I need to create a vector of sine signal.

So I'm trying to figure out what the structure of the function that describe sine signal ?

For example the function $\sin 2x$ meets the requirements?

If the answer is no, then what is the reason?


1 Answer 1


I think you posted similar question 3 days ago, regarding your teacher claiming that $\sin 2x$ is not a sinusoidal function. Nevertheless this function is definitely sinusoidal. Otherwise how come we can say that signals can be decomposed into sums of sinusoids (and cosinusoids)? You have plenty of orthogonal waves: $\sin x, \ \sin 2x, \ \sin 3x, \ldots$, and all of them are sinusoids.

The only thing that is coming into my mind, is that:

Discrete-time sinusoid is periodic only if its fundamental frequency $f_0$ is a rational number

For a sinusoid with frequency $f_0$ to be periodic, we should have:

$$ \sin[2\pi f_0 (N+n)+\theta] = \sin[2\pi f_0 n + \theta]$$

This relations is true if and only if there exists an integer $k$ such that:

$$2\pi f_0N=2\pi k $$ or, equivalently:

$$f_0 = \dfrac{k}{N} $$

To determine the fundamental frequency of a periodic sinusoid, we express its frequency as above and cancel common factors so that $k$ and $N$ have no common divisors. Then the fundamental period of the sinusoid is equal to N.

So for example:

  • $f_0 = \dfrac{31}{60}$ implies that $N=60$
  • $f_0 = \dfrac{30}{60}$ implies that $N=2$

Thus if you define your discrete sinusoid to be:

  • $\sin[2 f_0 t]$, or

  • $\sin[\sqrt{2} f_0 t]$

then these are no longer periodic in digital domain. Why? Think of that in following way: each sample at start of the new period will be shifted slightly - it's not a rational multiple of $\pi$. Below is some plot for two signals:

  • $\sin [2 \pi t]$
  • $\sin [2 \pi \sqrt{2} t]$

Sampling frequency is $10 \ \mathtt{Hz}$ and upper time limit is $8 \ \mathtt{s} $

enter image description here

You can see a first sinusoid is periodic - every 10 samples (1 second) you get repeating pattern. On the contrary second one, of which fundamental frequency cannot be represented as a rational number (no way to decompose $\sqrt{2}$) periods are not the same - thus your signal has no period. Check the figure below for overlay of 11 periods:

enter image description here

  • $\begingroup$ Answer updated with some plots as promised. $\endgroup$
    – jojeck
    Jun 22, 2014 at 15:54

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