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author | Nao Pross <np@0hm.ch> | 2021-11-01 00:22:02 +0100 |
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committer | Nao Pross <np@0hm.ch> | 2021-11-01 01:18:13 +0100 |
commit | 5d395dc3b2c8e9c94cf72f235f63586cde84903a (patch) | |
tree | ab2ee738b2f904d0e4e5f1ddcbf308c6e1f68032 /doc/thesis/chapters | |
parent | Fix notation and typos, add placeholder titlepage (diff) | |
download | Fading-5d395dc3b2c8e9c94cf72f235f63586cde84903a.tar.gz Fading-5d395dc3b2c8e9c94cf72f235f63586cde84903a.zip |
On the FT of h(t,tau)
Diffstat (limited to 'doc/thesis/chapters')
-rw-r--r-- | doc/thesis/chapters/theory.tex | 18 |
1 files changed, 12 insertions, 6 deletions
diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex index 4176ea4..2c4687b 100644 --- a/doc/thesis/chapters/theory.tex +++ b/doc/thesis/chapters/theory.tex @@ -56,7 +56,7 @@ i.e. a pulse function\footnote{Typically a root raised cosine to optimize for ba \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 \(\omega_c = 2\pi / T\). Why this is possible is explained in the next section. +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 \(\omega_c = 2\pi / T\). How this is possible is explained in the next section. \subsection{Orthogonality of carrier signals} @@ -162,7 +162,7 @@ The simplest way to understand the multipath fading, is to consider it from a ge } \end{figure} -The problem is that, as is geometrically evident, some paths are longer than others. Thus the signal is received by the receiver multiple times with different phase shifts \cite{Gallager,Messier}. To mathematically model this effect, we describe the received signal \(r(t)\) as a linear combination of delayed copies of the sent signal \(s(t)\), each with a different attenuation \(c_k\) and phase shift \(\tau_k\): +The problem is that, as is geometrically evident, some paths are longer than others. Because of this fact, the signal is seen by the receiver multiple times with different phase shifts \cite{Gallager,Messier}. To mathematically model this effect, we describe the received signal \(r(t)\) as a linear combination of delayed copies of the sent signal \(s(t)\), each with a different attenuation \(c_k\) and phase shift \(\tau_k\): \begin{equation} \label{eqn:geom-multipath-rx} r(t) = \sum_k c_k s(t - \tau_k). \end{equation} @@ -174,13 +174,13 @@ A further complication arises, when one end (or both) is not stationary. In that We have thus observed that the arrangement can be modelled as a linear time-\emph{varying} system (LTV), if the sender or the receiver (or anything else in the channel) is moving, and as a linear time \emph{invariant} (LTI) system if the geometry is stationary. Regardless of which of the two cases, linearity alone is sufficient to approximate the channel as finite impulse response (FIR) filter \cite{Messier}. Mathematically we can rewrite LTV version of equation \eqref{eqn:geom-multipath-rx} using a convolution product as following: \begin{align*} r(t) = \sum_k c_k(t) s(t - \tau_k(t)) &= \sum_k c_k(t) \int_\mathbb{R} s(\tau) \delta(\tau - \tau_k(t)) \,d\tau \\ - &= \int_\mathbb{R} s(\tau) \sum_k c_k(t) \delta(\tau - \tau_k(t)) \,dt' = s(\tau) * h(t, \tau), + &= \int_\mathbb{R} s(\tau) \sum_k c_k(t) \delta(\tau - \tau_k(t)) \,d\tau = s(\tau) * h(\tau, t), \end{align*} obtaining a new function \begin{equation} \label{eqn:multipath-impulse-response} - h(t,\tau) = \sum_k c_k(t) \delta(\tau - \tau_k(t)), + h(\tau, t) = \sum_k c_k(t) \delta(\tau - \tau_k(t)), \end{equation} -that describes the \emph{impulse response} of the channel. This function is dependant on two time parameters: actual time \(t\) and convolution time \(\tau\). To better understand \(h(t,\tau)\), consider an example in shown in figure \ref{fig:multipath-impulse-response}. Each stem represents a weighted Dirac delta, so each series of stems of the same color, along the convolution time \(\tau\) axis, is a channel response at some specific time \(t\). Along the other \(t\) axis we see how the entire channel response changes over time\footnote{In the figure only a finite number of stems was drawn, but actually \(h(t,\tau)\) is continuous in \(t\), i.e. the weights \(c_k(t)\) of the Dirac deltas change continuously.}. Notice that the stems are not quite aligned to the \(\tau\) time raster (dotted lines), that is because in equation \eqref{eqn:multipath-impulse-response} not only the weights \(c_k\) but also the delays \(\tau_k\) are time dependent. +that describes the \emph{impulse response} of the channel. This function is dependant on two time parameters: actual time \(t\) and convolution time \(\tau\). To better understand \(h(\tau, t)\), consider an example in shown in figure \ref{fig:multipath-impulse-response}. Each stem represents a weighted Dirac delta, so each series of stems of the same color, along the convolution time \(\tau\) axis, is a channel response at some specific time \(t\). Along the other \(t\) axis we see how the entire channel response changes over time\footnote{In the figure only a finite number of stems was drawn, but actually \(h(\tau, t)\) is continuous in \(t\), i.e. the weights \(c_k(t)\) of the Dirac deltas change continuously.}. Notice that the stems are not quite aligned to the \(\tau\) time raster (dotted lines), that is because in equation \eqref{eqn:multipath-impulse-response} not only the weights \(c_k\) but also the delays \(\tau_k\) are time dependent. \begin{figure} \centering @@ -193,7 +193,13 @@ that describes the \emph{impulse response} of the channel. This function is depe \subsection{Spectrum of a multipath fading channel} -With a continuous time channel model we can now discuss the spectral properties of a fading channel. For this section, we will assume a LTI channel impulse response \(h(\tau)\) and consider a simple geometry. +With a continuous time channel model we can now discuss the spectral properties of a fading channel, since the frequency response is the Fourier transform of the impulse response, mathematically \(H(f, t) = \fourier h(\tau, t)\). In this case however \(h(\tau, t)\) depends on two time variables, but that is actually not an issue, it just means that the frequency response is also changing with time. Hence we perform the Fourier transform with respect to the channel (convolution) time variable \(\tau\) to obtain +\begin{equation} \label{eqn:multipath-frequency-response} + H(f, t) = \int_\mathbb{R} \sum_k c_k(t) \delta(\tau - \tau_k(t)) e^{-2\pi jf\tau} \, d\tau + = \sum_k c_k(t) e^{-2\pi jf \tau_k(t)}. +\end{equation} + +Equation \eqref{eqn:multipath-frequency-response} indicates that the frequency response is has a periodic (complex) sinusoidal shape, which has some important implications. A series of plots of the magnitude of the frequency response is shown in figure \ref{fig:multipath-frequency-response-plots}. \subsection{Discrete-time model} |