**1. Introduction**

As wireless technologies evolve beyond 5G [1–3], there is a growing need to attain peak data rates of about gigabits per second per user, which is required for high definition video, remote surgery, autonomous vehicles, gaming and so on, while at the same time consuming minimum transmit power. This can only be achieved by using multiple antennas at the transmitter and receiver [4–8], small constellations like quadrature shift keying (QPSK) and powerful error correcting codes like turbo or low density parity check (LDPC) codes. Having a large number of antennas in the mobile handset is feasible in mm-wave frequencies [9–12] (30–300 GHz) due to the small antenna size. The main concern about mm wave communications has been its rather

high attenuation in outdoor environments with rain and snow [13]. Therefore, at least in the initial stages, mm wave could be deployed indoors. The second issue relates to the poor penetration characteristics of mm wave through walls, doors, windows and other materials. This points towards to usage of mm wave [9] in a single room, say a big auditorium or underground parking and so on. Reconfigurable intelligent surface (RIS) [11–17] could be used to boost the propagation of mm waves, both indoors and outdoors. Most of the massive MIMO systems discussed in the literature are multiuser (MU) [18–26], that is, the base station has a large number of antennas and the mobile handset has only a single antenna (*Nt* ¼ 1). A large number of users are served simultaneously by the base station. A comparison between MU-MMIMO and SU-MMIMO is given in **Table 1** [27, 28].

The base station in MU-MMIMO uses beamforming to improve the signal-to-noise ratio at the mobile handset. On the other hand, SU-MMIMO uses spatial multiplexing to improve the spectral efficiency in the downlink and uplink. The comparison between beamforming and spatial multiplexing is given in **Table 2** [27, 28]. The total transmit power of SU-MMIMO using uncoded QPSK versus MU-MMIMO using *M*ary QAM is shown in **Table 3**. The minimum Euclidean distance between symbols of all constellations is taken to be 2. The peak-to-average power ratio (PAPR) for SU-MMIMO using QPSK is compared with MU-MMIMO using *M*-ary QAM in **Table 4** [27]. Of course in the case of frequency selective fading channels, OFDM needs to be used, which would result in PAPR greater than 0 dB even for QPSK signaling. It is clear from **Tables 1**–**4** that technologies that use SU-MMIMO have a lot to gain.

Moreover, since all transmit antennas use the same carrier frequency, there is no increase in bandwidth. SU-MMIMO with equal number of transmit and receive


#### **Table 1.**

*Comparison of MU-MMIMO and SU-MMIMO.*


#### **Table 2.**

*Comparison of beamforming and spatial multiplexing.*

*New Results on Single User Massive MIMO DOI: http://dx.doi.org/10.5772/intechopen.112469*


#### **Table 3.**

*SU-MMIMO using QPSK vs. MU-MMIMO using* M*-ary.*


#### **Table 4.**

*PAPR of SU-MMIMO using QPSK vs. MU-MMIMO using* M*-ary.*

antennas is given in [29, 30]. The probability of erasure in MIMO-OFDM is presented in [31]. A practical SU-MMIMO receiver with estimated channel, carrier frequency offset and timing is described in [32, 33]. SU-MMIMO with unequal number of transmit and receive antennas and precoding is discussed in [34, 35] and the case without precoding in [36, 37]. All the earlier research on SU-MMIMO involved the use of a parallel concatenated turbo code (PCTC) and uncorrelated channel. In this work, we investigate the performance of SU-MMIMO using (a) serial concatenated turbo code (SCTC) in uncorrelated channel and (b) PCTC in correlated channel. Throughout this article we assume that the channel is known perfectly at the receiver. Perfect carrier and timing synchronization is also assumed.

This work is organized as follows. Section II discusses SU-MMIMO with SCTC in uncorrelated channel, the procedure for bit-error-rate (BER) estimation and computer simulation results. Section III deals with SU-MMIMO using PCTC in correlated channel with and without precoding along with computer simulation results. Section IV presents the conclusions and scope for future work.

