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:
$$x(n)\stackrel{\mathrm{DTFT}}{\leftrightarrow}X(\omega )$$$$x(t)\stackrel{\mathrm{CTFT}}{\leftrightarrow}X(\Omega )$$ where
$\omega $ and
$\Omega $ 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).
$X(\omega )=\sum_{n=0}^{N-1} x(n)e^{-i\omega n}$
where
$X(\omega )$ is the continuous function that is indexed by thereal-valued parameter
$-\pi \le \omega \le \pi $ . The other function,
$x(n)$ , is a discrete function that is indexed by
integers.
We want to work with
$X(\omega )$ on a computer. Why not just
sample$X(\omega )$ ?
In
we sampled at
$\omega =\frac{2\pi}{N}k$ where
$k=\{0, 1, \dots , N-1\}$ and
$X(k)$ for
$k=\{0, \dots , N-1\}$ is called the
Discrete Fourier Transform (DFT) of
$x(n)$ .
The DTFT of the image in
is written as follows:
$X(\omega )=\sum_{n=0}^{N-1} x(n)e^{-i\omega n}$
where
$\omega $ is any
$2\pi $ -interval, for example
$-\pi \le \omega \le \pi $ .
where again we sampled at
$\omega =\frac{2\pi}{N}k$ where
$k=\{0, 1, \dots , M-1\}$ . 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
$X(\omega )$ in Matlab.)
Choosing m
Case 1
Given
$N$ (length of
$x(n)$ ), choose
$\gg (M, N)$ 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
$$\forall n, n=\{0, \dots , N-1\}\colon x(n)$$ Let's concentrate on
$M=N$ :
$$x(n)\stackrel{\mathrm{DFT}}{\leftrightarrow}X(k)$$ for
$n=\{0, \dots , N-1\}$ and
$k=\{0, \dots , N-1\}$$$\mathrm{numbers}\leftrightarrow \mathrm{N\; numbers}$$
Discrete fourier transform (dft)
Define
$X(k)\equiv X(\frac{2\pi k}{N})$
where
$N=\mathrm{length}(x(n))$ and
$k=\{0, \dots , N-1\}$ . In this case,
$M=N$ .
Represent
$x(n)$ in terms of a sum of
$N$complex sinusoids of amplitudes
$X(k)$ and frequencies
$$\forall k, k\in \{0, \dots , N-1\}\colon {\omega}_{k}=\frac{2\pi k}{N}$$
Fourier Series with fundamental frequency
$\frac{2\pi}{N}$
Remark 1
IDFT treats
$x(n)$ as though it were
$N$ -periodic.
When we deal with the DFT, we need to remember that, in
effect, this treats the signal as an
$N$ -periodic sequence.
A sampling perspective
Think of sampling the continuous function
$X(\omega )$ , as depicted in
.
$S(\omega )$ will represent the sampling function applied to
$X(\omega )$ 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(Ω)
$\sum $∞∞δω2kN
Given the above equation, we can take the DTFT and get thefollowing equation:
$N\sum $∞∞δnmNSn
Why does
equal
$S(n)$ ?
$S(n)$ is
$N$ -periodic,
so it has the following
Fourier Series :
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