Orthogonal frequency division multiplexing (OFDM) with offset quadrature amplitude modulation (OQAM) has been widely discussed in the literature and is considered a popular waveform for 5th generation (5G) wireless telecommunications and beyond. In this work, we show that OFDM-OQAM can be generated using the Hilbert transform and is equivalent to single sideband modulation (SSB), that has roots in analog telecommunications. The transmit filter for OFDM-OQAM is complex valued whose real part is given by the pulse corresponding to the root raised cosine spectrum and the imaginary part is the Hilbert transform of the real part. The real-valued digital information (message) are passed through the transmit filter and frequency division multiplexed on orthogonal subcarriers. The message bandwidth corresponding to each subcarrier is assumed to be narrow enough so that the channel can be considered ideal. Therefore, at the receiver, a matched filter can used to recover the message. Turbo coding is used to achieve bit-error-rate (BER) as low as \(10^{-5}\) at an average signal-to-noise ratio (SNR) per bit close to 0 db. The system has been simulated in discrete time.

INTRODUCTION↩︎

Orthogonal frequency division multiplexing (OFDM) [1]–[3] and OFDM offset quadrature amplitude modulation (OFDM-OQAM) [4] are the preferred modulation techniques for transmission of digital information over frequency selective channels, both fading and non-fading. The variants of OFDM-OQAM [5]–[9] are known as filter bank multicarrier (FBMC) [4]–[7] and universal filtered multicarrier (UFMC) [10], [11] and generalized frequency division multiplexing (GFDM) [12]–[14] in the literature.

One of the key advantages of OFDM-OQAM/FBMC/UFMC over OFDM is its immunity against carrier frequency offsets (CFO). In other words, it may not be necessary for an OFDM-OQAM/FBMC/UFMC system to estimate and cancel the CFO, unlike OFDM. However, it has been shown in [1]–[3] that it is possible to estimate and cancel the CFO very accurately with large scope for parallel processing. The other important feature of OFDM-OQAM is the spectral containment of each subcarrier using a transmit filter, which is absent in OFDM. However, OFDM is more attractive than OFDM-OQAM in terms of implementation simplicity.

In this work, we demonstrate that OFDM-OQAM/FBMC/UFMC can be efficiently implemented using the Hilbert transform at the transmitter and a matched filter at the receiver [15]. It must be noted that using OQAM would improve the symbol density in time-frequency space, at the cost of introducing intersymbol interference (ISI) at the matched filter output and increasing the receiver complexity. Therefore, in this work, we do not use OQAM and yet have a symbol density in time frequency space greater than unity. We however retain the nomenclature “OFDM-OQAM/FBMC/UFMC” since this work deals with multicarrier communications.

We use the following notation. Complex quantities are denoted by a tilde e.g., \(\tilde{p}(t)\), estimates are denoted by hat e.g., \(\hat{b}\) and convolution is denoted by star e.g., \(p(t)\star g(t)\). This article is organized as follows. We first discuss the theory of OFDM-OQAM, followed by derivation of Hilbert transform of the pulse corresponding to root-raised cosine spectrum. The discrete-time implementation issues are discussed next, motivating the need for a modified Hilbert transform. The discrete-time system model used for computer simulations is presented, followed by results and conclusions.

THEORY↩︎

At the outset, we note that multicarrier communication is equivalent to frequency division multiplexing. The time-frequency representation of a multicarrier communication system is shown in Figure 1 [4]. The symbol-rate is \(1/T\) baud and the subcarrier spacing is \(\mathscr{B}\). The red colour dots denote symbols. The subcarrier frequency can be positive or negative, as will be apparent later.