### **2. SU-MMIMO with SCTC**

#### **2.1 System model**

Consider the block diagram in **Figure 1** [36, 38]. The input bits *ai*, 1≤*i* ≤*Ld*<sup>1</sup> is passed through an outer rate-1*=*2 recursive systematic convolutional (RSC) encoder to obtain the coded bit stream *bi*, 1≤*i*≤ *Ld*, where

**Figure 1.** *SU-MMIMO with serially concatenated turbo code.*

$$L\_d = \mathfrak{L}\_{d1}.\tag{1}$$

Now *bi* is input to an interleaver to generate *ci*, 1≤*i* ≤*Ld*. Next *ci* is passed through an inner rate-1*=*2 RSC encoder and mapped to symbols *Si*, 1≤*i* ≤*Ld*, in a quadrature phase shift keyed (QPSK) constellation having symbol coordinates �1 � j, where <sup>j</sup> <sup>¼</sup> ffiffiffiffiffiffi �<sup>1</sup> <sup>p</sup> . Throughout this article we assume that bit "0" maps to <sup>þ</sup>1 and bit "1" maps to �1. The set of *Ld* QPSK symbols constitute a "frame" and are transmitted using *Nt* antennas. We assume that

$$\frac{L\_d}{N\_t} = \text{an integer}\tag{2}$$

so that all symbols in the frame are transmitted using *Nt* antennas. The set of QPSK symbols transmitted simultaneously using *Nt* antennas constitute a "block". The generator matrix for both the inner and outer rate-1*=*2 RSC encoder is given by

$$\mathbf{G}(D) = \left[ \mathbf{1} \, \frac{\mathbf{1} + D^2}{\mathbf{1} + D + D^2} \right]. \tag{3}$$

Hence, both encoders have *SE* ¼ 4 states in the trellis. Assuming uncorrelated Rayleigh flat fading, the received signal for the *<sup>k</sup>th* re-transmission (0<sup>≤</sup> *<sup>k</sup>*<sup>≤</sup> *Nrt* � 1, *<sup>k</sup>* is an integer) is given by (2) of [36], which is repeated here for convenience

$$
\tilde{\mathbf{R}}\_k = \tilde{\mathbf{H}}\_k \mathbf{S} + \tilde{\mathbf{W}}\_k \tag{4}
$$

where **S**∈ *Nt*�<sup>1</sup> whose elements are drawn from the QPSK constellation, **H**~ *<sup>k</sup>* ∈ *Nr*�*Nt* whose elements are mutually independent and CN 0, 2*σ*<sup>2</sup> *H* � � and

**W**~ *<sup>k</sup>* ∈ *Nr*�<sup>1</sup> is the additive white Gaussian noise (AWGN) vector whose elements are mutually independent and CN 0, 2*σ*<sup>2</sup> *W* � �. Note that *σ*<sup>2</sup> *<sup>H</sup>*, *σ*<sup>2</sup> *<sup>W</sup>* denote the variance per dimension (real part or imaginary part) and *Nr* is the number of receive antennas. We assume that **H**~ *<sup>k</sup>* and **W**~ *<sup>k</sup>* are independent across blocks and re-transmissions, hence (4) in [29] is valid with *N* replaced by *Nt*. Recall that (see also (16) of [36])

$$N\_{\text{tot}} = N\_t + N\_r. \tag{5}$$

Following the procedure given in Section 4 of [36] we get (see (36) of [36])

$$
\tilde{Y}\_i = F\_i \mathbb{S}\_i + \tilde{U}\_i \qquad \text{for } \mathbb{1} \le i \le N\_t. \tag{6}
$$

After concatenation over blocks, *Y*~*<sup>i</sup>* in (6) for 1≤*i* ≤*Ld* is sent to the turbo decoder (see also the sentence after (25) in [29]). For the sake of consistency with earlier work [38], we re-index *i* as 0 ≤*i* ≤*Ld* � 1 and use the same index *i* for *ai*, *bi*, *ci* and *Yi* without any ambiguity. In the next subsection, we discuss the turbo decoding (BCJR) algorithm [39, 40] for the inner code.

