Question

Generate plots for the function in Example 8.1 with sample intervals of 0.5 sec, 1 sec, and 10 sec. Explain the results in light of the sampling theorem. Example 8.1 Sampling Theorem and Aliasing Consider the function FO sin( at)+sin br) Using a trigonometric identity for the sum of two sinusoidal functions, we can rewrite FO as the following product: a+ b If frequencies a and b are close in value, the bracketed term has a very low frequency in comparison to the sinusoidaal term on the right. Therefore, the bracketed term modulates the amplitude of the higher frequency sinusoidal term. The resulting waveform exhibits what is called a beat frequency that is common in optics, mechanics, and acoustics when twoInternet Link waves close in frequency add. For more information and an audio example of beat frequency see Intemet Link 8.1 and Video Demo 8.3 8.1 Beat frequency To ilustrate aliasing associated with improper sampling, the waveform is plotted in the following figures using two differeat sampling frequencies. If a and b are chosen as rad b0.9a sec then to sample the signal F) property, the sampling rate must be more than twice the high- est frequency in the signal: Video Demo 8.3 Beat fraquency from Therefore, the time interval between samples (plotted points) must be mbding signals of similar trequency The first data set is plotted with a time interval of 0.01 sec (100 Hz sampling rate), providing an adequate representation of the wavefom. The second data set is plotted with a time interval of 0.75 sec (1.33 Hz sampling rate), which is less than twice the maximum fre- quency of the waveform (2 Hz). Therefore, the signal is undersampled, and allasing results The sampled waveform is an incorrect representation, and its observed maximum frequency appears to be appeoximately 0.4 Hz because there are approximately four cycles over 10 sec.

Answer 1

As shown in example 8.1, let's first consider sampling frequencies > 2Hz Fs 100 Hz Fs 10 Hz Fs 5 Hz 1.5 1.5 0.5 0.5 0.5 0.5 0.5 0.5 -1 1.5 0 12345 67810 Time (s) 0 123 5 67810 0 1234 5 67810 Time (s) Time (s) As it can be seen from the above graphs, the shape of the signal is largely preserved when using these frequencies which means that sampling frequencies are sufficiently high to capture all the information in these signals to successfully reproduce them. Going forward let's consider the 100 Hz sampling rate as our reference since it does the best job of capturing all the information correctly

fs 100Hz 1.5 0.5 0.5 -1 -2 0 1 23 45 6 789 Time (s) The above graph shows the signal sampled at 100 Hz and 2 Hz. It can be seen that at 2 Hz, while the amplitude information is largely incorrect, the frequency of the signal is still captured somewhat accurately (i.e., both signals have about 10 cycles in 10s). For the given signal, 2Hz is the Nyquist frequency, which means that any lower a sampling rate and the information captured from the signal will be inaccurate and thus insufficient to successfully reproduce the signal which will be shown by the graphs below.

1.5 0.5 0.5 -1 -2 1 2 3 45678 Time (s) The above graph shows the signal sampled at 100 Hz, 2 Hz and 1 Hz. It can clearly be observed that when the same signal is sampled at 1 Hz, all information regarding the signal is lost and it appears as though the frequency of the original signal is 0.1 Hz (1 cycle in 10 seconds)

10 Sample Interval 10s fs 100Hz 1.5 s 0.1 Hz 0.2 -0.4 0.5 -0.5 0.8 -1 1.5 1.2 0 02 3 456789 0 Time (s) 23 4 5 6789 10 Time (s) The above graph shows the signal sampled at 100 Hz, 2 Hz and 0.1 Hz. When the signal is sampled at 0.1 Hz, it appears as though there is no information to be captured in 10s of the signal. However, there is a change in the signal but the change in magnitude is very low as shown in the second image (image on the right)

Graphs of signal sampled at 2Hz and 1Hz (for reference) Sample Interval = 0.5s Sample interval = 1s 1.5 0.8 0.6 0.4 0.2 0.5 0 0.2 0.4 0.6 0.8 0.5 0 2 4 10 10 5 Time (s) 0 2 4 5 Time (s)