Following this answer , I tried like this $$h(t)=\frac{u(t)}{t+1}$$

where $u(t)$ is unit step function.

I can write the input-output relation like this:

\begin{align} y(t)&=\int_{-\infty}^{+\infty}x(t-\tau)h(\tau)d\tau\\ &=\int_{-\infty}^{+\infty}x(t-\tau)\frac{u(\tau)}{(\tau+1)}d\tau\\ &=\int_{0}^{+\infty}\frac{x(t-\tau)}{(\tau+1)}d\tau \end{align}

$$y_1(t)=\int_{0}^{+\infty}\frac{x(t-t_o-\tau)}{(\tau+1)}d\tau$$ \begin{align} y(t-t_0)&=\int_{0}^{+\infty}\frac{x(t-t_o-\tau)}{(\tau+1)}d\tau\\ &=y_1(t) \end{align}

So the system is time Invariant but the given answer is time variant. Is there any mistake I am doing here? Or the answer is wrong?

In general is there any process to check directly "time variant" or "invariant" of an impulse response without doing this long process?


The actual question is this

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  • $\begingroup$ Are the solutions that you linked to from a book or an accompanying solutions manual? $\endgroup$ – Matt L. Nov 13 '17 at 20:52
  • $\begingroup$ @MattL. No, its not from any standard book...its from a coaching institute they provide previous year exam question's solution. $\endgroup$ – Rohit Nov 14 '17 at 1:04
  • $\begingroup$ Why there is downvote in the question can some one give me the reason please? $\endgroup$ – Rohit Nov 14 '17 at 1:05

The question whether a system described by an impulse response $h(t)$ is time-varying or not is pointless. If a system can be characterized by an impulse response $h(t)$ then it must be linear and time-invariant (LTI).


Now that you've added the original exam question, it turns out that you need to find a system that is time-invariant AND causal AND stable. Clearly, all systems in the problem are time-invariant (because they all are described by an impulse response), but not all of them are causal and stable. So you just need to check causality and stability of the given impulse responses.

EDIT 2 (following up on a discussion in the comments):

If a given impulse response is literally just the system's response to an impulse and it is unsure whether or not this impulse response completely characterizes the system's behavior, nothing can be said about time variance. That is because the given function could be the impulse response of an LTI system, but it might just as well be the response of a time-varying system to an impulse, which doesn't say much about the system. In the latter case, the only thing that could be concluded if $h(t)\neq 0$ for $t<0$ is that the system is non-causal. Note, however, that in the time-varying case $h(t)=0$ for $t<0$ does not imply causality.

The analyses of the four systems given in the solutions linked to in the question are wrong. They claim to derive whether or not a system is time-varying given its response to an impulse. E.g., the system with $h_2(t)=u(t)$ is "shown" to be time-invariant. That could be the case, but if we assume that $h_2(t)$ is just the system's response to an input signal $x(t)=\delta(t)$, we might as well assume that its response to $x(t)=\delta(t-1)$ is $y(t)=\sin(\omega_0t)$ or whatever else we fancy. Consequently, it's impossible to say if a system is time-varying or not if we only know its response to an impulse at $t=0$.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Peter K. Nov 13 '17 at 17:29

Here is how I do those questions:

Time invariant means you will get identical output signals if you 1) shift the input signal first, then put it through a system vs 2) put the input into system, then shift the entire output waveform.

To test this, you first delay the signal. To do so, you simply insert "-T" into the input signal. In your case the input is u(t). Note that (t+1) is not your input signal function, therefore you don't have to delay it. Going back, after delaying you get h=u(t-T)/(t+1).

Now, you first put the signal through a system, then delay the entire waveform. To do so, you replace every instance of t with (t-T). You get h=u(t-T)/(t-T+1).

Finally, you compare the two. In you case, they look different (the denominator is different), therefore the system is time variant. You don't get the same output if you first delay the input vs you delay the output waveform. If the two were to match, you would get time invariant.

Maybe this is not the most "math backed" answer, but this method gets the job done for me.

  • $\begingroup$ You misunderstood the question. "In your case the input is u(t)". No, $u(t)$ is the step function, and it's part of the impulse response of the system, not the input signal. Your conclusion that the system is time-varying is wrong. $\endgroup$ – Matt L. Nov 13 '17 at 9:17

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