This module will take the ideas of sampling CT signals further by examining how such operations can be performed in the frequency domain and by using a computer.
Introduction
We just covered ideal (and non-ideal) (time)
sampling of CT signals .
This enabled DT signal processing solutions for CTapplications (
):
Much of the theoretical analysis of such systems relied on
frequency domain representations. How do we carry out thesefrequency domain analysis on the computer? Recall the
following relationships:
$$x(n)\stackrel{\mathrm{DTFT}}{}X()$$$$x(t)\stackrel{\mathrm{CTFT}}{}X()$$ where
$$ and
$$ are continuous frequency
variables.
Sampling dtft
Consider the DTFT of a discrete-time (DT) signal
$x(n)$ . Assume
$x(n)$ is of finite duration
$N$ (
i.e. , an
$N$ -point signal).
$X()=\sum_{n=0}^{N-1} x(n)e^{-in}$
where
$X()$ is the continuous function that is indexed by thereal-valued parameter
$-\pi \le \le \pi $ . The other function,
$x(n)$ , is a discrete function that is indexed by
integers.
We want to work with
$X()$ on a computer. Why not just
sample$X()$ ?
In
we sampled at
$=\frac{2\pi}{N}k$ where
$k=\{0, 1, , N-1\}$ and
$X(k)$ for
$k=\{0, , N-1\}$ is called the
Discrete Fourier Transform (DFT) of
$x(n)$ .
The DTFT of the image in
is written as follows:
$X()=\sum_{n=0}^{N-1} x(n)e^{-in}$
where
$$ is any
$2\pi $ -interval, for example
$-\pi \le \le \pi $ .
where again we sampled at
$=\frac{2\pi}{N}k$ where
$k=\{0, 1, , M-1\}$ . For example, we take
$$M=10$$ . In the
following section we will discuss in
more detail how we should choose
$M$ , the number of samples in
the
$2\pi $ interval.
(This is precisely how we would plot
$X()$ in Matlab.)
Given
$N$ (length of
$x(n)$ ), choose
$(M, N)$ to obtain a dense sampling of the DTFT (
):
Case 2
Choose
$M$ as small as
possible (to minimize the amount of computation).
In general, we require
$M\ge N$ in order to represent all information in
$$\forall n, n=\{0, , N-1\}\colon x(n)$$ Let's concentrate on
$M=N$ :
$$x(n)\stackrel{\mathrm{DFT}}{}X(k)$$ for
$n=\{0, , N-1\}$ and
$k=\{0, , N-1\}$$$\mathrm{numbers}\mathrm{Nnumbers}$$
Discrete fourier transform (dft)
Define
$X(k)\equiv X(\frac{2\pi k}{N})$
where
$N=\mathrm{length}(x(n))$ and
$k=\{0, , N-1\}$ . In this case,
$M=N$ .
Represent
$x(n)$ in terms of a sum of
$N$complex sinusoids of amplitudes
$X(k)$ and frequencies
$$\forall k, k\in \{0, , N-1\}\colon {}_{k}=\frac{2\pi k}{N}$$
Fourier Series with fundamental frequency
$\frac{2\pi}{N}$
Remark 1
IDFT treats
$x(n)$ as though it were
$N$ -periodic.
Think of sampling the continuous function
$X()$ , as depicted in
.
$S()$ will represent the sampling function applied to
$X()$ and is illustrated in
as well. This will result in our
discrete-time sequence,
$X(k)$ .
Remember the multiplication in the frequency domain is equal
to convolution in the time domain!
Inverse dtft of s()
$\sum $∞∞2kN
Given the above equation, we can take the DTFT and get thefollowing equation:
$N\sum $∞∞nmNSn
Why does
equal
$S(n)$ ?
$S(n)$ is
$N$ -periodic,
so it has the following
Fourier Series :
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Azam
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Prasenjit
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Damian
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.