Abstract

Single-user massive multiple-input multiple-output (MIMO) systems have a large number of antennas at the transmitter and receiver. This results in a large overall throughput (bit-rate), of the order of tens of gigabits per second, which is the main objective of the recent fifth-generation (5G) wireless standard. It is feasible to have a large number of antennas in mm-wave frequencies, due to the small size of the antennas. This chapter deals with the coherent detection of orthogonal frequency division multiplexed (OFDM) signals transmitted through frequencyselective Rayleigh fading MIMO wireless channels. Low complexity, discrete-time algorithms are developed for channel estimation, carrier and timing synchronization, and finally turbo decoding of the data at the receiver. Computer simulation results are presented to validate the theory.

Keywords: 5G, channel capacity, channel estimation, single-user massive MIMO, OFDM, spatial multiplexing, retransmissions, synchronization, turbo codes

### 1. Introduction

The main objective of the fifth-generation [1–15] wireless communication standard is to provide peak data rates of 10 gigabit per second (Gbps) for each user, ultralow latency (the time duration between transmission of information and getting a response) of less than 1 ms, and, last but not the least, very low bit error rates (BER) ( < 1010). High data rates are essential for streaming ultrahigh definition (4k) video. Low latency is required for future driverless cars and remote surgeries. An important feature of the 5G network is that it involves not only people but also smart devices. For example, it may be possible to control a microwave oven or geyser located in the home, from the office. High data rates are feasible by using a large number of transmitting antennas. For example, if each transmit antenna transmits at a rate of 100 megabits per second (Mbps), then using 100 transmit antennas would result in an overall bit-rate of 10 Gbps. This technique of increasing the overall bit-rate by using a large number of transmit antennas is also known as spatial multiplexing (not to be confused with spatial modulation [16–20], wherein not all the transmit antennas are simultaneously active). This is illustrated in Figure 1, where the i th transmit antenna sends Ci bits of information and each of the receive antennas gets C=N bits

Figure 1. Illustration of spatial multiplexing for N � N MIMO.

of information, in each transmission (see Proposition A.1 and A.2 in [21]). It must be noted that a large array of transmit antennas can also be used for beamforming [22, 23] and beam steering (the ability to focus the transmitted signal in a particular direction, without moving the antenna), which is not the topic of this chapter. In fact, the basic idea used in this chapter is captured in the following proposition.

Proposition 1.1 Signals transmitted and received by antennas separated by at least λ=2 (λ ¼ c=ν where c is the velocity of light and ν is the carrier frequency) undergo independent fading.

A typical massive MIMO antenna array is shown in Figure 2. The black dots denote the antennas, and the circles denote obstructions used to prevent mutual coupling between the antennas. While spatial multiplexing is a big advantage in massive MIMO, the main problem lies in the high complexity of data detection at the receiver. To understand this issue, consider the signal model:

$$
\tilde{\mathbf{R}} = \tilde{\mathbf{H}} \mathbf{S} + \tilde{\mathbf{W}} \tag{1}
$$

additive white Gaussian noise (AWGN) vector. Here C denotes the set of complex numbers. Due to Proposition 1.1, the elements of H~ are statistically independent. Moreover, if there is no line-of-sight (LOS) path between the transmitter and receiver, the elements of H~ are zero-mean Gaussian. The elements of W~ are also assumed to be independent. The real and imaginary parts of the elements of H~ and W~ are also assumed to be independent. Now, the problem statement is find S given R~ . There are several methods of solving this problem, assuming that H~ is known.

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

1. Perform an exhaustive search over all the MN possibilities of S. This is known as the maximum likelihood (ML) approach, which has got an exponential

2. Pre-multiply R~ with H~ 1. This is known as the zero-forcing approach and has a complexity of the order of 2N<sup>3</sup> (N<sup>3</sup> complexity for computing the inverse and another N<sup>3</sup> for matrix multiplication). This approach usually leads to noise

3. The third approach, known as sphere decoding, has polynomial complexity ( <sup>C</sup><sup>0</sup> <sup>N</sup><sup>C</sup><sup>1</sup> , where <sup>C</sup>0>1 and <sup>C</sup>1>3) and has been widely studied in the

Data detection in single-user massive MIMO systems using retransmissions,

been proposed [34], where it was assumed that H~ is known at the receiver. In this work, which is an extension of [34], we present a coherent receiver for massive MIMO systems, where not only H~ but also the carrier frequency offset and timing are estimated. Moreover, the signal model in Eq. (1) is valid for flat fading channels. When the channel is frequency selective (the length of the discrete-time channel impulse response is greater than unity), orthogonal frequency division multiplexing needs to be used, since OFDM converts a frequency-selective channel into a flat fading channel (length of the discrete-time channel impulse response is equal to unity) [35]. To this end, the channel estimation and carrier and timing synchronization algorithms developed in [36] for single-input single-output (SISO) OFDM, [37, 38] for single-input multiple-output (SIMO) OFDM, and [21, 39] for multipleinput multiple-output (MIMO) OFDM are used in this work. In [40], a linear prediction-based detection of serially concatenated QPSK is presented, which does not require any preamble. The prospect of using superimposed training [41] in the context of massive MIMO looks quite intimidating, since the signal at each receive antenna is already a superposition of the signals from a large number of transmit

This work is organized as follows. Section 2 presents the system model. The discrete-time receiver algorithms are presented in Section 3. The computer simulation results are discussed in Section 4, and the chapter concludes with Section 5.

The transmitted frame structure is shown in Figure 3(a). The signal in the blue boxes is sent from transmit antenna nt. The signal in the red boxes is sent from other antennas. Note that in the preamble phase, only one transmit antenna is active at a time, whereas in the data phase, all transmit antennas are active simultaneously. In

, where Nrt is the number of retransmissions, has

enhancement and a poor symbol error rate (SER) performance.

complexity.

DOI: http://dx.doi.org/10.5772/intechopen.85893

literature [24–33].

antennas.

49

2. System model

having a complexity of Nrt <sup>N</sup><sup>3</sup>

where R~ ∈ C<sup>N</sup>�<sup>1</sup> is the received vector, H~ ∈ C<sup>N</sup>�<sup>N</sup> is the channel matrix, S∈ C<sup>N</sup>�<sup>1</sup> is the symbol vector drawn from an M-ary 2D constellation, and W~ ∈ C<sup>N</sup>�<sup>1</sup> is the

Figure 2. A massive MIMO antenna array.

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

additive white Gaussian noise (AWGN) vector. Here C denotes the set of complex numbers. Due to Proposition 1.1, the elements of H~ are statistically independent. Moreover, if there is no line-of-sight (LOS) path between the transmitter and receiver, the elements of H~ are zero-mean Gaussian. The elements of W~ are also assumed to be independent. The real and imaginary parts of the elements of H~ and W~ are also assumed to be independent. Now, the problem statement is find S given R~ . There are several methods of solving this problem, assuming that H~ is known.


Data detection in single-user massive MIMO systems using retransmissions, having a complexity of Nrt <sup>N</sup><sup>3</sup> , where Nrt is the number of retransmissions, has been proposed [34], where it was assumed that H~ is known at the receiver. In this work, which is an extension of [34], we present a coherent receiver for massive MIMO systems, where not only H~ but also the carrier frequency offset and timing are estimated. Moreover, the signal model in Eq. (1) is valid for flat fading channels. When the channel is frequency selective (the length of the discrete-time channel impulse response is greater than unity), orthogonal frequency division multiplexing needs to be used, since OFDM converts a frequency-selective channel into a flat fading channel (length of the discrete-time channel impulse response is equal to unity) [35]. To this end, the channel estimation and carrier and timing synchronization algorithms developed in [36] for single-input single-output (SISO) OFDM, [37, 38] for single-input multiple-output (SIMO) OFDM, and [21, 39] for multipleinput multiple-output (MIMO) OFDM are used in this work. In [40], a linear prediction-based detection of serially concatenated QPSK is presented, which does not require any preamble. The prospect of using superimposed training [41] in the context of massive MIMO looks quite intimidating, since the signal at each receive antenna is already a superposition of the signals from a large number of transmit antennas.

This work is organized as follows. Section 2 presents the system model. The discrete-time receiver algorithms are presented in Section 3. The computer simulation results are discussed in Section 4, and the chapter concludes with Section 5.

### 2. System model

The transmitted frame structure is shown in Figure 3(a). The signal in the blue boxes is sent from transmit antenna nt. The signal in the red boxes is sent from other antennas. Note that in the preamble phase, only one transmit antenna is active at a time, whereas in the data phase, all transmit antennas are active simultaneously. In

of information, in each transmission (see Proposition A.1 and A.2 in [21]). It must be noted that a large array of transmit antennas can also be used for beamforming [22, 23] and beam steering (the ability to focus the transmitted signal in a particular direction, without moving the antenna), which is not the topic of this chapter. In fact,

Proposition 1.1 Signals transmitted and received by antennas separated by at least

A typical massive MIMO antenna array is shown in Figure 2. The black dots denote the antennas, and the circles denote obstructions used to prevent mutual coupling between the antennas. While spatial multiplexing is a big advantage in massive MIMO, the main problem lies in the high complexity of data detection at

where R~ ∈ C<sup>N</sup>�<sup>1</sup> is the received vector, H~ ∈ C<sup>N</sup>�<sup>N</sup> is the channel matrix, S∈ C<sup>N</sup>�<sup>1</sup> is the symbol vector drawn from an M-ary 2D constellation, and W~ ∈ C<sup>N</sup>�<sup>1</sup> is the

<sup>R</sup><sup>~</sup> <sup>¼</sup> HS <sup>~</sup> <sup>þ</sup> <sup>W</sup><sup>~</sup> (1)

the basic idea used in this chapter is captured in the following proposition.

the receiver. To understand this issue, consider the signal model:

independent fading.

Illustration of spatial multiplexing for N � N MIMO.

Figure 1.

Multiplexing

Figure 2.

48

A massive MIMO antenna array.

