In the post on Soft Input Viterbi decoder, we had discussed BPSK modulation with convolutional coding and soft input Viterbi decoding in AWGN channel. Let us know discuss the derivation of soft bits for 16QAM modulation scheme with Gray coded bit mapping. The channel is assumed to be AWGN alone.

Gray Mapped 16-QAM constellation

In the past, we had discussed BER for 16QAM in AWGN modulation. The 4 bits in each constellation point can be considered as two bits each on independent 4-PAM modulation on I-axis and Q-axis respectively.

Table: Gray coded constellation mapping for 16-QAM

Figure: 16QAM constellation plot with Gray coded mapping

Channel Model

The received coded sequence is

, where

is the modulated coded sequence taking values in the alphabet

.

is the Additive White Gaussian Noise following the probability distribution function,

with mean and variance .

Demodulation

For demodulation, we would want to maximize the probability that the bit was transmitted given we received i.e . This criterion is called Maximum a posteriori probability (MAP).

Using Bayes rule,

.

Note: The probability that all constellation points occur are equally likely, so maximizing is equivalent to maximizing .

Soft bit for b0

The bit mapping for the bit b0 with 16QAM Gray coded mapping is shown below. We can see that when b0 toggles from 0 to 1, only the real part of the constellation is affected.

Figure: Bit b0 for 16QAM Gray coded mapping

When the b0 is 0, the real part of the QAM constellation takes values -3 or -1. The conditional probability of the received signal given b0 is 0 is,

.

When the bit0 is 1, the real part of the QAM constellation takes values +1 or +3. The conditional probability given b0 is zero is,

.

We can define a likelihood ratio that if

.

The likelihood ratio for b0 is,

.

Region #1 ()

When , then we can assume that relative contribution by constellation +3 in the numerator and -1 in the denominator is less and can be ignored. So the likelihood ratio reduces to,

.

Taking logarithm on both sides,

.

Region #2 (), Region #3 ()

When or , then we can assume that relative contribution by constellation +3 in the numerator and -3 in the denominator is less and can be ignored. So the likelihood ratio reduces to,

.

Taking logarithm on both sides,

.

Region #4 ()

If , then we can assume that relative contribution by constellation +1 in the numerator and -3 in the denominator is less and can be ignored. So the likelihood ratio reduces to,

.

Taking logarithm on both sides,

.

Soft bit for b1

The bit mapping for the bit b1 with 16QAM Gray coded mapping is shown below. We can see that when b0 toggles from 0 to 1, only the real part of the constellation is affected.

Figure: Bit b1 for 16QAM Gray coded mapping

When the b1 is zero, the real part of the QAM constellation takes values -3 or +3. The conditional probability given b0 is zero is,

.

When the bit0 is 1, the real part of the QAM constellation takes values -1 or +1. The conditional probability given b0 is zero is,

.

We can define a likelihood ratio that if

.

The likelihood ratio for b1 is,

.

Region #1 (), Region#2 ()

When or , then we can assume that relative contribution by constellation +1 in the numerator and +3 in the denominator is less and can be ignored. So the likelihood ratio reduces to,

.

Taking logarithm on both sides,

.

Region #3 (), Region #4 ()

If or , then we can assume that relative contribution by constellation -1 in the numerator and -3 in the denominator is less and can be ignored. So the likelihood ratio reduces to,

.

Taking logarithm on both sides,

.

Summary

Note: As the factor  is common to all the terms, it can be removed.

The softbit for bit b0 is,

.

The softbit for bit b1 is,

.

The softbit for bit b1 can be simplified to,

.

It is easy to observe that the softbits for bits b2, b3 are identical to softbits for b0, b1 respectively except that the decisions are based on the imaginary component of the received vector .

The softbit for bit b2 is,

.

The softbit for bit b3 is,

.

As described in the paper, Simplified Soft-Output Demapper for Binary Interleaved COFDM with Application to HIPERLAN/2, Filippo Tosato1, Paola Bisaglia, HPL-2001-246 October 10th , 2001, the term

and

,

This simplification avoids the need for having a threshold check in the receiver for sofbits b0 and b2 respectively.

