Lecture 6: Sampling Theorem

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Schedule :: Sampling Theorem :: Issues for Sampling :: Some Applications :: Spectral Sampling

Due to the increased use of computers in all engineering applications, including signal processing, it is important to spend some more time examining issues of sampling. In this chapter we will look at sampling both in the time domain and the frequency domain.

We have already encountered the sampling theorem and, arguing purely from a trigonometric-identity point of view, have established the Nyquist sampling criterion for sinusoidal signals. However, we have not fully addressed the sampling of more general signals, nor provided a general proof. Nor have we indicated how to reconstruct a signal from its samples. With the tools of Fourier transforms and Fourier series available to us we are now ready to finish the job that was started months ago.

To begin with, suppose we have a signal which we wish to sample. Let us suppose further that the signal is bandlimited to Hz. This means that its Fourier transform is nonzero for $-2\pi B < \omega < 2\pi B$ < \omega < 2\pi B$" align="middle" border="0" height="30" width="129" />. Plot spectrum.
We will model the sampling process as multiplication of by the ``picket fence'' function

$\begin{displaymath}\delta_{T}(t) = \sum_{n} \delta(t-nT). \end{displaymath}$

We encountered this periodic function when we studied Fourier series. Recall that by its Fourier series representation we can write

$\begin{displaymath}\delta_{T}(t) = \frac{1}{T}\sum_{n} e^{jn\omega_s t} \end{displaymath}$

where $\omega_s = \frac{2\pi}{T}$ . The frequency $f_s = \omega_s/(2\pi) = 1/T$ is the sampling frequency in samples/sec. Suppose that the sampling frequency is chosen so that $f_s > 2B$

2B$" align="middle" border="0" height="30" width="60" />, or equivalently, $\omega_s > 4\pi B$ 4\pi B$" align="middle" border="0" height="30" width="71" />.

The sampled output is denoted as $\xbar(t)$ , where

$\begin{displaymath}\xbar(t) = x(t)\delta_{T}(t) \end{displaymath}$

Using the F.S. representation we get

$\begin{displaymath}\xbar(t) = x(t)\frac{1}{T}\sum_{n} e^{jn\omega_s t} \end{displaymath}$

Now lets look at the spectrum of the transformed signal. Using the convolution property,

$\begin{displaymath}\Xbar(\omega) = \frac{1}{2\pi}\frac{1}{T}X(\omega)*\sum_{n}2\... ...omega - n\omega_s) = \frac{1}{T}\sum_{n}X(\omega - n\omega_s). \end{displaymath}$

Plot the spectrum of the sampled signal with both $\omega$ frequency and $f$

frequency. Observe the following:

The spectrum is periodic, with period $2\pi$ , because of the multiple copies of the spectrum.
The spectrum is scaled down by a factor of .
Note that in this case there is no overlap between the images of the spectrum.

Now consider the effect of reducing the sampling rate to $f_s < 2B$ < 2B$" align="middle" border="0" height="30" width="60" />. In this case, the duplicates of the spectrum overlap each other. The overlap of the spectrum is aliasing.

This demonstration more-or-less proves the sampling theorem for general signals. Provided that we sample fast enough, the signal spectrum is not distorted by the sampling process. If we don't sample fast enough, there will be distortion. The next question is: given a set of samples, how do we get the signal back? From the spectrum, the answer is to filter the signal with a lowpass filter with cutoff $\omega_c \geq 2\pi B$ . This cuts out the images and leaves us with the original spectrum. This is a sort of idealized point of view, because it assumes that we are filtering a continuous-time function $\xbar(t)$ , which is a sequence of weighted delta functions. In practice, we have numbers representing the value of the function . How can we recover the time function from this?

Theorem 1 (The sampling theorem) If is bandlimited to Hz then it can be recovered from signals taken at a sampling rate 2B$" align="middle" border="0" height="30" width="60" />. The recovery formula is

$\begin{displaymath}x(t) = \sum_n x(nT) g(t-nT) \end{displaymath}$

where

$\begin{displaymath}g(t) = \frac{\sin(\pi f_s t)}{\pi f_s t} = \sinc(\pi f_s t).\end{displaymath}$

Show what the formula means: we are interpolating in time between samples using the $\sinc$ function.

We will prove this theorem. Because we are actually lacking a few theoretical tools, it will take a bit of work. What makes this interesting is we will end up using in a very essential way most of the transform ideas we have talked about.