The complex envelope of a linearly modulated digital signal is given by [16] \[\begin{align} \label{Eq:FLOGEN95Eq1} \tilde{s}(t) & = \sum_{k=-\infty}^{\infty} S_k \, \tilde{p}(t-kT) \nonumber \\ & = s_I(t) + \mathrm{j}\, s_Q(t) \qquad (say) \end{align}\tag{1}\] where \(S_k\) denotes complex-valued symbols drawn from an \(M\)-ary constellation, \(\tilde{p}(t)\) denotes the (possibly complex-valued) impulse response of the transmit filter, the subscripts “\(I,\, Q\)” denote in-phase (real) and quadrature (imaginary) components respectively. When the symbols \(S_k\) are uncorrelated, the power spectral density of \(\tilde{s}(t)\) in (1 ) is [16] \[\label{Eq:FLOGEN95Eq1951} S_{\tilde{s}}(F) = \frac{P_{\mathrm{av}}}{2T} \left| \tilde{P}(F) \right|^2\tag{2}\] where \(\tilde{P}(F)\) is the Fourier transform of \(\tilde{p}(t)\) and \(P_{\mathrm{av}}\) denotes the average power of the \(M\)-ary constellation. If the transmit filter \(\tilde{p}(t)=p(t)\) is real-valued [16] having a root raised cosine (RRC) spectrum with roll-off factor \(\rho\), \(0<\rho \le 1\), then the minimum spacing between subcarriers for no aliasing would be \[\label{Eq:FLOGEN95Eq1952} \mathscr{B} = \frac{2(1+\rho)}{2T}\tag{3}\] which is essentially the two-sided bandwidth of \(\tilde{P}(F)\). Therefore, the symbol density in time-frequency space is the inverse of the area of each rectangle in Figure 1 [4]: \[\label{Eq:FLOGEN95Eq2} \frac{1}{\mathscr{B}T} = \frac{1}{(1+\rho)} < 1.\tag{4}\] In this article, we propose a complex-valued transmit filter given by \[\label{Eq:FLOGEN95Eq3} \tilde{p}(t) = p(t) + \mathrm{j}\, \hat{p}(t)\tag{5}\] where \(p(t)\) has an RRC spectrum and \(\hat{p}(t)\) is the Hilbert transform of \(p(t)\). The Fourier transform of \(\tilde{p}(t)\) in (5 ) is \[\begin{align} \label{Eq:FLOGEN95Eq4} \tilde{P}(F) & = P(F) + \mathrm{j}\, (-\mathrm{j}\mathrm{sgn}(F)) P(F) \nonumber \\ & = \left \{ \begin{array}{ll} 2P(F) & for F>0\\ P(0) & for F=0\\ 0 & for F<0 \end{array} \right. \end{align}\tag{6}\] where \(P(F)\) is the Fourier transform of \(p(t)\) and \(\mathrm{sgn}(\cdot)\) is the signum function [17], [18]. The minimum spacing between subcarriers in this case is \[\label{Eq:FLOGEN95Eq5} \mathscr{B} = \frac{1+\rho}{2T}\tag{7}\] resulting in symbol density in time-frequency space equal to \[\label{Eq:FLOGEN95Eq6} \frac{1}{\mathscr{B}T} = \frac{2}{(1+\rho)} \ge 1.