Reference

Simplified Soft-Output Demapper for Binary Interleaved COFDM with Application to HIPERLAN/2, Filippo Tosato1, Paola Bisaglia, HPL-2001-246 October 10th , 2001

My friend Mr. Balaji volunteers for Vibha, a non-profit  organization whose mission is to ensure that every underprivileged child attains his or her right to education, health and opportunity.

Vibha, which was founded in 1991 has a volunteer network of 825 members spread across Atlanta, Austin, Bay Area, Boston, Chicago, Dallas, Houston, Jacksonville, Los Angeles, Milwaukee, Twin Cities – Minnesota, New York, Philadelphia, Sacramento, Washington DC and several other cities across the US and India. Till date, we have supported about 190 projects in India and the US. You may read more about their projects here.

To support Vibha’s cause, Mr Balaji has set up a donation page – The Dream Mile … A few miles for a million dreams . The Dream Mile 5k/10k Run/Walk is Vibha’s flagship event for increasing awareness and raising funds to help underprivileged children. Vibha has chosen  My New Red Shoes as one of the beneficiaries of this event.

To find out more about My New Red Shoes, please watch the video below.

To find out more about Alamb, a project by Vibha in India, please watch the video below.

If you like some of the work done by Vibha, please consider making a donation to the cause. As Mr. Balaji has mentioned:

$120 can provide education to a child for a whole year

$50 can facilitate the rehabilitation of a mentally handicapped child for a year

$40 can sponsor the non-formal education of a child for a whole year.

To donate for this cause, please visit the donation page The Dream Mile … A few miles for a million dreams

Thanks again for your support.

In a post on Minimum Shift Keying (MSK), we had discussed that MSK uses two frequencies which are separated by and phase discontinuity is avoided in symbol boundaries. In that post, we had discussed MSK as a continuous phase transmit signal and showed that phase changes through 0, 90, 180 and 270 degrees. In this post, we will discuss MSK transmission as a variant of offset-QPSK technique. Further, we will discuss the receiver structure and show that bit error rate with coherent demodulation of MSK (using time) is equivalent to that of BPSK modulation. The channel assumed is AWGN.

MSK transmitter

Using bit-sequence and the explanation provided in the excellent paper  Minimum shift keying: A spectrally efficient modulation Subbarayan Pasupathy, IEEE Communications Magazine, July 1979 as a reference,  let us proceed as follows:

a) Consider a 8 bit sequence which is BPSK modulated as [+1, +1, -1, -1, -1, +1, +1 , +1] having symbol duration of .

b) The same sequence can be sent over QPSK modulation, where even bits are send on I-arm and the odd bits are send over Q-arm. To keep the same data rate, the bits on the I/Q arm are send for symbol periods.

Figure: Transmission of BPSK/QPSK modulation

c) Now, a variant of QPSK called offset-QPSK (O-QPSK) can be generated by having a relative delay between the I-arm and Q-arm by symbol period .

Note: The advantage of O-QPSK is that phase of the signal can jump at a maximum of only 90 degrees (when compared to 180 degrees in QPSK). Having a smaller phase jump ensures that the spectrum of the waveform is cleaner even when there are distortions in the transmitter.

d) Researchers have found that they can make the phase transitions zero, if  rectangular pulse shapes used in O-QPSK are replaced by sinusoidal pulse shapes i.e by using and on the I and Q arm respectively. This is also known as Minimum Shift Keying (MSK) !

Figure: Transmission of O-QPSK/MSK modulation

Assuming that the carrier frequency is , the MSK transmission can be written as

,

where

are the even pulse sequence (send on the I arm) and

are the odd pulse sequence (send on the Q arm).

Using trigonometric identities, the above equation can also be written as,

,

where

and

.

We can see that MSK can indeed be visualized as a form of frequency shift keying (FSK), where the two frequencies are and .

