Is \$\sin(-t)\$ LTI or linear time variant? It seems like linear time variant but we know \$\sin(-t) = \sin(t)\$. Then \$\sin(t)\$ is linear time invariant. So is \$\sin(-t)\$ linear time variant or linear time invariant?

  • 3
    \$\begingroup\$ Doesn't sin(-t) = -sin(t)? \$\endgroup\$
    – Andy aka
    Feb 15 at 14:04
  • \$\begingroup\$ yep. cos(-t) = cos(t). sin(-t) = -sin(t). Sine is a vertical coordinate on a unit circle. If you go t or -t, you get opposite vertical coordinates, but the same horizontal (cos). \$\endgroup\$
    – Ilya
    Feb 15 at 14:14
  • \$\begingroup\$ You are only trying to reason (wrongly) about time invariance, but totally forgot about linearity. \$\endgroup\$
    – Eugene Sh.
    Feb 15 at 15:22

1 Answer 1


LTI is a description used for systems.

Systems are usually described by their transfer function \$H(s)\$ which is the laplace-transform of the system's impulse response \$h(t)\$, but they can also be described by their differential equation. Assuming that \$h(t) = \sin(-t)\$ we can see the following.

$$H(s) = \frac{Y(s)}{X(s)} =\mathcal{L} \{ \sin(-t) \} = -\frac{1}{s^2+1} $$

The differential equation for the system becomes

$$\ddot{y}(t)+y(t)=-x(t) $$

  1. A system is linear, if the terms in the differential equation themselves are linear. That is, there are no \$y^2(t)\$ or something like that.
  2. A system is time-invariant, if the coefficients of the differential equation are constants. That is, they don't depend on time.

I will leave it to you to draw the conclusion whether or not the system is linear and time-invariant.

  • \$\begingroup\$ I got you. Thank you. I've forgot this point. I was directly finding the Time variant or invariant. Thank you for the explanation. \$\endgroup\$ Feb 16 at 5:18
  • 1
    \$\begingroup\$ @AkhilBasoya You are welcome. If you feel your question has been answered, you could (and should, for the record) hit the checkmark next to one of the answers to complete the question-and-answering of this post. \$\endgroup\$
    – Carl
    Feb 16 at 19:44
  • \$\begingroup\$ Not yet I am still confused. I think I should post the image for more clearification. \$\endgroup\$ Feb 16 at 22:34

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.