# Why is the digital frequency highest at $\pi$ and what does it mean in terms of analog frequency? [duplicate]

I can't seem to comprehend why the digital frequency is highest at $$\pi$$. I know the relationship between the digital and analog frequency is given by: $$\omega = 2\pi\frac{F}{F_s}=2\pi f$$ Where $$F$$ is the analog frequency and $$F_s$$ is the sampling frequency.
Why does the digital frequency increase from $$0$$ to $$\pi$$ and then decrease from $$\pi$$ to $$2\pi$$?

Check this figure. Consider the signal $$x[n]=A$$ which is constant but can be written as $$x[n]=A\cos(0n)$$, so its frequency is zero. There is no oscillation.

Let's increase the frequency. For the signal $$x[n]=A\cos(\pi n/4)$$, it oscillates at frequency $$\omega_0=\pi/4$$. The oscillation frequency increases.

For the frequency $$\omega_0=\pi$$, the signal is $$x[n]=A\cos(\pi n)=A(-1)^n$$, thus the signal oscillates between $$A$$ and $$-A$$ for each pair of consequtive samples. You can see that the signal cannot oscillate at a higher frequency, there is no faster oscillation than this.

For frequency $$\omega_0=7\pi/4$$, the signal can be written as $$x[n]=A\cos\Big(\frac{7\pi n}{4}\Big) = A\cos\Big(2\pi n-\frac{\pi n}{4}\Big)=A\cos\Big(\frac{\pi n}{4}\Big)$$ which oscillates at a lower frequency.

Finally, for $$\omega_0=2\pi$$, you have a constant signal again, no oscillation.

It is the nature of discrete time that introduces this effect.

• It makes much more sense now. However, I still don't understand the relationship between the digital and analog frequency. If you increase the digital frequency from $0$ to $\pi$ for example, what happens to the analog frequency? Does it change or does the sampling frequency change?
– A6EE
Feb 9, 2020 at 13:22
• The analog frequency $f$ is related to the digital one by $$f=\omega F_s/2\pi$$ During sampling, you can uniquely represent analog frequencies up to $F_s/2$ in the digital domain.
– GKH
Feb 9, 2020 at 13:30
• Oh, I see, so, the change in the digital frequency happens due to change in the analog frequency. However, since the sampling frequency is $F_s$, if we increase the analog frequency to higher than $\frac{F_s}{2}$, we start to get aliases in the digital domain. Is that correct?
– A6EE
Feb 9, 2020 at 13:36
• Absolutely. I have also added a figure to support my answer.
– GKH
Feb 9, 2020 at 14:43

The digital frequency represented with $$\pi$$ is the Normalized Angular Frequency, and has units of frequency given as radians/sample. This signal can be represented as a rotating phasor on the complex IQ plane: $$Ae^{j\omega n}$$, where $$A$$ is the amplitude and $$\omega$$ is the normalized angular frequency. With this representation the normalized frequency is very clear in that the frequency represented is the number of radians that have rotated in one sample.

Consider the analog complex signal given in the plot below where the analog radian frequency $$\Omega_o$$ is given in units of radians/sec. Once sampled, if we divide the analog radian frequency $$\Omega_o$$ by the sampling rate to get the normalized radian frequency $$\omega_o$$, we can see that the signal will rotate $$\omega_o$$ radians for each sample. If the signal with frequency $$\omega_o$$ rotates past $$\pi$$ radians on a single sample, we cannot distinguish it from another signal that has rotated in the other direction $$2\pi -\omega_o$$.

• Someone downvoted this, please explain why so I can understand the error if there is something wrong. Thank you! Feb 9, 2020 at 16:30
• Hi. Because your digital frequency is $\omega_o$ measured in radians/sample, your above description seems correct to me. Feb 9, 2020 at 18:36
• Thanks for weighing in @RichardLyons. Someone sees an issue so perhaps it is hidden in there somewhere and we get a chance to clear it up. Feb 9, 2020 at 18:39