The first step is to notice that the spectrum of the sampled signal,

$\begin{displaymath}\Xbar(\omega) = \frac{1}{T} \sum_n X(\omega - n\omega_s) \end{displaymath}$

is periodic and hence has a Fourier series. The period of the function in frequency is $\omega_s$ , and the fundamental frequency is

$\begin{displaymath}p_0 = \frac{2\pi}{\omega_s} = \frac{1}{f_s} = T. \end{displaymath}$

By the F.S. we can write

$\begin{displaymath}\Xbar(\omega) = \sum_n c_n e^{jn\omega T} \end{displaymath}$

where the are the F.S. coefficients

$\begin{displaymath}c_n = \frac{1}{\omega_s}\int_{\omega_s} \Xbar(\omega)e^{-jn\o... ...{-\infty}^\infty \frac{1}{T}X(\omega)e^{-jn\omega T} d\omega. \end{displaymath}$

But the integral is just the inverse F.T., evaluated at :

$\begin{displaymath}c_n = \frac{1}{T} \left.\frac{2\pi}{\omega_s} x(t)\right\vert _{t=-nT} = x(-nT), \end{displaymath}$

so

$\begin{displaymath}\Xbar(\omega) = \sum_n x(-nt) e^{jn\omega T} = \sum_n x(nt) e^{-jn\omega T}. \end{displaymath}$
Let $g(t) = \sinc(\pi f_s t)$ . Then

$\begin{displaymath}g(t) \Leftrightarrow T\rect\left(\frac{\omega}{2\pi f_s}\right). \end{displaymath}$
Let

$\begin{displaymath}y(t) = \sum_n x(nT) g(t-nT). \end{displaymath}$
We will show that by showing that $Y(\omega) = X(\omega)$ . We can compute the F.T. of using linearity and the shifting property:

$\begin{displaymath}Y(\omega) = \sum_n x(nT) T\rect\left(\frac{\omega}{2\pi f_s}\... ...ft(\frac{\omega}{2\pi f_s}\right) \sum_n x(nT) e^{-j\omega nT} \end{displaymath}$

Observe that the summation on the right is the same as the F.S. we derived in step 1:

$\begin{displaymath}Y(\omega) = T\rect\frac{\omega}{2\pi f_s} \Xbar(\omega) . \end{displaymath}$

Now substituting in the spectrum of the sampled signal (derived above)

$\begin{displaymath}Y(\omega) = T\rect\left(\frac{\omega}{2\pi f_s}\right)\left( \frac{1}{T} \sum_n X(\omega-n\omega_s)\right) = X(\omega) \end{displaymath}$

since is bandlimited to $-\pi f_s < \omega < \pi f_s$ < \omega < \pi f_s$" align="middle" border="0" height="30" width="117" /> or < f < f_s/2$" align="middle" border="0" height="32" width="128" />.

Notice that the reconstruction filter is based upon a sinc function, whose transform is a rect function: we are really just doing the filtering implied by our initial intuition.

In practice, of course, we want to sample at a frequency higher than just twice the bandwidth to allow room for filter rolloff.

Copyright 2008, by the Contributing Authors. Cite/attribute Resource. admin. (2006, June 12). Lecture 6: Sampling Theorem. Retrieved November 23, 2009, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/6_1node1.html. This work is licensed under a Creative Commons License.