#### **2.2 BCJR for the inner code**

Let D*<sup>n</sup>* denote the set of states that diverge from state *n* in the trellis [38, 40]. Similarly, let C*<sup>n</sup>* denote the set of states that converge to state *n*. Let *α<sup>i</sup>*,*<sup>n</sup>* denote the forward sum-of-products (SOP) at time *i*, 0≤*i* ≤*Ld* � 2, at state *n*, 0≤ *n*≤*SE* � 1. Then the forward SOP can be recursively computed as follows (see also (30) of [38]):

$$a'\_{i+1,n} = \sum\_{m \in \mathcal{L}\_n} a\_{i,m} \chi\_{i,m,n} P(c\_{i,m,n}); \quad a\_{0,n} = \mathbf{1}; \quad a\_{i+1,n} = a'\_{i+1,n} \left(\sum\_{n=0}^{s\_{\overline{E}^{-1}}} a\_n\right) \tag{7}$$

where *P c*ð Þ *<sup>i</sup>*,*m*,*<sup>n</sup>* denotes the *a priori* probability of the systematic bit corresponding to the transition from encoder state *m* to *n*, at time *i* (this is set to 0.5 at the beginning of the first iteration). The last equation in (7) is required to prevent numerical instabilities [40]. We have

$$\gamma\_{i,m,n} = \exp\left(-\frac{\left(\tilde{Y}\_i - S\_{m,n}\right)^2}{2\sigma\_U^2}\right) \tag{8}$$

where *Y*~*<sup>i</sup>* is given by (6), *Sm*,*<sup>n</sup>* is the QPSK symbol corresponding to the transition from encoder state *m* to *n* and *σ*<sup>2</sup> *<sup>U</sup>* is given by (38) of [36] which is repeated here for convenience:

$$E\left[\left|\tilde{U}\_i\right|^2\right] = \frac{8\sigma\_H^4 N\_r (N\_t - 1) + 4\sigma\_W^2 \sigma\_H^2 N\_r}{N\_{rt}} \triangleq \sigma\_U^2. \tag{9}$$

Robust turbo decoding (see Section 4.2 of [41]) can be employed to compute *γ<sup>i</sup>*,*m*,*<sup>n</sup>* in (8). Similarly, let *β<sup>i</sup>*,*<sup>m</sup>* denote the backward SOP at time *i*, 1≤*i* ≤*Ld* � 1, at state *m*, 0≤ *m* ≤*SE* � 1. Then the backward SOP can be recursively computed as (see also (33) of [38]):

$$\boldsymbol{\beta}'\_{i,m} = \sum\_{n \in \mathcal{D}\_m} \beta\_{i+1,n} \boldsymbol{\gamma}\_{i,m,n} \mathbf{P}(c\_{i,m,n});\\\boldsymbol{\beta}\_{L,n} = \mathbf{1};\\\boldsymbol{\beta}\_{i,m} = \boldsymbol{\beta}'\_{i,n} \boldsymbol{\gamma} \left(\sum\_{m=0}^{s\_{\mathbb{E}^{-1}}} \boldsymbol{\frac{\boldsymbol{\beta}}{\boldsymbol{\beta}}}\right). \tag{10}$$