Note: This equation is comparable to that described in the post Simulating MSK transmission

The MSK transmitter block diagram is shown below

Figure: Block diagram of MSK transmitter

MSK Receiver

The receiver for the MSK transmission can be constructed as follows:

a) Down convert the RF signal to I and Q arms by I/Q down conversion

b) Multiply the I and Q arms by and respectively

c) Integrate over the a period of

d) On I arm, perform hard decision decoding on the integrator output at every to get the even bits 

e) On Q arm, delay by time , perform hard decision decoding on the integrator output at every to get the odd bits

The block diagram is shown below:

Figure: MSK receiver block diagram

Simulation Model

The Matlab/Octave script performs the following

(a) Generate random binary sequence of +1’s and -1’s.

(b) Group them into even and odd symbols

(c) Perform rectangular pulse shaping on the even and odd symbols, delay the odd symbols by

(d) Multiply the even and odd symbols by and respectively and transmit

(e) Add additive white Gaussian noise (AWGN) for the given value of

(f) Multiply the I and Q arm by  and respectively and integrate over period

(g) Using the I arm, recover even bits by performing hard decision decoding on integrator output every time

(h) Using the Q arm, recover even bits by performing hard decision decoding on integrator output delayed by time every time

(i) Count the bit errors

(j) Repeat for multiple values of and plot the simulation and theoretical results.

Click here to download Matlab/Octave script for computing BER with MSK transmission and reception in AWGN channel

Figure: BER plot for MSK transmission/reception in AWGN channel

Observations

1./ The BER with MSK is identical to BER with BPSK modulation. This is because MSK demodulation in the above simulation is performed by integrating the received symbol over  time .

2./ Note that it is indeed possible to demodulate MSK by observing only over time and demodulate it as a FSK with two carriers at and . If such a demodulation is performed, then the BER with MSK will be 3dB poorer (comparable to FSK demodulation) when compared to BPSK modulation.

Reference

Minimum shift keying: A spectrally efficient modulation Subbarayan Pasupathy, IEEE Communications Magazine, July 1979

Using the services of a new author ‘RV’, we are starting a new series of articles in the blog. Typically in India, many of the competitive examinations pertaining to Engineering (GATE, IES) and rectuitment by private and public sector companies (ISRO, BSNL, BEL, BHEL) uses examination with objective questions for the first level screening. We are hoping that this Objective Question Answer series, mostly discussing topics pertaining to Electronics and Communication Engineering, will help those preparing for those examination. Kindly do give your feedback via comment and/or email. Thanks, Krishna

Q#1: SQNR with pulse code modulation

If the number of bits per sample in a Pulse Coded Modulation (PCM) system is increased from 5 bits to 6 bits,the improvement in signal to quantization noise ratio will be (a) 3 dB (b) 6 dB (c) 2pi dB (d) 0 dB

Explanation

In a pulse coded modulation system, the analog voltage is quantized into discrete voltage. In the post on Signal to Quantization Noise Ratio (SNR) for a quantized sinusoidal, we have derived that the

, where is the number of bits in the pulse coded modulation system.

So if the number of bits in a pulse coded modulation system (aka ADC) is increased by 1, the improvement in SQNR is around 6dB. Right answer (b).

Q#2 Image frequency in superheterodyne receiver

A superheterodyne radio receiver with an intermediate frequency of 455 kHz is tuned to a station operating at 1200 KHz. The associated image frequency is ________ kHz

(a) 1200 (b) 2110 (c) 1100 (d) 455

Explanation

The simple structure of an RF mixer used in a super hetero-dyne receiver is shown below.

Figure: RF mixer in a superheterodyne receiver

Lets assume the following:

is the RF frequency,

is the local oscillator frequency and

is the intermediate frequency.

We know, from basic trigonometric identities that the output at the mixer will consist of spectrum at and .

Note:

We can safely assume that the the is filtered away. The output of the mixer is a real signal and hence have both +ve and -ve spectrum. Please refer to the post on Negative frequency for bit more details on spectrum of real and complex sinusoidals.

