Time-invariant and Time-varying Systems

Question

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

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

Solution:

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 $y(t,T)$

ie,

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

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

Answer 1

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.

Answer 2

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.