Let *ρ*þð Þ *n* denote the state that is reached from encoder state *n* when the input symbol is þ1. Similarly let *ρ*�ð Þ *n* denote the state that can be reached from encoder state *n* when the input symbol is �1. Then for 0≤ *i*≤*Ld* � 1 we compute

$$\mathbf{C}\_{i+} = \sum\_{n=0}^{\mathbb{S}-1} a\_{i,n} \boldsymbol{\gamma}\_{i, \mathfrak{n}, \boldsymbol{\rho}^+(n)} \boldsymbol{\beta}\_{i+1, \boldsymbol{\rho}^+(n)}; \quad \mathbf{C}\_{i-} = \sum\_{n=0}^{\mathbb{S}-1} a\_{i,n} \boldsymbol{\gamma}\_{i, \mathfrak{n}, \boldsymbol{\rho}^-(n)} \boldsymbol{\beta}\_{i+1, \boldsymbol{\rho}^-(n)}.\tag{11}$$

Finally, the extrinsic information that is fed to the BCJR algorithm for the outer code is computed as, for 0 ≤*i* ≤*Ld* � 1, (see (36) of [38]):

$$E(\mathbf{c}\_i = +1) = \mathbf{^{C}\_{i^+} (\_{(\mathbf{C}\_{i^+} + \mathbf{C}\_{i^-})} ; E(\mathbf{c}\_i = -1) = \mathbf{C}\_{i-} / (\mathbf{C}\_{i+} + \mathbf{C}\_{i-}).\tag{12}$$

Next, we describe the BCJR for the outer code.

#### **2.3 BCJR for the outer code**

Let *α<sup>i</sup>*,*<sup>n</sup>* denote the forward SOP at time *i*, 0≤*i* ≤*Ld*<sup>1</sup> � 2, at state *n*, 0≤*n*≤ *SE* � 1. Then the forward SOP is recursively computed as follows:

$$a'\_{i+1,n} = \sum\_{m \in \mathcal{C}\_n} a\_{i,m} \chi\_{sp,i,m,n} \chi\_{par,i,m,n} P(a\_{i,m,n}); \\ a\_{0,n} = 1; \\ a\_{i+1,n} = a'\_{i+1,n} \left(\sum\_{n=0}^{s\_{\mathbb{Z}^{-1}}} a'\_{i+1,n}\right) \tag{13}$$

where *P a*ð Þ *<sup>i</sup>*,*m*,*<sup>n</sup>* denotes the *a priori* probability of the systematic bit corresponding to the transition from state *m* to state *n*, at time *i*. In the absence of any other information, we assume ð Þ¼ *ai*,*m*,*<sup>n</sup>* 0*:*5 [42]. We also have for 0≤*i* ≤*Ld*<sup>1</sup> � 1 (similar to (38) of [38])

$$\gamma\_{\text{sys},i,m,\boldsymbol{\mu}} = \begin{cases} E(c\_{\pi(2i)} = +1) & \text{if } \mathcal{H}\_1 \\ E(c\_{\pi(2i)} = -1) & \text{if } \mathcal{H}\_2 \end{cases}; \quad \gamma\_{\text{par},i,m,\boldsymbol{\mu}} = \begin{cases} E(c\_{\pi(2i+1)} = +1) & \text{if } \mathcal{H}\_3 \\ E(c\_{\pi(2i+1)} = -1) & \text{if } \mathcal{H}\_4 \end{cases} \tag{14}$$

where *π*ð Þ� denotes the interleaver map and

H<sup>1</sup> : systematic bit from state *m* to *n* is þ 1; H<sup>2</sup> : systematic bit from state *m* to *n* is � 1

$$\mathcal{H}\_3: \text{parity bit from state } m \text{ to } n \text{ is } +1; \\ \mathcal{H}\_4: \text{parity bit from state } m \text{ to } n \text{ is } -1. \tag{15}$$

