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This module builds on the intuition developed in the sampling module to discuss the Nyquist-Shannon sampling theorem, including a full statement and a proof.

Introduction

With the introduction of the concept of signal sampling, which produces a discrete time signal by selecting the values of the continuous time signal at evenly spaced points in time, it is now possible to discuss one of the most important results in signal processing, the Nyquist-Shannon sampling theorem. Often simply called the sampling theorem, this theorem concerns signals, known as bandlimited signals, with spectra that are zero for all frequencies with absolute value greater than or equal to a certain level. The theorem implies that there is a sufficiently high sampling rate at which a bandlimited signal can be recovered exactly from its samples, which is an important step in the processing of continuous time signals using the tools of discrete time signal processing.

Nyquist-shannon sampling theorem

Statement of the sampling theorem

The Nyquist-Shannon sampling theorem concerns signals with continuous time Fourier transforms that are only nonzero on the interval ( - B , B ) for some constant B . Such a function is said to be bandlimited to ( - B , B ) . Essentially, the sampling theorem has already been implicitly introduced in the previous module concerning sampling. Given a continuous time signals x with continuous time Fourier transform X , recall that the spectrum X s of sampled signal x s with sampling period T s is given by

X s ( ω ) = 1 T s k = - X ω - 2 π k T s .

It had previously been noted that if x is bandlimited to ( - π / T s , π / T s ) , the period of X s centered about the origin has the same form as X scaled in frequency since no aliasing occurs. This is illustrated in [link] . Hence, if any two ( - π / T s , π / T s ) bandlimited continuous time signals sampled to the same signal, they would have the same continuous time Fourier transform and thus be identical. Thus, for each discrete time signal there is a unique ( - π / T s , π / T s ) bandlimited continuous time signal that samples to the discrete time signal with sampling period T s . Therefore, this ( - π / T s , π / T s ) bandlimited signal can be found from the samples by inverting this bijection.

This is the essence of the sampling theorem. More formally, the sampling theorem states the following. If a signal x is bandlimited to ( - B , B ) , it is completely determined by its samples with sampling rate ω s = 2 B . That is to say, x can be reconstructed exactly from its samples x s with sampling rate ω s = 2 B . The angular frequency 2 B is often called the angular Nyquist rate. Equivalently, this can be stated in terms of the sampling period T s = 2 π / ω s . If a signal x is bandlimited to ( - B , B ) , it is completely determined by its samples with sampling period T s = π / B . That is to say, x can be reconstructed exactly from its samples x s with sampling period T s .

The spectrum of a bandlimited signals is shown as well as the spectra of its samples at rates above and below the Nyquist frequency. As is shown, no aliasing occurs above the Nyquist frequency, and the period of the samples spectrum centered about the origin has the same form as the spectrum of the original signal scaled in frequency. Below the Nyquist frequency, aliasing can occur and causes the spectrum to take a different than the original spectrum.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Ece 454 and ece 554 supplemental reading. OpenStax CNX. Apr 02, 2012 Download for free at http://cnx.org/content/col11416/1.1
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