2

ሜP = 1 γ 0

− 1 γ 2

= 1.13 − 1

83.333 = 1.118

P γ 3

ሜP = 1 γ 0

− 1 γ 3

= 1.13 − 1

333.33 = 1.127

Water-filling algorithm application

γ 1 = 0.8333, p 1 = 0.1

γ 2 = 83.333, p 2 = 0.5

γ 3 = 333.33, p 3 = 0.4

List the r = 3 SNRs from low to high: γ 1 = 0.833 < γ 2 = 83.333 < γ 3 = 333.33


1

=

γ 0

σ3

i=2

1

p γ i

3 p γ

1 + σ i

i=2 = 1

1 + 0.5

+ 0.4

= 1.12 γ

γ i 0.5+0.4 83.333 333.33

0 = 0.89

1 γ 0 obtained above, solve for the relative power, P γ i Τ ሜP, for

2. Using the value of Τ

the ith SNR using the following equation:

P γ i

ሜP

= 1 γ 0

− 1 γ i

i = 2,3

P γ 1

ሜP = 0

P γ 2

ሜP = 1 − 1 = 1.12 − 1

γ 0 γ 2 83.333 = 1.108

P γ 3

ሜP = 1 − 1 = 1.12 − 1

γ 0 γ 3 333.33 = 1.117



with the iteration count k incremented by 1.



Interpretation and capacity

Transmitted power

◦ When the transmitter knows that the SNR at the receiver is γ 1 = 0.833 don’t bother

transmitting any power P γ 1 = 0.

◦ When the transmitter knows that the SNR at the receiver is γ 2 = 83.333 transmit at P γ 2

= 1.108 ሜP.

◦ When the transmitter knows that the SNR at the receiver is γ 3 = 333.33 transmit at P γ 3

= 1.117 ሜP.

◦ Average transmitted power = p γ 1 ∙ P γ 1 + p γ 2 ∙ P γ 2 + p γ 3 ∙ P γ 3 = 0.1 ∙ 0 + 0.5

∙ 1.108 ሜP + 0.4 ∙ 1.117 ሜP = ሜP

Capacity

∞

C = න B log 2

γ 0

γ

γ 0

p γ dγ

γ i

3

C = B log 2

i=2 γ 0

p γ i

= 30000 0.5 log 2 83.333 ⋅ 1.12 + 0.4 log 2 333.33 ⋅ 1.12

= 200.82 kbps


Capacity comparison

Observations:

1. AWGN capacity always better than CSIR

only

2. At high SNR, CSIT/R not better than CSIR

only

◦ When all is good…

Rayleigh fading channel


Capacity of Frequency-Selective Fading Channels

Up to now, we have considered flat fading channels.

We will now look at frequency selective fading channels where the channel

frequency response is not flat along the bandwidth of the channel.

Our previous lecture on OFDM should give us a hint on how to handle the

capacity computation for a frequency selective channel.

We consider two cases:

◦ Time-invariant channels – where the frequency response of the channel H(f) remains

constant and we assume that it is known at both the transmitter and receiver.

◦ Time-varying channels – where the frequency response of the channel H(f) = H(f,i), i.e., the

channel varies over both frequency and time.

time index i


Time-Invariant Channels

Assumptions:

◦H(f) constant and known at both the transmitter and receiver

◦Total transmit power constraint P

◦H(f) is block-fading in frequency: frequency divided into “subchannels” of

bandwidth B, where H(f) = H j is constant over each block (“subchannel”).

The frequency-selective fading channel consists of a set

of parallel AWGN channels with SNR |H j | 2 P j /(N 0 B)

where P j is the power transmitted in “subchannel” j

(which may be different for different “subchannels”)


46

Time-Invariant Channels

The frequency-selective fading channel consists of a set

of parallel AWGN channels with SNR |H j | 2 P j /(N 0 B)

The capacity of this parallel set is the sum of rates

associated with each channel with power optimally

allocated over all channels, constrained to the total

power

constrained

This is the “discrete” version of the

constrained optimization we had for

the flat fading channel with CSI known

at transmitter and receiver


47

Time-Invariant Channels (cont.)

The answer is found using Lagrange and leads to a “water-filling” solution

for some cutoff γ 0 and with

that must satisfy

and the optimal capacity is


Water-filling in frequency

It is the same case as the flat fading channel

but now in the frequency domain.

The optimal capacity is achieved by sending at

different rates and powers over each

subchannel.

Multicarrier modulation can be used with

“adaptive loading” (adjust rate and power in

each “subchannel”).

send

nothing

here


Water-filling freq. domain example

Consider a time-invariant frequency-selective block fading channel consisting of

three subchannels of bandwidth B = 1 MHz each. The frequency response

associated with each channel is H a = 2, H b = 1 and H c = 3.

