24 avr 2020 · Solution: Waveform g(t) is unlimited in time, so it has a hard bandlimited frequency response We have a bandwidth of 50 Hz, so the Nyquist rate is 100 Hz ⇐ The Nyquist interval is the reciprocal of the Nyquist rate = 1/100 sec = 10 ms ⇐ Note: Therefore, 2B = 100 Hz, so B = 50 Hz
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24 avr 2020 · Solution: Waveform g(t) is unlimited in time, so it has a hard bandlimited frequency response We have a bandwidth of 50 Hz, so the Nyquist rate is 100 Hz ⇐ The Nyquist interval is the reciprocal of the Nyquist rate = 1/100 sec = 10 ms ⇐ Note: Therefore, 2B = 100 Hz, so B = 50 Hz
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Find the Nyquist rate and Nyquist interval for the continuous-time the message signal from its samples, sampling frequency or sampling rate should be greater
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Nyquist rate is 200 Hz, and the Nyquist interval is 1/200 seconds domain, we find that the signal g(t) has a bandwidth equal to twice that of the sinc pulse
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1 Find the minimum sampling rate required to avoid aliasing 2 If , What is the discrete-time signal after sampling?
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19 avr 2012 · ((F) The Nyquist rate for the derivative of a signal is greater than the Determine the respective values of wi and W2, both in terms of W,
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This rate is lower than the Nyquist rate, and we will get aliasing The shifted triangles of width 2 will now be spaced one unit apart in the frequency domain,
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components, we must find the largest sampling time interval (Δts), or lowest sampling frequency (fs) that will allow recovery of the signal without error ATMS 320
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10 oct 2000 · From this example, we can see the reason for the term aliasing That is, the signal now takes on a different \persona," or a false presentation, due
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Determine the Nyquist sampling rate and the Nyquist sampling interval for the signals When f0 = 40, we start to see the aliasing effect in the frequency domain
[PDF] Sampling - MIT OpenCourseWare
22 nov 2011 · To determine the effect of sampling, compare the original signal x(t) The minimum sampling frequency, 2ωm, is called the “Nyquist rate ” 21
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