# Reason of effect of continuity and discreteness on periodicity of signal?

We know that a sine signal in continuous time is a periodic signal with a period $$2\pi$$ whereas the same sine signal in discrete time is aperiodic. My question is how changing just a type of signal (by sampling) can affect the signal's periodicity and why this happens whereas we have just omitted some real values between samples and how can we make use of this fact practically?

• What makes you think the discrete signal is aperiodic in general? For example, if I sample a sine wave with frequency 1 Hz at 4 samples per second, the result is periodic.
– MBaz
Feb 26 '19 at 3:22
• @MBaz, it's a weird esoteric thing that doesn't really matter, but sampling a continuous-time periodic signal that has a period equal to an irrational number of sampling periods, that sampled discrete-time signal will not ever satistfy $$x[n+P] = x[n] \qquad \forall n \in \mathbb{Z}$$ for any integer $P$. Feb 26 '19 at 19:59
• @robertbristow-johnson That's my point; you have to really go out of your way to have an aperiodic discrete signal when sampling a periodic signal.
– MBaz
Feb 27 '19 at 0:00
• @MBaz My question is still there !! Sine signal is just one example. there are many other examples showing above-mentioned phenomenon and there must be some practical significance of this fact. Feb 27 '19 at 7:13

To claim that "sampling changes the continuity and periodicity of a signal", implies that given a signal $$x(t)$$ and its sampled version $$q(n) = x(n \cdot T), n \in \mathbb{N}, T = \frac{1}{Fs}$$ there is a way to "transform" $$x(t)$$ in an aperiodic $$q(n)$$ just by periodic sampling.

This is impossible, especially for linear systems that cannot go into oscillations out of their own accord.

The only thing that you do with sampling is "tuning in" to a part of a signal's spectrum. Here is why, given a periodic signal at some frequency:

$$x(t) = sin(2 \pi f t)$$

We are going to sample it with a periodic train of pulses:

$$q(t) = \text{III}_T(t) \cdot x(t) \implies \frac{1}{T} \sum_{n=-\infty}^{n=\infty}e^{i 2 \pi n \frac{t}{T}} \cdot x(t) \implies$$

$$\frac{1}{T} \sum_{n=-\infty}^{n=\infty}\left( \cos\left(2 \pi n \frac{t}{T}\right) +i \cdot \sin\left(2 \pi n \frac{t}{T}\right) \right) \cdot x(t)$$

This expression by itself, does not guarantee that $$q(t)$$ is not going to be aliased and this is why $$x(t)$$ is first going through a low pass anti-aliasing filter. After this operation, sampling, is putting a "canvas" over that antialiased spectrum (discretises it).

How it is then possible that the operation described above can generate an aperiodic $$q(t)$$ for a periodic $$x(t)$$?

The only thing that this operation does is sample from the already existing components of $$x(t)$$ which are periodic functions. If $$x(t)$$ contained discontinuities, so would $$q(t)$$. If $$x(t)$$ was aperiodic, so would $$q(t)$$. At the end of the day, what you do here is a straightforward mapping. You don't change any "type of signal".

The case that robert-bristow-johnson mentions is a result from number theory, transposed to periods of sinusoids. You cannot compose an irrational number out of integers and this tiny little difference that remains after each period presents itself as a "running phase" of the sinusoid.

There are however cases where sampling can have an effect on the signal, BUT that is entirely resonable. In the integration of non-linear differential equations by naive schemes such as Euler's method, the "sampling" that is effected by choosing a fixed step size can lead to large errors in the reconstructed waveform and more often than not in cases like this, the sampling becomes irregular. More dense around areas of large errors and vice versa. But even in that case, periodic regions (in the input) would remain periodic in the output.

Hope this helps.