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	Info Lecture 6: Sampling Theorem Document Actions Schedule :: Sampling Theorem :: Issues for Sampling :: Some Applications :: Spectral Sampling Due to the increased use of computers in all engineering applications, including signal processing, it is important to spend some more time examining issues of sampling. In this chapter we will look at sampling both in the time domain and the frequency domain. We have already encountered the sampling theorem and, arguing purely from a trigonometric-identity point of view, have established the Nyquist sampling criterion for sinusoidal signals. However, we have not fully addressed the sampling of more general signals, nor provided a general proof. Nor have we indicated how to reconstruct a signal from its samples. With the tools of Fourier transforms and Fourier series available to us we are now ready to finish the job that was started months ago. To begin with, suppose we have a signal which we wish to sample. Let us suppose further that the signal is bandlimited to Hz. This means that its Fourier transform is nonzero for $-2\pi B < \omega < 2\pi B$ < \omega < 2\pi B$" align="middle" border="0" height="30" width="129" />. Plot spectrum. We will model the sampling process as multiplication of by the ``picket fence'' function $\begin{displaymath}\delta_{T}(t) = \sum_{n} \delta(t-nT). \end{displaymath}$ We encountered this periodic function when we studied Fourier series. Recall that by its Fourier series representation we can write $\begin{displaymath}\delta_{T}(t) = \frac{1}{T}\sum_{n} e^{jn\omega_s t} \end{displaymath}$ where $\omega_s = \frac{2\pi}{T}$ . The frequency $f_s = \omega_s/(2\pi) = 1/T$ is the sampling frequency in samples/sec. Suppose that the sampling frequency is chosen so that 2B$" align="middle" border="0" height="30" width="60" />, or equivalently, $\omega_s > 4\pi B$ 4\pi B$" align="middle" border="0" height="30" width="71" />. The sampled output is denoted as $\xbar(t)$ , where $\begin{displaymath}\xbar(t) = x(t)\delta_{T}(t) \end{displaymath}$ Using the F.S. representation we get $\begin{displaymath}\xbar(t) = x(t)\frac{1}{T}\sum_{n} e^{jn\omega_s t} \end{displaymath}$ Now lets look at the spectrum of the transformed signal. Using the convolution property, $\begin{displaymath}\Xbar(\omega) = \frac{1}{2\pi}\frac{1}{T}X(\omega)\sum_{n}2\... ...omega - n\omega_s) = \frac{1}{T}\sum_{n}X(\omega - n\omega_s). \end{displaymath}$ Plot the spectrum of the sampled signal with both $\omega$ frequency and frequency. Observe the following: The spectrum is periodic, with period $2\pi$ , because of the multiple copies of the spectrum. The spectrum is scaled down by a factor of . Note that in this case* there is no overlap between the images of the spectrum. Now consider the effect of reducing the sampling rate to < 2B$" align="middle" border="0" height="30" width="60" />. In this case, the duplicates of the spectrum overlap each other. The overlap of the spectrum is aliasing. This demonstration more-or-less proves the sampling theorem for general signals. Provided that we sample fast enough, the signal spectrum is not distorted by the sampling process. If we don't sample fast enough, there will be distortion. The next question is: given a set of samples, how do we get the signal back? From the spectrum, the answer is to filter the signal with a lowpass filter with cutoff $\omega_c \geq 2\pi B$ . This cuts out the images and leaves us with the original spectrum. This is a sort of idealized point of view, because it assumes that we are filtering a continuous-time function $\xbar(t)$ , which is a sequence of weighted delta functions. In practice, we have numbers representing the value of the function . How can we recover the time function from this? Theorem 1 (The sampling theorem) If is bandlimited to Hz then it can be recovered from signals taken at a sampling rate 2B$" align="middle" border="0" height="30" width="60" />. The recovery formula is $\begin{displaymath}x(t) = \sum_n x(nT) g(t-nT) \end{displaymath}$ where $\begin{displaymath}g(t) = \frac{\sin(\pi f_s t)}{\pi f_s t} = \sinc(\pi f_s t).\end{displaymath}$ Show what the formula means: we are interpolating in time between samples using the $\sinc$ function. We will prove this theorem. Because we are actually lacking a few theoretical tools, it will take a bit of work. What makes this interesting is we will end up using in a very essential way most of the transform ideas we have talked about. The first step is to notice that the spectrum of the sampled signal, $\begin{displaymath}\Xbar(\omega) = \frac{1}{T} \sum_n X(\omega - n\omega_s) \end{displaymath}$ is periodic and hence has a Fourier series. The period of the function in frequency is $\omega_s$ , and the fundamental frequency is $\begin{displaymath}p_0 = \frac{2\pi}{\omega_s} = \frac{1}{f_s} = T. \end{displaymath}$ By the F.S. we can write $\begin{displaymath}\Xbar(\omega) = \sum_n c_n e^{jn\omega T} \end{displaymath}$ where the are the F.S. coefficients $\begin{displaymath}c_n = \frac{1}{\omega_s}\int_{\omega_s} \Xbar(\omega)e^{-jn\o... ...{-\infty}^\infty \frac{1}{T}X(\omega)e^{-jn\omega T} d\omega. \end{displaymath}$ But the integral is just the inverse F.T., evaluated at : $\begin{displaymath}c_n = \frac{1}{T} \left.\frac{2\pi}{\omega_s} x(t)\right\vert _{t=-nT} = x(-nT), \end{displaymath}$ so $\begin{displaymath}\Xbar(\omega) = \sum_n x(-nt) e^{jn\omega T} = \sum_n x(nt) e^{-jn\omega T}. \end{displaymath}$ Let $g(t) = \sinc(\pi f_s t)$ . Then $\begin{displaymath}g(t) \Leftrightarrow T\rect\left(\frac{\omega}{2\pi f_s}\right). \end{displaymath}$ Let $\begin{displaymath}y(t) = \sum_n x(nT) g(t-nT). \end{displaymath}$ We will show that by showing that $Y(\omega) = X(\omega)$ . We can compute the F.T. of using linearity and the shifting property: $\begin{displaymath}Y(\omega) = \sum_n x(nT) T\rect\left(\frac{\omega}{2\pi f_s}\... ...ft(\frac{\omega}{2\pi f_s}\right) \sum_n x(nT) e^{-j\omega nT} \end{displaymath}$ Observe that the summation on the right is the same as the F.S. we derived in step 1: $\begin{displaymath}Y(\omega) = T\rect\frac{\omega}{2\pi f_s} \Xbar(\omega) . \end{displaymath}$ Now substituting in the spectrum of the sampled signal (derived above) $\begin{displaymath}Y(\omega) = T\rect\left(\frac{\omega}{2\pi f_s}\right)\left( \frac{1}{T} \sum_n X(\omega-n\omega_s)\right) = X(\omega) \end{displaymath}$ since is bandlimited to $-\pi f_s < \omega < \pi f_s$ < \omega < \pi f_s$" align="middle" border="0" height="30" width="117" /> or < f < f_s/2$" align="middle" border="0" height="32" width="128" />. Notice that the reconstruction filter is based upon a sinc function, whose transform is a rect function: we are really just doing the filtering implied by our initial intuition. In practice, of course, we want to sample at a frequency higher than just twice the bandwidth to allow room for filter rolloff. Copyright 2008, by the Contributing Authors. Cite/attribute Resource. admin. (2006, June 12). Lecture 6: Sampling Theorem. Retrieved November 23, 2009, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/6_1node1.html. This work is licensed under a Creative Commons License.	Reuse Course Download this course

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Lecture 6: Sampling Theorem

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