Relaying MIMO channel capacity with imperfectchannel knowledge at the receiver

Triantafyllos Kanakis, Predrag B. RapajicSchool of Engineering, the University of Greenwich, UK, ME4 4TB,

Email: {t.kanakis, p.rapajic}@gre.ac.uk

Abstract— In this paper, we study a scenario of multiple inputmultiple output system using relays in an angle showing howthe topology of the system can play its role in the situation. Weassume perfect channel knowledge at the transmitter and therelay(s) which is represented by a matrix of zero mean circularlysymmetric complex Gaussian (ZMCSCG) random variables withknown covariances, but imperfect channel state information(CSI) at the receiver. The imperfect CSI at the receiver causeserrors to the received signal and as a consequence the channelcapacity saturates. We derive the channel capacity formulas fortwo different transmission algorithms; uniform and waterfilling.Finally we simulate and compare the results.

I. INTRODUCTION

In this paper we will implement a simple scenario forrelaying MIMO channel. We set a transmitter an a receiverwith a relay put in an angle between the transmitter and thereceiver and we show that the angles between the transmitterand the relay and between the relay and the receiver should notbe larger than 60◦. We assume a perfect channel knowledgeat the transmitter and the relay but imperfect channel stateinformation at the receiver which will cause errors in thecommunication. We will derive the channel capacity formulasfor our scenario using uniform and waterfilling transmissionalgorithms and we will simulate them to show the results. Atlow SNRs we expect to see waterfilling to perform better thanuniform as it handles power in a smarter way than uniformdoes. At high SNRs we expect to see waterfilling and uniformtechniques to perform equally well because there is plenty ofpower to distribute over the channel elements in equal portions.

II. SYSTEM MODEL

For the analysis shown in this paper a wireless systemwith t transmit and r receive antennas is employed. Thechannel fading is assumed to be Rayleigh and the channelstate information (CSI) at the receiver is imperfect. The isrepresented in the following general form

y = Hx + n (1)

where H is the channel matrix with independent identicallydistributed (iid) Gaussian elements with ZMCSCG indepen-dent real and imaginary parts each with variance 1/2. Thetotal transmitted power is limited to P

tr(E ∣∣xxH∣∣) ≤ P (2)

Denoted by n is a t× 1 noise vector with iid ZMCSCG noisesamples with variance σ2

n.

A. Capacity

The maximum mutual information lower bound with Gaus-sian input signalling of a MIMO system is referred as channelcapacity and is given by [1], [2] and [3] to be

I(X;Y ) ≥ log2 det

(Ir +

HQHH

σ2n + σ2

etr(RtQ)Rr

)(3)

where H is the estimation of the channel matrix at the receivesides of each individual link, Q is the input covariance matrixwhere Q = E ∣∣xxH∣∣, σ2

e and σ2n are the variance of error and

the variance of noise respectively. Rr is the r × r receiveantenna correlation matrix while Rt is the t × t transmitantenna correlation matrix and they effect the system onlyif the system is spatially correlated. For simplicity we assumethat the antenna elements are not spatially correlated and sothe we correlation matrices are unit matrices. So the channelcapacity is calculated by

I(X;Y ) ≥ log2 det

(Ir +

HQHH

σ2n + σ2

etr(Q)

)(4)

III. GENERAL PHYSICAL MODEL

In urban areas, buildings interrupt or scatter the radio signalsand also the physical ground is not flat. For these reasons themost communication systems are not placed in a straight linebut the antenna arrays are placed in angles between them. Weassume that the distances between the relay and the transmitterand between the relay and the receiver are equal. So anisosceles triangle is formed.

Fig. 1. Antenna arrays positioned into the corners of an isosceles trianglewith 30◦ base angles and 120◦ top angle. Denoted by lower case letters arethe distances between the corners and by capital hat letters are the angles.

From the illustration provided by Fig. 1 we can realize thatthe larger side is ”c”, the one between points A and B. When

801-4244-0957-8/07/$25.00 ©2007 IEEE.

the base angles are larger than 60◦ then the top angle will besmaller than 60◦ in order to satisfy the sum of 180◦. In thiscase distance ”c” between points A and B is going to be thesmaller side of the triangle.

For the case where the top angle is smaller than the baseangles the pathloss of the path without the relay will be smallerthan the path with the relay due to increased distance. In thiscase the relay will act as scatterer of the signal and will notsignificantly gain to the system. Hence when the top angle islarger than the base angles, then the relay will offer a SNRgain to the system.

