Here Its says Hilbert transform is a non-causal,linear,and time-invariant system

How can I prove it mathematically?

wikipedia says the input output relation like this $$\boxed{y(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau)}{t-\tau}d\tau}$$

so from this relation it showing time varying nature because

for $X(t-t_o)$,$y(t)$ is $${y(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau-t_o)}{t-\tau}d\tau}$$

and $${y(t-t_o)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau)}{(t-t_o)-\tau}d\tau}$$

So both are not same so its Time variant

  • 2
    $\begingroup$ If you browse the link you gave, you can find information on how to prove things, for instance Showing Linearity and Time Invariance, or Not. I suggest you provide your insight, and where you actually require help $\endgroup$ Nov 12 '17 at 10:44
  • $\begingroup$ @LaurentDuval Sir please check now $\endgroup$
    – Rohit
    Nov 12 '17 at 13:43


  1. Prove that $$y(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau)}{t-\tau}d\tau=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(t-\tau)}{\tau}d\tau\tag{1}$$
  2. Use the right-most expression in $(1)$ to show that the system is time-invariant.

Equivalently, you can directly show that the original input-output relation is a convolution integral, from which it follows that the system must be LTI. Then find the expression for the impulse response $h(t)$ and show that $h(t)\neq 0$ for $t<0$, which means that the system is not causal.


The same variable substitution that can be used to show $(1)$ can also be used to directly show the equivalence of the response to $x(t-t_0)$ and the delayed response to $x(t)$:



  • $\begingroup$ Sir why do I need to change the integration like that in equation 1...why its giving time varying without changing it into the other integration form? $\endgroup$
    – Rohit
    Nov 12 '17 at 15:29
  • $\begingroup$ @Rohit: You don't need to change it; it must makes it easier to see that the system is time-invariant. $\endgroup$
    – Matt L.
    Nov 12 '17 at 17:01
  • $\begingroup$ But Why am I seeing it Time varying in my answer sir? $\endgroup$
    – Rohit
    Nov 12 '17 at 17:03
  • $\begingroup$ @Rohit: That's the mistake. You only think you see time variance. Prove that the last two expressions in your question are the same by variable substitution. $\endgroup$
    – Matt L.
    Nov 12 '17 at 17:52

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