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

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?

 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?

• sorry, no idea what you're talking about. You'll have to give us a concrete formula / problem. $2\pi$ is just so ubiquitous in signal processing. Please Edit your question to include a clear case where you're confused about its usage. – Marcus Müller Jun 19 at 21:43
• Hi @Marcus Müller, sry for long delay. I added an example. – Vitali Pom Jun 19 at 22:09
• It's useful to consider that the units of $2\pi$ are radians per cycle. – Cedron Dawg Jun 20 at 1:05

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 $$\frac{1}{f_{sampling}} = 3sec$$.

• So when I have pi in function declaration I move to liniar radians and when there is no pi it is already liniar? – Vitali Pom Jun 19 at 23:09
• No, cos(t) is $2\pi$ periodic always, because cos(t + $2\pi$) is cos(t). So, after every $2\pi$sec the signal repeats. If cosine' argument changes to $\frac{\pi t}{3}$, then the period changes to $\frac{2\pi}{\pi/3}$. I donot recommend using the word linear or non linear here. It's just that cosines repeat after the value inside ( ) of cos() becomes $2\pi$ – DSP Rookie Jun 20 at 0:15
• Note: $$\omega = 2\pi f$$ In units: $$\frac{Radians}{Second} = \frac{Radians}{Cycle} \cdot \frac{Cycles}{Second}$$ I would argue though that samples are much more important than seconds when measuring rotations in discrete (which most our computation is) cases. $$\frac{Radians}{Sample} = \frac{Radians}{Cycle} \cdot \frac{Cycles}{Sample} = \frac{\frac{Radians}{Cycle}}{\frac{Samples}{Frame}} \cdot \frac{Cycles}{Frame}$$ Which is conventionally represented in variables as $$\omega = \frac{2\pi}{N} \cdot k$$ – Cedron Dawg Jun 20 at 3:54
• So every Fourier Transform represents in radians form? – Vitali Pom Jun 20 at 10:00
• @VitaliPom Unless stated otherwise, you should always assume that radians are the units for angles in higher mathematics and engineering. It is the "natural" measure, and the one that works correctly for Calculus. – Cedron Dawg Jun 20 at 11:37

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 $$T=\frac{1}{f_s}$$.

• Hi, sry I apollgies, but don't we have a delta on $\pi/3$ in frequency domain? How does that turn into 1/6? – Vitali Pom Jun 19 at 22:34
• Hi: Because, since $\omega = 2 \pi f$ so $f = \frac{\omega }{2 \pi}$. So, you divide $\frac{\pi}{3}$ by $2 \pi$. – mark leeds Jun 20 at 3:14