In the case where an obstacle is between the transmitter andthe receiver, the pathloss is even greater and hence the powerof the signal that arrives at the receive side of the system isvery weak. In order to tackle this problem the most effectivesolution is to use a relay on the top of the obstacle. In thiscase the top angles can be smaller than 60◦ and the relay willstill play a significant role in the improvement of the SNR andas a consequence the improvement of the channel capacity ofthe system, as it is shown in the rest of this paper.

A. Analytical Results

As shown in Fig. 1 distance c is the actual transmitter-receiver distance, a is the distance between the relay and b isthe distance between the relay and the receiver. The SNR ofthe system without the relay is calculated by:

SNR1 =Pr1

N(5)

Pr1 = Pt1Kc−γ (6)

Where Pt is the total power transmitted, Pr is the totalpower received, K is the total gains and losses of the systemand γ is the pathloss index which normally varies between2 and 6. We Assume that the climate conditions are uniformeverywhere within the system and same equipment for all thenodes of the system is used. Assuming that the SNR of thesystem with the relay will transmit by using half of the powerto the first hope and the other half to the second hop.

SNR2 =Pr2

N(7)

Pt2 =Pt1

2=⇒ Pr2 =

Pt1

2Ka−γ (8)

By assuming that identical equipment is used, we comparethe two SNRs.

SNR2

SNR1=

12Pt1Ka−γ

Pt1Kc−γ=⇒ SNR2

SNR1=

12

(a

c

)−γ

=⇒ (9)

SNR2

SNR1=

12

( c

a

)γ

(10)

c2 = a2 + b2 − 2ab cos C (11)

but in the case where no obstacles bothers the communication,C is always larger than 60◦, so c2 ≥ a2 therefore the ratio c

ais always a gain to the system as long as angle C is larger than60◦. In the case where an obstacle interrupts the transmissionthen any c

a will be beneficial as we do not have more effectivedirect components.

IV. NUMERICAL RESULTS

The channels examined in this paper will either use thewaterfilling or uniform transmission algorithms. The reasonof selecting those two algorithms is to show our results intwo widely known and examined transmission algorithms. Itis also useful to check what is the capacity gain of the systemwhen we use a simple relay system according to the scenarioshown in the previous section.

A. Numerical Results - The gain of the relay

As it is shown in section III by placing a relay in an anglefrom the transmitter and the receiver our SNR will gain fromthe use of the relay if the angle is not larger than 60◦. If theangle is larger than 60◦ then the SNR will gain if an obstacleis between the transmitter and the receiver. For calculationpurposes, in this paper we assume that the angle between thetransmitter and the relay is and the relay and the receiver is30◦ and also we assume a single relay as shown in figure (1).

c2 = a2 + b2 − 2ab cos C

but C = 120◦ and cos C = − 12

c2 = 3a2 =⇒ c =√

3a2 =⇒ c = 1.732a

Having done this simple geometry calculations the SNRgain of the system can be calculated. The system without therelay has the following SNR

SNR1 =Pr1

N

Pr1 = Pt1Kc−γ =⇒ Pr1 = Pt1K(1.732a)−γ

where Pt is the total power transmitted, Pr the total powerreceived and K the total gains and losses of the system. Thepathloss index γ normally varies from 2 to 6 where 2 isthe open space pathloss index and 6 is the pathloss index ofurban environments full with scatterers and shadows. For ourcalculations we use a pathloss index γ = 3 as it is a mediancase and represents a sub-urban environment.

The system with the relay has the following SNR assumingthat half of the total power is used to transmit from thetransmitter to the relay and the second half is used to transmitfrom the relay to the receiver.

SNR2 =Pr2

NPr2 =

Pt12

Ka−γ

By comparing the two SNRs found above we can see thatSNR2 = 2.59 × SNR1. In practise though we transmitin two time slots and so we use double channel resources.Hence the channel capacity will be half of the minimum

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among the capacities of the two sub-channels (hops), but forthe simulations we use half the gain in order to provide fairapproach, so SNR2 = 1.295 × SNR1. It is also possible totransmit in one time slot but noise will be coupled at the relayand carried over the receiver, so the noise level is expected toincrease, but this is beyond the research of this paper.