λ=2 (λ ¼ c=ν where c is the velocity of light and ν is the carrier frequency) undergo

The term i in the above equations denotes the i

DOI: http://dx.doi.org/10.5772/intechopen.85893

kth retransmission are CN 0; 2σ<sup>2</sup>

1 2 E ~ hk,n,nr,nt

1 2 E ~ hk,n,nr,nt

1 2 E ~ hk,n,nr,nt

1 2 E ~ hk,n,nr,nt

quasi-static, that is, ~

CN 0; 2σ<sup>2</sup>

where

51

w � � and satisfy

neously from all the transmit antennas. The channel coefficients ~

f

Gaussian random variable) and satisfy the following relations [21]:

~ h ∗ k,m,nr,nt h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

~ h ∗ k,n,mr,nt h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

~ h ∗ k,n,nr,mt h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

~ h ∗ i,n,nr,nt h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

The channel span assumed by the receiver is [21, 36, 39].

The length of the cyclic prefix or suffix is [21, 36, 39].

E w~ k,n,nr

E w~ k,n,nr

E w~ k,n,nr

w~ <sup>∗</sup> k,m,nr h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

w~ <sup>∗</sup> k,n,mr h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

w~ <sup>∗</sup> i,m,nr h i <sup>¼</sup> <sup>σ</sup><sup>2</sup>

The noise and channel coefficients are assumed to be independent. The frequency offset ω<sup>0</sup> is uniformly distributed over ½ � �0:03; 0:03 radians, and the ML frequency offset estimator searches in the range ½ � �ω0,max;ω0,max radians [42]

1 2

1 2

1 2

index, and 1≤ nt ≤ N is the index to the transmit antenna. Note that in this work, the same preamble is transmitted one after the other by each of the transmit antennas, as shown in Figure 3(a). In [21], different preambles are transmitted simulta-

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

with the receive antenna nr (1≤ nr ≤ N) and transmit antenna nt (1≤ nt ≤ N) for the

where "\*" denotes complex conjugate and δKð Þ� is the Kronecker delta function. Observe that Eq. (3) implies a uniform power delay profile. Even though an exponential power delay profile is more realistic, we have used a uniform power delay profile, since it is expected to give the worst-case BER performance, as all the multipath components have the same power [21]. The channel is assumed to be

The length of all the N<sup>2</sup> channel impulse responses is assumed to be Lh, which is proportional to the difference between the longest and shortest multipath [21].

The length of the preamble is Lp, and the length of the data is Ld. The AWGN noise samples <sup>w</sup><sup>~</sup> k,n,nr for the <sup>k</sup>th retransmission at time <sup>n</sup> and receive antenna nr are

� � (CN ð Þ� denotes a circularly symmetric

<sup>f</sup> δKð Þ n � m

<sup>f</sup> δKð Þ nr � mr

<sup>f</sup> δKð Þ nt � mt

Lhr ¼ 2Lh � 1: (4)

Lcp ¼ Lhr � 1: (5)

<sup>w</sup>δKð Þ n � m

<sup>w</sup>δKð Þ nr � mr

ω0,max ¼ 0:04 radian: (7)

<sup>w</sup>δKð Þ k � i :

<sup>f</sup> δKð Þ k � i

hk,n,nr,nt is time-invariant over one frame (retransmission).

th subcarrier, n denotes the time

hk,n,nr,nt associated

(3)

(6)

practice, each transmit antenna could use a different preamble. However, in this work, we assume that all transmit antennas use the same preamble. The signals in Figure 3(a) are defined as follows (similar to [21]):

$$\begin{aligned} \tilde{s}\_{1,n} &= \frac{1}{L\_p} \sum\_{i=0}^{L\_p - 1} \mathbf{S}\_{1,i} \mathbf{e}^{\mathbf{j} \cdot 2\pi i/L\_p} \quad \text{for} \quad 0 \le n \le L\_p - 1 \\\\ \tilde{s}\_{3,n,n\_t} &= \frac{1}{L\_d} \sum\_{i=0}^{L\_d - 1} \mathbf{S}\_{3,i,n\_t} \mathbf{e}^{\mathbf{j} \cdot 2\pi i/L\_d} \quad \text{for} \quad 0 \le n \le L\_d - 1 \\\\ \tilde{s}\_{2,n,n\_t} &= \tilde{s}\_{3,L\_d - L\_{cp} + n,n\_t} \quad \text{for} \quad 0 \le n \le L\_{cp} - 1 \\\\ \tilde{s}\_{4,n} &= \tilde{s}\_{1,n} \quad \text{for} \quad 0 \le n \le L\_{cp} - 1. \end{aligned} \tag{2}$$

### Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

The term i in the above equations denotes the i th subcarrier, n denotes the time index, and 1≤ nt ≤ N is the index to the transmit antenna. Note that in this work, the same preamble is transmitted one after the other by each of the transmit antennas, as shown in Figure 3(a). In [21], different preambles are transmitted simultaneously from all the transmit antennas. The channel coefficients ~ hk,n,nr,nt associated with the receive antenna nr (1≤ nr ≤ N) and transmit antenna nt (1≤ nt ≤ N) for the kth retransmission are CN 0; 2σ<sup>2</sup> f � � (CN ð Þ� denotes a circularly symmetric Gaussian random variable) and satisfy the following relations [21]:

$$\begin{aligned} \frac{1}{2} \mathcal{E} \left[ \ddot{h}\_{k,n,n,n\_{r},n\_{t}} \ddot{h}\_{k,m,n\_{r},n\_{t}}^{\*} \right] &= \ \sigma\_{f}^{2} \delta\_{K}(n-m) \\\\ \frac{1}{2} \mathcal{E} \left[ \ddot{h}\_{k,n,n\_{r},n\_{t}} \ddot{h}\_{k,n,m,n\_{r},n\_{t}}^{\*} \right] &= \ \sigma\_{f}^{2} \delta\_{K}(n\_{r}-m\_{r}) \\\\ \frac{1}{2} \mathcal{E} \left[ \ddot{h}\_{k,n,n\_{r},n\_{t}} \ddot{h}\_{k,n,n\_{r},m\_{t}}^{\*} \right] &= \ \sigma\_{f}^{2} \delta\_{K}(n\_{t}-m\_{t}) \end{aligned} \tag{3}$$
 
$$\frac{1}{2} \mathcal{E} \left[ \ddot{h}\_{k,n,n\_{r},n\_{t}} \ddot{h}\_{i,n,n\_{r},n\_{t}}^{\*} \right] &= \sigma\_{f}^{2} \delta\_{K}(k-i)$$

where "\*" denotes complex conjugate and δKð Þ� is the Kronecker delta function. Observe that Eq. (3) implies a uniform power delay profile. Even though an exponential power delay profile is more realistic, we have used a uniform power delay profile, since it is expected to give the worst-case BER performance, as all the multipath components have the same power [21]. The channel is assumed to be quasi-static, that is, ~ hk,n,nr,nt is time-invariant over one frame (retransmission). The length of all the N<sup>2</sup> channel impulse responses is assumed to be Lh, which is proportional to the difference between the longest and shortest multipath [21]. The channel span assumed by the receiver is [21, 36, 39].

$$L\_{hr} = 2L\_h - 1.\tag{4}$$

The length of the cyclic prefix or suffix is [21, 36, 39].

$$L\_{cp} = L\_{hr} - \mathbf{1}.\tag{5}$$

The length of the preamble is Lp, and the length of the data is Ld. The AWGN noise samples <sup>w</sup><sup>~</sup> k,n,nr for the <sup>k</sup>th retransmission at time <sup>n</sup> and receive antenna nr are CN 0; 2σ<sup>2</sup> w � � and satisfy

$$\frac{1}{2}E\left[\tilde{w}\_{k,n,n\_r}\tilde{w}\_{k,m,n\_r}^\*\right] = \sigma\_w^2 \delta\_K(n-m)$$

$$\frac{1}{2}E\left[\tilde{w}\_{k,n,n\_r}\tilde{w}\_{k,n,m\_r}^\*\right] = \sigma\_w^2 \delta\_K(n\_r - m\_r) \tag{6}$$

$$\frac{1}{2}E\left[\tilde{w}\_{k,n,n\_r}\tilde{w}\_{i,m,n\_r}^\*\right] = \sigma\_w^2 \delta\_K(k-i).$$

The noise and channel coefficients are assumed to be independent. The frequency offset ω<sup>0</sup> is uniformly distributed over ½ � �0:03; 0:03 radians, and the ML frequency offset estimator searches in the range ½ � �ω0,max;ω0,max radians [42] where

$$
\rho\_{0,\text{max}} = 0.04 \,\text{radian}.\tag{7}
$$

practice, each transmit antenna could use a different preamble. However, in this work, we assume that all transmit antennas use the same preamble. The signals in

(a) Frame structure for kth retransmission. (b) Signal from transmit antenna nt. (c) Receiver for the data phase.

~s2,n,nt ¼ ~s3,Ld�Lcpþn,nt for 0≤n ≤Lcp � 1

~s4,n ¼ ~s1,n for 0≤ n≤Lcp � 1:

<sup>S</sup>1,iej 2πni=Lp for 0 <sup>≤</sup>n<sup>≤</sup> Lp � <sup>1</sup>

<sup>S</sup>3,i,ntej 2<sup>π</sup>ni=Ld for 0 <sup>≤</sup><sup>n</sup> <sup>≤</sup>Ld � <sup>1</sup>

(2)

Figure 3(a) are defined as follows (similar to [21]):

Ld ∑ Ld�1 i¼0

<sup>~</sup>s1,n <sup>¼</sup> <sup>1</sup> Lp ∑ Lp�1 i¼0

Figure 3.

Multiplexing

50

<sup>~</sup>s3,n,nt <sup>¼</sup> <sup>1</sup>

For convenience, and without loss of generality, we assume that ω<sup>0</sup> is constant over Nrt retransmissions.

