# Periodic or not Periodic signal

I am trying to solve following problem which I try to find the signal is periodic or not.

However, as it is different from traditional examples I do not know how to solve it. Can you please give me clues or solve it?

• Do you know the (formal) definition of a periodic signal? Commented Oct 26, 2017 at 16:17
• Yes I know the definition. Commented Oct 26, 2017 at 16:34

If a signal is periodic then its derivative will also be periodic

So $$x(t)=sin((\sqrt{2t}+5)+sin(\pi t))-1$$ its derivative is $$x'(t)=\frac{cos((\sqrt{2t}+sin(\pi t)+5))}{\sqrt{2t}}+\pi cos(\pi t)cos((\sqrt2t+sin(\pi t)+5))$$

$$x'(t)=x_1(t)+x_2(t)$$ where $x_1(t)=\frac{cos((\sqrt{2t}+sin(\pi t)+5))}{\sqrt{2t}}$ and $x_2(t)=\pi cos(\pi t)cos((\sqrt2t+sin(\pi t)+5))$ Now $x_1(t)$ is aperiodic because wolframalpha result $$\lim_{t \to \infty}\frac{cos((\sqrt{2t}+sin(\pi t)+5))}{\sqrt{2t}}=0$$

Now $$x_2(t)=\pi cos(\pi t)cos((\sqrt2t+sin(\pi t)+5))$$ $$x_2(t)=\frac{\pi}{2}{[cos((\pi t-\sqrt2t-sin(\pi t)-5))+cos((\pi t+\sqrt2t+sin(\pi t)+5))]}$$

Now $x_2(t)$ can also be checked with above property of derivative individually that will come as aperiodic signal

So $x'(t)$ is sum of two aperiodic signal that will be a aperiodic signal.

Hence $x(t)$ is aperiodic signal

• Thank you very much. Can I ask you another little question? Commented Oct 26, 2017 at 16:35
• @NihadAzimli I think you can't do with that method because what $T$ you will take for $sin((\sqrt{2t}+5)$ this signal itself is aperiodic.same like asking what will be the period($T$) of $e^t$ Commented Oct 26, 2017 at 17:02