B. Numerical Results - Waterfilling

Assuming a system operating using the modal design,waterfilling is the technique that assigns each mode a fractionof the total power proportional to the strength of the mode.In other words, strong modes will get a larger amount ofthe total power than the weakest one. The input covariancematrix Q is responsible to distribute the total power over thetransmitting elements. The input covariance matrix Q can beeigen-decomposed and written as

Q = UHΛQUHH

where ΛQ is a non negative diagonal matrix and is of the form

ΛQ =

SNR1 0 . . . 0

0 SNR2 . . ....

......

. . ....

0 · · · · · · SNRt

(12)

The channel capacity of a system using a waterfillingtransmission algorithm is determined by the equation (13)

C = E[log2

∣∣∣∣∣Ir +HQHH

σ2n + σ2

etr(Q)

∣∣∣∣∣]

(13)

From equation (12) it is shown that the tr(Q) = SNR soequation (13) becomes

C = E[log2

∣∣∣∣∣Ir +HQHH

σ2n + σ2

e(SNR)

∣∣∣∣∣]

(14)

C. Numerical Results - Uniform

An extension to the modal design referred in the Waterfillingsection, the concept of the uniform transmission algorithm isno matter what the modal quality is, the system assigns toeach of the modes SNR

t fractions of the total power.

Q =

SNRt 0 . . . 0

0 SNRt . . .

......

.... . .

...0 · · · · · · SNR

t

(15)

The capacity of a channel transmitting using the uniformtechnique will be determined by

C = E[log2

∣∣∣∣∣Ir +SNR

M HHH

σ2n + σ2

eSNR

∣∣∣∣∣]

(16)

V. SIMULATION RESULTS

Relaying communication systems can be composed of mul-tiple MIMO systems. Each MIMO transmitter-receiver systemis called a hop. The system of figure (1) is showing a two hopsMIMO relaying system. In order to determine the channelcapacity of a relaying system we need to determine thecapacity of each hop independently. Then, the channel capacityof the system is the smaller of all. That makes sense as wecannot use higher data-rate than this our system can afford.

In our simulations we use the scenario stated in section III,we assume the analogy between the SNR without relay andwith relay to be 2.5 (SNR1 = 2.5 × SNR2), we assumethe noise variance to be unit (σ2

n = 1) and the estimationof the channel matrix at the receiver H to be result ofthe white channel matrix Hω affected by the

√1 − σ2

e rule(H = Hω ×√1 − σ2

e

). Also we assume a perfect CSI at

the transmitter and the relay and imperfect CSI only at thereceiver.

Finally we need to note that the relay operates as a propertransmitter/receiver which, receives the signal, analyzes it andit sends to the receiver as a new message. And also, we needto note that the equipments of the transmitter, the relay andthe receiver are identical and so the gains and the losses areequal.

A. Waterfilling

Figure (2) shows how the imperfect CSI at the receivereffects the system operating the waterfilling transmission strat-egy. It is shown that an error with variance of 0.1

(σ2

e = 0.1)

can cause our system to saturate at about 20dB. The saturationin this case is that the channel capacity will not increasesignificantly after this point. If we keep increasing the powerfrom this point on, it will only lead to a waste of energy as wewill not gain in capacity. It is also shown that the use of relayimproves the system slightly and that the system saturates abit later than a simple MIMO channel.

By increasing the error variance to 0.2 we can see thatalthough the system behaves similarly with the smaller errorvariance at the low SNRs, it does not increase with thesame rate as the previous case and also the channel capacitysaturates a lot sooner. Finally by increasing the error varianceto 0.5 the system’s capacity does not improve much from theinitial state. It saturates at about 5 to 6dB and we can noticea bit of a degrade at high SNRs (after 13-14dB).

B. Uniform

Figure (3) shows how the imperfect CSI at the receivereffects the system operating the uniform transmission strategy.An error variance of 0.1

(σ2

e = 0.1)

causes the system tosaturate a bit later than 20dB. For higher SNRs it is expectedto merge with the waterfilling transmission. That makes senseif we take into account that waterfilling transmission strategybehaves just like uniform in high SNRs because all the modestake almost same amount of power. The difference in powerbetween the nodes is negligible.