In our example, we have and with . The possible choices of LO frequency are

(a) , if

(b) , if

Now, if we assume that

(a) , its easy to guess that both and can result in

(b) , its easy to guess that both and can result in .

So the image frequencies are and if .

Given that we do not have 290kHz as a choice, its reasonable to assume that the questionnaire assumed and hence . Right answer : (b)

Q#3 Bandwidth of wide band FM

A 10 MHz carrier is frequency modulated with a sinusoidal signal at 500Hz, the maximum frequency deviation being 50 KHz. The bandwidth required as given by the Carson’s rule is

(a) 101kHz (b) 50kHz (c) 105 kHz (d) 100kHz

Explanation

In an FM transmission if we send a a sinusoidal at frequency is

, where is the maximum frequency deviation caused by the message signal.

With some sophisticated mathematical tricks, the signal can be represented as, ,

where is called the modulation index, and

is the Bessel function of first kind for the value .

Strictly, the spectrum of the FM modulated signal has components at frequencies where is an integer from 0 to .

However, the value of tends to 0 as .

Narrow band FM

If , then , , and

Then the FM signal can be approximated as,

.

This type of FM signal, also called as narrow band FM requires bandwidth of .

Wideband FM

In a wideband FM, the modulation index i.e . Its intuitiveve to guess that we require at-least as the bandwidth. Further, according to Carson’s rule bandwidth of FM in the scenario is given by .

So, for our question, the bandwidth is 2*(50+0.5) = 101kHz. Right answer (a)

Reference : Frequency Modulation, Phase Modulation and FM spectra

Q#4 Aliases with sampling

A 1.0 kHz signal is sampled at the rate of 1.8 kHz and the samples are applied to an ideal rectangular LPF with cut-off frequency of 1.1 kHz, then the output of the filter contains

(a) Only 800 Hz component (b) 800 Hz and 900 Hz components

(c) 800 Hz and 100 Hz components (d) 800Hz, 900 and 100 Hz components

Explanation

According to Nyquist sampling theorem, a bandlimited signal should be sampled atleast greater than the twice of the maximum signal. If this condition is not satisfied aliasing will take place. Figure: Aliasing caused due to under sampling

In the above question, 1.0 kHz signal should be sampled at least at the rate of 2kHz. In this example the sampling was performed at 1.8kHz, whih means the frequency range which does not cause aliasing is from -0.9kHz to +0.9kHz. Any frequency outside this range will get aliased to this range.

In general, if the frequency and the sampling frequency is , the aliased signal will be at .

In this example, the frequency at 1kHz will get aliased to 1-1.8 = -0.8kHz and the frequency at -1kHz will get aliased to -1+1.8 = 0.8kHz. After low pass filtering by 1.1kHz, only 0.8kHz component remains. Right answer (a)

Q#5: Sampling Double sideband suppressed carrier signal

A 4 GHz carrier is DSB Sc modulated by a low pass message signal with maximum frequency of 2 MHz. The resultant signal is to be ideally sampled. The minimum frequency of the sampling in train should be

(a) 4 MHz (b) 8 MHz (c) 8 GHz (d) 8.004 GHz

Explanation

According to Nyquist sampling theorem, a bandlimited signal should be sampled equal to greater than the twice of the maximum frequency component of the signal. In this example, the spectrum after up-conversion has a maximum frequency of 4000MHz + 2MHz = 4002MHz. So the sampling frequency required to prevent aliasing 8004MHz = 8.004GHz. Right answer (d)

In the post on transmit pulse shaping filter, we had discussed pulse shaping using rectangular and sinc. In this post we will discuss about optimal receiver structure when pulse shaping is used at the transmitter. The receiver structure is also called as matched filter. For the discussion, we will assume rectangular pulse shaping, the channel is AWGN only and the modulation is BPSK.

Transmitter

The simplified transmitter structure is as shown below:

Transmit pulse shaping filter

Figure: Transmit block diagram

The sequence of operation are as follows:

a). The random binary data comes as input to the the BPSK modulator circuit.

b). The BPSK modulator maps bits to symbols i.e bit0 to -1 and bit1 to +1

The sequence of transmit symbols maybe visualized as follows.

