Filter-Bank Modulation Based Signal Design and Transmission Techniques for Intensity Modulated MIMO Visible Light Communication Systems

Abstract

Current research in wireless communication undoubtedly points towards the tremendous advantages of using visible light as a spectrum for significantly boosting the capacities and capabilities of communication systems. The intensity of light can be rapidly modulated to obtain a communication scheme which can transport much larger volumes of data at higher rates and lesser latency compared to current radio frequency methods. In this paper, we have developed two novel filter bank based multicarrier techniques for communication using visible light. We propose a new signal design that employs pulse shaping with square root raised cosine filtering and frequency translation to generate orthogonal multicarrier signals. Based on this, we propose a filter bank modulation technique for multiple-input–multiple-output visible light communication. Further, we develop another filter bank modulation method using spatial indexing which further improves the data rate and spectral efficiency. The advantages of the proposed methods are thoroughly investigated through exhaustive simulation studies. The methods are found to yield very good bit error rate performance and high data rates.

Appendix: Proof for Orthogonality Conditions in (4), (5) and (6)

In this section we prove the orthogonality conditions mentioned in (4), (5) and (6). In order to achieve this task, we first provide the magnitude response of FT of SRRC pulse denoted by $\vert H_{Ts}(f)\vert$ as

$$\begin{aligned} \vert H_{T_s}(f) \vert ={\left\{ \begin{array}{ll} 1 &{}\quad when \quad 0\le f \le \frac{1-\beta }{2T_s}\\ \sqrt{\frac{1}{2}\left( 1- sin\frac{\pi \left( 2fT_s-1\right) }{2\beta } \right) } &{}\quad when \quad \frac{1-\beta }{2T_s}\le f \le \frac{1+\beta }{2T_s}\\ 0 &{}\quad otherwise \end{array}\right. } \end{aligned}$$

(33)

It is evident from (33) that the magnitude of $H_{Ts}(f)$ is zero for $\vert f \vert > \frac{1+\beta }{2T_s}$. Let us define $f_k=\frac{k}{\eta T_s}$ for convenience. Then the FT of (1) and (2) is given, respectively, as

$$\begin{aligned} {\mathcal {F}}\left\{ g_k(t) \right\} =G_k(f)=\frac{H_{T_s}(f-f_k)+H_{T_s}(f+f_k)}{2} \end{aligned}$$

(34)

and

$$\begin{aligned} {\mathcal {F}}\left\{ {\hat{g}}_k(t) \right\} ={\hat{G}}_k(f)=\frac{H_{T_s}(f-f_k)-H_{T_s}(f+f_k)}{2j} \end{aligned}$$

(35)

Note that the operator ${\mathcal {F}}\left\{ \cdot \right\}$ is the FT operator. Also, $H_{T_s}(f)= {\mathcal {F}}\left\{ h_{T_s}(t) \right\}$. Since the $\vert H_{T_s}(f) \vert =0$ for $\vert f \vert > \frac{1+\beta }{2T_s}$, we obtain

$$\begin{aligned} \vert H_{T_s}(f-f_k) \vert =0 \quad for {\left\{ \begin{array}{ll} f > f_k+ \frac{1+\beta }{2T_s} \\ f < f_k- \frac{1+\beta }{2T_s} \end{array}\right. } \end{aligned}$$

(36)

Similarly we obtain

$$\begin{aligned} \vert H_{T_s}(f+f_k) \vert =0 \quad for {\left\{ \begin{array}{ll} f > -f_k+ \frac{1+\beta }{2T_s} \\ f < -f_k- \frac{1+\beta }{2T_s} \end{array}\right. } \end{aligned}$$

(37)

Since the magnitude spectrum of $\vert H_{T_s}(f-f_k) \vert$ and $\vert H_{T_s}(f+f_k) \vert$ satisfy the conditions in (36) and (37), the magnitude spectrum of $G_k(f)$ and ${\hat{G}}_k(f)$ also satisfy the same zero conditions mentioned in both (36) and (37). With these notions in mind we will prove the orthogonality conditions mentioned in (4), (5) and (6).

