Problem 1
Express the sampling theorem in words.
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Solution 1
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Problem 2
Theoretically, why is the sinc-function so important for reconstruction?
Sketch a sinc(t). What are the values for integer values of t?
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Solution 2
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Problem 3
Argue that the sampling rate for CD should be over 40KHz.
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Solution 3
The human ear can hear frequencies up to 20 KHz, so according to the sampling theorem
we should sample at a rate equal to or exceeding 40KHz. In practice we always have to sample
at more than the double rate, partly due to finite precision.
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Problem 4: (By Don Johnson)
What is the simplest bandlimited signal? Using this
signal, convince yourself that less than two
samples/period will not suffice to specify it. If the
sampling rate
1Ts1Ts
is not high enough, what signal would your resulting undersampled signal become?
Hint: Try the
aliasing applet.
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Solution 4
The simplest bandlimited signal is the sine wave. At the
Nyquist frequency, exactly two samples/period would
occur. Reducing the sampling rate would result in fewer
samples/period, and these samples would appear to have
arisen from a lower frequency sinusoid.
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Problem 5
Are the filter h(t) described by the sinc function the only filter
we can use as a perfect reconstruction filter? If not what are the condition that
would allow us to use another filter?
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Solution 5
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Problem 6
If you found that it is possible to use another filter in
5
specify such a filter. Hint: Try using the domain which usually simplifies things...
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Solution 6
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Problem 7
What are the difficulties introduced when we want to apply the results of this
chapter in practice?
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Solution 7
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Problem 8
If a real signal has frequency content up to
F1F1.
What is then the bandwith of the signal?
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Solution 8
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Problem 9
If a real signal has frequency content confined in the interval
-F1F1
F1
F1
.
What is then the bandwith of the signal?
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Solution 9
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Problem 10
What can be said in general for the spectrum of a discrete signal which
is the result of sampling an analog signal that is NOT bandlimited?
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Solution 10
The spectrum will ALWAYS overlap,there will always be aliasing.
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Exercises related to the Aliasing applet
Link to the
aliasing applet
(Right click if you want to open it in a new window).
In the following problems, as in the aliasing applet, we are studying a
sinusoidal signal,
xt=sin2πft
x
t
2
f
t
, which is sampled at
Fs=8000
Fs
8000
.
Problem 11
What is the frequency limitation of an analog sinusoidal
signal if we want to avoid aliasing, given
Fs=8000
Fs
8000
?
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Solution 11
With a sampling frequency of 8000 Hz, the maximum frequency
of the analog signal is 4000 Hz, as given by
the sampling theorem.
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Problem 12
Describe with words the type of signal we "reconstruct" from the samples
when the input frequency (of the sinusoidal signal) is higher than the sample rate can deal with?
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Solution 12
The signal we "reconstruct" is a sinusoidal signal with a frequency
that is lower than the original because of aliasing.
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Problem 13
Find an expression the signal we "reconstruct" from the samples
when the input frequency is 6000 Hz.
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Solution 13
When the input frequency is 6000 Hz, a sampling frequency of
8000 Hz is to low, i.e aliasing will occur. The sampled signal will
have frequency components at +6000 Hz and -6000 Hz plus some new frequency components
as a result of aliasing.
We know from the
proof of the sampling theorem
that the sampled signal is periodic with
Fs=8000
Fs
8000
. Thus a frequency component at 6000 Hz
implies frequencies at -2000 Hz, -10000 Hz, 14000 Hz and so on.
Similarly a frequency component at -6000 Hz give rise to(among others) a 2000 Hz component.
Looking only at the positive frequencies the "reconstructed" signal will only have a 2000
Hz frequency component. The removal of the 6000 Hz and above frequencies are due to the reconstruction
filter. The filter is designed based on a maximum input signal frequency of 4000 Hz.
Thus the "reconstructed" signal can be written as:
sin2π2000t
2
2000
t
.
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Problem 14
Explain the "strange" sample points when the input input frequency is 4000 Hz.
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Solution 14
The sampled signal can be written as
xsn=sin2π4000n8000=sinπn=0
xs
n
2
4000
n
8000
n
0
.
Thus all the samples are zero-valued.
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Problem 15
Explain the "strange" sample points when the input input frequency is 8000 Hz.
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Solution 15
The sampled signal can be written as
xsn=sin2π8000n8000=sin2πn=0
xs
n
2
8000
n
8000
2
n
0
.
Thus all the samples are zero-valued.
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Problem 16
Find an expression for the signal we can reconstruct from the samples
when the input frequency is 4000 Hz.
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Solution 16
As shown in problem 14, the samples are zero valued.
A reconstructing filter cannot distinguish this from the all zero
signal so the reconstructed signal will be the all zero signal.
Note that a small change in the sinusoidal signals phase would produce
samples that are not only zero-valued. The "reconstructed" signal will
then be a equal to the original signal. This problem illustrates that
sampling twice the signals highest frequency component does
not always guarantee perfect recontstruction. If we could increase
the sampling frequency to, say,
Fs=8000.00001
Fs
8000.00001
, we could reconstruct the original signal.
I.e sampling at a rate greater than
twice the highest frequency component yields the desired
reconstruction.
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Problem 17
Find an expression for the "reconstructed" signal from the samples
when the input frequency is 8000 Hz.
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Solution 17
As shown in problem 15, the samples are zero valued.
A reconstructing filter cannot distinguish this from the all zero
signal so the reconstructed signal will be the all zero signal.
Note that a small change in the sinusoidal signals phase would produce
samples that are not only zero-valued. The "reconstructed" signal will
then be a signal with aliased components.
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