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 $2\pi$ ?


4 Answers 4


Take a complex exponential


Let's assume that $\omega=\pi$. This gives


(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


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:


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.


A continuous-time sinusoid $x(t) = \sin(2 \pi f_0 t) = \sin(\omega_0 t) $ is periodic with period $T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0}$ where $f_0$ is frequency in Hz, and $\omega_0$ is frequency in radians per second. Since $f_0$, $\omega_0$, and $T_0$ are all continuous variables, there is no difficulty on their interpretation: larger the $f_0$ shorter will $T_0$ be.

For the discrete-time case, however, a fundamental difference exists: the period $N$ is an integer. And the smallest integer period is $N_{min} = 1$ which corresponds to a DC signal, and the minimim period for a sinusoid is effectively $N_{min} = 2$. Furthermore, discrete-time frequency $\omega$ is effectively periodic with $2\pi$, which limits the analysis to the range $\omega \in [0,2\pi)$.

The integer period $N$ (if it exists) of the sinuodid $x[n] = \sin(\omega_0 n)$ is found from the relation $$x[n] = x[n+N]$$ which yields: $$\sin(\omega_0 n) = \sin(\omega_0 (n+N)) = \sin(\omega_0 n + \omega_0 N) = \sin(\omega_0 n + 2\pi m) $$ 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 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 begin from $\omega=0$ at the minimum, and reach up to a maximum of $\omega = \pi$, and then decrease again to a minimum into $\omega = 2\pi$, which must be a minimum by the observation that $\omega$ is periodic with $2\pi$. Hence if $\omega = 0$ is a minimum, then $\omega = 0 + 2\pi = 2\pi$ is also an effective minimum, and so on.


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 $\pi/T_s$.


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