During the preamble phase, the signal at receive antenna nr, for the kth retransmission, can be written as (for 0≤n≤ Lp þ Lcp þ Lh � 2)

$$\begin{aligned} \tilde{\boldsymbol{r}}\_{k,n,n\_{r},n\_{t},p} &= \left( \tilde{\boldsymbol{s}}\_{5,n} \star \tilde{\boldsymbol{h}}\_{k,n,n\_{r},n\_{t}} \right) \mathbf{e}^{\mathbf{j}\boldsymbol{\alpha}\boldsymbol{\eta}\mathbf{n}} + \tilde{\boldsymbol{w}}\_{k,n,n\_{r},n\_{t},p} \\ &= \tilde{\boldsymbol{y}}\_{k,n,n\_{r},n\_{t},p} \mathbf{e}^{\mathbf{j}\boldsymbol{\alpha}\boldsymbol{\eta}\mathbf{n}} + \tilde{\boldsymbol{w}}\_{k,n,n\_{r},n\_{t},p} \end{aligned} \tag{8}$$

where "⋆" denotes linear convolution, <sup>~</sup>s5,n is depicted in Figure 3(a), <sup>~</sup> hk,n,nr,nt denotes the channel impulse response between transmit antenna nt and receive antenna nr for the kth retransmission, and

$$
\tilde{\mathcal{Y}}\_{k,n,n\_r,n\_t,p} = \tilde{\mathfrak{s}}\_{\mathsf{S},n} \star \tilde{h}\_{k,n,n\_r,n\_t}.\tag{9}
$$

SNRav,b,p ¼

DOI: http://dx.doi.org/10.5772/intechopen.85893

where [36]

(see Proposition A.2 in [21]).

FFT for the i

53

retransmission, is

can be written as (for 0≤n ≤Ld þ Lcp þ Lh � 2)

~rk,n,nr,d ¼ ∑

where ~s6,n,nt is depicted in Figure 3(a) and

N nt¼1

~yk,n,nr,d ¼ ∑

<sup>R</sup>~k,i,nr,d <sup>¼</sup> <sup>∑</sup>

N nt¼1

where <sup>H</sup><sup>~</sup> k,i,nr,nt and <sup>W</sup><sup>~</sup> k,i,nr,d denote the Ld-point FFT of <sup>~</sup>

respectively. The average SNR per bit corresponding to Eq. (18) is

~s6,n,nt ⋆ ~

<sup>¼</sup> <sup>~</sup>yk,n,nr,de<sup>j</sup>ω0<sup>n</sup> <sup>þ</sup> <sup>w</sup><sup>~</sup> k,n,nr,d

N nt¼1

¼

<sup>¼</sup> <sup>A</sup><sup>2</sup>

E H~ k,i,nr,nt � � � � <sup>2</sup> h i <sup>¼</sup> <sup>2</sup>Lhσ<sup>2</sup>

E W~ k,i,nr,nt,p �

� �

E 2A<sup>2</sup> H~ k,i,nr,nt � � � �

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

E W~ k,i,nr,nt,p �

2A<sup>2</sup> 2Lhσ<sup>2</sup>

Lhσ<sup>2</sup> <sup>f</sup> NNrt

> Lpσ<sup>2</sup> w

� <sup>2</sup> h i <sup>¼</sup> <sup>2</sup>Lpσ<sup>2</sup>

E S1,i j j<sup>2</sup> h i <sup>¼</sup>

where A is a constant to be determined and it is assumed that each sample of each receive antenna gets 2=ð Þ NNrt bits of information during the preamble phase

During the data phase, the signal for the kth retransmission at receive antenna nr

hk,n,nr,nt � �ej<sup>ω</sup>0<sup>n</sup> <sup>þ</sup> <sup>w</sup><sup>~</sup> k,n,nr,d

<sup>~</sup>s6,n,nt <sup>⋆</sup> <sup>~</sup>

th (0 <sup>≤</sup>i<sup>≤</sup> Ld � 1) subcarrier and receive antenna nr, during the <sup>k</sup>th

The subscript "d" in Eqs. (16) and (17) denotes data. Assuming perfect carrier and timing synchronization at the receiver, the signal at the output of the Ld-point

<sup>H</sup><sup>~</sup> k,i,nr,nt

hk,n,nr,nt

2Lpσ<sup>2</sup> w � � � <sup>2</sup>

2 h i

� �

f � �NNrt

� <sup>2</sup> h i �

f

w

<sup>Δ</sup> 2A<sup>2</sup>

¼ Es

NNrt 2

(14)

(15)

(16)

: (17)

hk,n,nr,nt and w~ k,n,nr,d,

<sup>S</sup>3,i,nt <sup>þ</sup> <sup>W</sup><sup>~</sup> k,i,nr,d (18)

The subscript "p" in Eqs. (8) and (9) denotes the preamble. Note that any random carrier phase can be absorbed in the channel impulse response. We have

$$\begin{aligned} \left| \tilde{s}\_{1,n} \oslash\_{L\_p} \tilde{s}\_{1,-n}^\* = E\_i \delta\_K(n) \\\\ \overset{L\_p}{\rightleftharpoons} \left| \tilde{S}\_{1,i} \right|^2 \\\\ = \text{a constant for } 0 \le i \le L\_p - 1 \end{aligned} \tag{10}$$

where " ⊙ Lp " denotes an Lp-point circular convolution, "⇌ Lp " denotes the Lp-point discrete Fourier transform (DFT) or the fast Fourier transform (FFT), and

$$E\_s = \sum\_{n=0}^{L\_p - 1} |\tilde{s}\_{1,n}|^2. \tag{11}$$

Due to the presence of the cyclic suffix, we have

$$
\bar{s}\_{5,n} \star \bar{s}\_{1,-n}^\* \begin{cases} = 0 & \text{for } 1 \le n \le L\_{hr} - 1 \\\\ = E\_\circ & \text{for } n = 0 \\\\ \ll E\_\circ & \text{otherwise.} \end{cases} \tag{12}
$$

Assuming perfect carrier and timing synchronization (ω<sup>0</sup> is perfectly canceled and the frame boundaries are perfectly known) at the receiver, the signal at the output of the Lp-point FFT for the i th (0 <sup>≤</sup><sup>i</sup> <sup>≤</sup>Lp � 1) subcarrier and receive antenna nr, due to the preamble sent from transmit antenna nt during the kth retransmission, is

$$
\tilde{R}\_{k,i,n\_r,n\_t,p} = \tilde{H}\_{k,i,n\_r,n\_t} \mathbf{S}\_{\mathbf{1},i} + \tilde{W}\_{k,i,n\_r,n\_t,p} \tag{13}
$$

where <sup>H</sup><sup>~</sup> k,i,nr,nt and <sup>W</sup><sup>~</sup> k,i,nr,nt,p denote the Lp-point FFT of <sup>~</sup> hk,n,nr,nt and w~ k,n,nr,nt,p, respectively. The average SNR per bit corresponding to Eq. (13) is Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

$$\begin{split} \text{SNR}\_{\text{av},b,p} &= \frac{E\left[2A^2 \left| \bar{H}\_{k,i,n,n\_t,n} \right|^2 \right]}{E\left[\left| \bar{W}\_{k,i,n,n,p} \right|^2 \right]} \times \frac{\text{NN}\_{rt}}{2} \\ &= \frac{2A^2 \left(2L\_h \sigma\_f^2 \right) \text{NN}\_{rt}}{(2L\_p \sigma\_w^2) \times 2} \\ &= \frac{A^2 L\_h \sigma\_f^2 \text{NN}\_{rt}}{L\_p \sigma\_w^2} \end{split} \tag{14}$$

where [36]

For convenience, and without loss of generality, we assume that ω<sup>0</sup> is constant

hk,n,nr,nt � �

<sup>¼</sup> <sup>~</sup>yk,n,nr,nt,pej<sup>ω</sup>0<sup>n</sup> <sup>þ</sup> <sup>w</sup><sup>~</sup> k,n,nr,nt,p

where "⋆" denotes linear convolution, <sup>~</sup>s5,n is depicted in Figure 3(a), <sup>~</sup>

denotes the channel impulse response between transmit antenna nt and receive

<sup>~</sup>yk,n,nr,nt,p <sup>¼</sup> <sup>~</sup>s5,n <sup>⋆</sup><sup>~</sup>

The subscript "p" in Eqs. (8) and (9) denotes the preamble. Note that any random carrier phase can be absorbed in the channel impulse response.

<sup>1</sup>,�<sup>n</sup> <sup>¼</sup> EsδKð Þ <sup>n</sup>

⇌ Lp ~ S1,i � � � � 2

where " ⊙ Lp " denotes an Lp-point circular convolution, "⇌

Due to the presence of the cyclic suffix, we have

1,�n

8 >><

>>:

~s5,n ⋆~s <sup>∗</sup>

output of the Lp-point FFT for the i

retransmission, is

52

<sup>e</sup>jω0<sup>n</sup> <sup>þ</sup> <sup>w</sup><sup>~</sup> k,n,nr,nt,p

hk,n,nr,nt

¼ a constant for 0≤ i ≤ Lp � 1

Lp-point discrete Fourier transform (DFT) or the fast Fourier transform (FFT), and

¼ Es for n ¼ 0 ≪ Es otherwise:

Assuming perfect carrier and timing synchronization (ω<sup>0</sup> is perfectly canceled and the frame boundaries are perfectly known) at the receiver, the signal at the

antenna nr, due to the preamble sent from transmit antenna nt during the kth

w~ k,n,nr,nt,p, respectively. The average SNR per bit corresponding to Eq. (13) is

<sup>R</sup>~k,i,nr,nt,p <sup>¼</sup> <sup>H</sup><sup>~</sup> k,i,nr,nt

where <sup>H</sup><sup>~</sup> k,i,nr,nt and <sup>W</sup><sup>~</sup> k,i,nr,nt,p denote the Lp-point FFT of <sup>~</sup>

~s1,nj 2 : �

¼ 0 for 1 ≤ n ≤ Lhr � 1

th (0 <sup>≤</sup><sup>i</sup> <sup>≤</sup>Lp � 1) subcarrier and receive

Es ¼ ∑ Lp�1 n¼0

(8)

hk,n,nr,nt

(10)

(12)

: (9)

Lp

� (11)

<sup>S</sup>1,i <sup>þ</sup> <sup>W</sup><sup>~</sup> k,i,nr,nt,p (13)

hk,n,nr,nt and

" denotes the

During the preamble phase, the signal at receive antenna nr, for the kth

retransmission, can be written as (for 0≤n≤ Lp þ Lcp þ Lh � 2)

<sup>~</sup>rk,n,nr,nt,p <sup>¼</sup> <sup>~</sup>s5,n <sup>⋆</sup><sup>~</sup>

antenna nr for the kth retransmission, and

~s1,n ⊙ Lp ~s <sup>∗</sup>

over Nrt retransmissions.