1.1 Proof for (4)

The inner product of $G_k(f)$ and ${\hat{G}}_k(f)$ is given by

$$\begin{aligned} {<}G_k(f),G_p(f){>}=\int _{-\infty }^{+\infty }G_k(f)G_p^*(f) df \end{aligned}$$

(38)

Let us consider the case where $k \ne p$. In such a scenario, the function very nearer to $G_k(f)$ is $G_{k+1}(f)$ or $G_{k-1}(f)$, where $\vert k-p \vert =1$. We consider $p-k=1$ and the same is true for $k-p=1$. Also, we define the frequency difference as

$$\begin{aligned} \delta f_{k-p}=f_p-f_k=\frac{(k-p)}{\eta T_s} \end{aligned}$$

(39)

It is evident from (39) that the frequency difference will be minimum, when $\eta$ is maximum which is equal to 0.5 and $\vert k-p \vert =1$. In such a scenario, we obtain

$$\begin{aligned} \delta f_{1}=\frac{2}{ T_s} \end{aligned}$$

(40)

Since $G_k(f)$ and $G_p(f)$ satisfies the conditions mentioned in (36) and (37), it is obvious that the maximum bandwidth of $G_k(f)$ (when $\beta =1$) is $\frac{2}{T_s}$. Since $G_k(f)$ is symmetric function centered at $f_k$, the magnitude response $\vert G_k(f)\vert =0$, $\forall f > f_k+\frac{1}{T_s}$. Similarly, for $\vert {\hat{G}}_p(f)\vert =0$, $\forall f< f_p-\frac{1}{T_s}$. Since $G_k(f)$ and $G_p(f)$ are separated by a frequency difference of $\delta f_{1}$, we can conclude that

$$\begin{aligned} <G_k(f),G_p(f)>=0 \quad if \; k\ne p \end{aligned}$$

(41)

Since it is true for minimum frequency separation, it is obvious that it is true for all other possible values of $\eta$ where $\eta \le 0.5$. However, if $\eta > 0.5$, then the $\vert G_k(f)\vert \ne 0$ and $\vert G_p(f)\vert \ne 0$ will become non-zero in the specified range and the orthogonality condition is not satisfied.

However, for the situation where $k=p$, we have two cases where ${\bar{m}}=0$ and ${\bar{m}} \ne 0$. We first consider the case where ${\bar{m}}=0$. At the sampling points ${\bar{m}}T_s$, we obtain

$$\begin{aligned} \int _{-\infty }^{+\infty }g_k(t)g_k^*(t-{\bar{m}}T_s) dt \left| _{{\bar{m}}=0} = \int _{-\infty }^{+\infty }\left| g_k(t) \right| ^2 dt \right. \end{aligned}$$

(42)

Since the function $|g_k(t)|^2$ is a bounded function, we obtain

$$\begin{aligned} \int _{-\infty }^{+\infty }|g_k(t)|^2 dt=C_{G} \end{aligned}$$

(43)

where $C_G$ is the area under the curve $|g_k(t)|^2$. Next we consider the case where ${\bar{m}} \ne 0$. Then we obtain

$$\begin{aligned} \int _{-\infty }^{+\infty }g_k(t)g_k^*(t-{\bar{m}}T_s) dt \left| _{{\bar{m}}\ne 0}=\int _{ f_k- \frac{1+\beta }{2T_s}}^{+ f_k+ \frac{1+\beta }{2T_s}}|G_k(f)|^{2}e^{+j2\pi f {\bar{m}} T_s} df \left| _{{\bar{m}}\ne 0} \right. \right. \end{aligned}$$

(44)

Since $e^{+j2\pi f {\bar{m}} T_s}$ is a periodic function and the function $|G_k(f)|^{2}$ is bounded in frequency and magnitude, also symmetric and even, the above integral in (44) becomes zero.

1.2 Proof for (5)

It is evident from (35) that both $G_k(f)$ and ${\hat{G}}_k(f)$ share the same magnitude spectrum. Hence the same orthogonality proof mentioned in “Proof for (4)” appendix section is true for orthogonality condition mentioned in (5). Hence a separate description is not provided.

