1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
|
% vim: set ts=2 sw=2 noet spell:
\chapter{Theory}
\begin{figure}
\centering
\input{figures/tikz/overview}
\caption{
Block diagram for a general wireless communication system with annotated signal names. Frequency domain representations of signals use the uppercase symbol of their respective time domain name.
\label{fig:notation}
}
\end{figure}
In this section we will briefly give the mathematical background required by the modulation schemes used in the project. The notation used is summarised in figure \ref{fig:notation}. For conciseness encoding schemes and (digital) signal processing calculations are left out and discussed later. Thus for this section \(m_e = m\).
%% TODO: Par on notation m(n) = m(nT) = discrete time
\section{Quadrature amplitude modulation (\(M\)-ary QAM)}
Quadrature amplitude modulation is a family of modern digital modulation methods, that use an analog carrier signal. The simple yet effective idea behind QAM is to encode extra information into an orthogonal carrier signal, thus increasing the number of bits sent per unit of time. A diagram showing the process is found in figure \ref{fig:quadrature-modulation}.
%% TODO: Quick par on "we will dicusss M-Ary QAM"
\paragraph{Bit splitter}
As mentioned earlier, quadrature modulation allows sending more than one bit per unit time. The first step to do it is to use a so called bit splitter, converts the continuous data stream \(m(n)\) into pairs of chunks of \(\sqrt{M}\) bits. The two bit vectors of length \(\sqrt{M}\), denoted by \(\vec{m}_i\) and \(\vec{m}_q\), are called in-phase and quadrature component respectively. The reason will become more clear later.
\paragraph{Binary to level converter}
%% TODO: explain why gray code
Both bit vectors \(\vec{m}_i, \vec{m}_q \in \{0,1\}^{\sqrt{M}}\) are sent through a binary to level converter. It's purpose is to reinterpret the bit vector as a number, usually in gray code, and to convert them into an analog waveform, which we will denote with \(m_i(t)\) and \(m_q(t)\) respectively. Mathematically the binary to level converter can be described as:
\begin{equation}
m_i(t) = \text{Level}(\vec{m}_i) \cdot p(t),
\end{equation}
i.e. with a pulse function \(p(t)\) (typically a root raised cosine to optimize for bandwidth) scaled by the interpreted binary value, which we will write here with a ``Level'' function. So at this point the analog waveform is already encoding \(\sqrt{M}\) bits per unit time, but actually it is possible to do better.
\paragraph{Mixer}
Having analog level signals, it is this now possible to mix them with radio frequency carriers. Because there are two waveforms, one might expect that two carrier frequencies are necessary, however this is not the case.
The two component \(m_i(t)\) and \(m_q(t)\) are mixed with two different periodic signals \(\phi_i(t)\) and \(\phi_q(t)\) that have the same frequency \(\omega_c = 2\pi / T\). Now the clever part: the carrier functions are picked to be \emph{orthonormal}, mathematically this is expressed by the conditions
\begin{subequations}
\begin{align}
\langle \phi_i | \phi_q \rangle
&= \int_T \phi_i^* \phi_q \, dt = \int_T \phi_i \phi_q^* \, dt
= 0, \text{ and } \\
\langle \phi_k | \phi_k \rangle
&= \int_T \phi_k^* \phi_k \,dt = 1,
\text{ where } k \text{ can be either } i \text{ or } q.
\end{align}
\end{subequations}
These rather abstract conditions remarkably allow for something very special. By defining a new signal
\begin{equation}
s = m_i\phi_i + m_q\phi_q,
\end{equation}
%% TODO: is this assumption correct?
notice that assuming \(m_i\) and \(m_q\) are constant over the period carrier's period \(T\),
\begin{align*}
\langle s | \phi_i \rangle = \int_T s^* \phi_i \,dt
&= \int m_i \phi_i^* \phi_i + m_q \phi_q^* \phi_i \,dt \\
&= m_i \underbrace{\int_T \phi_i^* \phi_i \,dt}_{1}
+ m_q \underbrace{\int_T \phi_q^* \phi_i \,dt}_{0} = m_i,
\end{align*}
which effectively means that it is possible to isolate a single component of the signal out of \(s\). The same of course works with \(\phi_q\) as well resulting in \(\langle s | \phi_q \rangle = m_q\).
% This formulation is rather abstract, in practice we usually pick \(\phi_i(t) = \cos(\omega_c t)\) and \(\phi_q(t) = j\sin(\omega_c t)\).
% \begin{figure}
% \centering
% \input{figures/tikz/qpks-constellation}
% \caption{
% % TODO: write caption
% \label{fig:qpks-constellation}
% }
% \end{figure}
\subsection{Phase Shift Keying (PSK)}
PSK is a popular modulation type for data transmission\cite{Meyer2011}. With a bipolar binary signal, the amplitude remains constant and only the phase will be changed with phase jumps of 180 degrees, which can be seen as a multiplication of the carrier signal with $\pm$ 1. That is alow known as binary phase shift keying.
% \begin{figure}
% % TODO: Better Image
% % https://sites.google.com/site/billmahroukelec675/bipolar-phase-shift-keying
% \includegraphics[width=5cm]{./image/BPSK2.png}
% \end{figure}
\subsection{Quadrature Phase Shift Keying (QPSK)}
Two bits are modulated at ones with the same bandwidth as a 2-PSK so more informations are transmitted at the same time. \cite{Meyer2011}
%TODO: Image Signal Raum
Most times there is noise and the points on the constellation diagram become a surface.
If the surfaces overlap there will be a problem with decoding.
\section{Fading}
\subsection{Geometric Model}
\subsection{Statistical Model}
%% TODO: write about advantage of statistical model instead of geometric
\paragraph{Continuous time model}
Continuous time small scale fading channel response.
time varying channel impulse response:
\begin{equation}
h(t, \tau) = \sum_k c_k (t) \delta(\tau - \tau_k(t))
\end{equation}
received signal \(y = h * x\), i.e. convolution with channel model.
\subsection{Time discretization of the model}
%% TODO: explain why
Assume \(x\) is a time discrete signal with and bandwidth \(W\), thus the pulse is sinc shaped
\begin{equation}
x(t) = \sum_n x[n] \sinc(t/T - n)
\end{equation}
Ideal sampling at rate \(2W\) of \(y\) gives
|