Figure: Transmit symbols for baseband PAM transmission

c). The upsampling block (the block with upward arrow) insert zeros between the samples based on the oversampling factor. For example, if the oversampling factor is 4, the block inserts 3 zeros between samples.

d). The last block in the chain is the filter, where the upsampled sequence is convolved with the filter. In this example, we have assumued rectangular filtering (also known as zero order hold), i.e. to repeat the current symbol till the next symbol arrives and so on. Mathematically, the filter can be represented as,

.

The filtered waveform can be as shown below.

Figure: Baseband PAM transmisison with rectangular filtering

The transmit spectrum assume that the symbol duration is shown below.

Figure: Transmit spectrum for rectangular matched filter

Matched filter Receiver

The filter having an impulse response is called a matched filter of .

Matched filter response for rectangular pulse

Figure: Matched filter response for a rectangular pulse shaping filter

In this example, as is a rectangular function, the mtched filter is also rectangular.

We pass the received signal through the matched filter . The output of the convolution peaks at time and we sample the at that time instant to perform the demodulation.

Further, if the filtered transmit signal is corrupted by noise, the filter with impulse response matched to maximizes the output Signal to Noise Ratio. The proof for this claim is detailed in Chapter 5.1.2 Matched Filter Demodulator in Digital Communications by John Proakis.

Simulation Model

The Matlab/Octave script performs the following

(a) Generate random binary sequence of +1’s and -1’s.

(b) Upsample by inserting zeros and the convolve with rectangular filter

(c) Add white Gaussian noise.

(d) Convolve the received samples with matched filter, and extract output at time T

(e) Perform hard decision decoding and count the bit errors

(f) Repeat for multiple values of and plot the simulation and theoretical results.

Click here to download Matlab/Octave script for computing the BER for BPSK modulation in AWGN only with using rectangular pulse shaping at the transmitter and corresponding matched filter

Figure: BER plot for BPSK with rectangular pulse shaping and matched filtering

Observations

Can see that the simulated results are in good agreement with the theoretical BER results for BPSK in AWGN in AWGN.

References

Digital Communications by John Proakis.

In the past, we had discussed several posts on two transmit two receive MIMO communication, where the transmission was based on V-BLAST. The details about V-BLAST can be read from the landmark paper V-BLAST: An architeture for realizing very high data rates over the rich scattering wireless channel – P. W. Wolniansky, G. J. Foschini, G. D. Golden, R. A. Valenzuela. We will assume that the channel is a flat fading Rayleigh multipath channel and the modulation is BPSK.

Figure: 2 transmit 2 receive MIMO channel

V-BLAST transmission for 2×2 MIMO channel

In a 2×2 MIMO channel, probable usage of the available 2 transmit antennas can be as follows:

1. Consider that we have a transmission sequence, for example

2. In normal transmission, we will be sending in the first time slot, in the second time slot, and so on.

3. However, as we now have 2 transmit antennas, we may group the symbols into groups of two. In the first time slot, send and from the first and second antenna. In second time slot, send and from the first and second antenna, send and in the third time slot and so on.

4. Notice that as we are grouping two symbols and sending them in one time slot, we need only time slots to complete the transmission – data rate is doubled !

5. This forms the simple explanation of a probable MIMO transmission scheme with 2 transmit antennas and 2 receive antennas.

Other Assumptions

1. The channel is flat fading – In simple terms, it means that the multipath channel has only one tap. So, the convolution operation reduces to a simple multiplication. For a more rigorous discussion on flat fading and frequency selective fading, may I urge you to review Chapter 15.3 Signal Time-Spreading from [DIGITAL COMMUNICATIONS: SKLAR]

2. The channel experience by each transmit antenna is independent from the channel experienced by other transmit antennas.

3. For the transmit antenna to receive antenna, each transmitted symbol gets multiplied by a randomly varying complex number . As the channel under consideration is a Rayleigh channel, the real and imaginary parts of are Gaussian distributed having mean and variance .

