A continuous-time sinusoidal wave is given by
x(t) = Acos(2⇡Ft + “). (1)
For A = 5, F = 10, ” = 0, and 0 < t < 1,
(a) Define and plot x(t) and |X(j!)|. For 0  t < 1, use a step size of 1
256 , and for −128  ! < 127 a
step size of 1. To approximate |X(j!)|, use abs(fftshift(fft(x))).
(b) Determine the maximum frequency of x(t) from the graph of |X(j!)|. Since the plot of x(t) goes from
0 to 1, you can also count the number of peaks from the plot of x(t). Verify that you get a frequency of
10 Hz as indicated in the question. Use the maximum frequency to determine the Nyquist sampling
frequency and period. Matlab deals with vectors and matrices. Therefore, to sample, we need to
know how many elements to skip before taking a sample. We will call this the Nyquist step size. It
can be calculated from the Nyquist period by multiplying the period with the step size 1

Verify
that the Nyquist step size is 12.8. Meaning, we need to take a sample every 12 elements of the vector
x. Any more skipping and we violate the Nyquist condition.
(c) Oversample x(t) to obtain x[n] and plot it. To oversample, use a step size of 2 (much smaller than 12.8).
Reconstruct x(t) from x[n] by calling the provided sinc interpolator function (i.e., sincinterp()) to
obtain xr(t). Since you want the reconstructed signal to have the same size as x(t), pass N = 256
to sincinterp(). Find and plot |Xr(j!)| along with x[n] and xr(t). Find the maximum frequency
of the reconstructed signal from the graphs of x(t) and |X(j!)|. Is it still 10 Hz? If not, what is the
new frequency? Is this expected? Explain.
(d) Critically sample x(t) by repeating part (c) with a step size of 12. Note that it is very close to the
Nyquist step size (related to the Nyquist period).
(e) Undersample x(t) and repeat part (c) with a step size of 15. Note that it is larger than the Nyquist
step size. Verify that the frequency of the reconstructed signal is smaller than the original signal.
What is this phenomenon called?