Multiplexing

We have

$$E\left[\left|\ddot{H}\_{k,i,n\_r,n\_t}\right|^2\right] = 2L\_k \sigma\_f^2$$

$$E\left[\left|\ddot{W}\_{k,i,n\_r,n\_t,p}\right|^2\right] = 2L\_p \sigma\_w^2 \tag{15}$$

$$E\left[\left|S\_{1,i}\right|^2\right] \stackrel{\Delta}{=} 2A^2$$

$$= \quad E\_s$$

where A is a constant to be determined and it is assumed that each sample of each receive antenna gets 2=ð Þ NNrt bits of information during the preamble phase (see Proposition A.2 in [21]).

During the data phase, the signal for the kth retransmission at receive antenna nr can be written as (for 0≤n ≤Ld þ Lcp þ Lh � 2)

$$\begin{aligned} \tilde{r}\_{k,n,n\_r,d} &= \sum\_{n\_t=1}^N \left( \tilde{s}\_{k,n,n\_t} \star \tilde{h}\_{k,n,n\_r,n\_t} \right) \mathbf{e}^{\mathrm{i}a \otimes n} + \tilde{w}\_{k,n,n\_r,d} \\\\ &= \tilde{\mathcal{Y}}\_{k,n,n\_r,d} \mathbf{e}^{\mathrm{i}a \otimes n} + \tilde{w}\_{k,n,n\_r,d} \end{aligned} \tag{16}$$

where ~s6,n,nt is depicted in Figure 3(a) and

$$
\tilde{\mathcal{Y}}\_{k,n,n\_r,d} = \sum\_{n\_t=1}^N \tilde{\varsigma}\_{\mathfrak{G},n,n\_t} \star \tilde{h}\_{k,n,n\_r,n\_t}.\tag{17}
$$

The subscript "d" in Eqs. (16) and (17) denotes data. Assuming perfect carrier and timing synchronization at the receiver, the signal at the output of the Ld-point FFT for the i th (0 <sup>≤</sup>i<sup>≤</sup> Ld � 1) subcarrier and receive antenna nr, during the <sup>k</sup>th retransmission, is

$$
\tilde{R}\_{k,i,n\_r,d} = \sum\_{n\_t=1}^{N} \tilde{H}\_{k,i,n\_r,n\_t} \mathbf{S}\_{\mathbf{\tilde{3}},i,n\_t} + \tilde{W}\_{k,i,n\_r,d} \tag{18}
$$

where <sup>H</sup><sup>~</sup> k,i,nr,nt and <sup>W</sup><sup>~</sup> k,i,nr,d denote the Ld-point FFT of <sup>~</sup> hk,n,nr,nt and w~ k,n,nr,d, respectively. The average SNR per bit corresponding to Eq. (18) is

Multiplexing

$$\begin{split} \text{SNR}\_{\text{av},b,d} &= \underbrace{\left[\left|\sum\_{n=1}^{N} \bar{H}\_{k,i,n\_r,n\_t} \text{S}\_{\text{3},i,n\_t}\right|^2\right]}\_{E\left[\left|\bar{W}\_{k,i,n\_r,d}\right|^2\right]} \times 2N\_{rt} \\ &= \frac{2\left(2L\_h \sigma\_f^2\right) N(2N\_{rt})}{2L\_d \sigma\_w^2} \\ &= \frac{4L\_h \sigma\_f^2 N N\_{rt}}{L\_d \sigma\_w^2} \end{split} \tag{19}$$

Therefore, the radio frequency (RF) amplifiers at the transmitter must have a dynamic range of at least (note that the RF amplifiers have to also deal with the

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

1

Let us now consider the case where the preamble power is equal to the data

¼ 2 Ld

> ffiffiffiffiffi Lp Ld r :

A ¼ 10 log <sup>10</sup>ð Þ 4 dB

Lhσ<sup>2</sup> <sup>f</sup> NNrt

<sup>¼</sup> Lhσ<sup>2</sup>

¼ 6 dB:

Lpσ<sup>2</sup> w

<sup>f</sup> NNrt Ldσ<sup>2</sup> w

<sup>f</sup> NNrt Ldσ<sup>2</sup> w

¼ 10 log <sup>10</sup>ð Þ 4 dB:

(24)

(25)

(26)

¼ 6 dB:

<sup>E</sup> j j <sup>~</sup>s1,n <sup>2</sup> h i

<sup>E</sup> <sup>~</sup>s3,n,nt j j<sup>2</sup> h i

power at each transmit antenna. From Eqs. (22) and (23), we have [36]

2A<sup>2</sup> Lp

) A ¼

Substituting for A from Eq. (25), we obtain the average SNR per bit of the

SNRav,b,p <sup>¼</sup> <sup>A</sup><sup>2</sup>

SNRav,b,d <sup>¼</sup> <sup>4</sup>Lhσ<sup>2</sup>

SNRav,b,d SNRav,b,p � �

than that of the data phase by 6 dB. In what follows, we assume that A is

In other words, the average SNR per bit of the preamble phase would be less

The receiver algorithms have been adapted from [21, 36, 37, 39] and will be

The first task of the receiver is to detect the presence of a valid signal, that is, the start of frame (SoF). The SoF detection and coarse frequency offset estimation are performed for each receive antenna 1≤nr ≤ N, transmit antenna 1≤ nt ≤ N, and retransmission 1≤k≤ Nrt as given by the following rule (similar to Eq. (17) in [21]:

peak-to-average power ratio (PAPR) problem [43–49])

0 @

10 log <sup>10</sup>

DOI: http://dx.doi.org/10.5772/intechopen.85893

preamble phase and the data phase as

given by (21).

55

3. Receiver algorithms

) 10 log <sup>10</sup>

briefly described in the following subsections.

3.1 Start of frame and frequency offset estimation

choose that value of m^ <sup>k</sup> nr ð Þ ; nt and ν^<sup>k</sup> nr ð Þ ; nt which maximizes

where [36].

$$\begin{aligned} E\left[\left|\tilde{\boldsymbol{W}}\_{k,i,n\_r,d}\right|^2\right] &= 2\boldsymbol{L}\_d \sigma\_w^2\\ E\left[\left|\mathbf{S}\_{3,i,n\_t}\right|^2\right] &\stackrel{\Delta}{=} 2 \end{aligned} \tag{20}$$

and it is assumed that each receive antenna gets 1=ð Þ 2Nrt bits of information in each transmission [34]. We impose the constraint that

$$\text{SNR}\_{\text{av},b,p} = \text{SNR}\_{\text{av},b,d}$$

$$\Rightarrow \frac{A^2 L\_h \sigma\_f^2 \text{NN}\_{\text{rt}}}{L\_p \sigma\_w^2} = \frac{4 L\_h \sigma\_f^2 \text{NN}\_{\text{rt}}}{L\_d \sigma\_w^2} \tag{21}$$

$$\Rightarrow A = \sqrt{\frac{4 L\_p}{L\_d}}.$$

Let us now compare the average power of the preamble with that of the data, at the transmitter. The average power of the preamble in the time domain is

$$\begin{split} E\left[\left|\bar{\mathbf{s}}\_{1,n}\right|^{2}\right] &= \frac{1}{L\_{p}^{2}} E\left[\sum\_{i=0}^{L\_{p}-1} \mathbf{S}\_{1,i} \mathbf{e}^{\left[2\pi n i/L\_{p}\right]} \sum\_{l=0}^{L\_{p}-1} \mathbf{S}\_{1,l}^{\*} \mathbf{e}^{-j2\pi nl/L\_{p}}\right] \\ &= \frac{1}{L\_{p}^{2}} \sum\_{i=0}^{L\_{p}-1} \sum\_{l=0}^{L\_{p}-1} E\left[\mathbf{S}\_{1,i} \mathbf{S}\_{1,l}^{\*}\right] \times \mathbf{e}^{-j2\pi n(i-l)/L\_{p}} \\ &= \frac{1}{L\_{p}^{2}} \sum\_{i=0}^{L\_{p}-1} \sum\_{l=0}^{L\_{p}-1} 2A^{2} \delta\_{K}(i-l) \times \mathbf{e}^{-j2\pi n(i-l)/L\_{p}} \\ &= \frac{2A^{2}}{L\_{p}} \\ &= \frac{8}{L\_{d}} \end{split} \tag{22}$$

where A is defined in Eqs. (15) and (21). Similarly, the average power of the data in the time domain is

$$E\left[\left|\check{s}\_{3,n,n\_t}\right|^2\right] = \frac{2}{L\_d}.\tag{23}$$

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

Therefore, the radio frequency (RF) amplifiers at the transmitter must have a dynamic range of at least (note that the RF amplifiers have to also deal with the peak-to-average power ratio (PAPR) problem [43–49])

$$\mathbf{10}\log\_{10}\left(\frac{E\left[\left|\check{s}\_{1,n}\right|^2\right]}{E\left[\left|\check{s}\_{3,n,n\_t}\right|^2\right]}\right) = \mathbf{10}\log\_{10}(4)\text{ dB}\tag{24}$$

$$= \mathbf{6}\text{ dB}.$$

Let us now consider the case where the preamble power is equal to the data power at each transmit antenna. From Eqs. (22) and (23), we have [36]

$$\begin{split} \frac{2A^2}{L\_p} &= \frac{2}{L\_d} \\ \Rightarrow A &= \sqrt{\frac{L\_p}{L\_d}}. \end{split} \tag{25}$$

Substituting for A from Eq. (25), we obtain the average SNR per bit of the preamble phase and the data phase as

$$\text{SNR}\_{\text{av},b,p} = \frac{A^2 L\_h \sigma\_f^2 N N\_{rt}}{L\_p \sigma\_w^2}$$

$$= \frac{L\_h \sigma\_f^2 N N\_{rt}}{L\_d \sigma\_w^2}$$

$$\text{SNR}\_{\text{av},b,d} = \frac{4 L\_h \sigma\_f^2 N N\_{rt}}{L\_d \sigma\_w^2} \tag{26}$$

$$\Rightarrow \quad \text{10} \log\_{10} \left( \frac{\text{SNR}\_{\text{av},b,d}}{\text{SNR}\_{\text{av},b,p}} \right) = \text{10} \log\_{10}(4) \text{ dB}.$$

$$= \text{6 dB}.$$

In other words, the average SNR per bit of the preamble phase would be less than that of the data phase by 6 dB. In what follows, we assume that A is given by (21).

### 3. Receiver algorithms

The receiver algorithms have been adapted from [21, 36, 37, 39] and will be briefly described in the following subsections.

