0
$\begingroup$

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 $2$ ?

| improve this question | |
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$$ \frac{d}{dt} u(t) = \delta(t) $$ $$ \frac{d}{dt} u(-t) = -\delta(t) $$ $$ \frac{d}{dt} ( u(t) + u(-t))=0$$ 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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user28715 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. $\endgroup$ – user28715 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
  • 1
    $\begingroup$ I'll concede that. $\endgroup$ – user28715 Jun 20 '17 at 15:08
0
$\begingroup$

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 $t$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $1$.

| improve this answer | |
$\endgroup$
-1
$\begingroup$

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.