Signal Let s sample the signal at a time interval o Dr. Christopher M. Godrey University o North Carolina at Asheville Photo: C. Godrey Let s sample the signal at a time interval o Reconstruct the curve rom the sampled points Samples at Samples at High-requency components in the curve are sampled poorly Given that there can be sampling problems, what sampling interval should be selected in order to ully deine the given input signal? Low-requency components in the curve are sampled well Samples at We ll look at a version o the sampling theorem to answer this but irst, let s review some properties o sines and cosines 1
Fourier Analysis Fourier analysis shows that any analog signal can be represented perectly by an ininite sum o sines and cosines Fourier Analysis D1 Great attributes o sine unctions It s possible to deine a sine wave, with unity amplitude, using only two points (more is better)! For a given signal with multiple requency components, we must ind the largest sampling time interval ( s ), or lowest sampling requency ( s ) that will allow recovery o the signal without error Note that = 1/() Think o as the # complete cycles per unit time I we know max (i.e., the highest requency o the wave), we can appropriately sample the signal or error-ree recovery! We can igure out max using Fourier analysis Theorem Let s sample a signal periodically every s s is the sampling time interval s = 1/ s is the sampling requency Then each o the p samples strung together uniquely deine the sampled signal, where p(t)=p(n s ) and n is an integer s D2 Theorem For a limited bandwidth signal (bandwidth reers to the range o requencies contained within a signal) with maximum requency max, the sampling requency s must be greater than twice the maximum requency max in order to uniquely reconstruct the signal without aliasing: s > 2 max The requency 2 max is called the Nyquist rate. Used in this context, the Nyquist rate is the lower bound or the sampling rate necessary or alias-ree sampling. The Nyquist rate is a property o the signal. The Nyquist Frequency The Nyquist requency is hal the sampling requency: N = s /2=1/(2 s ) I the sampled signal contains max > N, aliasing (requency olding) will occur. The sampled wave will appear to have a requency that is olded to a lower requency. To completely reconstruct a wave without error, N must be greater than max. 2
The Nyquist Frequency The Nyquist rate is a property o a continuous signal I you want to sample a signal without aliasing, you must sample it with a sampling requency that is greater than or equal to the Nyquist rate or that signal. The Nyquist requency is a property o a discrete sampling system Any given sampling interval s chosen or a particular sampling system (e.g., radar) corresponds with a particular Nyquist requency, regardless o the properties o the sampled signal. Example Assume that there is some signal with many requency components associated with it We know that the maximum requency component o the signal is 5 khz To completely reconstruct the signal, we would need to sample the signal at what requency and time interval? Answer: We would need to sample the wave at the Nyquist rate, or s = 2 max = 10 khz. This corresponds with a s o 0.0001 s. The Nyquist requency in this case is s /2 = 5 khz, so no aliasing will occur. Another question: What happens i we sample at a requency that is less than the Nyquist rate? Recovery o the original signal is impossible Funny things happen with the reconstructed signal I we sample close to the actual requency o the input signal, the output will appear olded back to a lower requency. Let s sample a simple sine wave I we sample with s < 2 max, those original requencies that are above N will appear as alse low-requency components o the sampled signal. 3
I we sample it once per cycle ( s = max ), we might think it is constant! I we sample it 1.5 times per cycle ( s = 1.5 max ), the reconstructed wave looks like a lower-requency sine wave Why? Because N = ½ s = ¾ max < max The original wave has a higher requency than our Nyquist requency. Sampled at twice the requency o the input wave (i.e., s = 2 max ), the reconstruction approximates the original sine wave (i we sample in the right spot). Oversampling results in an even better approximation to the original input signal Why? Because N = ½ s = max In general, the aliased output requency is Frequency olding ± aliased = input 2m N where: aliased is the aliased output requency input is the input requency o the original wave m is an integer N is the Nyquist requency The x-axis shows the input requency (i.e., the requency o the original signal) in terms o the Nyquist requency (which is determined by the sampling interval). ± aliased = input 2m N 4
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