Observe that in (14) and (15) it is assumed that after the parallel-to-serial conversion in **Figure 1**, *b*2*<sup>i</sup>* corresponds to the systematic (data) bits and *b*2*i*þ<sup>1</sup> corresponds to the parity bits for 0≤*i* ≤*Ld*<sup>1</sup> � 1. Similarly, let *β<sup>i</sup>*,*<sup>m</sup>* denote the backward SOP at time *i*, 1≤*i* ≤*Ld*<sup>1</sup> � 1, at state *m*, 0 ≤ *m* ≤*SE* � 1. Then the backward SOP can be recursively computed as:

$$\beta\_{i,m}' = \sum\_{n \in \mathcal{D}\_m} \beta\_{i+1,n} \chi\_{\mathcal{Y}s,i,m,n} \chi\_{par,i,m,n} P(a\_{i,m,n});\\\beta\_{L\_{i1,n}} = \mathbf{1};\\\beta\_{i,m} = \beta\_{i,n}' \left(\sum\_{n=0}^{\S\_E - 1} \rho\_{i,n}'\right). \tag{16}$$

Next, for 0 ≤*i* ≤*Ld*<sup>1</sup> � 1 we compute

$$B\_{2i+} = \sum\_{n=0}^{\mathcal{S}\_{\mathbb{E}}-1} a\_{i,n} \gamma\_{\text{par},i,n,\rho^+(n)} \beta\_{i+1,\rho^+(n)}; \\ B\_{2i-} = \sum\_{n=0}^{\mathcal{S}\_{\mathbb{E}}-1} a\_{i,n} \gamma\_{\text{par},i,n,\rho^-(n)} \beta\_{i+1,\rho^-(n)}.\tag{17}$$

Let *μ*þð Þ *n* , and *μ*�ð Þ *n* denote the states that are reached from state *n* when the parity bit is þ1 and �1respectively. Similarly for 0≤ *i*≤*Ld*<sup>1</sup> � 1 compute

$$B\_{2i+1+} = \sum\_{n=0}^{\mathbb{S}-1} a\_{i,n} \chi\_{\text{sys},i,n,\mu^+(n)} \beta\_{i+1,\mu^+(n)};\\B\_{2i+1-} = \sum\_{n=0}^{\mathbb{S}-1} a\_{i,n} \chi\_{\text{sys},i,n,\mu^-(n)} \beta\_{i+1,\mu^-(n)}.\tag{18}$$

The extrinsic information that is sent to the inner decoder for 0 ≤*i*≤ *Ld* � 1 is computed as

$$E(b\_i = +1) = {}^{B\_{i+}}\!/\_{(B\_{i+} + B\_{i-})} {}^{\*}E(b\_i = -1) = B\_{i-} / (B\_{i+} + B\_{i-}) \tag{19}$$

where *Bi*þ, *Bi*� are given by (17)or (18) depending on whether *i* is even or odd respectively. Note that *P c*ð Þ *<sup>i</sup>*,*m*,*<sup>n</sup>* for 0 ≤*i* ≤*Ld* � 1 in (7) and (10) is equal to

$$P(c\_{i,m,n}) = \begin{cases} E(b\_{\pi^{-1}(i)} = +1) & \text{if } \mathcal{H}\_1 \\ E(b\_{\pi^{-1}(i)} = -1) & \text{if } \mathcal{H}\_2 \end{cases} \tag{20}$$

where *<sup>π</sup>*�<sup>1</sup>ð Þ� denotes the inverse interleaver map. Note that *ci*,*m*,*<sup>n</sup>* are the systematic (data) bits for the inner encoder.

After the convergence of the BCJR algorithm in the last iteration, the final *a posteriori* probabilities of *ai* for 0≤ *i*≤*Ld*<sup>1</sup> � 1 is given by

$$P(\mathfrak{a}\_{i}=+\mathbf{1}) = E(b\_{2i}=+\mathbf{1})E(\mathfrak{c}\_{\pi(2i)}=+\mathbf{1}); P(\mathfrak{a}\_{i}=-\mathbf{1}) = E(b\_{2i}=-\mathbf{1})E(\mathfrak{c}\_{\pi(2i)}=-\mathbf{1})\tag{21}$$

where *E c*ð Þ *<sup>i</sup>* ¼ �1 and *E b*ð Þ *<sup>i</sup>* ¼ �1 are given by (12) and (19) respectively. Finally note that for 0≤*i* ≤*Ld*<sup>1</sup> � 1

$$
\mathfrak{a}\_i = \mathfrak{b}\_{2i} = \mathfrak{c}\_{\pi(2i)}.\tag{22}
$$

In the next section we present the estimation of the bit-error-rate (BER) of the SCTC.