1.3 Proof for (6)

In this section, we prove the orthogonality of $g_k(t)$ and ${\hat{g}}_p(t)$. Note that the inner product of $g_k(t)$ and ${\hat{g}}_p(t)$ is equal to the inner product of $G_k(f)$ and ${\hat{G}}_p(f)$ according to the power theorem (or Parseval’s theorem) of FT. Thus, we obtain

$$\begin{aligned}&<G_k(f),{\hat{G}}_p(f)>=\frac{j}{4}\left( <H_{T_s}(f-f_k),H_{T_s}(f-f_p)>\right. \end{aligned}$$

(45)

$$\begin{aligned}&\quad -<H_{T_s}(f-f_k),H_{T_s}(f+f_p)>+<H_{T_s}(f+f_k),H_{T_s}(f-f_p)>\end{aligned}$$

(46)

$$\begin{aligned}&\quad \left. -<H_{T_s}(f+f_k),H_{T_s}(f+f_p)>\right) \end{aligned}$$

(47)

Since $H_{T_s}(f-f_k)$ and $H_{T_s}(f+f_p)$ are the frequency shifted versions of $H_{T_s}(f)$ that are centered at $f_k$ and $-f_p$ respectively, the frequency difference between these center frequencies is given by $f_k+f_p=\frac{k+p}{\eta T_s}$. Hence, the magnitude of $H_{T_s}(f-f_k)$ in the frequency range of $H_{T_s}(f+f_p)$ is always zero for any value of k and p. Thus (45) can be rewritten as

$$\begin{aligned}<G_k(f),{\hat{G}}_p(f)>&=\frac{j}{4}\left(<H_{T_s}(f-f_k) ,H_{T_s}(f-f_p)>\right. \nonumber \\&-\left. <H_{T_s}(f+f_k),H_{T_s}(f+f_p)>\right) \end{aligned}$$

(48)

By using (6), we have

$$\begin{aligned} \int _{-\infty }^{+ \infty }g_k(t){\hat{g}}_p^{*}(t-{\bar{m}}T_s) dt= \int _{-\infty }^{+ \infty } G_k(f){\hat{G}}_p^{*}(f)e^{+j2\pi f{\bar{m}}T_s}df \end{aligned}$$

(49)

We define

$$\begin{aligned} \int _{-\infty }^{+ \infty } G_k(f){\hat{G}}_p^{*}(f)e^{+j2\pi f{\bar{m}}T_s} df=I \end{aligned}$$

(50)

where I is result of the integral. Using (45), we obtain

$$\begin{aligned}&I=\int _{-\infty }^{+ \infty } \frac{je^{+j2\pi f{\bar{m}}T_s}}{4}\left( H_{T_s}(f-f_k)H_{T_s}^{*}(f-f_p) \right. \nonumber \\&\quad \left. -H_{T_s}(f+f_k)H_{T_s}^{*}(f+f_p)\right) df \end{aligned}$$

(51)

Here we have two possible cases, where $k=p$ and $k \ne p$. Let us consider the first case with $k=p$. Defining $P_{rc}(f)=H_{T_s}(f)H_{T_s}^{*}(f)$, where $P_{rc}(f)$ follows the magnitude response of raised cosine filter. Also, by interchanging the order of summation and integration, we obtain

$$\begin{aligned} I=\int _{-\infty }^{+ \infty } \frac{1}{4}e^{+j(2\pi f{\bar{m}}T_s+\frac{\pi }{2})} \left( P_{rc}(f-f_k) -P_{rc}(f+f_k) \right) df \end{aligned}$$

(52)

However, the area under $P_{rc}(f-f_k)$ and $P_{rc}(f+f_k)$ is finite and are same as the area under $P_{rc}(f)$ which is given by

$$\begin{aligned} \int _{-\infty }^{+ \infty } P_{rc}(f)df=A_{P_{rc}} \end{aligned}$$

(53)

It is because these filters are frequency shifted versions of $P_{rc}(f)$. Hence, by using the product rule for integration, we obtain

$$\begin{aligned} I&=\frac{1}{4}e^{+j(2\pi f{\bar{m}}T_s+\frac{\pi }{2})}\left( A_{P_{rc}} -A_{P_{rc}}\right) \nonumber \\&\quad - \int _{-\infty }^{+ \infty } 2\pi {\bar{m}} T_s e^{+j(2\pi f{\bar{m}}T_s+\frac{\pi }{2})}\left( A_{P_{rc}} -A_{P_{rc}}\right) df=0 \end{aligned}$$

(54)

Hence the proof. Similarly, for $k \ne p$, we have the minimum possible frequency difference of $\delta f_1$. As explained in “Proof for (4)” appendix section, we obtain the inner product of $G_k(f)$ and ${\hat{G}}_p(f)$ as zero because they don’t overlap in frequency domain. Hence it is obvious that they are orthogonal.