4.7 Exercises

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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.

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.

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The spectrum will ALWAYS overlap,there will always be aliasing.

With a sampling frequency of 8000 Hz, the maximum frequency of the analog signal is 4000 Hz, as given by the sampling theorem .

The signal we "reconstruct" is a sinusoidal signal with a frequency that is lower than the original because of aliasing.

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: $\sin (2\pi \times 2000t)$ .

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.

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.

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.

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.