Transcript Document 7810540

Examining the Signal
• Examine the signal using a very high-speed system, for example, a 50 MHz
digital oscilloscope.
Setting the Sampling Conditions
• In most circumstances, as when using computers, sampling is DIGITAL.
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
Digital Sampling
• The analog signal, y(t), is sampled every dt seconds, N times for a period
of T seconds, yielding the digital signal y(rdt), where r = 1, 2, …, N.
• For this situation:
Figure 12.1
Digital Sampling Errors
• When is signal is digitally sampled, erroneous results occur if either one
of the following occur:
Digital Sampling Errors
• To avoid amplitude ambiguity, set the sample period equal to the least
common (integer) multiple of all of the signal’s contributory periods.
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)
Illustration of Correct Sampling
y(t) = 5sin(2pt)
→ f = 1 Hz
with fs = 8 Hz
Figure 12.7
Illustration of Aliasing
y(t) = sin(20pt)
>> f = 10 Hz
with
fs = 12 Hz
The Folding Diagram
To determine the
aliased frequency, fa:
Example: f = 10 Hz; fs = 12 Hz
Figures 12.8 and 12.9
Aliasing of sin(20pt)
y(t) = sin(20pt)
→ f = 10 Hz
with
fs = 12 Hz
Aliasing of sin(20pt)
y(t) = 5sin(2pt)
→ f = 1 Hz
fs = 1.33 Hz
Figure 12.13
In-Class Example
• At what cyclic frequency will y(t) = 3sin(4pt) appear if
fs = 6 Hz?
fs = 4 Hz ?
fs = 2 Hz ?
fs = 1.5 Hz ?
Correct Sample Time Period
y(t) =
3.61sin(4pt+0.59)
+ 5sin(8pt)
Figure 12.16
Sampling with Aliasing
y(t) = 5sin(2pt)
→ f = 1 Hz
fs = 1.33 Hz
Figure 12.13
Sampling with Amplitude Ambiguity
y(t) = 5sin(2pt)
→ f = 1 Hz
fs = 3.33 Hz
Figure 12.12
In-Class Example
y(t) = 6 + 2sin(pt/2) + 3cos(pt/5) +4sin(pt/5 + p) – 7sin(pt/12)
fi (Hz):
Ti (s):
Smallest sample period that contains all integer multiples of the Ti’s:
Smallest sampling to avoid aliasing (Hz):