I am always confused by this, please make it clearer for me here:
1)If I have a system which has Impulse response, which means \$H(j\omega)\$, so my system is LTI.
Am I right?
2) Same with transfer function \$H(s)\$, which is basically stronger.

If it is false, I have to find if it is LTI by usual stuff? with \$y(t)\$

For example: $$y'\left(t\right)+\frac{1}{RC}y\left(t\right)=\frac{1}{RC}x\left(t\right)$$ $$H\left(j\omega \right)=\frac{1}{RC}\left(\frac{1}{j\omega +\frac{1}{RC}}\right)$$ and the impulse signal: $$h\left(t\right)=\frac{1}{RC}\cdot e^{-\frac{t}{RC}}$$

can I know if it LTI based on the impulse response, or I must find \$y(t)\$ and then like the usual long way.


2 Answers 2


It is perfectly reasonable to talk about a linear time-varying system having an impulse response. It's just than instead of the impulse response being \$h(\tau)\$, it is \$h(t, \tau)\$, with \$t\$ being the time and \$\tau\$ being the time lag since the impulse.

When this is the case, \$H(j\omega)\$ (and \$H(s)\$) does not exist.

Howevever, \$H(j\omega)\$ and \$H(s)\$ do imply that the system in question has a time-invariant impulse response and is, therefor, time-invariant itself.

In your example:


Each one of the operations in this example are LTI:

  • I see one addition (two if you bring everything over to one side or another). Addition is linear and time invariant.
  • I see a time derivative. The time derivative is linear and time invariant.
  • I see multiplication by constants. Multiplication by a constant is both linear and time invariant.

If you have an assemblage of systems that are each LTI, and the assembly is done using operations that don't violate linearity, then the result is linear and time invariant.

So you can stop right there and demonstrate that the system is LTI -- then you can go on to use Fourier* or Laplace analysis, knowing that you're "legal".

* I really hesitate to add this because at your stage this will probably be confusing. However, honesty compels me: you can use Fourier analysis to usefully analyze some linear time varying systems. It's not as easy, and it's mostly useful for systems that are or resemble heterodyne radio systems (i.e., the only time-varying part is multiplication by a sine or other periodic wave someplace). But it can be done.

  • \$\begingroup\$ There is depth and breadth, despite the brevity, and it squarely addresses questions (1) and (2). So +1. (I'm just worried the questioner isn't quite ready given the added concrete 'for example' they provided [my reading tea leaves from that.]) \$\endgroup\$ Aug 13 at 0:38
  • \$\begingroup\$ So I cant use (1) and (2). I guess If I need to find linear and time variant, I have to do it the usual way, right? Thanks :) \$\endgroup\$
    – user323806
    Aug 13 at 6:40
  • 1
    \$\begingroup\$ If your real question is how to unequivocally find out if a system is nonlinear or time varying without working out its entire response, then you may want to ask just that in another question. \$\endgroup\$
    – TimWescott
    Aug 13 at 14:42

Not quite. You have it backwards. Every system will have an impulse response. If a system happens to be an LTI system, then that impulse response completely describes the system, and if you know the impulse response of that LTI system, you can calculate the output for any given system.

  • \$\begingroup\$ Then I have to find if it is linear by using usuall way? with the sum of inputs and time variant also? \$\endgroup\$
    – user323806
    Aug 13 at 6:38
  • 1
    \$\begingroup\$ "Every system will have an impulse response". I don't see how that can be true, or at least I can't see unique and meaningful impulse responses for any arbitrary system. Can you please back that up with examples, particularly for systems that have higher-order (i.e. \$x^2\$ or \$e^x\$) response to the input or in the states? \$\endgroup\$
    – TimWescott
    Aug 13 at 14:40
  • \$\begingroup\$ @TimWescott. "Impulse Response" defined as "output of a system in response to the input of an impulse". It doesn't need to be bounded, it doesn't matter if it goes to infinity. It doesn't matter if the response isn't the same every time you put an impulse into it. \$\endgroup\$ Aug 13 at 16:51
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    \$\begingroup\$ Yes, I know what the definition means. AFAIK it is an utterly useless concept to use with a nonlinear system because the output will either be zero, stuck at infinity, or otherwise useless to predict the system behavior. So -- could you please back up your assertion that an impulse response is a useful way to analyze a nonlinear system without linearizing the system first. \$\endgroup\$
    – TimWescott
    Aug 13 at 23:28
  • \$\begingroup\$ I get in more than enough trouble over things I do assert. \$\endgroup\$ Aug 13 at 23:46

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