\tag{8}\] It is emphasized here that when \(\tilde{p}(t)\) is given by (5 ), the symbols \(S_k\) have to be real-valued. The reason is as follows. The complex envelope in (1 ) can be written as \[\label{Eq:FLOGEN95Eq7} \tilde{s}(t) = \tilde{s}_1(t) \star \tilde{p}(t)\tag{9}\] where “\(\star\)” denotes convolution and \[\begin{align} \label{Eq:FLOGEN95Eq8} \tilde{s}_1(t) & = \sum_{k=-\infty}^{\infty} S_k \, \delta_D(t-kT) \nonumber \\ & = s_{1,\, I}(t) + \mathrm{j}\, s_{1,\, Q}(t) \qquad (say) \end{align}\tag{10}\] where \(\delta_D(\cdot)\) denotes the Dirac-delta function [17], [18]. At the receiver, we use a matched filter given by \(\tilde{p}^*(-t)\). The matched filter output would be given by [16] \[\label{Eq:FLOGEN95Eq9} \tilde{y}(t) = \tilde{s}_1(t) \star \tilde{p}(t) \star \tilde{p}^*(-t).\tag{11}\] Now \[\begin{align} \label{Eq:FLOGEN95Eq10} \tilde{p}(t) \star \tilde{p}^*(-t) & = \left[ p(t) + \mathrm{j}\, \hat{p}(t) \right] \star \left[ p(-t) - \mathrm{j}\, \hat{p}(-t) \right] \nonumber \\ & = p(t) \star p(-t) + \hat{p}(t) \star \hat{p}(-t) + \mathrm{j}\, \left[ \hat{p}(t) \star p(-t) - p(t) \star \hat{p}(-t) \right]. \end{align}\tag{12}\] It can be shown that for real-valued \(p(t)\) [17], [18] \[\begin{align} \label{Eq:FLOGEN95Eq11} \hat{p}(t) \star \hat{p}(-t) & = p(t) \star p(-t) \nonumber \\ & = R_{pp}(t) \nonumber \\ \hat{p}(t) \star p(-t) & = R_{\hat{p}p}(t) \nonumber \\ & = -p(t) \star \hat{p}(-t) \nonumber \\ & = -R_{p\hat{p}}(t). \end{align}\tag{13}\] Substituting (12 ) and (13 ) in (11 ) we obtain the matched filter output as \[\label{Eq:FLOGEN95Eq12} \tilde{y}(t) = \sum_{k=-\infty}^{\infty} 2S_k \left[ R_{pp}(t-kT) + \mathrm{j}\, R_{\hat{p}p}(t-kT) \right].\tag{14}\] The condition for zero ISI is \[\label{Eq:FLOGEN95Eq13} R_{pp}(mT) = \delta_K(mT)\tag{15}\] where \(\delta_K(\cdot)\) is the Kronecker delta function [16]. However, \(R_{\hat{p}p}(t)\) does not satisfy the zero ISI condition. Hence using (15 ), the symbol-rate sampler at the matched filter output would yield \[\begin{align} \label{Eq:FLOGEN95Eq14} \tilde{y}(nT) & = \sum_{k=-\infty}^{\infty} 2S_k \left( R_{pp}(nT-kT) + \mathrm{j}\, R_{\hat{p}p}(nT-kT) \right) \nonumber \\ & = 2S_n + 2\mathrm{j}\, \sum_{k=-\infty}^{\infty} S_k R_{\hat{p}p}(nT-kT). \end{align}\tag{16}\] It is clear from (16 ) that implementing the matched filter as given in (11 ) and (12 ) with complex-valued symbols \(S_k\) is not possible, due to crosstalk (interference between in-phase and quadrature components). Therefore, the symbols \(S_k\) have to be real-valued. One possible implementation of the proposed OFDM-OQAM/FBMC/UFMC transmitter is shown in Figure 2.

