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Problems related to the Sampling Theorem module.
Express the sampling theorem in words.
Fill in the solution here...
Theoretically, why is the sinc-function so important for reconstruction? Sketch a sinc(t). What are the values for integer values of t?
Fill in the solution here...
Argue that the sampling rate for CD should be over 40KHz.
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 sampleat more than the double rate, partly due to finite precision.
What is the simplest bandlimited signal? Using this signal, convince yourself that less than twosamples/period will not suffice to specify it. If the sampling rate $\frac{1}{{T}_{s}}$ is not high enough, what signal would your resulting undersampled signal become? Hint: Try the aliasing applet .
The simplest bandlimited signal is the sine wave. At the Nyquist frequency, exactly two samples/period wouldoccur. Reducing the sampling rate would result in fewer samples/period, and these samples would appear to havearisen from a lower frequency sinusoid.
Are the filter h(t) described by the sinc function the only filterwe can use as a perfect reconstruction filter? If not what are the condition that would allow us to use another filter?
Fill in a solution here
If you found that it is possible to use another filter in specify such a filter. Hint: Try using the domain which usually simplifies things...
Fill in a solution here
What are the difficulties introduced when we want to apply the results of thischapter in practice?
Fill in a solution here
If a real signal has frequency content up to ${F}_{1}$ . What is then the bandwith of the signal?
Fill in a solution here
If a real signal has frequency content confined in the interval $\left[-{F}_{1} , {F}_{1}\right]$ . What is then the bandwith of the signal?
Fill in a solution here
What can be said in general for the spectrum of a discrete signal whichis the result of sampling an analog signal that is NOT bandlimited?
The spectrum will ALWAYS overlap,there will always be aliasing.
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, $x(t)=\sin (2\pi ft)$ , which is sampled at ${F}_{s}=8000$ .
What is the frequency limitation of an analog sinusoidalsignal if we want to avoid aliasing, given ${F}_{s}=8000$ ?
With a sampling frequency of 8000 Hz, the maximum frequency of the analog signal is 4000 Hz, as given by the sampling theorem .
Describe with words the type of signal we "reconstruct" from the sampleswhen the input frequency (of the sinusoidal signal) is higher than the sample rate can deal with?
The signal we "reconstruct" is a sinusoidal signal with a frequency that is lower than the original because of aliasing.
Find an expression the signal we "reconstruct" from the sampleswhen the input frequency is 6000 Hz.
When the input frequency is 6000 Hz, a sampling frequency of 8000 Hz is to low, i.e aliasing will occur. The sampled signal willhave 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 ${F}_{s}=8000$ . 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 2000Hz 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: $\sin (2\pi \times 2000t)$ .
Explain the "strange" sample points when the input input frequency is 4000 Hz.
The sampled signal can be written as ${x}_{s}(n)=\sin (2\pi \times 4000\frac{n}{8000})=\sin (\pi n)=0$ . Thus all the samples are zero-valued.
Explain the "strange" sample points when the input input frequency is 8000 Hz.
The sampled signal can be written as ${x}_{s}(n)=\sin (2\pi \times 8000\frac{n}{8000})=\sin (2\pi n)=0$ . Thus all the samples are zero-valued.
Find an expression for the signal we can reconstruct from the sampleswhen the input frequency is 4000 Hz.
As shown in problem 14, the samples are zero valued. A reconstructing filter cannot distinguish this from the all zerosignal 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 willthen be a equal to the original signal. This problem illustrates that sampling twice the signals highest frequency component doesnot always guarantee perfect recontstruction. If we could increase the sampling frequency to, say, ${F}_{s}=8000.00001$ , we could reconstruct the original signal. I.e sampling at a rate greater than twice the highest frequency component yields the desiredreconstruction.
Find an expression for the "reconstructed" signal from the sampleswhen the input frequency is 8000 Hz.
As shown in problem 15, the samples are zero valued. A reconstructing filter cannot distinguish this from the all zerosignal 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 willthen be a signal with aliased components.
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