4. The channel experienced between each transmit to the receive antenna is independent and randomly varying in time.

5. On the receive antenna, the noise has the Gaussian probability density function with

with and .

7. The channel is known at the receiver.

System Model

The received signal on the first receive antenna is,

.

The received signal on the second receive antenna is,

.

where

, are the received symbol on the first and second antenna respectively,

is the channel from transmit antenna to receive antenna,

is the channel from transmit antenna to receive antenna,

is the channel from transmit antenna to receive antenna,

is the channel from transmit antenna to receive antenna,

, are the transmitted symbols and

is the noise on receive antennas.

We assume that the receiver knows , , and . The receiver also knows and . The unknown s are and . Two equations and two unknowns. Can we solve it? Answer is YES.

For convenience, the above equation can be represented in matrix notation as follows:

.

Equivalently,

Receiver structures

MIMO with Zero Forcing Equalization

The zero forcing approach tries to find a matrix which satisfies . The Zero Forcing (ZF) linear detector for meeting this constraint is given by,

.

MIMO with MMSE Equalization

The Minimum Mean Square Error (MMSE) approach tries to find a coefficient which minimizes the criterion,

.

Solving,

.

Zero Forcing Equalization with Successive Interference Cancellation

Using the Zero Forcing (ZF) equalization approach described above, the receiver can obtain an estimate of the two transmitted symbols , , i.e.

.

Take one of the estimated symbols (for example ) and subtract its effect from the received vector and , i.e.

.

Expressing in matrix notation,

,

The above equation is same as equation obtained for receive diversity case. Optimal way of combining the information from multiple copies of the received symbols in receive diversity case is to apply Maximal Ratio Combining (MRC).

The equalized symbol is,

.

This forms the simple explanation for Zero Forcing Equalizer with Successive Interference Cancellation (ZF-SIC) approach.

Zero Forcing Equalization with Optimally ordered Successive Interference Cancellation

In classical Successive Interference Cancellation, the receiver arbitrarily takes one of the estimated symbols, and subtract its effect from the received symbol and . However, we can have more intelligence in choosing whether we should subtract the effect of first or first. To make that decision, let us find out the transmit symbol (after multiplication with the channel) which came at higher power at the receiver. The received power at the both the antennas corresponding to the transmitted symbol is,

.

The received power at the both the antennas corresponding to the transmitted symbol is,

.

If then the receiver decides to remove the effect of from the received vector and and then re-estimate . Else if the receiver decides to subtract effect of from the received vector and , and then re-estimate .

MMSE equalization with optimaly ordered Successive Interference Cancellation

Using the Minimum Mean Square Error (MMSE) equalization, the receiver can obtain an estimate of the two transmitted symbols , , i.e.

.

If then the receiver decides to remove the effect of from the received vector and . Else if the receiver decides to subtract effect of from the received vector and , and then re-estimate .

Once the effect of either or is removed, the new channel becomes a one transmit antenna, 2 receive antenna case and the symbol on the other spatial dimension can be optimally equalized by Maximal Ratio Combining (MRC).

MIMO with ML equalization

The Maximum Likelihood receiver tries to find which minimizes,

Since the modulation is BPSK, the possible values of is +1 or -1 Similarly also take values +1 or -1. So, to find the Maximum Likelihood solution, we need to find the minimum from the all four combinations of and .

The estimate of the transmit symbol is chosen based on the minimum value from the above four values i.e

if the minimum is ,

if the minimum is ,

if the minimum is and
if the minimum is .

Simulation Results

The plot embedded below captures BER for 2 transmit 2 receive MIMO V-BLAST ttansmission/reception for BPSK modulation in a flat fading independent Rayleigh channel, for the different equalizer structures discussed above.

Figure: BER plot for 2 transmit 2 receive MIMO channel for BPSK modulation

Observations

1. The BER curve with ZF equalization for 2×2 MIMO channel is identical to BER plot for 1 transmit 1 receive system

2. Ordered variant of successive interference cancellation shows better performance than the simple successive interference cancellation

3. MMSE equalization with ordered successive interference cancellation provides performance which is slightly poorer than ML.