The transmit power constraint is P = 10 mW and the noise PSD is N 0 = 10 −9 W/Hz.

Find the capacity of this channel and the optimal power allocation that achieves

this capacity.

Solution

We first find γ j = |H j | 2 P/(N 0 B) for each subchannel:

◦ γ a = 2 2 ‧(10‧10 -3 )/(10 -9 ‧10 6 ) =40,

◦ γ b = 1 2 ‧(10‧10 -3 )/(10 -9 ‧10 6 ) = 10,

◦ γ c = 3 2 ‧(10‧10 -3 )/(10 -9 ‧10 6 ) =90

3

2

1

H j

a b c

1 2 3

MHz


Water-filling general algorithm for freq. domain

List the r linear SNR values from low to high: γ 1 < γ 2 < ⋯ < γ r


1

= 1

1 + σr 1

γ 0 r−k+1 i=k γ i

2. Using the value of 1Τγ 0 obtained above, solve for the relative power, P i ΤP, for the

ith SNR using the following equation:

P i

= 1 − 1 i = k, . . . , r − 1, r

P γ 0 γ i


if P k ΤP< 0), discard that channel by settingP k ΤP= 0 and rerun the algorithm with the

iteration count k incremented by 1.




γ a = 40

γ b = 10

γ c = 90

List the 3 linear SNR values from low to high: γ 1

= γ b = 10 < γ 2 = γ a = 40 < γ 3 = γ c = 90

1. Set the iteration count k to 1 and compute 1Τγ 0 by solving the following equation:

1

= 1 1 + σ3 1

γ 0 3 i=1 = 1 1 + 1 + 1 + 1 = 0.378 γ γ i 3 10 40 90

0 = 2.64

2. Using the value of 1Τγ 0 obtained above, solve for the relative power, P i ΤP, for the ith

SNR using the following equation:

P i

= 1 − 1 i = 1,2,3

P γ 0 γ i

P 1

P = 1 − 1 γ 0 γ 1

= 0.378 − 1 10 = 0.278

P 2

P = 1 − 1 γ 0 γ 2

= 0.378 − 1 40 = 0.353

P 3

P = 1 − 1 γ 0 γ 3

= 0.378 − 1 90 = 0.366


ifP i ΤP< 0), discard that channel by settingP i ΤP= 0 and rerun the algorithm with the

iteration count k incremented by 1.




The powers allocated to each subchannel are

The resulting capacity is

P 1

P = 1 − 1 = 0.378 − 1

γ 0 γ 1 10 = 0.278 ⟹ P 1 = P b = 10mW ∙ 0.278 = 2.78mW

P 2

P = 1 γ 0

− 1 γ 2

= 0.378 − 1

40 = 0.353 ⟹ P 2 = P a = 10mW ∙ 0.353 = 3.53mW

P 3

P = 1 γ 0

− 1 γ 3

= 0.378 − 1

90 = 0.366 ⟹ P 3 = P c = 10mW ∙ 0.366 = 3.66mW

γ 1 = 10

γ 2 = 40

γ 3 = 90

γ j

C = σ j:γj ≥γ 0

B log 2

γ 0

= 1,000,000 log 2 10 ⋅ 0.378 + log 2 40 ⋅ 0.378 + log 2 90 ⋅ 0.378

= 10.93 Mbps


Time-Varying channels

The capacity of time-varying frequency-selective fading

channels is, in general, unknown, however some

approximations exist.

One approximation is to take the channel bandwidth B

of interest and divide it up into subchannels of size

equal to the channel coherence bandwidth B c , then

assume that each resulting sub-channel is

independent, time-varying, and flat-fading with H(f, i)

= H j [i] on the j-th subchannel.

Assuming transmitter and receiver channel CSI, the

optimal power allocation results in a two-dimensional

water-filling solution both in time and frequency.

Power

frequency

Power allocation

time


Main points

Shannon’s theorem defines the capacity of a channel as the maximum rate that

can be transmitted over that channel with negligible probability of error.

C = B log 2 1 + γ

The capacity of a flat fading channel with CSIR is worst than that of an AWGN

channel with same average SNR

The capacity of a flat fading channel with CSIT and CSIR can sometimes be better

than an AWGN channel with same average SNR (very low).

Water-filling: shovel as many bits as you can into the channel when the channel

is good, don’t waste you time when the channel is bad (principles of adaptive

loading and frequency selective scheduling)

◦ Used in flat fading and in frequency selective channels with CSIR/CSIT. More soon…


07 - CAPACITY OF WIRELESS CHANNELS v2 (3)

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