# Very basic question about how we define frequency in signal processing

When talking about general periodic continuous-time signals for which $$x(t + T_0) = x(t)$$ where $T_0$ is the fundamental period we define the fundamental frequency $\omega_0$ as $\omega_0 = 2\pi/T_0$.

The way I interpret this is that $1/T_0$ is the frequency of the signal in cycles per second and there are $2\pi$ radians in one cycle, therefore the angular frequency is $2\pi/T_0$ radians per second. But are there really $2\pi$ radians in one cycle in general?

Take $x(t) = \tan(t)$ for example. Its fundamental period is $\pi$ and using the definition of fundamental frequency above, its fundamental frequency is 2 radians per second. But in the case of the tangent function, aren't there $\pi$ radians in one cycle, as opposed to $2\pi$ radians? This would give us a definition of frequency as $\omega_0 = \pi / T_0$. Or do we refer to the unit circle by convention when talking about "cycles"?

Apologies if this question sounds very basic.

For the tangent function a cycle is indeed done in $\pi$ radians, because you want

\begin{align}\tan(x) &= \tan(x+T)\\ \implies \frac{\sin(x)}{\cos(x)} &= \frac{\sin(x+T)}{\cos(x+T)} \end{align} $$\implies \sin(x+T)\cos(x) - \sin(x)\cos(x+T) =0$$ $$\implies \sin\big[\left(x + T\right)-x\big] =0\equiv \sin\left(T\right)=0$$ $$\implies T = \pi k, \quad \text{with}\quad k\in \mathbb Z$$

Smallest strictly positive $k$ (i.e. $k= 1$) gives us $T_0 = \pi$. Anyway, we know that.

Regarding the interpretations, let's use $\Omega \ [\text{rad/sec}]$ to avoid ambiguity with $\omega \ [\text{rad/sample}]$ mostly used in the discrete case. From the definition of the angular frequency $\Omega = 2\pi F$ with the fundamental period $T_0 = 1/F$, as you have already described, we have $$\Omega_0 = \frac{2\pi}{T_0}\tag{1}$$ The interpretation should go from $(1)$ and read as the number of cycles of the periodic signal in $\mathbf{2\pi}$ radians. As examples, look at the plot below, As seen above,

• $\tan(x)$ has $T_0 = \pi$ implying $\Omega_0 = 2\pi/\pi=2 \ \text{rad/sec}$ or 2 cycles in $\mathbf{2\pi}$ radians (or equivalently one cycle in $\pi$ radians)
• $\sin(x)$ has $T_0 = 2\pi$ implying $\Omega_0 = 2\pi/2\pi=1 \ \text{rad/sec}$ or 1 cycle in $\mathbf{2\pi}$ radians
• $\sin(2\pi x)$ has $T_0 = 1$ (see $x$-axis) implying $\Omega_0 = 2\pi/1=2\pi \ \text{rad/sec}$ or $\mathbf{2\pi\approx 6.2832}$ (see $x$-axis) cycles in $\mathbf{2\pi}$ radians.

See it this way, the ratio in equation $(1)$ is simply stating how many $T_0$ (i.e. fundamental periods or full cycles) you have in $2\pi$. You have $T_0 = 2\pi \implies \Omega_0 = 1 \ [\text{rad/sec}]$ meaning you have $1$ fundamental period (full cycle) in $2\pi$. Similarly, when $T_0 = 1 \implies \Omega_0 = 2\pi \ [\text{rad/sec}]$ meaning you have $2\pi$ fundamental periods (full cycles) in $2\pi$. Etc. for other cases.

• how is $\Omega_0$ number of cycles in $2\pi$ radians? Isn't it $2\pi$ radians per "time of one cycle" (the period)? – user33568 Feb 11 '18 at 0:46
• I'm also not very convinced since you have used "2 rad/sec" to refer to two cycles in $2\pi$ radians, which somehow implies that a radian is equivalent to a cycle? I'm a bit confused (especially from a dimensional analysis point of view). – user33568 Feb 11 '18 at 0:50
• @0MW, please see my edit in the last paragraph. – Gilles Feb 11 '18 at 9:10
• @0MW any tick on this? – Gilles Mar 15 '18 at 6:25

My guess is that you're thinking about it in a way that makes things confusing.( as I do often enough also so I can relate. ). The periodicity of the sine function is $2 \pi$ because it takes $2 \pi$ radians to get back to where it began. So the function $\sin(2 \pi t)$ is zero at $t = 0$ and zero at $t = 1$ so the fundamental frequency in this case would be $\frac{2\pi}{1} = 2 \pi$. The function is also zero at $t = \frac{1}{2}$ but that doesn't count as the end of the cycle because there are still more values for the function to "see" before it begins again. I hope that helps.

• The textbook that I am using defines the fundamental frequency for any periodic signal as $2\pi /T_0$ though, not just sinusoids. – user33568 Feb 7 '18 at 20:47
• Note that the funfamental frequency of a function can be changed just by multiplying the function. For example $\sin(4 \pi t)$ has a period of $\frac{1}{2}$. – mark leeds Feb 7 '18 at 20:57
• could you name your textbook – user28715 Feb 7 '18 at 22:30
• Signals and Systems by Alan Oppenheim. – user33568 Feb 7 '18 at 22:41
• you’re right, that’s how they define it there. In Fundamentals of Physics, by Haliday and Resnick, 1974 printing, on page 230, frequency is 1/T and 2pi/T is angular frequency. I don’t have Oppenheim and Schafer’s book with me to see if they made the same claim. – user28715 Feb 7 '18 at 23:10

a fundamental frequency is $$f_0=\frac{1}{T_0}$$ We consider phase in radians if we wish to express the periodicity in terms of a linear combination of sines and cosines.

Given this definition, the fundamental frequency is consistent .

• Is it wrong to say that $\tan (t)$ has a fundamental frequency of 2 radians per second then? The tangent function has no Fourier Series expansion. – user33568 Feb 7 '18 at 19:42
• I stay away from words used to characterize moral choices. A frequency is the inverse of a period. I say use radians when meaningful. (-1)^n is periodic. Are radians essential? In the frequency domain, yes, as characterizing a fundamental freq? No – user28715 Feb 7 '18 at 20:18
• Do you mean that the choice of using $2\pi$ radians to "represent" one cycle is conventional? – user33568 Feb 7 '18 at 20:33
• If 360 degrees = 2 pi radians, why wouldn’t 360/T be any less a “fundamental” frequency than what you posted? 1/T works regardless of basis function. Your original question is about frequency. – user28715 Feb 7 '18 at 20:59
• I came to the conclusion that it doesn't really matter how we relate the fundamental frequency to the period so long as it is inversely proportional to the period (the periodicity of the signal is uniquely characterized by $T_0$, and even $\omega$ for that matter as long as we are consistent when comparing signals). So we might as well have defined $\omega = 5\pi / T_0$: it doesn't really make a difference in the math since the period can still be recovered from that definition. It's just that for a sinusoid $\cos \omega t$ using the definition with $2\pi$ allows us to say that $\omega$... – user33568 Feb 7 '18 at 21:17

the $\tan(x)$ function is not only periodic with period $\pi$, but is also periodic with period $2\pi$. (or any other integer times $\pi$.)