# Discrete-time signals increase in frequency and period

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 $$\omega_0 = \pi/2$$.

• – leonbloy Jan 3 at 18:10
• The period of the discrete-time sinusoids always decrease as the frequency increases if certain conditions are met. Please see my answer below. – Gilles Jan 6 at 8:25

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.

• I encountered the same problem in the past with the meaning of "frequency" here. What's the mathematical definition for the frequency in this case? – S.H.W Jan 3 at 12:34
• Thanks for the answer. I don’t know why 999pi/1000 cannot be the “frequency “ here? I agree pi is the fastest frequency but I don’t understand why the frequency close to pi is also fast. 999pi/1000 is very close to pi but it’s slow – Ziyuan Ning Jan 3 at 17:08
• @ZiyuanNing: It's not slow, the samples are alternating, aren't they? I tried to explain in my answer that it's not about the period of the discrete-time signal. – Matt L. Jan 3 at 17:10
• I understand your point now. Thank you, that makes sense. – Ziyuan Ning Jan 3 at 18:19
• @S.H.W: One way to think about it is "what would be the frequency of the analog waveform that results at the output of an ideal DAC when fed with this input sequence?" – Matt L. Jan 3 at 18:22

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.

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 $$\omega_0 = \pi/2$$.

• What do you mean by a period of $2.0202$ for a discrete-time signal? A discrete-time sequence $x[n]$ is periodic if it satisfies $x[n]=x[n+N]$ for some integer $N$. – Matt L. Jan 3 at 12:06
• Indeed, a discrete-time sinusoid $x[n]$ of frequency $f_0$ is periodic with period $N$ $\iff f_0 = \frac kN$, where $k$ and $N$ are relatively prime integers. – Gilles Jan 3 at 12:35