Question 9.14: Consider the following four sinusoids: x1[n] = sin(0.1πn), x......
Consider the following four sinusoids:
x_1[n] = sin(0.1πn), x_2[n] = sin(0.2πn), x_3[n] = sin(0.6πn), x_4[n] = sin(0.7πn) −∞ < n < ∞.
Find if they are periodic and if so find their fundamental periods. Are these signals harmonically related? Use MATLAB to plot these signals from n = 0, ··· , 40. Comment on which of these signals resemble sampled analog sinusoids and indicate why some do not.
To find if they are periodic, we rewrite the given signals as
x_{1}[n]=\sin(0.1\pi n)=\sin\left({\frac{2\pi}{20}}n\right),\;\;x_{2}[n]=\sin(0.2\pi n)=\sin\left({\frac{2\pi}{20}}2n\right),
x_{3}[n]=\sin(0.6\pi n)=\sin\left({\frac{2\pi}{20}}6n\right),\;\;x_{4}[n]=\sin(0.7\pi n)=\sin\left({\frac{2\pi}{20}}7n\right)
indicating the signals are periodic of fundamental periods 20, with frequencies harmonically related. When plotting these signals using MATLAB the first two resemble analog sinusoids but not the other two. See Fig. 9.6.
One might think that x_3[n] and x_4[n] look like that because of aliasing, but that is not the case. To obtain cos(ω_0n) we could sample an analog sinusoid cos(Ω_0t) using a sampling period T_s = 1 so that according to the Nyquist condition
T_{s}=1\leq{\frac{\pi}{\Omega_{0}}}
where π/Ω_0 is the maximum value permitted for the sampling period for no aliasing. Thus x_3[n] = sin(0.6πn) = sin(0.6πt)|_{t=n T_s=n} when T_s = 1 and in this case
T_{s}=1\leq{\frac{\pi}{0.6\pi}}\approx1.66.
Comparing this with x_2[n] = \sin(0.2πn) = \sin(0.2πt)|_{t=nT_s=n} we get
T_{s}=1\leq{\frac{\pi}{0.2\pi}}=5.
Thus when obtaining x_2[n] from sin(0.2πt) we are oversampling more than when we obtain x_3[n] from sin(0.6πt) using the same sampling period, as such x_2[n] resembles more an analog sinusoid than x_3[n], but no aliasing is present. Similarly for x_4[n].
