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# Exercises

Module by: Anders Gjendemsjø. E-mail the author

Summary: Exercises to TTT4110: Information and Signal Theory, Sampling theorem.

Problems related to the Sampling Theorem module.

## Exercise 1

Express the sampling theorem in words.

### Solution

Fill in the solution here...

## Exercise 2

Theoretically, why is the sinc-function so important for reconstruction? Sketch a sinc(t). What are the values for integer values of t?

### Solution

Fill in the solution here...

## Exercise 3

Argue that the sampling rate for CD should be over 40KHz.

### Solution

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.

## Exercise 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.

### Solution

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.

## Exercise 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?

### Solution

Fill in a solution here

## Exercise 6

If you found that it is possible to use another filter in Exercise 5 specify such a filter. Hint: Try using the domain which usually simplifies things...

### Solution

Fill in a solution here

## Exercise 7

What are the difficulties introduced when we want to apply the results of this chapter in practice?

### Solution

Fill in a solution here

## Exercise 8

If a real signal has frequency content up to F1F1. What is then the bandwith of the signal?

### Solution

Fill in a solution here

## Exercise 9

If a real signal has frequency content confined in the interval F1 F1 F1 F1 . What is then the bandwith of the signal?

### Solution

Fill in a solution here

## Exercise 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?

### Solution

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

## 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 .

### Exercise 11

What is the frequency limitation of an analog sinusoidal signal if we want to avoid aliasing, given Fs=8000 Fs 8000 ?

#### Solution

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

### Exercise 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?

#### Solution

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

### Exercise 13

Find an expression the signal we "reconstruct" from the samples when the input frequency is 6000 Hz.

#### Solution

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 .

### Exercise 14

Explain the "strange" sample points when the input input frequency is 4000 Hz.

#### Solution

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.

### Exercise 15

Explain the "strange" sample points when the input input frequency is 8000 Hz.

#### Solution

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.

### Exercise 16

Find an expression for the signal we can reconstruct from the samples when the input frequency is 4000 Hz.

#### Solution

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.

### Exercise 17

Find an expression for the "reconstructed" signal from the samples when the input frequency is 8000 Hz.

#### Solution

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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