From 7cd6547ac35ecc3812b27241106a6910ac6e22c1 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Thu, 14 Oct 2021 19:08:34 +0200 Subject: Spectrum replication --- DigSig1.tex | 59 +++++++++++++++++++++++++++++++++++------------------------ 1 file changed, 35 insertions(+), 24 deletions(-) diff --git a/DigSig1.tex b/DigSig1.tex index 10379ae..6056b6e 100644 --- a/DigSig1.tex +++ b/DigSig1.tex @@ -67,8 +67,7 @@ \subsection{Random variables} -A \emph{random variable} (RV) is a function \(x : \Omega \to \mathbb{R}\). -The \emph{distribution function} of a RV is a function \(F_x : \mathbb{R} \to [0,1]\) that is always monotonically increasing and given by +A \emph{random variable} (RV) is a function \(x : \Omega \to \mathbb{R}\). The \emph{distribution function} of a RV is a function \(F_x : \mathbb{R} \to [0,1]\) that is always monotonically increasing and given by \[ F_x(\alpha) = \Pr{x \leq \alpha}. \] @@ -98,8 +97,7 @@ The \emph{variance} of a RV is \[ \Var{x} = \sigma^2 = \E{(x - \E{x})^2} = \E{x^2} - \E{x}^2, \] -where \(\sigma\) is called the \emph{standard deviation}. -The variance is sometimes also called the \emph{second moment} of a RV, the \emph{\(n\)-th moment} of a RV is \(\E{x^n}\). +where \(\sigma\) is called the \emph{standard deviation}. The variance is sometimes also called the \emph{second moment} of a RV, the \emph{\(n\)-th moment} of a RV is \(\E{x^n}\). \subsection{Jointly distributed RVs} @@ -122,9 +120,7 @@ Recall the three important operations for the analysis of analog signals. The Laplace transform reduces to the Fourier transform under the substitution \(s = j\Omega\). \subsection{Linear Systems} -Recall that superposition holds. -Thus the system is characterized completely by the impulse response function \(h(t)\). -The output in the time domain \(y(t)\) is given by the convolution product +Recall that superposition holds. Thus the system is characterized completely by the impulse response function \(h(t)\). The output in the time domain \(y(t)\) is given by the convolution product \[ y(t) = h(t) * x(t) = \int_\mathbb{R} h(t - t') x(t') \,dt', \] @@ -135,15 +131,11 @@ and in the frequency domain \(Y(\Omega) = H(\Omega) X(\Omega)\), where \(H(\Omeg \section{Sampling and reconstruction} -To sample a signal \(x(t)\) it means to measure (take) the value at a periodic interval every \(T\) seconds. \(T\) is thus called the \emph{sample interval} and \(f_s =1/T\) is the \emph{sampling frequency}. We will introduce the notation -\[ - x[n] = x(nT) -\] -to indicate that a signal is a set of discrete samples. +To sample a signal \(x(t)\) it means to measure (take) the value at a periodic interval every \(T\) seconds. \(T\) is thus called the \emph{sample interval} and \(f_s =1/T\) is the \emph{sampling frequency}. \subsection{Sampling theorem} -To represent a signal \(x(t)\) by its samples \(x[n]\) two conditions must be met: +To represent a signal \(x(t)\) by its samples \(\hat{x}(nT)\) two conditions must be met: \begin{enumerate} \item \(x(t)\) must be \emph{bandlimited}, i.e. there must be a frequency \(f_\text{max}\) after which the spectrum of \(x(t)\) is always zero. \item The sampling rate \(f_s\) must be chosen so that @@ -151,28 +143,47 @@ To represent a signal \(x(t)\) by its samples \(x[n]\) two conditions must be me f_s \geq 2 f_\text{max}. \] \end{enumerate} -In other words you need at least 2 samples / period to reconstruct a signal. -When \(f_s = 2 f_\text{max}\), the edge case, the sampling rate is called \emph{Nyquist rate}. -The interval \(\left[-f_s / 2, f_2 / 2\right]\), and its multiples are called \emph{Nyquist intervals}, as they are bounded by the Nyquist frequencies. -It would be good to have an arbitrarily high sampling frequency but in reality there is upper limit given by processing time \(T_\text{proc}\). Thus \(2f_\text{max} \leq f_s \leq f_\text{proc}\). +In other words you need at least 2 samples / period to reconstruct a signal. When \(f_s = 2 f_\text{max}\), the edge case, the sampling rate is called \emph{Nyquist rate}. The interval \(\left[-f_s / 2, f_2 / 2\right]\), and its multiples are called \emph{Nyquist intervals}, as they are bounded by the Nyquist frequencies. It would be good to have an arbitrarily high sampling frequency but in reality there is upper limit given by processing time \(T_\text{proc}\). Thus \(2f_\text{max} \leq f_s \leq f_\text{proc}\). \subsection{Discrete-Time Fourier Transform} -Mathematically speaking, to sample a signal is equivalent multiplying a function with the \emph{impulse train distribution}\footnote{Sometimes it is also called the Dirac comb.} +Mathematically speaking, to sample a signal is equivalent multiplying a function with the the so called \emph{impulse train distribution} (aka Dirac comb). +\[ + s(t) = \sum_{n = -\infty}^{\infty} \delta(t - nT), +\] +so we write \(\hat{x}(t) = s(t)\cdot x(t)\) to represent a sampled signal. Because of the special propertie of the Dirac delta, The spectrum of a sampled function \(\hat{x}\) is \[ - \Comb_T (t) = \sum_{n = -\infty}^{\infty} \delta(t - nT), + \hat{X}(f) = \sum_{n = -\infty}^{\infty} x(nT) e^{-2\pi jfTn}. \] -so \(x[n] = \Comb_T(t)\, x(t)\). -Interestingly the impulse train is periodic, and has thus a Fourier series with all coefficients equal to \(1/T\). -So the Fourier transform of a comb is also a comb, i.e. +This can be thought as a numerical approximation of the real spectrum \(X(f)\) which gets better as \(T \to 0\), i.e. \[ - \Comb_T(t) \leftrightarrow \Comb_{1/T}(\Omega). + X(f) = \lim_{T \to 0} T\hat{X}(f). \] - +If we have a finite number \(L\) of samples to work with, we will repeat them periodically and obtain what is known as the \emph{Discrete-Time Fourier Transform} (DTFT), i.e. +\[ + \hat{X}(f) \approx \hat{X}_L(f) = \sum_{n = 0}^{L -1} x(nT) e^{-2\pi jfTn}. +\] + \subsection{Spectrum replication and aliasing} + +Notice that the impulse train is periodic, and has thus a Fourier series, whose coefficients all equal to \(1/T\) (\(= f_s\), the sampling rate). So the Fourier transform of a comb is also a comb. The consequence is that, because the Fourier of the product \(x(t)\cdot s(t)\) in the time domain becomes a convolution \(X(f) * S(f)\) where \(S(f)\) is an impulse train of Dirac deltas spaced \(1/T = f_s\) apart, what is called \emph{spectrum replication} happens, mathematically +\[ + \hat{X}(f) + = \sum_{n = -\infty}^{\infty} x(nT) e^{2\pi jfTn} + = \frac{1}{T}\sum_{m = -\infty}^\infty X(f - mf_s). +\] +In other words, the modulation of the property of the Fourier transform copies the baseband spectrum into integer multiples of the sampling frequency. This is why \(f_s \geq 2f_\text{max}\). The important result is that +\[ + X(f) = T \hat{X}(f), \quad + \text{for} \quad -\frac{f_2}{2} \leq f \leq \frac{f_s}{2}, +\] +and if the sampling theorem is satisfied the exact original spectrum can be recovered with a low pass filter. + + % Alias frequency \(f_a = f \pmod{f_s}\). % Anti-aliasing: analog LP prefilter cutoff \@ \(f_s/2\) +\section{Quantization} \end{document} -- cgit v1.2.1