Discrete-time signals increase in frequency and period

Question

I am studying Discrete-Time Signal Processing by Alan V. Oppenheim and Ronald W. Schafer. I am confused when it says

periodic discrete signal oscillates more rapidly when its frequency is increased from $0$ to $\pi$

and

frequencies in the vicinity of $\pi$ are referred to as high frequencies

But I don't think it's monotonically increasing, for example, $\omega_0 = 999\pi/1000$ has a way longer period than $\omega_0 = \pi/2$.

Answer 1

You're right that the periods of the two signals are very different. The period of a sinusoidal sequence

$$x[n]=\cos(n\omega_0+\phi)\tag{1}$$

with $\omega_0=\pi/2$ equals $N=4$, whereas for $\omega_0=999\pi/1000$ it equals $N=2000$.

But this is not what "frequency" refers to here. Note that in general the sequence $x[n]$ as defined in $(1)$ isn't even periodic. It is only periodic if $\omega_0$ satisfies

$$\omega_0=\frac{2\pi p}{q},\qquad p,q\in\mathbb{Z}\tag{2}$$

Maximum frequency for a discrete-time signal is achieved if the samples alternate in sign. This is the case for $\omega_0=\pi$, and it is achieved for almost all samples if $\omega_0$ is close to $\pi$. For $\omega_0=\pi/2$ there are always two samples with the same sign before the sign changes, so the frequency is half of the frequency of a sequence for which consecutive samples have alternating signs.

Answer 2

On when does the period of the discrete-time sinusoid decrease as the frequency increase.

For a discrete-time sinusoid $x[n]$ of frequency $f_0$ $$ x[n] = x[n + N] \iff f_0 = \frac kN\tag{1} $$ where $k$ and $N$ are relatively prime integers. The condition in Equation $(1)$ is tantamount to $$ \omega_0 N = 2\pi k \tag{2} $$ What to note down is that discrete-time signals are defined only for integer indices $n$. And that $k$ in Equation $(1)$ and $(2)$ correspond to an integer number of periods (i.e. $kT_s$) of the corresponding continuous-time sinusoid.

With that in mind; if the frequency of a discrete-time sinusoid is varied from $\omega_1$ with period $N_1$ to $\omega_2$ with period $N_2$, and they both fit in exactly one period of their corresponding continuous-time equivalents, meaning $$ \frac{2\pi}{\omega_1}\in \mathbb N\quad\text{and}\quad\frac{2\pi}{\omega_2}\in \mathbb N\tag{3} $$ then it can be seen that $$ \omega_2 > \omega_1 \implies N_2 < N_1\tag{4} $$

A quick check can be made using the values in the table below. The values of $2\pi/\omega_0$ for which the discrete-time sinusoid fit in exactly one period of their corresponding continuous-time equivalent are highlighted in blue.

$$ \begin{array}{cccc} \hline \omega_0 & f_0 & N & \frac{2\pi}{\omega_0}\\ \hline 0 & \infty & 0 & \infty\\ \frac{3\pi}{32} & \frac{3}{64} & 64 & \frac{64}{3}\\ \frac{\pi}{8} & \frac{1}{16} & 16 & \color{blue}{16}\\ \frac{\pi}{4} & \frac{1}{8} & 8 & \color{blue}{8}\\ \frac{3\pi}{8} & \frac{3}{16} & 16 & \frac{16}{3}\\ \frac{\pi}{2} & \frac{1}{4} & 4 & \color{blue}{4}\\ \frac{29\pi}{30} & \frac{29}{60} & 60 & \frac{60}{29}\\ \frac{999\pi}{1000} & \frac{999}{2000} & 2000 & \frac{2000}{999}\\ \pi & \frac{1}{2} & 2 & \color{blue}{2}\\\hline \end{array} $$

In all other cases, increasing the value of $\omega_0$ does not necessarily decrease the period of the signal.

Answer 3

The period of $\omega_0 = 999\pi/1000$ is 2.0202. The period $\omega_0 = \pi/2$ is 4. So the period of $\omega_0 = 999\pi/1000$ is roughly half the period $\omega_0 = \pi/2$.