### 3.1 Start of frame and frequency offset estimation

The first task of the receiver is to detect the presence of a valid signal, that is, the start of frame (SoF). The SoF detection and coarse frequency offset estimation are performed for each receive antenna 1≤nr ≤ N, transmit antenna 1≤ nt ≤ N, and retransmission 1≤k≤ Nrt as given by the following rule (similar to Eq. (17) in [21]: choose that value of m^ <sup>k</sup> nr ð Þ ; nt and ν^<sup>k</sup> nr ð Þ ; nt which maximizes

SNRav,b,d ¼

where [36].

Multiplexing

¼

each transmission [34]. We impose the constraint that

) A2 Lhσ<sup>2</sup> <sup>f</sup> NNrt

¼ 1 L2 p E ∑ Lp�1 i¼0

¼ 1 L2 p ∑ Lp�1 i¼0 ∑ Lp�1

¼ 1 L2 p ∑ Lp�1 i¼0 ∑ Lp�1

<sup>¼</sup> <sup>2</sup>A<sup>2</sup> Lp

¼ 8 Ld

<sup>E</sup> j j <sup>~</sup>s1,n <sup>2</sup> h i

in the time domain is

54

E ∑ N nt¼1

4

2 2Lhσ<sup>2</sup> f � �

<sup>¼</sup> <sup>4</sup>Lhσ<sup>2</sup>

E W~ k,i,nr,d �

> Lpσ<sup>2</sup> w

the transmitter. The average power of the preamble in the time domain is

l¼0

l¼0 2A<sup>2</sup>

<sup>E</sup> <sup>~</sup>s3,n,nt j j<sup>2</sup> h i

� �

E S3,i,nt j j<sup>2</sup> h i

� <sup>2</sup> h i

and it is assumed that each receive antenna gets 1=ð Þ 2Nrt bits of information in

SNRav,b,p ¼ SNRav,b,d

) A ¼

Let us now compare the average power of the preamble with that of the data, at

S1,iej 2<sup>π</sup>ni=Lp ∑

E S1,iS <sup>∗</sup> 1,l

where A is defined in Eqs. (15) and (21). Similarly, the average power of the data

¼ 2 Ld

<sup>¼</sup> <sup>4</sup>Lhσ<sup>2</sup>

s

ffiffiffiffiffiffiffiffi 4Lp Ld

:

Lp�1

" #

e�j 2πnl=Lp

: (23)

l¼0 S ∗ 1,l

� � � <sup>e</sup>�j 2πn ið Þ �<sup>l</sup> <sup>=</sup>Lp

<sup>δ</sup>Kð Þ� <sup>i</sup> � <sup>l</sup> <sup>e</sup>�j 2πn ið Þ �<sup>l</sup> <sup>=</sup>Lp

<sup>f</sup> NNrt Ldσ<sup>2</sup> w

2Ldσ<sup>2</sup> w

<sup>f</sup> NNrt Ldσ<sup>2</sup> w

� � � � �

H~ k,i,nr,nt

E W~ k,i,nr,d �

� �

Nð Þ 2Nrt

2 2

S3,i,nt

� <sup>2</sup> h i � <sup>2</sup>Nrt

> <sup>¼</sup> <sup>2</sup>Ldσ<sup>2</sup> w

¼ <sup>Δ</sup> 2 � � � � � 3 5

(19)

(20)

(21)

(22)

$$\left| \left( \tilde{r}\_{k,m,n\_r,n\_t,p} \mathbf{e}^{-j\hat{\boldsymbol{\nu}}\_k}(\boldsymbol{n}\_r,\boldsymbol{n}\_t) \boldsymbol{m} \right) \star \tilde{s}\_{1,L\_p-1-m,n\_t}^\* \right| \tag{27}$$

where ILhr is an Lhr � Lhr identity matrix. The estimate of the channel is [21, 36, 39]

~sH

<sup>5</sup> ~rk,m1,nr,nt,p: (35)

<sup>5</sup> w~ k,m1,nr,nt,p: (36)

k,nr,nt

ω<sup>r</sup> ¼ ω<sup>0</sup> � ω^ <sup>0</sup> (39)

ωrLd < 0:1 radians (40)

<sup>m</sup><sup>2</sup> <sup>¼</sup> <sup>m</sup><sup>1</sup> <sup>þ</sup> N Lp <sup>þ</sup> Lcp � � (41)

T

Ld�1

N�1

Ld�<sup>1</sup> (42)

(43)

(44)

<sup>u</sup>ILhr : (37)

(38)

<sup>5</sup> ~s<sup>5</sup> � ��<sup>1</sup>

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

To see the effect of noise on the channel estimate in Eq. (35), consider

~s H

wLd 4Lp

<sup>~</sup>rk,m1,nr,nt,p � <sup>~</sup>s5h^

In this section, we assume that the residual frequency offset given by

so that the effect of inter carrier interference (ICI) is negligible. Let

where m<sup>1</sup> is defined in Eq. (31). Note that m<sup>2</sup> is the starting point of the data phase. Define the FFT input in the time domain for the kth retransmission and

~rk,m2,nr,d ¼ ½ � ~rk,m2,nr,d … ~rk,m2þLd�1,nr,d

<sup>R</sup><sup>~</sup> k,nr,d <sup>¼</sup> <sup>R</sup><sup>~</sup> k,0,nr,d … <sup>R</sup>~k,Ld�1,nr,d

<sup>R</sup><sup>~</sup> k,i,d <sup>¼</sup> <sup>R</sup><sup>~</sup> k,i, <sup>1</sup>,d … <sup>R</sup>~k,i,N,d

where <sup>R</sup>~k,i,nr,d is given by Eq. (18). Construct a matrix:

for 0≤ i≤ Ld � 1. Note that from Eq. (18)

where we have followed the notation in Eq. (16). The Ld-point FFT of Eq. (42) is

h i<sup>T</sup>

h i

� �:

ILhr ¼ <sup>Δ</sup> 2σ<sup>2</sup>

<sup>~</sup>rk,m1,nr,nt,p � <sup>~</sup>s5h^

k,nr,nt

� �<sup>H</sup>

h

It can be shown that [21, 39]

DOI: http://dx.doi.org/10.5772/intechopen.85893

3.3 Noise variance estimation

σ^2

3.4 Post-FFT operations

is such that

receive antenna nr as

57

<sup>w</sup> <sup>¼</sup> <sup>1</sup> 2LpN<sup>2</sup>

^k,nr,nt <sup>¼</sup> <sup>~</sup>s<sup>H</sup>

<sup>u</sup><sup>~</sup> <sup>¼</sup> <sup>~</sup>s<sup>H</sup> <sup>5</sup> ~s<sup>5</sup> � ��<sup>1</sup>

<sup>E</sup> <sup>u</sup>~u~<sup>H</sup> � � <sup>¼</sup> <sup>σ</sup><sup>2</sup>

The noise variance per dimension is estimated as

Nrt ∑ Nrt k¼1 ∑ N nt¼1 ∑ N nr¼1

where ~rk,m,nr,nt,p is given in Eq. (8) and

$$\hat{\nu}\_k(n\_r, n\_t) \in \left\{-\alpha\_{0,\max} + \frac{2lo\_{0,\max}}{B\_1} \right\} \tag{28}$$

for 0≤ l≤ B1, where l and B<sup>1</sup> [21] are positive integers and ω0,max is given in Eq. (7). Observe that m^ <sup>k</sup> nr ð Þ ; nt satisfies Eqs. (18) and (19) in [21]. The average value of the frequency offset estimate is given by

$$
\hat{\boldsymbol{\alpha}}\_{0} = \frac{\sum\_{k=1}^{N\_{rt}} \sum\_{n\_r=1}^{N} \sum\_{n\_t=1}^{N} \hat{\nu}\_k(n\_r, n\_t)}{N^2 N\_{rt}}.\tag{29}
$$

### 3.2 Channel estimation

We assume that the SoF has been estimated using Eq. (27) with outcome m<sup>0</sup> given by (assuming the condition (19) in [21] is satisfied for all k, nr, and nt)

$$m\_0 = \hat{m}\_1(1, 1) - L\_p + 1 \quad 0 \le m\_0 \le L\_h \quad -1 \tag{30}$$

and the frequency offset has been perfectly canceled [36, 38]. Observe that any value of k, nr, and nt can be used in the computation of Eq. (30). We have taken k ¼ nr ¼ nt ¼ 1. Define [21, 36, 39].

$$m\_1 = m\_0 + L\_h - 1.\tag{31}$$

The steady-state, preamble part of the received signal for the kth retransmission and receive antenna nr can be written as [21, 36, 39]

$$
\tilde{\mathbf{r}}\_{k,m\_1,n\_r,n\_t,p} = \tilde{\mathbf{s}}\_{\mathsf{S}} \tilde{\mathbf{h}}\_{k,n\_r,n\_t} + \tilde{\mathbf{w}}\_{k,m\_1,n\_t,n\_r,n\_t,p} \tag{32}
$$

where

$$
\begin{bmatrix}
\tilde{\mathbf{r}}\_{k,m\_{1},n\_{r},n\_{t},p} \\
\tilde{\mathbf{w}}\_{k,m\_{1},n\_{r},n\_{t},p}
\end{bmatrix} = \begin{bmatrix}
\tilde{r}\_{k,m\_{1},n\_{r},n\_{t},p} & \dots & \tilde{r}\_{k,m\_{1}+L\_{p}-1,n\_{r},n\_{t},p}
\end{bmatrix}\_{L\_{p}\times 1}^{T}
$$

$$
\tilde{\mathbf{w}}\_{k,m\_{1},n\_{r},n\_{t},p} = \begin{bmatrix}
\tilde{\omega}\boldsymbol{\upmu}\_{k,m\_{1},n\_{r},n\_{t},p} & \dots & \tilde{\omega}\boldsymbol{\upmu}\_{k,m\_{1}+L\_{p}-1,n\_{r},n\_{t},p}
\end{bmatrix}^{T}\_{L\_{p}\times 1}
$$

$$
\tilde{\mathbf{h}}\_{k,n\_{r},n\_{t}} = \begin{bmatrix}
\tilde{h}\_{k,0,m\_{r},n\_{t}} & \dots & \tilde{h}\_{k,L\_{d}-1,n\_{r},n\_{t}}
\end{bmatrix}^{T}\_{L\_{dr}\times 1}
\tag{33}
$$

$$
\tilde{\mathbf{s}}\_{5} = \begin{bmatrix}
&\tilde{\boldsymbol{s}}\_{5,L\_{dr}} - 1 & \dots & \tilde{\boldsymbol{s}}\_{5,0}
\end{bmatrix}\_{L\_{p}\times L\_{dr}}
$$

$$
\tilde{\mathbf{s}}\_{5} = \begin{bmatrix}
&\vdots & \dots & &\vdots \\
&\vdots & \dots & &\vdots \\
&\vdots & \dots & &\tilde{\mathbf{s}}\_{5,L\_{dr}-1}
\end{bmatrix}\_{L\_{p}\times L\_{dr}}
$$