The corresponding receiver is shown in Figure 3. In the next section, we derive \(\hat{p}(t)\) when \(p(t)\) has an RRC spectrum.

HILBERT TRANSFORM OF RRC SPECTRUM↩︎

The RRC spectrum which is the Fourier transform of \(p(t)\) is given by [16] \[\label{Eq:FLOGEN95Eq15} P(F) = \left \{ \begin{array}{ll} \frac{1}{\sqrt{2B}} & for -F_1 \le F \le F_1 \\ \frac{1}{\sqrt{2B}} \cos \left( \frac{\pi(|F|-F_1)}{4B-4F_1} \right) & for F_1 \le |F| \le 2B-F_1 \\ 0 & elsewhere \end{array} \right.\tag{17}\] where \[\begin{align} \label{Eq:FLOGEN95Eq16} 2B & \stackrel{\Delta}{=} \frac{1}{T} \nonumber \\ \rho & \stackrel{\Delta}{=} 1- \frac{F_1}{B}. \end{align}\tag{18}\] Therefore \[\label{Eq:FLOGEN95Eq17} \hat{p}(t) = \int_{F=-\infty}^{\infty} -\mathrm{j}\, \mathrm{sgn}(F) P(F) \mathrm{e}^{\mathrm{j}\, 2\pi Ft} \, dF\tag{19}\] where \(P(F)\) is given by (17 ) and (18 ). We have \[\begin{align} \label{Eq:FLOGEN95Eq18} I_1 & = \int_{F=-F_1}^{F_1} - \mathrm{j}\, \mathrm{sgn}(F) \frac{1}{\sqrt{2B}} \mathrm{e}^{\mathrm{j}\, 2\pi Ft} \, dF \nonumber \\ & = \frac{1}{\pi t\sqrt{2B}} \left[ 1 - \cos(2\pi F_1 t) \right]. \end{align}\tag{20}\] Similarly \[\begin{align} \label{Eq:FLOGEN95Eq19} I_2 & = \int_{F=F_1}^{2B-F_1} - \mathrm{j}\, \frac{1}{\sqrt{2B}} \cos \left( \frac{\pi(F-F_1)}{4B-4F_1} \right) \mathrm{e}^{\mathrm{j}\, 2\pi Ft} \, dF \nonumber \\ & \qquad + \int_{F=-(2B-F_1)}^{-F_1} \mathrm{j}\, \frac{1}{\sqrt{2B}} \cos \left( \frac{\pi(-F-F_1)}{4B-4F_1} \right) \mathrm{e}^{\mathrm{j}\, 2\pi Ft} \, dF \nonumber \\ & = \frac{1}{\sqrt{2B}} \int_{F=B(1-\rho)}^{B(1+\rho)} \left\{ \sin \left[ 2\pi Ft + \frac{\pi(F-F_1)}{4B-4F_1} \right] + \sin \left[ 2\pi Ft - \frac{\pi(F-F_1)}{4B-4F_1} \right] \right\} \, dF \nonumber \\ & = \frac{1}{\sqrt{2B}} \int_{F=B(1-\rho)}^{B(1+\rho)} \left\{ \sin \left[ \alpha F - \beta \right] + \sin \left[ \beta - \gamma F \right] \right\} \, dF \nonumber \\ & = \frac{1}{\gamma\sqrt{2B}} \left\{ \cos \left[ \beta - \gamma B(1+\rho) \right] - \cos \left[ \beta - \gamma B(1-\rho) \right] \right\} \nonumber \\ & \qquad - \frac{1}{\alpha\sqrt{2B}} \left\{ \cos \left[ \alpha B(1+\rho) - \beta \right] - \cos \left[ \alpha B(1-\rho) - \beta \right] \right\} \end{align}\tag{21}\] where \[\begin{align} \label{Eq:FLOGEN95Eq20} F_1 & = B(1-\rho) \nonumber \\ 2B-F_1 & = B(1+\rho) \nonumber \\ 4B-4F_1 & = 4B\rho \nonumber \\ \alpha & = \frac{\pi(1+8Bt\rho)}{4B\rho} \nonumber \\ \beta & = \frac{\pi F_1}{4B\rho} \nonumber \\ \gamma & = \frac{\pi(1-8Bt\rho)}{4B\rho}. \end{align}\tag{22}\] Now from (22 ) \[\begin{align} \label{Eq:FLOGEN95Eq21} \beta - \gamma B & = \frac{-\pi}{4} + 2\pi Bt \nonumber \\ \gamma B\rho & = \frac{\pi}{4} - 2\pi Bt\rho \nonumber \\ \alpha B - \beta & = \frac{\pi}{4} + 2\pi Bt \nonumber \\ \alpha B\rho & = \frac{\pi}{4} + 2\pi Bt\rho. \end{align}\tag{23}\] Substituting (23 ) in (21 ) we obtain \[\begin{align} \label{Eq:FLOGEN95Eq22} I_2 & = \frac{1}{\sqrt{2B}} \sin(2\pi Bt(1+\rho)) \left\{ \frac{1}{\gamma} + \frac{1}{\alpha} \right\} \nonumber \\ & \qquad - \frac{1}{\sqrt{2B}} \cos(2\pi Bt(1-\rho)) \left\{ \frac{1}{\gamma} - \frac{1}{\alpha} \right\}. \end{align}\tag{24}\] Now \[\begin{align} \label{Eq:FLOGEN95Eq23} \frac{1}{\gamma} - \frac{1}{\alpha} & = \frac{64B^2\rho^2 t}{\pi(1-64B^2t^2\rho^2)} \nonumber \\ \frac{1}{\gamma} + \frac{1}{\alpha} & = \frac{8B\rho}{\pi(1-64B^2t^2\rho^2)}. \end{align}\tag{25}\] Substituting (25 ) in (24 ) we get \[\begin{align} \label{Eq:FLOGEN95Eq24} I_2 & = \frac{1}{\sqrt{2B}} \sin(2\pi Bt(1+\rho)) \left\{ \frac{8B\rho}{\pi(1-64B^2t^2\rho^2)} \right\} \nonumber \\ & \qquad - \frac{1}{\sqrt{2B}} \cos(2\pi Bt(1-\rho)) \left\{ \frac{64B^2\rho^2 t}{\pi(1-64B^2t^2\rho^2)} \right\}. \end{align}\tag{26}\] Finally \(\hat{p}(t)\) in (19 ) is given by \[\label{Eq:FLOGEN95Eq25} \hat{p}(t) = I_1 + I_2\tag{27}\] where \(I_1,\, I_2\) are given by (20 ) and (26 ) respectively.

