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According to Oppenheim's signal and systems solution book and other solutions like this

$$ Ev(\cos(4\pi t)u(t)) = 1/2 \times ( \cos(4\pi t)u(t) + \cos(4\pi t)u(-t)) = 1/2(\cos(4\pi t)) $$

$$ \textrm{for }-\infty < t < \infty $$

is periodic and its period $1/2$.

But I think it is not equal $ 1/2(\cos 4\pi t) $ at $ t = 0 $ so it is not periodic because at $ t = 0 $ it is equal to $ 1 $ different from other multiples of period.

I have also curious about the value of $ u(t)+u(-t) $. Is it $0$, $1$, or $2$ ?

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  • $\begingroup$ What is the meaning of $Ev(\cdot)$? $\endgroup$ – MBaz Mar 12 '15 at 22:13
  • $\begingroup$ $ Ev(f(t)) $ is even part of function f(t). $ Ev(f(t)) = 1/2(f(t) + f(-t)) $ $\endgroup$ – verdery Mar 12 '15 at 22:20
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It is a matter of convention. The answer depends on the definition of the step function $u(t)$. In signal processing, it is common to define $u(0)=1/2$. Under this definition, we have that $Ev(\cos(4\pi t)u(t)) = 1/2(0.5+0.5)=1/2$, so the signal is indeed periodic.

If $u(0)$ is defined as either 0 or 1, then you'd be correct and the signal wouldn't be periodic.

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  • $\begingroup$ Could you give me a resource(book, link) that is explain this convention. I have not seen this before. $\endgroup$ – verdery Mar 12 '15 at 22:37
  • $\begingroup$ @verdery, see en.wikipedia.org/wiki/Heaviside_step_function , in particular the section "Zero argument". $\endgroup$ – MBaz Mar 12 '15 at 22:38
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$$ \frac{d}{dt} u(t) = \delta(t) $$ $$ \frac{d}{dt} u(-t) = -\delta(t) $$ $$ \frac{d}{dt} ( u(t) + u(-t))=0$$ And this is true for all derivatives, Therefore $$ ( u(t) + u(-t)) = 1$$ and is continuous for all $t$ including $t=0$ so the function in question is periodic.

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  • $\begingroup$ I think this derivation proves that the function $u(t) + u(-t)$ has a limit as $t$ goes to $0$, but not necessarily continuous at $t=0$ and therefore does not prove the fact that $u(0)+u(-0)=1$. For the question, which looks for the periodicity of the even part of $\cos(4\pi t) u(t)$ the value at the origin is important. $\endgroup$ – Fat32 Jun 20 '17 at 4:49
  • $\begingroup$ If you accept that all the derivatives exist and are convergent which in this case they are zero, the function is analytic. If you don't accept that all the derivatives exist I really haven't proved anything. $\endgroup$ – Stanley Pawlukiewicz Jun 20 '17 at 6:03
  • $\begingroup$ You need to establish the I think part before you can move to the therefore part, so your therefore is an opinion. I have no problem with opinions but it is not a proof regardless of how many reputation points you have. $\endgroup$ – Stanley Pawlukiewicz Jun 20 '17 at 6:13
  • $\begingroup$ ok. The formal derivative, based on the classical calculus of analytic functions and not on Lebesgue measurable sense, of the discontinuous function $u(t)$ does not exist at the discontinuity point $t=0$ hence all of your derivations indeed is formally incorrect at the discontinuity point $t=0$... Therefore I don't know what to unprove now in fact :-) $\endgroup$ – Fat32 Jun 20 '17 at 12:38
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    $\begingroup$ I'll concede that. $\endgroup$ – Stanley Pawlukiewicz Jun 20 '17 at 15:08
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So, let's check.

For a signal to be periodic with period $P$, we need:

$$ x(t) = x(t+P) $$

for all $t$.

Here $$ x(t) = 1/2(\cos(4\pi t))$$ and $$ x(t+P) = 1/2(\cos(4\pi [t+P]))$$ so for such a $P$ to exist we need $$ \cos(4\pi t) = \cos(4\pi [t+P]) = \cos(4\pi t) \cos(4\pi P) - \sin(4\pi t) \sin(4\pi P) $$ by applying the $\cos(A+B)$ identity.

We know that $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$, so $P=\frac{1}{2}$ satisfies the equation for all $t$.

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  • $\begingroup$ Peter, I think the question is not whether $\cos(4\pi t)$ is periodic, but whether the even function at $t=0$ is equal to $1/2$. $\endgroup$ – MBaz Mar 12 '15 at 22:34
  • $\begingroup$ @MBaz: Ah, you may be right! Oh well. :-) $\endgroup$ – Peter K. Mar 13 '15 at 1:40
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Some authors say periodic signals should exist alike on positive and negative time axis.

$\cos(4\pi t)$ is periodic but, when multiplied by $u(t)$, $\cos(4\pi t)u(t)$ exists on positive axis only, so it's not periodic.

Regarding the step function, Simon Haykin states $u(0)$ is undefined for CTS due to abrupt transition from $0$ to $1$.

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x(t)=1/2(cos 4 pi t u(t)+cos 4pi(-t) u(-t)) ... even function cos(-x) =cosx but u(-t) =o as unit step function exists when t>=0 so u(-t) =0 u(t) =1 for all even values x(t)= 1/2 cos 4 pi t ... u(t)=1 for all even values t= 2 pi / omega omega = 4 t= 2pi/4 =1/2 sec [ periodic function]

if cos is replaced by sin then sin function gets 0 for even values t=0 u(t) =1 but sin 0 =0 .. x(t) =0 non periodic

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  • $\begingroup$ Could you please edit your maths to make your answer clearer? I tried to do so to help out, e.g. $x(t)=1/2(\cos(4 \pi t)u(t) + cos(4\pi(-t)) u(-t))$, but when I got to '...for all even values x(t)' it got too terse for me to unambiguously understand the meaning of the rest. $\endgroup$ – Thomas Arildsen Jun 20 '17 at 12:12

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