Observe that ~s<sup>5</sup> is independent of m<sup>1</sup> and due to the relations in Eqs. (10), (15), and (21), we have

$$\tilde{\mathbf{s}}\_{\ $}^{H}\tilde{\mathbf{s}}\_{\$ } = \frac{8L\_p}{L\_d} \mathbf{I}\_{L\_{\&r}}.\tag{34}$$

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

where ILhr is an Lhr � Lhr identity matrix. The estimate of the channel is [21, 36, 39]

$$
\hat{\mathbf{h}}\_{k,n\_r,n\_t} = \left(\tilde{\mathbf{s}}\_5^H \tilde{\mathbf{s}}\_5\right)^{-1} \tilde{\mathbf{s}}\_5^H \tilde{\mathbf{r}}\_{k,m\_1,n\_r,n\_t,p}.\tag{35}
$$

To see the effect of noise on the channel estimate in Eq. (35), consider

$$
\tilde{\mathbf{u}} = \left(\tilde{\mathbf{s}}\_{\mathfrak{F}}^H \tilde{\mathbf{s}}\_{\mathfrak{F}}\right)^{-1} \tilde{\mathbf{s}}\_{\mathfrak{F}}^H \tilde{\mathbf{w}}\_{k, m\_1, n\_r, n\_t, p}. \tag{36}
$$

It can be shown that [21, 39]

<sup>~</sup>rk,m,nr,nt,p <sup>e</sup>�<sup>j</sup> <sup>ν</sup>^<sup>k</sup> nr ð Þ ; nt <sup>m</sup> � �

ν^<sup>k</sup> nr ð Þ ; nt ∈ �ω0,max þ

<sup>k</sup>¼<sup>1</sup>∑<sup>N</sup>

for 0≤ l≤ B1, where l and B<sup>1</sup> [21] are positive integers and ω0,max is given in Eq. (7). Observe that m^ <sup>k</sup> nr ð Þ ; nt satisfies Eqs. (18) and (19) in [21]. The average

nr¼∑<sup>N</sup>

N2 Nrt

We assume that the SoF has been estimated using Eq. (27) with outcome m<sup>0</sup> given by (assuming the condition (19) in [21] is satisfied for all k, nr, and nt)

and the frequency offset has been perfectly canceled [36, 38]. Observe that any value of k, nr, and nt can be used in the computation of Eq. (30). We have taken

The steady-state, preamble part of the received signal for the kth retransmission

h i<sup>T</sup>

� �<sup>T</sup>

� �<sup>T</sup>

~s5,Lhr�<sup>1</sup> … ~s5,<sup>0</sup>

~s5,LpþLhr�<sup>2</sup> … ~s5,Lp�<sup>1</sup>

Observe that ~s<sup>5</sup> is independent of m<sup>1</sup> and due to the relations in Eqs. (10), (15),

<sup>5</sup> <sup>~</sup>s<sup>5</sup> <sup>¼</sup> <sup>8</sup>Lp Ld

⋮ … ⋮

hk,Lhr�1,nr,nt

~rk,m1,nr,nt,p ¼ ~rk,m1,nr,nt,p … ~rk,m1þLp�1,nr,nt,p

<sup>w</sup><sup>~</sup> k,m1,nr,nt,p <sup>¼</sup> <sup>w</sup><sup>~</sup> k,m1,nr,nt,p … <sup>w</sup><sup>~</sup> k,m1þLp�1,nr,nt,p

hk,0,nr,nt … <sup>~</sup>

~sH

� � �

where ~rk,m,nr,nt,p is given in Eq. (8) and

value of the frequency offset estimate is given by

3.2 Channel estimation

Multiplexing

where

and (21), we have

56

k ¼ nr ¼ nt ¼ 1. Define [21, 36, 39].

and receive antenna nr can be written as [21, 36, 39]

<sup>h</sup><sup>~</sup> k,nr,nt <sup>¼</sup> <sup>~</sup>

~s<sup>5</sup> ¼

<sup>~</sup>rk,m1,nr,nt,p <sup>¼</sup> <sup>~</sup>s5h<sup>~</sup>

<sup>ω</sup>^ <sup>0</sup> <sup>¼</sup> <sup>∑</sup>Nrt

⋆~s <sup>∗</sup>

2lω0,max B1

nt¼1ν^<sup>k</sup> nr ð Þ ; nt

m<sup>0</sup> ¼ m^ <sup>1</sup>ð Þ� 1; 1 Lp þ 1 0 ≤ m<sup>0</sup> ≤ Lh � 1 (30)

m<sup>1</sup> ¼ m<sup>0</sup> þ Lh � 1: (31)

k,nr,nt þ w~ k,m1,nr,nt,p (32)

Lhr�1

Lp�Lhr

:

ILhr : (34)

Lp�1

Lp�1

(33)

� �

1,Lp�1�m,nt

� �

� (27)

: (29)

(28)

$$E\left[\tilde{\mathbf{u}}\,\tilde{\mathbf{u}}^H\right] = \frac{\sigma\_w^2 L\_d}{4L\_p} \mathbf{I}\_{L\_{dr}} \stackrel{\Delta}{=} 2\sigma\_u^2 \mathbf{I}\_{L\_{dr}}.\tag{37}$$

### 3.3 Noise variance estimation

The noise variance per dimension is estimated as

$$\begin{split} \hat{\sigma}\_{w}^{2} = \frac{1}{2L\_{p}N^{2}N\_{rt}} \sum\_{k=1}^{N\_{rt}} \sum\_{n=1}^{N} \sum\_{n\_{r}=1}^{N} \left( \tilde{\mathbf{r}}\_{k,m\_{1},n\_{r},n\_{t},p} - \tilde{\mathbf{s}}\_{5} \hat{\mathbf{h}}\_{k,n\_{r},n\_{t}} \right)^{H} \\ \qquad \left( \tilde{\mathbf{r}}\_{k,m\_{1},n\_{r},n\_{t},p} - \tilde{\mathbf{s}}\_{5} \hat{\mathbf{h}}\_{k,n\_{r},n\_{t}} \right). \end{split} \tag{38}$$

### 3.4 Post-FFT operations

In this section, we assume that the residual frequency offset given by

$$a\rho\_r = a\rho\_0 - \hat{a}\rho\_0 \tag{39}$$

is such that

$$
\alpha\_r L\_d < 0.1 \quad \text{radians} \tag{40}
$$

so that the effect of inter carrier interference (ICI) is negligible. Let

$$m\_2 = m\_1 + N(L\_p + L\_{cp})\tag{41}$$

where m<sup>1</sup> is defined in Eq. (31). Note that m<sup>2</sup> is the starting point of the data phase. Define the FFT input in the time domain for the kth retransmission and receive antenna nr as

$$
\tilde{\mathbf{r}}\_{k,m\_{2\star},n\_r,d} = \begin{bmatrix}
\tilde{r}\_{k,m\_{2\star},n\_r,d} & \dots & \tilde{r}\_{k,m\_2+L\_d-1,n\_r,d}
\end{bmatrix}^T\_{L\_d \times \mathbf{1}} \tag{42}
$$

where we have followed the notation in Eq. (16). The Ld-point FFT of Eq. (42) is

$$
\tilde{\mathbf{R}}\_{k,n\_r,d} = \begin{bmatrix}
\tilde{R}\_{k,0,n\_r,d} & \dots & \tilde{R}\_{k,L\_d-1,n\_r,d}
\end{bmatrix}\_{L\_d \times 1}^T \tag{43}
$$

where <sup>R</sup>~k,i,nr,d is given by Eq. (18). Construct a matrix:

$$
\tilde{\mathbf{R}}\_{k,i,d} = \begin{bmatrix}
\tilde{\mathbf{R}}\_{k,i,1,d} & \dots & \tilde{\mathbf{R}}\_{k,i,N,d}
\end{bmatrix}\_{N \times 1} \tag{44}
$$

for 0≤ i≤ Ld � 1. Note that from Eq. (18)

Multiplexing

$$
\ddot{\mathbf{R}}\_{k,i,d} = \ddot{\mathbf{H}}\_{k,i} \mathbf{S}\_{3,i} + \dot{\mathbf{W}}\_{k,i,d} \tag{45}
$$

where σ^<sup>2</sup>

where

efficiency is

Lp = 512

Lcp = 18 Nrt = 2

Lp = 4096

Lcp = 18 Nrt = 2

Throughput for various simulation parameters.

Table 1.

59

FFT of (35). The term Fi,nt in Eq. (51) is given by

DOI: http://dx.doi.org/10.5772/intechopen.85893

where again nt is an odd integer.

3.5 Throughput and spectral efficiency

<sup>w</sup> is given by Eq. (38) and <sup>H</sup>^ k,i,nr,nt is obtained by taking the Ld-point

H^ k,i,nr,nt � � � � 2

� �

" #

2σ<sup>2</sup> U

Nrt N Lp <sup>þ</sup> Lcp � � <sup>þ</sup> Ld <sup>þ</sup> Lcp � � : (57)

N = 8 38.77%

N = 8 39.84%

Fk,i,nt (54)

� 2

: (55)

. The equations for

(56)

Fi,nt <sup>¼</sup> <sup>1</sup> Nrt ∑ Nrt k¼1

Fk,i,nt ¼ ∑

Eqs. (32) and (33) of [34], with γ1,i,n, <sup>ρ</sup>þð Þ <sup>n</sup> replaced by γ1,i,n, <sup>ρ</sup>þð Þ <sup>n</sup> ,nt

N nr¼1

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

The extrinsic information from decoder 1 to decoder 2 is computed using

decoder 2 are similar, except that γ2,i,m,n in Eq. (34) of [34] should be replaced by

<sup>γ</sup>2,i,m,n,ntþ<sup>1</sup> <sup>¼</sup> exp � <sup>Y</sup><sup>~</sup> i,ntþ<sup>1</sup> � Fi,ntþ<sup>1</sup>Sm,n �

Recall from Figure 3(a) that during the preamble phase, only one transmit antenna is active at a time, whereas during the data phase, all the transmit antennas

The numerator of Eq. (57) denotes the total number of data bits transmitted, and the denominator represents the total number of QPSK symbol durations over Nrt retransmissions. The symbol rate during the preamble phase and data phase is the same. In the data phase, we are transmitting coded QPSK, that is, in each data bit duration, two coded QPSK symbols are sent simultaneously from two transmit antennas (see Figure 3(b)). Thus, during the data phase, each transmit antenna sends half a bit of information in each transmission. Therefore, the spectral

Simulation parameters Throughput T

Ld = 1024 N = 4 32.38%

Ld = 8192 N = 4 33.21%

are simultaneously active. Thus the throughput can be defined as [36, 37].

