When do we divide by $2\pi$ while solving Nyquist problems?

Question

I slightly lost the tempo in the course for a couple of days, so I apologies for the noob question if seemed so.

Why do we divide by $2\pi$ in Nyquist problems sometimes and sometimes not? I saw there is a frequency and a radial frequency. But I don't understand when do we divide max frequency by $2\pi$ and when do we not?

[Edit] for instance:

We have $cos(t\pi/3)$ for what T can we sample it with a dirac train, such as that we can recover the original signal afterwards?

On my lecture the max T is 3. I see the 3 but how the rest of $2\pi/3$ is gone?

Answer 1

Well! all signals in this world are made up of sum of different rotations(sinusoidals) - different in three senses:

a. how big is the amplitude (A)

b. how fast is the rotation ($\omega$)

c. where is the starting point of the rotation (phase $\phi$)

Fourier made this very clear. How do we measure rapidness of the rotating signals(sinusoidals) : by their angular velocity $\omega$ which is given in radians/seconds. This is the correct and most appropriate unit for measuring rotations - how much of angle is being covered per unit time! If a constituent signal is rotating by $200\pi$ radians in 1 second, we say $\omega = 200\pi$ rad/sec.

But we can also measure the rapidness of these rotating signals as number of rotations per unit time. That is $f$ expressed in $Hertz$ : number of rotations per second.

How are $f$ and $\omega$ related? One complete rotation covers a full circle meaning an angle of $360^o = 2\pi \ radians$. That means if the signal is making $200$ rotations per second then it is covering an angle of $200*2\pi = 400\pi$ per second. So, the relation between $f$ and $\omega$ is basically: $$\omega = 2\pi f$$

That is why you divide $\omega$ by $2\pi$ when you want to express the frequency in $Hz$.

The example you have given is $\cos{\frac{\pi t}{3}} = \cos{\frac{2\pi t}{6}}$. The period of this sinusoidal is $T = 6$sec, therefore, the frequency in $Hz$ will be $f = \frac{1}{6}$ and in radians/sec will be $\omega = 2\pi f = \frac{2\pi}{6}$ rad/sec.

(Think about why the period of this sinusoidal is $6sec$. Figure out that the sinusoidal will repeat after $t = 6sec$. Figure out that the sinusoidal makes 1 full rotation of angle $2\pi$radians in $6sec$.)

And the sampling frequency is twice the frequency of the sinusoidal giving : $$f_{sampling} = \frac{2}{6} = \frac{1}{3}sec$$ Meaning Sampling period is $\frac{1}{f_{sampling}} = 3sec$.

Answer 2

Why do we divide by $2\pi$ in Nyquist problems sometimes and sometimes not? I saw there is a frequency and a radial frequency. But I don't understand when do we divide max frequency by $2\pi$ and when do we not?

The relation between frequency and radian frequency is $2\pi f=\omega$, where $f$ is the frequency in units of Hertz and $\omega$ is the radian frequency is units of radians per second. They are communicating the same information and the necessary units will be apparent from the question or application.

We have $\text{cos}(tπ/3)$ for what $T$ can we sample it with a dirac train, such as that we can recover the original signal afterwards?

From this you can say that the signal has frequency $f=\frac{1}{6}$ and so Nyquist says that you need to sample the signal at $f_s > \frac{1}{3}$ (twice the frequency). This is where the $T=3$ comes from because the sample period is $T=\frac{1}{f_s}$.