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−5 0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

10

11

SNR [dB]

Nor

mal

ised

Cap

acity

[nat

s/s/

Hz]

123456

Fig. 2. Performance analysis for Waterfilling transmission strategy1. 4 × 4 MIMO system with variance of error σ2

e = 0.12. 4 × 4 × 4 Relaying MIMO system with variance of error σ2

e = 0.13. 4 × 4 MIMO system with variance of error σ2

e = 0.24. 4 × 4 × 4 Relaying MIMO system with variance of error σ2

e = 0.25. 4 × 4 MIMO system with variance of error σ2

e = 0.56. 4 × 4 × 4 Relaying MIMO system with variance of error σ2

e = 0.5

By increasing the error variance to 0.2 causes the system toperform worse that with error variance of 0.1. The behavior issimilar to the waterfilling behavior but the capacity achievedthroughout all the SNRs is lower than waterfilling’s. Finally aterror variance of 0.5 we can only notice a smoother behaviorbut still the capacity levels remain the same. It is interestingto see that waterfilling degrades at high SNRs and it seemsthat it does merge with uniform in this case too.

C. Comparison

Figure (4) is a comparison between two equal size systemsusing different transmission algorithms. On the one hand, thereis a uniform 4× 4× 4 relaying MIMO system and a uniform4 × 4 simple MIMO system. On the other hand we have awaterfilling 4×4×4 relaying MIMO system and a waterfilling4×4 MIMO system. All the systems are having error varianceof 0.2. This comparison graph is used to graphically representwhat is theoretically expected and stated above; waterfillingperforms better than uniform in low SNRs but it merges onhigh SNRs. Waterfilling even degrades at some stage and itmerges with uniform at very high SNRs.

Although waterfilling systems increase their capacity rapidlyor at least much faster than the uniform systems do, they alsosaturate much faster too. In the particular case of figure (4) thewaterfilling seems to saturate at about 14dB while the uniformsystem seems to saturate at about 22dB. Waterfilling systemsseem to have a very sharp saturation while uniform have amuch smoother saturation.

−5 0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

10

11

SNR [dB]

Nor

mal

ised

Cap

acity

[nat

s/s/

Hz]

123456

Fig. 3. Performance analysis for Uniform transmission strategy1. 4 × 4 MIMO system with variance of error σ2

e = 0.12. 4 × 4 × 4 Relaying MIMO system with variance of error σ2

e = 0.13. 4 × 4 MIMO system with variance of error σ2

e = 0.24. 4 × 4 × 4 Relaying MIMO system with variance of error σ2

e = 0.25. 4 × 4 MIMO system with variance of error σ2

e = 0.56. 4 × 4 × 4 Relaying MIMO system with variance of error σ2

e = 0.5

−5 0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

10

11

SNR [dB]

Nor

mal

ised

Cap

acity

[nat

s/s/

Hz]

1234

Fig. 4. A performance comparison of waterfilling versus uniform transmis-sion algorithms.1. Uniform transmission strategy - 4×4 MIMO system with variance of errorσ2

e = 0.12. Uniform transmission strategy - 4 × 4 × 4 Relaying MIMO system withvariance of error σ2

e = 0.13. Waterfilling transmission strategy - 4 × 4 MIMO system with variance oferror σ2

e = 0.14. Waterfilling transmission strategy - 4×4×4 Relaying MIMO system withvariance of error σ2

e = 0.1

VI. CONCLUSIONS

In this paper we examined a simple relaying MIMO sce-nario. We put a relay into an angle from the transmitter and

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the receiver and we showed what the power gain of a relayingMIMO system is over a simple MIMO system. We derivedthe capacity formula for uniform and waterfilling transmissionstrategies by assuming imperfect CSI at the receiver and weused those formulas in simulations. The simulations showedhow the channels saturate when imperfect CSI exists at thereceiver.

From the simulations we proved that waterfilling trans-mission strategy performs better than uniform transmissionstrategy capacity-wise in both MIMO and Relaying MIMOsystems for low SNRs. Only in higher SNRs we have shownthat the capacity-wise performance of the two algorithmsmerge. So in the comparison between the two different trans-mission algorithms we have to accept a trade of complexityfor performance.

REFERENCES

[1] Emre Telatar, ”Capacity of Multi-Antenna Gaussian Channels”, Euro-pean Transaction on Telecommunications EET, vol.10 Nov/Dec 1999

[2] L. Musavian, M. Dohler, M. R. Nakai and A. H. Aghvami ”TransmitterDesign in Partial Coherent Multiple Antenna Systems”, IEEE Intern.Conf. on Communic., vol.4, pp.2261-2265, May 2005

[3] A. Paulraj, R. Nabar, D. Gore, ”Introduction to Space-Time WirelessCommunications”, Cambridge University Press, May 2003

[4] B. Wang, J. Zhang; Host-Madsen, ”On the capacity of MIMO relaychannels”, IEEE Trans. Inform. Theory, vol.21, pp. 29-43, Jan 2005

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