#### **2.4 Estimation of BER**

The estimation of BER of SCTC is based on the following propositions:

**Proposition1.** *The extrinsic information as computed in (12) and (19) lies in the range* ½ � 0, 1 *(0 and 1 included). The extrinsic information in the range* ð Þ 0, 1 *, 0 and 1 excluded, is Gaussian distributed* [43] *for each frame.*

This is illustrated in **Figure 2** for different values of the frame length *Ld*1, over many frames (*F*). We find that for large values of *Ld*1, the histogram better approximates the Gaussian characteristic. It may be noted that the extrinsic information at the output of one decoder is equal to the *a priori* probabilities for the other decoder.

**Figure 2.**

*Normalized histogram for Ntot = 1024, Nt = 512, Nrt = 2 (a) Ld1 = 1024, SNRav, b = 1.25 dB, F = 105 frames (b) Ld1 = 50,176, SNRav, b = 0.3 dB, F = 2000 frames (c) expanded view of (around) r3,i = 0 and (d) Ld1 = 50,176, SNRav, b = 0.5 dB, F = 2000 frames.*

**Proposition 2.** *After convergence of the BCJR algorithm in the final iteration, the extrinsic information at a decoder output has the same mean and variance as that of the a priori probability at its input.*

**Proposition 3.** *The mean and variance of the Gaussian distribution may vary from frame to frame.*

This is illustrated in **Figure 3** over two frames, that is, *F* ¼ 2.

$$P(e) = \frac{1}{2} \text{erfc}\left(\sqrt{\frac{A^2}{\sigma^2}}\right). \tag{23}$$

Based on *Propositions 1 & 2* and (22), after convergence of the BCJR algorithm, we can write for 0≤ *i*≤ *Ld*<sup>1</sup> � 1

$$E(b\_{2i} = +1) = \frac{1}{\sigma\sqrt{2\pi}}\mathbf{e}^{-\left(\mathbf{r}\_{1i} - \mathbf{A}\right)^2/\left(2\sigma^2\right)};\\E\left(\mathbf{c}\_{\pi\left(2i\right)} = +1\right) = \frac{1}{\sigma\sqrt{2\pi}}\mathbf{e}^{-\left(\mathbf{r}\_{2i} - \mathbf{A}\right)^2/\left(2\sigma^2\right)}\tag{24}$$

where it is assumed that bit "0" maps to *A* and bit "1" maps to �*A* and

**Figure 3.**

*Normalized histogram over two frames (F = 2) for Ntot = 1024, Nt = 512, Nrt = 2 (a) Ld1 = 1024, SNRav, <sup>b</sup> = 1.25 dB and (d) Ld1 = 50,176, SNRav, b = 0.5 dB.*

$$r\_{1,i} = \pm A + w\_{1,i}; r\_{2,i} = \pm A + w\_{2,i} \tag{25}$$

where *w*1,*<sup>i</sup>*, *w*2,*<sup>i</sup>* denote real-valued samples of zero-mean additive white Gaussian noise (AWGN) with variance *σ*2. Similarly we have

$$E(b\_{2i} = -1) = \frac{1}{\sigma \sqrt{2\pi}} \mathbf{e}^{-\left(\mathbf{r}\_{1i} + \mathbf{A}\right)^2 / \left(2\sigma^2\right)};\\E\left(c\_{\pi(2i)} = -\mathbf{1}\right) = \frac{1}{\sigma \sqrt{2\pi}} \mathbf{e}^{-\left(\mathbf{r}\_{2i} + \mathbf{A}\right)^2 / \left(2\sigma^2\right)}.\tag{26}$$