DISCRETE TIME IMPLEMENTATION↩︎

Theoretically when \(p(t)\) has an RRC spectrum, both \(p(t),\, \hat{p}(t)\) have an infinite time span. In practice they have to be truncated, in which case (15 ) is only approximately valid. Moreover, the transmitter and receiver in Figures 2 and 3 have to be implemented in discrete time. In this section, we explore these issues.

The magnitude response of \(p(mT_s)\) (RRC) and \(\tilde{p}(mT_s)\) (HT-RRC) given by \[\label{Eq:FLOGEN95Eq25951} \tilde{p}(mT_s)=p(mT_s)+\mathrm{j}\,\hat{p}(mT_s)\tag{28}\] is shown in Figure 4 for \(T=1\) sec, \(\rho=0.161\) and sampling frequency \(F_s=1/T_s=5\) Hz. Both \(p(mT_s)\) and \(\hat{p}(mT_s)\) lie in the range \(-M\le m \le M\) for \(M=100\). The parameter \(M\) is referred to as the one-sided window length. Note that in Figure 4, \(\tilde{P}(\cdot)\) is the discrete Fourier transform of \(\tilde{p}(\cdot)\) given in (28 ).

The ratio of signal-to-interference (SIR) power is defined as \[\label{Eq:WFLOGEN95Eq26} SIR = 10 \log_{10} \left[ \frac{R_{gg}^2(0)}{\sum_{\substack{n\\n\ne 0}}R_{gg}^2(nT)} \right] \quad db.\tag{29}\] where \(g(\cdot)\) is \(p(\cdot)\) or \(\hat{p}(\cdot)\). Note that \[\begin{align} \label{Eq:FLOGEN95Eq27} R_{gg}(mT_s) & = g(mT_s) \star g(-mT_s) \nonumber \\ R_{gg}(nT) & = R_{gg}(nIT_s) \nonumber \\ I & = \frac{T}{T_s} \end{align}\tag{30}\] where “\(\star\)” denotes (discrete-time) linear convolution and \(I\) is the interpolation factor.

The SIR in decibels is shown in Table [Tbl:FLOGEN202395SIR1] for different values of the one-sided window length \(M\) when \(T=1\) sec, \(\rho=0.161\), \(F_s=5=1/T_s\) Hz. We find from Table [Tbl:FLOGEN202395SIR1] that the SIR of \(R_{\hat{p}\hat{p}}(t)\) is much lower than \(R_{pp}(t)\) for the same values of \(M\). This is probably because the signum function in the Hilbert transform (see (19 )) has a discontinuity at \(F=0\). In the next section, we discuss the modified Hilbert transform that avoids this discontinuity.

MODIFIED HILBERT TRANSFORM↩︎

Consider the modified Hilbert transform given by [15] \[\label{Eq:FLOGEN95Eq28} \tilde{H}(F) = \left\{ \begin{array}{ll} \mathrm{e}^{\,\mathrm{j}[\pi(F+aF_1)/(2aF_1)+\pi/2]} & for |F|\le aF_1 \\ - \mathrm{j} & for F\ge aF_1 \\ \mathrm{j} & for F\le -aF_1 \end{array} \right.\tag{31}\] for \(0<a\le 1\), which is plotted in Figure 5.

