Determine whether the following system is time-invariant or not:

$y(t)=x(t)\sin 10\pi t$


Given: $y(t)=x(t)\sin 10\pi t$

$y(t)=T[x(t)]=x(t)\sin 10\pi t$

The output due to input delayed by $T$ sec is:

$y(t,T)=T[x(t-T)]=y(t)|_{x(t)=x(t-T)}=x(t-T)\sin 10\pi t$

The output delayed by T sec is:

$y(t-T)=y(t)|_{t=t-T}=x(t-T)\sin 10\pi (t-T)$

$y(t,T) \ne y(t-T)$

Conclusion: The delayed output is not equal to the output due to delayed input. Therefore, the system is time invariant.

My query/doubt:

When the output delayed by T sec :

$y(t-T)=y(t)|_{t=t-T}=x(t-T)\sin 10\pi (t-T) =x(t-T)\sin (10\pi t-10\pi T)$

In case the value of $T$ is $\frac{2n}{10}$; $n\ge 0$, then the above equation becomes equal to $y(t,T)$


$y(t,T) = y(t-T)$ , When $T=\frac{2n}{10}$; $n\ge 0$

So can we say that the system is time-invariant when $T=\frac{2n}{10}$; $n\ge 0$ ?

  • 1
    $\begingroup$ What crappy notation where $T$ is used to denote the linear Transformation as well as the delay! and the conclusion that the system is time-invariant is a bald-faced contradiction of the previous sentence which proves exactly the opposite. $\endgroup$ Apr 9, 2018 at 3:27
  • $\begingroup$ The "sin10πt" makes the system time variant. I mean just look at it, clearly the output of the system will vary even if your input is held constant. It appears that you've dived too far into the analysis without stepping back and looking at what logically makes sense. $\endgroup$
    – Izzo
    Jun 4, 2019 at 14:33
  • $\begingroup$ In your post you wrote "Conclusion: The delayed output is not equal to the output due to delayed input." Which is correct - This makes the system time-variant. In your post, it is only the next sentence that is incorrect. $\endgroup$
    – David
    Jun 4, 2019 at 16:19
  • $\begingroup$ This looks like homework to be, in which case the tag should be added $\endgroup$ Jun 4, 2019 at 23:03

2 Answers 2


Your conclusion is wrong: the system is time variant.

Regarding your question: for a system to be time invariant, it needs to be time invariant for any delay.

Even if there are specific delays when the system behaves as time-invariant, in general the system still is classified as time-variant.

  • $\begingroup$ OK, and what if $t\in \frac{2}{10}\mathbb{Z}$? Why do many believe that $t\in \mathbb{R}$ is $\endgroup$ Jun 4, 2019 at 22:44
  • 1
    $\begingroup$ @LaurentDuval I think the question assumes continuous time. If $t \in \frac{2}{10}\mathbb{Z}$ then indeed it becomes time-invariant. I don't know under what conditions this kind of (possibly non-uniform) "sampling" turns a continuous-time time-variant system into time-invariant. $\endgroup$
    – MBaz
    Jun 4, 2019 at 22:56
  • 1
    $\begingroup$ In a mood for laugh, and to insist on people being sufficient precise to avoid opening doors to twisted minds like mine. $\endgroup$ Jun 4, 2019 at 23:00

Just for the fun, and to emphasize on the importance of being precise about the domains one is talking about: if $t$ is not a real variable, but for instance $t$ belongs to an affine version of relative integers: $t=\frac{1}{5}k+b$, then the (now discrete) system rewrites $$y(k)=x(k)\sin\left(10\pi\left(\frac{1}{5}k+b\right)\right)$$ and it becomes a trivially time-invariant system: it takes the form

$$y(k)=\alpha x(k)$$

and is thus instantaneous.

If you want to play the "clever person" with your teacher/instructor, this counter-example could work... or be disastrous.

  • 1
    $\begingroup$ It's a valuable skill as a student to be able to tell what your professor will like and dislike :) $\endgroup$
    – MBaz
    Jun 5, 2019 at 0:13

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