<sup>T</sup> <sup>¼</sup> NLd=<sup>2</sup>

where

$$
\tilde{\mathbf{H}}\_{k,i} = \begin{bmatrix}
\tilde{H}\_{k,i,1,1} & \dots & \tilde{H}\_{k,i,1,N} \\
\vdots & \dots & \vdots \\
\tilde{H}\_{k,i,N,1} & \dots & \tilde{H}\_{k,i,N,N}
\end{bmatrix}\_{N \times N} \tag{46}
$$

$$
\mathbf{S}\_{3,i} = \begin{bmatrix}
\mathbf{S}\_{3,i,1} & \dots & \mathbf{S}\_{3,i,N}
\end{bmatrix}\_{N \times 1}^T
$$

$$
\tilde{\mathbf{W}}\_{k,i,d} = \begin{bmatrix}
\tilde{W}\_{k,i,1,d} & \dots & \tilde{W}\_{k,i,N,d}
\end{bmatrix}\_{N \times 1}^T.
$$

which is similar to Eq. (1) in [34]. Let

$$\tilde{\mathbf{Y}}\_{k,i} = \hat{\mathbf{H}}\_{k,i}^H \tilde{\mathbf{R}}\_{k,i,d} = \hat{\mathbf{H}}\_{k,i}^H \tilde{\mathbf{H}}\_{k,i} \mathbf{S}\_{\mathbf{\hat{3}},i} + \hat{\mathbf{H}}\_{k,i}^H \tilde{\mathbf{W}}\_{k,i,d} \tag{47}$$

where H^ k,i is constructed from the Ld-point FFT of h ^k,nr,nt in Eq. (35) and <sup>Y</sup>^ k,i is similar to <sup>Y</sup><sup>~</sup> <sup>k</sup> in Eq. (6) of [34]. The analysis when

$$
\hat{\mathbf{H}}\_{k,i} = \mathbf{\tilde{H}}\_{k,i} \tag{48}
$$

is given in [34]. Let

$$\tilde{\mathbf{Y}}\_{i} = \frac{\mathbf{1}}{N\_{rt}} \sum\_{k=1}^{N\_{rt}} \tilde{\mathbf{Y}}\_{k,i} \qquad \text{ for } \ 0 \le i \le L\_d - \mathbf{1}. \tag{49}$$

Note that <sup>Y</sup>~<sup>i</sup> is an <sup>N</sup> � 1 matrix, whose <sup>n</sup>th <sup>t</sup> element <sup>Y</sup>~i,nt is a noisy version of S3,i,nt in Eq. (18). The matrix

$$\tilde{\mathbf{Y}}\_{n\_{\ell}} = \left[ \tilde{\mathbf{Y}}\_{0,n\_{\ell}} \dots \tilde{\mathbf{Y}}\_{L\_{d}-1,n\_{\ell}} \right]\_{L\_{d} \times \mathbf{1}}^{T} \tag{50}$$

constructed from the elements of <sup>Y</sup>~<sup>i</sup> in Eq. (49) is fed to the turbo decoder. The forward (α) backward (β) recursions for decoder 1 of the turbo code is given by Eqs. (28) and (31) in [34]. The term γ1,i,m,n in Eq. (30) of [34] should be replaced by

$$\gamma\_{1,i,m,n,n\_t} = \exp\left[-\frac{\left|\tilde{Y}\_{i,n\_t} - F\_{i,n\_t}\mathbf{S}\_{m,n}\right|^2}{2\sigma\_U^2}\right] \tag{51}$$

where <sup>Y</sup>~i,nt is an element of <sup>Y</sup><sup>~</sup> nt in Eq. (50) and nt is an odd integer. The term <sup>σ</sup><sup>2</sup> U in Eq. (51) is given by Eq. (22) in [34] which is repeated here for convenience:

$$
\sigma\_U^2 = \frac{1}{N\_{rt}} \left( 4N\sigma\_H^2 \sigma\_W^2 + 8N(N-1)\sigma\_H^4 \right) \tag{52}
$$

with

$$\begin{aligned} \sigma\_W^2 &= \, \, L\_d \hat{\sigma}\_w^2\\ \sigma\_H^2 &= \frac{1}{2N^2 N\_{rt} L\_d} \sum\_{k=1}^{N\_{rt}} \sum\_{i=0}^{L\_d - 1} \sum\_{n\_r=1}^N \sum\_{n\_t=1}^N \left| \hat{H}\_{k, i, n\_r, n\_t} \right|^2 \end{aligned} \tag{53}$$

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

where σ^<sup>2</sup> <sup>w</sup> is given by Eq. (38) and <sup>H</sup>^ k,i,nr,nt is obtained by taking the Ld-point FFT of (35). The term Fi,nt in Eq. (51) is given by

$$F\_{i,n\_t} = \frac{1}{N\_{rt}} \sum\_{k=1}^{N\_{rt}} F\_{k,i,n\_t} \tag{54}$$

where

<sup>R</sup><sup>~</sup> k,i,d <sup>¼</sup> <sup>H</sup><sup>~</sup> k,i

S3,i ¼ ½ � S3,i, <sup>1</sup> … S3,i,N

<sup>R</sup><sup>~</sup> k,i,d <sup>¼</sup> <sup>H</sup>^ <sup>H</sup>

<sup>Y</sup><sup>~</sup> nt <sup>¼</sup> <sup>Y</sup>~0,nt … <sup>Y</sup>~Ld�1,nt

h i<sup>T</sup>

constructed from the elements of <sup>Y</sup>~<sup>i</sup> in Eq. (49) is fed to the turbo decoder. The forward (α) backward (β) recursions for decoder 1 of the turbo code is given by Eqs. (28) and (31) in [34]. The term γ1,i,m,n in Eq. (30) of [34] should be replaced by

> <sup>γ</sup>1,i,m,n,nt <sup>¼</sup> exp � <sup>Y</sup><sup>~</sup> i,nt � Fi,ntSm,n �

in Eq. (51) is given by Eq. (22) in [34] which is repeated here for convenience:

4Nσ<sup>2</sup> Hσ<sup>2</sup>

<sup>W</sup> <sup>¼</sup> Ldσ^ <sup>2</sup>

<sup>H</sup> <sup>¼</sup> <sup>1</sup> 2N<sup>2</sup>

σ2

σ2

where <sup>Y</sup>~i,nt is an element of <sup>Y</sup><sup>~</sup> nt in Eq. (50) and nt is an odd integer. The term <sup>σ</sup><sup>2</sup>

w

NrtLd

∑ Nrt k¼1 ∑ Ld�1 i¼0 ∑ N nr¼1 ∑ N nt¼1

<sup>W</sup><sup>~</sup> k,i,d <sup>¼</sup> <sup>W</sup><sup>~</sup> k,i, <sup>1</sup>,d … <sup>W</sup><sup>~</sup> k,i,N,d

H~ k,i, <sup>1</sup>, <sup>1</sup> … H~ k,i, <sup>1</sup>,N

H~ k,i,N, <sup>1</sup> … H~ k,i,N,N

� �<sup>T</sup>

k,i <sup>H</sup><sup>~</sup> k,i

⋮ … ⋮

T N�1

<sup>S</sup>3,i <sup>þ</sup> <sup>H</sup>^ <sup>H</sup> k,i

<sup>H</sup><sup>~</sup> k,i <sup>¼</sup>

which is similar to Eq. (1) in [34]. Let

<sup>Y</sup><sup>~</sup> k,i <sup>¼</sup> <sup>H</sup>^ <sup>H</sup>

similar to <sup>Y</sup><sup>~</sup> <sup>k</sup> in Eq. (6) of [34]. The analysis when

<sup>Y</sup>~<sup>i</sup> <sup>¼</sup> <sup>1</sup> Nrt ∑ Nrt k¼1

Note that <sup>Y</sup>~<sup>i</sup> is an <sup>N</sup> � 1 matrix, whose <sup>n</sup>th

σ2 <sup>U</sup> <sup>¼</sup> <sup>1</sup> Nrt

is given in [34]. Let

S3,i,nt in Eq. (18). The matrix

with

58

k,i

where H^ k,i is constructed from the Ld-point FFT of h

where

Multiplexing

<sup>S</sup>3,i <sup>þ</sup> <sup>W</sup><sup>~</sup> k,i,d (45)

(46)

(50)

(51)

<sup>2</sup> (53)

U

> N�1 :

<sup>H</sup>^ k,i <sup>¼</sup> <sup>H</sup><sup>~</sup> k,i (48)

<sup>Y</sup><sup>~</sup> k,i for 0 <sup>≤</sup>i<sup>≤</sup> Ld � <sup>1</sup>: (49)

� 2

H � � (52)

> H^ k,i,nr,nt � � � �

Ld�1

� �

" #

2σ<sup>2</sup> U

<sup>W</sup> <sup>þ</sup> <sup>8</sup>N Nð Þ � <sup>1</sup> <sup>σ</sup><sup>4</sup>

<sup>t</sup> element <sup>Y</sup>~i,nt is a noisy version of

<sup>W</sup><sup>~</sup> k,i,d (47)

^k,nr,nt in Eq. (35) and <sup>Y</sup>^ k,i is

$$F\_{k,i,n\_t} = \sum\_{n\_r=1}^{N} \left| \hat{H}\_{k,i,n\_r,n\_t} \right|^2. \tag{55}$$

The extrinsic information from decoder 1 to decoder 2 is computed using Eqs. (32) and (33) of [34], with γ1,i,n, <sup>ρ</sup>þð Þ <sup>n</sup> replaced by γ1,i,n, <sup>ρ</sup>þð Þ <sup>n</sup> ,nt . The equations for decoder 2 are similar, except that γ2,i,m,n in Eq. (34) of [34] should be replaced by

$$\gamma\_{2,i,m,n,n\_t+1} = \exp\left[-\frac{\left|\tilde{Y}\_{i,n\_t+1} - F\_{i,n\_t+1}\mathbf{S}\_{m,n}\right|^2}{2\sigma\_U^2}\right] \tag{56}$$

where again nt is an odd integer.