Clearly

$$\ln\left(\frac{E(b\_{2i}=+1)}{E(b\_{2i}=-1)}\right) = \frac{2A}{\sigma^2}r\_{1,i}; \quad \ln\left(\frac{E(c\_{\pi(2i)}=+1)}{E(c\_{\pi(2i)}=-1)}\right) = \frac{2A}{\sigma^2}r\_{2,i}.\tag{27}$$

From (21) and (26) we have for 0≤ *i*≤ *Ld*<sup>1</sup> � 1

$$\ln\left(\frac{P(a\_i = +1)}{P(a\_i = -1)}\right) = \frac{2A}{\sigma^2}(r\_{1,i} + r\_{2,i}) \triangleq \frac{2A}{\sigma^2}r\_{3,i}.\tag{28}$$

Consider the average

$$\mathcal{V} = \frac{2A}{\sigma^2 L\_{d2}} \sum\_{i=0}^{L\_{d2}-1} a\_i r\_{3,i} = \frac{4A^2}{\sigma^2} + \mathcal{Z} \tag{29}$$

where

$$Z = \frac{2A}{\sigma^2 L\_{d2}} \sum\_{i=0}^{L\_{d2}-1} a\_i (w\_{1,i} + w\_{2,i}); \quad L\_{d2} < L\_{d1} \tag{30}$$

Note that the average in (28) is done over less than *Ld*<sup>1</sup> terms to avoid situations like

$$P(a\_i = \pm \mathbf{1}) = \mathbf{1} \text{ or } \mathbf{0}.\tag{31}$$

In fact, only those time instants *i* have been considered in the summation of (28) for which

$$P(a\_i = \pm \mathbf{1}) > \mathbf{e}^{-500}.\tag{32}$$

Now

$$E[\mathcal{Z}] = 0; E[\mathcal{Z}^2] = \frac{4A^2}{\sigma^4 L\_{d2}^2} \sum\_{i=0}^{L\_{d2}-1} 2\sigma^2 = \frac{8A^2}{\sigma^2 L\_{d2}}\tag{33}$$

where we have used the fact that *w*1,*i*, *w*2,*<sup>i</sup>* are independent. Now, we know that the probability of error for the BPSK signal in (27), that is equal to [40].

$$r\_{3,i} = r\_{1,i} + r\_{2,i} = \pm 2A + w\_{1,i} + w\_{2,i} \tag{34}$$

Therefore from (28), (32) and (34) we have

$$P\_f(e) \approx \frac{1}{2} \text{erfc}\left(\sqrt{\frac{|\mathcal{Y}|}{4}}\right) \tag{35}$$

where *Pf*ð Þ*e* denotes the probability of bit error for frame "*f* " and

$$E\left[\mathcal{Z}^2\right] \to \mathbf{0} \qquad \text{for } L\_{d2} \gg 1. \tag{36}$$

Observe that it is necessary to take the absolute value of Y in (35) since there is a possibility that it can be negative. The average probability of bit error over *F* frames is given by

$$P(e) = \frac{1}{F} \sum\_{f=0}^{F-1} P\_f(e). \tag{37}$$

In the next section we present computer simulation results for SU-MMIMO using SCTC in uncorrelated channel.

#### **2.5 Simulation results**

The simulation parameters are given in **Table 5**. We can make the following observations from **Figures 4**–**6** [36]:

The theoretical prediction of BER closely matches with simulations.


In **Figure 4(c)** we observe that there is more than 1 dB improvement in SINR compared to **Figures 4**–**6(a, b)**. However, large values of *Ld*<sup>1</sup> may introduce more latency which is contrary to the requirements of 5G and beyond. In the next section we present SU-MMIMO using PCTC in correlated channel.

*New Results on Single User Massive MIMO DOI: http://dx.doi.org/10.5772/intechopen.112469*


**Table 5.**

*Simulation parameters for results in Figures 4–6.*

**Figure 4.** *Simulation results for* N*tot = 1024.*
