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

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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?

  • 1
    \$\begingroup\$ Do you know the (formal) definition of a periodic signal? \$\endgroup\$
    – Eugene Sh.
    Commented Oct 26, 2017 at 16:17
  • \$\begingroup\$ Yes I know the definition. \$\endgroup\$
    – fallen
    Commented Oct 26, 2017 at 16:34

1 Answer 1


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

  • \$\begingroup\$ Thank you very much. Can I ask you another little question? \$\endgroup\$
    – fallen
    Commented Oct 26, 2017 at 16:35
  • \$\begingroup\$ @NihadAzimli Yes Please \$\endgroup\$
    – Rohit
    Commented Oct 26, 2017 at 16:39
  • \$\begingroup\$ if our T1 and T2 is pi/something how can we find T0. Because LCM(T1,T2)/HCF(T1,T2) is not working for this example since we have to get integer for LCM and HCF \$\endgroup\$
    – fallen
    Commented Oct 26, 2017 at 16:47
  • \$\begingroup\$ @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\$ \$\endgroup\$
    – Rohit
    Commented Oct 26, 2017 at 17:02
  • \$\begingroup\$ Are you sure that this derivative is correct? The second sin is in the first sin. \$\endgroup\$ Commented Oct 26, 2017 at 18:24

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