### 3.5 Throughput and spectral efficiency

Recall from Figure 3(a) that during the preamble phase, only one transmit antenna is active at a time, whereas during the data phase, all the transmit antennas are simultaneously active. Thus the throughput can be defined as [36, 37].

$$\mathcal{P} = \frac{\text{NL}\_d/2}{N\_{\text{rt}} \left[ N \left( L\_p + L\_{\text{cp}} \right) + L\_d + L\_{\text{cp}} \right]}. \tag{57}$$

The numerator of Eq. (57) denotes the total number of data bits transmitted, and the denominator represents the total number of QPSK symbol durations over Nrt retransmissions. The symbol rate during the preamble phase and data phase is the same. In the data phase, we are transmitting coded QPSK, that is, in each data bit duration, two coded QPSK symbols are sent simultaneously from two transmit antennas (see Figure 3(b)). Thus, during the data phase, each transmit antenna sends half a bit of information in each transmission. Therefore, the spectral efficiency is


Table 1. Throughput for various simulation parameters.

$$\mathcal{G}' = \mathcal{N}/(2\mathcal{N}\_{\pi}) \quad \text{bits per transmission.} \tag{58}$$

The throughput for various simulation parameters is given in Table 1. Observe that when Lp ¼ Ld=2, Lcp ≪ Ld, and N ≫ 1, T ! 1=Nrt. In this work, we have used a rate-1=2 turbo code, that is, each data bit generates two coded QPSK symbols. The throughput can be doubled by using a rate-1 turbo code, obtained by puncturing.

### 4. Simulation results

The simulation parameters are given in Table 2. A "run" in Table 2 is defined as transmitting and receiving the frame in Figure 3(a) over Nrt retransmissions. The generating matrix of each of the constituent encoders of the turbo code is given by Eq. (49) in [21]. A question might arise: how does N ¼ 4, 8 correspond to a massive MIMO system, whereas in [34] N was as large as 512? The answer is in [34], an ideal massive MIMO was considered, wherein the channel, timing, and carrier frequency offset were assumed to be known, whereas in this work, the channel, timing, and carrier frequency offset are estimated. The estimation complexity and memory requirement increase as N<sup>2</sup> , for an N � N MIMO system. For example, the memory requirement of Eq. (27) when the number of frequency bins B<sup>1</sup> ¼ 1024 [21], preamble length Lp ¼ 4096, cyclic prefix length Lcp ¼ 18, channel length Lh ¼ 10, N ¼ 8 transmit and receive antennas, and Nrt ¼ 4 retransmissions is

$$\begin{aligned} \text{(memory requirement)} &= \left(L\_p + L\_{cp} + L\_h - \mathbf{1}\right) (B\_1 + \mathbf{1}) N^2 N\_{rt} \\ &= \mathbf{1081875200} \end{aligned} \tag{59}$$

double precision values. In fact Eq. (27) is implemented using multidimensional arrays in Scilab, instead of using for loops. Note that from Eq. (8), the length of the received signal during the preamble phase is Lp þ Lcp þ Lh � 1. If for loops are used, the memory requirement would be

$$\begin{aligned} \text{memory requirement} &= \left(L\_p + L\_{cp} + L\_h - 1\right) (B\_1 + 1) \\ &= 4226075 \end{aligned} \tag{60}$$

8 � 8 MIMO system? The answer is no. The solution lies in using multiple carrier frequencies as illustrated in Figure 4. Observe that with 8 � 8 MIMO and M carrier frequencies, we get an overall 8M � 8M MIMO system. The bit error rate results for a 4 � 4 MIMO system are shown in Figure 5. The bit error rate results for an 8 � 8 MIMO system are shown in Figure 6. The following observations can be made from

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

1. There is only 0.75 dB difference in performance between the ideal (id) and estimated (est) receiver, for Ld ¼ 1024, Nrt ¼ 2, and bit error rate equal to 10�4. On the other hand, there is hardly any performance difference between the ideal and estimated receiver for Ld ¼ 8192. This is because the noise

Bit error rate results for a 4 � 4 MIMO system. (a) Ld ¼ 1024 and Lp ¼ 512. (b) Ld ¼ 8192 and Lp ¼ 4096.

Figure 5:

Figure 5.

61

Figure 4.

Massive MIMO using multiple carrier frequencies.

DOI: http://dx.doi.org/10.5772/intechopen.85893

double precision values, which is much less than Eq. (59); however the simulations would run much slower. Does this mean that we cannot go higher than an


Table 2. Simulation parameters. Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893

Figure 4. Massive MIMO using multiple carrier frequencies.

8 � 8 MIMO system? The answer is no. The solution lies in using multiple carrier frequencies as illustrated in Figure 4. Observe that with 8 � 8 MIMO and M carrier frequencies, we get an overall 8M � 8M MIMO system. The bit error rate results for a 4 � 4 MIMO system are shown in Figure 5. The bit error rate results for an 8 � 8 MIMO system are shown in Figure 6. The following observations can be made from Figure 5:

1. There is only 0.75 dB difference in performance between the ideal (id) and estimated (est) receiver, for Ld ¼ 1024, Nrt ¼ 2, and bit error rate equal to 10�4. On the other hand, there is hardly any performance difference between the ideal and estimated receiver for Ld ¼ 8192. This is because the noise

Figure 5. Bit error rate results for a 4 � 4 MIMO system. (a) Ld ¼ 1024 and Lp ¼ 512. (b) Ld ¼ 8192 and Lp ¼ 4096.

S ¼ N=ð Þ 2Nrt bits per transmission: (58)

, for an N � N MIMO system. For example, the memory

Nrt

(59)

(60)

, 10<sup>3</sup>

The throughput for various simulation parameters is given in Table 1. Observe that when Lp ¼ Ld=2, Lcp ≪ Ld, and N ≫ 1, T ! 1=Nrt. In this work, we have used a rate-1=2 turbo code, that is, each data bit generates two coded QPSK symbols. The throughput can be doubled by using a rate-1 turbo code, obtained by puncturing.

The simulation parameters are given in Table 2. A "run" in Table 2 is defined as transmitting and receiving the frame in Figure 3(a) over Nrt retransmissions. The generating matrix of each of the constituent encoders of the turbo code is given by Eq. (49) in [21]. A question might arise: how does N ¼ 4, 8 correspond to a massive MIMO system, whereas in [34] N was as large as 512? The answer is in [34], an ideal massive MIMO was considered, wherein the channel, timing, and carrier frequency offset were assumed to be known, whereas in this work, the channel, timing, and carrier frequency offset are estimated. The estimation complexity and memory

requirement of Eq. (27) when the number of frequency bins B<sup>1</sup> ¼ 1024 [21], preamble length Lp ¼ 4096, cyclic prefix length Lcp ¼ 18, channel length Lh ¼ 10,

N ¼ 8 transmit and receive antennas, and Nrt ¼ 4 retransmissions is

memory requirement <sup>¼</sup> Lp <sup>þ</sup> Lcp <sup>þ</sup> Lh � <sup>1</sup> ð Þ <sup>B</sup><sup>1</sup> <sup>þ</sup> <sup>1</sup> <sup>N</sup><sup>2</sup>

¼ 1081875200

double precision values. In fact Eq. (27) is implemented using multidimensional arrays in Scilab, instead of using for loops. Note that from Eq. (8), the length of the received signal during the preamble phase is Lp þ Lcp þ Lh � 1. If for loops are used,

memory requirement <sup>¼</sup> Lp <sup>þ</sup> Lcp <sup>þ</sup> Lh � <sup>1</sup> ð Þ <sup>B</sup><sup>1</sup> <sup>þ</sup> <sup>1</sup>

¼ 4226075

double precision values, which is much less than Eq. (59); however the simulations would run much slower. Does this mean that we cannot go higher than an

Parameter Value Lp 512, 4096 Ld 1024, 8192 Lh 10 Lhr 19 Lcp 18 N 4, 8 Nrt 1, 2, 4

[21] 64, 1024 Runs 104

4. Simulation results

Multiplexing

requirement increase as N<sup>2</sup>

the memory requirement would be

B1

Table 2.

60

Simulation parameters.

Acknowledgements

Author details

India

63

K. Vasudevan\*, Shivani Singh and A. Phani Kumar Reddy

\*Address all correspondence to: vasu@iitk.ac.in

provided the original work is properly cited.

Department of Electrical Engineering, Indian Institute of Technology, Kanpur,

© 2019 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium,

at IIT Kanpur for running the computer simulations.

DOI: http://dx.doi.org/10.5772/intechopen.85893

The authors would like to thank the high-performance computing (HPC) facility

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions

Figure 6. Bit error rate results for an 8 � 8 MIMO system. (a) Ld ¼ 1024 and Lp ¼ 512. (b) Ld ¼ 8192 and Lp ¼ 4096.

variance (σ<sup>2</sup> <sup>w</sup>) decreases with increasing Lp, Ld, for a given average SNR per bit, as shown by Eq. (21).


### 5. Conclusions

This work describes the discrete-time algorithms for the implementation of a massive MIMO system. Due to the implementation complexity considerations, more than one carrier frequency is required to obtain a truly single-user massive MIMO system. Each carrier frequency needs to be associated with an 8 � 8 or 4 � 4 MIMO subsystem. The average SNR per bit has been used as a performance measure, which has not been done earlier in the literature. Perhaps the channel can also be estimated using Eq. (27), instead of using Eq. (35). This needs investigation.

Coherent Receiver for Turbo Coded Single-User Massive MIMO-OFDM with Retransmissions DOI: http://dx.doi.org/10.5772/intechopen.85893
