Property:
The frequency of oscillation of discrete time sinusoids sequence increases as $\omega$ increases from $0$ to $\pi$. If $\omega$ is increased from $\pi$ to $2\pi$ then frequency of oscillation decreases.
My question:
What is meant by "frequency of oscillation" ?
How can a frequency of a signal vary (increase) from $0$ to $\pi$ and (decrease) from $\pi$ to $2\pi$. Isn't it should be same from $0$ to $2\pi$ ?
Take a complex exponential
$$x[n]=e^{jn\omega}\tag{1}$$
Let's assume that $\omega=\pi$. This gives
$$x[n]=e^{jn\pi}=(-1)^n\tag{2}$$
(because $e^{j\pi}=-1$). Eq. $(2)$ shows that a signal with frequency $\pi$ is an alternating signal, so $\omega=\pi$ clearly is the maximum possible frequency of a discrete-time signal.
Now assume that $\omega>\pi$. Let's write $\omega=\pi+\Delta\omega$ with $0<\Delta\omega<\pi$. The signal $x[n]$ from $(1)$ can be written as
$$x[n]=e^{jn\omega}=e^{jn(\pi+\Delta\omega)}\tag{3}$$
Since the complex exponential function is $2\pi$-periodic, we can subtract a multiple of $2\pi$ ($2\pi n$, $n\in\mathbb Z$) from its argument without changing anything:
$$x[n]=e^{jn(\Delta\omega-\pi)}=e^{-jn(\pi-\Delta\omega)}\tag{4}$$
Eq. $(4)$ shows that increasing $\Delta\omega$ from $0$ to $\pi$ corresponds to decreasing the frequency from its maximum value $\pi$ down to zero.
I think it means the "apparent" frequency of oscillation. It's poorly worded.
What's happening is that the frequency is increased to the Nyquist rate in the first half and then above it, causing aliasing, in the second.
What looks like the frequency increases in the first half and decreases in the second because of aliasing.
I prefer the following approach to understand why the frequency of the discrete-time sinusoid behaves so. As the confusion arises because of the continuous-time analogy lets repeat it here:
A continuous-time sinusoid $x(t) = \sin(2 \pi f_0 t) = \sin(\omega_0 t) $ is said to be periodic in $T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0}$ where $f_0$ is frequency in Hz and $\omega_0$ is frequency in radians per second. It can be easily seen that since $f_0$, $\omega_0$ and $T_0$ are all continuous variables, there is no constraint on their interperetation: larger the $f_0$ (higher frequency) shorter will be $T_0$ (shorter period) for any $f_0$
But coming to the discrete-time case, a fundamental difference happens; the period $N$ of the discrete time sequences must be an integer and the smallest such integer period is $N_{min} = 1$ samples (in fact a period of $N=1$ samples corresponds to a DC signal and hence the minimim integer period for a sinusoid is effectively $N_{min} = 2$ samples.
Now what's the relation between the integer period and corresponding continuous frequency of the discrete-time sinuodid $x[n] = \sin(\omega_0 n)$ ? The answer comes from the relation $$x[n] = x[n+N]$$ which reveals $$\sin(\omega_0 n) = \sin(\omega_0 (n+N)) = \sin(\omega_0 n + \omega_0 N) = \sin(\omega_0 n + 2\pi m) $$ resulting in:
$$ \omega_0 N = 2\pi m \longrightarrow \omega_0 = \frac{ 2 \pi m}{ N} $$
Putting $m=1$ for $N=2$ gives the highest frequency as $$\omega_{max} = \frac{ 2 \pi m}{N_{min}} = \frac{2 \pi}{2} = \pi $$
On the other hand the minimum frequency goes to zero as the maximum (positive) integer period goes to infinity $N_{max} \rightarrow \infty$ hence $$ \omega_{min} = \frac{ 2\pi m}{N_{max}} \rightarrow 0 $$ for any finite $m$.
This way it can bee seen that discrete-time frequencies (which are fundamentally related to allowed range of discrete-time periods $N$) begin from $\omega=0$ as the minimum frequency and reach up to maximum frequency of $\omega = \pi$ for the minimum allowed integer period of $N_{min}=2$ samples.
The highest possible frequency that can be represented by $T_s$ sampling is given by time period $T = 2T_s$. Why? because oscillation frequency cannot go higher than sequence ${... 1, -1, 1, -1, ...}$. This corresponds to $\omega = 2\pi /2T_s = \pi/T_s$. Since in discrete index $T_s = 1$, therefore the frequency corresponds to $\pi$. Now let us say we want to sample a higher frequency, i.e., $f = 3F_s/4$. This means time period of our frequency of interest is $T = 4T_s/3$. Note that we can only capture multiples of $T_s$, which means we will be only sampling every $3$rd sample, which is same as if $f$ was $4T_s$, which corresponds to $\omega = \pi/4T_s$ a frequency lower than $\pi/T_s$.
