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Examining the Signal
• Examine the signal using a very high-speed system, for example, a 50 MHz digital oscilloscope.
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Setting the Sampling Conditions• In most circumstances, as when using computers, sampling is DIGITAL.
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The Number of Samples• The number of required samples depends upon what information is needed
→ there is not one specific formula for N..
• For example, consider two different signals
Solid: ‘normal’ (random) population with mean =3 and standard deviation = 0.5
Dotted: same as solid but with 0.001/s additional amplitude decrease
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Digital Sampling
Figure 12.1
• The analog signal, y(t), is sampled every t seconds, N times for a period of T seconds, yielding the digital signal y(rt), where r = 1, 2, …, N.
• For this situation:
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Digital Sampling Errors
• When is signal is digitally sampled, erroneous results occur if either one of the following occur:
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Digital Sampling Errors
The least common multiple or lowest common multiple or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple of both a and b. Since it is a multiple, a and b divide it without remainder. For example, the least common multiple of the numbers 4 and 6 is 12. (Ref: Wikipedia)
• To avoid amplitude ambiguity, set the sample period equal to the least common (integer) multiple of all of the signal’s contributory periods.
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Illustration of Correct Sampling
y(t) = 5sin(2t)
→ f = 1 Hz
with fs = 8 Hz
Figure 12.7
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y(t) = sin(20t)
>> f = 10 Hzwith
fs = 12 Hz
Illustration of Aliasing
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Figures 12.8 and 12.9
The Folding Diagram
Example: f = 10 Hz; fs = 12 Hz
To determine the aliased frequency, fa:
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y(t) = sin(20t)
→ f = 10 Hzwith
fs = 12 Hz
Aliasing of sin(20t)
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Aliasing of sin(20t)
y(t) = 5sin(2t)→ f = 1 Hzfs = 1.33 Hz
Figure 12.13
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In-Class Example
• At what cyclic frequency will y(t) = 3sin(4t) appear if fs = 6 Hz?
fs = 4 Hz ?
fs = 2 Hz ?
fs = 1.5 Hz ?
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Correct Sample Time Period
y(t) = 3.61sin(4t+0.59)
+ 5sin(8t)
Figure 12.16
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Sampling with Aliasing
y(t) = 5sin(2t)→ f = 1 Hzfs = 1.33 Hz
Figure 12.13
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Sampling with Amplitude Ambiguity
y(t) = 5sin(2t)→ f = 1 Hzfs = 3.33 Hz
Figure 12.12
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y(t) = 6 + 2sin(t/2) + 3cos(t/5) +4sin(t/5 + ) – 7sin(t/12)
Smallest sample period that contains all integer multiples of the Ti’s:
fi (Hz):
Ti (s):
Smallest sampling to avoid aliasing (Hz):
In-Class Example