Hence the modified Hilbert transform of \(p(t)\) can be written as \[\begin{align} \label{Eq:FLOGEN95Eq29} \hat{p}_{\mathrm{m}}(t) & = \int_{F=-\infty}^{\infty} \tilde{H}(F) P(F) \mathrm{e}^{\,\mathrm{j}\, 2\pi Ft} \, dF \nonumber \\ & = I_{11} + I_{12} + I_2 \end{align}\tag{32}\] where \(I_2\) is given by (26 ) and \[\begin{align} \label{Eq:FLOGEN95Eq30} I_{11} & = \int_{F=-aF_1}^{aF_1} \tilde{H}(F) P(F) \mathrm{e}^{\,\mathrm{j}\, 2\pi Ft} \, dF \nonumber \\ & = \frac{-1}{\sqrt{2B}} \int_{F=-aF_1}^{aF_1} \mathrm{e}^{\,\mathrm{j}\,\pi F/(2aF_1)} \mathrm{e}^{\,\mathrm{j}\, 2\pi Ft} \, dF \nonumber \\ & = \frac{-2aF_1}{\sqrt{2B}} sinc \left( aF_1A_1 \right) \end{align}\tag{33}\] where \[\begin{align} \label{Eq:FLOGEN95Eq31} A_1 & = 2t + \frac{1}{2aF_1} \nonumber \\ sinc(x) & = \frac{\sin(\pi x)}{\pi x}. \end{align}\tag{34}\] Similarly \(I_{12}\) in (32 ) is equal to \[\begin{align} \label{Eq:FLOGEN95Eq32} I_{12} & = \frac{-\mathrm{j}}{\sqrt{2B}} \int_{aF_1}^{F_1} \mathrm{e}^{\,\mathrm{j}2\pi Ft} \, dF + \frac{\mathrm{j}}{\sqrt{2B}} \int_{-F_1}^{-aF_1} \mathrm{e}^{\,\mathrm{j}2\pi Ft} \, dF \nonumber \\ & = \frac{2F_1(1+a)}{\sqrt{2B}} sinc(F_1 t(1+a)) \sin(\pi F_1 t(1-a)). \end{align}\tag{35}\]

The magnitude response of \(p(mT_s)\) (RRC) and \(\tilde{p}_{\mathrm{m}}(mT_s)\) (mHT-RRC) given by \[\label{Eq:FLOGEN95Eq33} \tilde{p}_{\mathrm{m}}(mT_s)=p(mT_s)+\mathrm{j}\,\hat{p}_{\mathrm{m}}(mT_s)\tag{36}\] is shown in Figure 6 for \(T=1\) sec, \(\rho=0.161\) and sampling frequency \(F_s=1/T_s=5\) Hz. Both \(p(mT_s)\) and \(\hat{p}_{\mathrm{m}}(mT_s)\) lie in the range \(-M\le m \le M\) for \(M=100\). Note that in Figure 6, \(\tilde{P}_{\mathrm{m}}(\cdot)\) is the discrete Fourier transform of \(\tilde{p}_{\mathrm{m}}(\cdot)\) given in (36 ).

We find from Table [Tbl:FLOGEN202395SIR2] that the SIR of \(R_{\hat{p}_{\mathrm{m}}\hat{p}_{\mathrm{m}}}(nT)\) is comparable to \(R_{pp}(nT)\).

DISCRETE TIME SYSTEM MODEL↩︎

The discrete time system model used for computer simulations is shown in Figure 7. The input \(b_k\) is arranged into frames of length \(L_{d1}\) bits and given to the rate-\(1/2\) turbo code [1]–[3]. The output of the turbo code of length \(L_d=2L_{d1}\) bits is mapped to binary phase shift keyed (BPSK) symbols \(S_{i,\, k}=\pm 1\), demultiplexed and transmitted simultaneously over \(N\) subcarriers. The frequency of the \(i^{th}\) subcarrier is given by \[\label{Eq:FLOGEN95Eq34} \omega_i = 2\pi i/N \quad radians \quad for 0\le i \le N-1\tag{37}\] where \(N=I\) is the total number of subcarriers and \(I\) is given by (30 ). The overall OFDM-OQAM signal is given by \[\label{Eq:FLOGEN95Eq34951951} \tilde{s}_o(mT_s) = \sum_{i=0}^{N-1} \tilde{s}_i(mT_s) \mathrm{e}^{\,\mathrm{j}\, \omega_i m}\tag{38}\] where \[\label{Eq:FLOGEN95Eq34951952} \tilde{s}_i(mT_s) = \sum_{k} S_{i,\, k} \tilde{p}_{\mathrm{m}}(mT_s - kT)\tag{39}\] where \(\tilde{p}_{\mathrm{m}}(\cdot)\) is given by (36 ).

