Question 9.4: What is true for analog sinusoids—that they are always perio......

The sampled signal x[n] = x(t)|_{t=nT_s} = cos(n + π/4) has a discrete frequency ω = 1 (rad) which cannot be expressed as 2πm/N for any integers m and N because π is an irrational number. So x[n] is not periodic.

Since the frequency of the continuous-time signal x(t) = cos(t + π/4) is Ω_0 = 1, the sampling period, according to the Nyquist sampling rate condition, should be

T_{s}\leq{\frac{\pi}{\Omega_{0}}}=\pi

and for the sampled signal x(t)|_{t=nT_s} = \cos(nT_s + π/4) to be periodic of fundamental period N or

cos((n + N )T_s + π/4) = cos(nT_s + π/4)     it is necessary that     NT_s = 2kπ

for an integer k, i.e., a multiple of 2π. Thus T_s = 2kπ/N ≤ π satisfies the Nyquist sampling condition at the same time that ensures the periodicity of the sampled signal. For instance, if we wish to have a discrete sinusoid with fundamental period N = 10, then T_s = kπ/5, for k chosen so that the Nyquist sampling rate condition is satisfied, i.e.,

0\lt T_{s}=k\pi/5\leq\pi\quad{\mathrm{so~that~}}\;0\lt k\leq5.

From these possible values for k we choose k = 1 and 3 so that N and k are not divisible by each other and we get the desired fundamental period N = 10 (the values k = 2 and 4 would give 5 as the period, and k = 5 would give a period of 2 instead of 10). Indeed, if we let k = 1 then T_s = 0.2π satisfies the Nyquist sampling rate condition, and we obtain the sampled signal

x[n]=x(t)|_{t=0.2\pi n}=\cos(0.2\pi n+\pi/4)=\cos\left({\frac{2\pi}{10}}n+{\frac{\pi}{4}}\right),

which according to its frequency is periodic of fundamental period 10. For k = 3, we have T_s = 0.6π < π and

x[n]=x(t)|_{t=0.6\pi n}=\cos(0.6\pi n+\pi/4)=\cos\left({\frac{2\pi\times3}{10}}n+{\frac{\pi}{4}}\right)

also of fundamental period N = 10.