4. With ML equalization, we come close to the performance of 1 transmit 2 receive MRC case. We gain both throughput gain and diversity gain.

Reference

[DIG-COMM-BARRY-LEE-MESSERSCHMITT] Digital Communication: Third Edition, by John R. Barry, Edward A. Lee, David G. Messerschmitt

V-BLAST: An architeture for realizing very high data rates over the rich scattering wireless channel – P. W. Wolniansky, G. J. Foschini, G. D. Golden, R. A. Valenzuela.

In this post lets discuss a closed-loop transmit diversity scheme, where the transmitter has the knowledge of the channel. As there is a feedback path required from the receiver, to communicate the channel seen by the receiver to the transmitter, the scheme is called closed-loop transmit diversity scheme. Recall that the transmit diversity using Space Time Coding (Alamouti STBC) does not require the knowledge of the channel. In this post, we will restrict our discussion to a 2 transmit, 1 receive case. We will assume that the channel is a flat fading Rayleigh multipath channel and the modulation is BPSK.

Channel Model

1. We have 1 receive antennas and two transmit antenna.

2. The channel is flat fading – In simple terms, it means that the multipath channel has only one tap. So, the convolution operation reduces to a simple multiplication. For a more rigorous discussion on flat fading and frequency selective fading, may I urge you to review Chapter 15.3 Signal Time-Spreading from [DIGITAL COMMUNICATIONS: SKLAR]

3. The channel experienced by each receive antenna is randomly varying in time. For the receive antenna, each transmitted symbol gets multiplied by a randomly varying complex number . As the channel under consideration is a Rayleigh channel, the real and imaginary parts of are Gaussian distributed having mean and variance .

4. The channel experience by each transmit antenna to receive antenna is independent from the channel experienced by other transmit antennas.

5. On each receive antenna, the noise has the Gaussian probability density function with

with and .

6. At each transmit antenna, the channel is known.

Figure: 2 transmit 1 receive beam steering

Transmit Beamforming

On the receive antenna, the received signal is,

where,

is the received symbol,
is the channel on the transmit antenna,
is the transmitted symbol and
is the noise on the receive antenna.

When transmit beamforming is applied, we multiply the symbol from each transmit antenna with a complex number corresponding to the inverse of the phase of the channel so as to ensure that the signals add constructively at the receiver. In this scenario, the received signal is,

,

where,

and

.

In this case, the signal at the receiver is,

.

For equalization, we need to divide the received symbol with the new effective channel, i.e,

.

BER Simulation Model

The Matlab/Octave script performs the following

(a) Generate random binary sequence of +1’s and -1’s.

(b) Multiply the symbols with the beam steering matrics – corresponding to the phase of the channel

(c) Perform equalization at the receiver

(d) Perform hard decision decoding and count the bit errors

(e) Repeat for multiple values of and plot the simulation and theoretical results.

Click here to download Matlab/Octave script for simulting BER for BPSK in flat fading Rayleigh channel with and without beamforming

Figure: BER plot for 2 transmit 1 receive beamforming for BPSK in Rayleigh channel

Observations

1. Sending the same information on multiple transmit antenna does not provide diversity gain. Intuituvely, this is due to the fact that the effective channel in a 2 transmit antenna case is again a Rayleigh channel; hence the BER performance is identical to 1 transmit 1 receive Rayleigh channel case.

2. If the transmit symbols are multiplied by a complex phase to ensure that the phases align at the receiver, there is diversity gain. However, the BER performance seems to be slighly poorer than the 1 transmit 2 receive MRC case. I guess, the noise is scaled by in the case of transmit beamforming, whereas the noise scaling is different in the case of Maximal Ratio Combining. I need to study bit more for a precise answer.

Reference

[DIG-COMM-BARRY-LEE-MESSERSCHMITT] Digital Communication: Third Edition, by John R. Barry, Edward A. Lee, David G. Messerschmitt