The spectrum of the complex-valued overall OFDM-OQAM signal \(\tilde{s}_o(mT_s)\) is shown in Figure 8, where \[\label{Eq:FLOGEN95Eq34951} \omega = 2\pi F T_s \quad radians\tag{40}\] where \(F\) is the frequency in Hz. Recall that the sampling frequency \(1/T_s\) Hz maps to \(2\pi\) in the digital frequency domain. The variance per dimension of complex-valued additive white Gaussian noise (AWGN) is \[\label{Eq:FLOGEN95Eq35} \frac{1}{2} E \left[ \left| \tilde{w}(mT_s) \right|^2 \right] = \sigma_w^2.\tag{41}\] The in-phase and quadrature components of \(\tilde{w}(mT_s)\) are assumed to be independent. At the receiver we have for the \(i^{th}\) subcarrier \[\label{Eq:FLOGEN95Eq36} x_{i,\, k} = S_{i,\, k} + z_{i,\, k}\tag{42}\] where \(z_{i,\, k}\) denotes real-valued samples of AWGN with variance \(\sigma_w^2/2\) [16]. We assume that both \(p(\cdot)\) and \(\hat{p}_{\mathrm{m}}(\cdot)\) have unit energy, that is \[\begin{align} \label{Eq:FLOGEN95Eq37} \sum_{m=-M}^{M} p^2(mT_s) & = 1 \nonumber \\ \sum_{m=-M}^{M} \hat{p}^2_{\mathrm{m}}(mT_s) & = 1. \end{align}\tag{43}\] Note that from (12 ), (13 ) and (16 ), \(x_{i,\, k,\, I},\, x_{i,\, k,\, Q}\) in Figure 7 have to be summed (or averaged). Since \(S_{i,\, k}\) carries half a bit of information, the average signal-to-noise ratio (SNR) per bit is defined as [16] \[\begin{align} \label{Eq:FLOGEN95Eq38} \mathrm{SNR}_{\mathrm{av},\, b} & = \frac{2E\left[S_{i,\, k}^2\right]}{2D noise variance} \nonumber \\ & = \frac{2\times 1}{2\times \sigma^2_w/2} \nonumber \\ & = \frac{2}{\sigma^2_w}. \end{align}\tag{44}\]

SIMULATION RESULTS↩︎

The discrete-time simulation parameters are given in Table [Tbl:FLOGEN95Sim95Param]. The transmit filters are given by \(p(mT_s-0.5T_s),\, \hat{p}_{\mathrm{m}}(mT_s-0.5T_s)\), for \(-M\le m\le M\), for integer \(m\) and \(M=8N,\, 16N\). This is because, we require the filter length to be an integer multiple of \(N\), for the sake of implementation simplicity.

The computer simulation results for the bit-error-rate (BER) vs the average SNR per bit (\(\mathrm{SNR}_{\mathrm{av},\, b}\)) are presented in Figure 9. The theoretical BER is obtained from [19]. The BER results for \(M=8N\) are denoted by “sim8” and “theory8”. The BER results for \(M=16N\) are denoted by “sim16” and “theory16”. Observe the close match between theory and simulations.

CONCLUSIONS↩︎

This article discusses the implementation of an OFDM-OQAM/FBMC/UFMC system in discrete-time, using Hilbert transform. A simple matched filter receiver is sufficient for detecting the symbols. In the present work, an AWGN channel is considered. Frequency selective fading channels along with carrier frequency offsets can be considered in future works.

References↩︎