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Intuitively, I feel like a time-variant system would necessarily have a time-dependent frequency response, and vice-versa. So, is the time-independence of the frequency response necessary and sufficient to say that a sytem is LTI ?

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  • $\begingroup$ Time-variant systems do not have a frequency response. They do have an impulse response, as all systems do. But the Fourier transform of the impulse response is not the Frequency response for time-variant systems. The concept of a time-dependent frequency response is, albeit being somewhat intuitive, not generally defined and assuming its existence can easily mislead you. Therefore, if a system has a frequency response, in the sense of its original definition, it is already LTI. Strictly speaking, the answers you have received here so far are incorrect. $\endgroup$ – Jazzmaniac Jan 25 at 16:30
  • $\begingroup$ good to see you back, @Jazzmaniac , but the fact is that Time-variant Linear Systems absolutely do have a frequency response (albeit a time-varying frequency response), and i spelled it out below. $\endgroup$ – robert bristow-johnson Jan 28 at 19:39
  • $\begingroup$ @robertbristow-johnson The frequency response of a system is defined only if the system can be diagonalised by the Fourier transform. That is not the case for time-dependent systems. What you are talking about is the Fourier transform of the impulse response, which is something entirely different. I give you that for a very slowly changing impulse response you can use most of the intuition of the frequency response, but quantifying that into a meaningful mathematical notion that deserves the name time-dependent frequency response does not work how you think it does. $\endgroup$ – Jazzmaniac Jan 28 at 23:18
  • $\begingroup$ Consider the system $y(t) = \exp(\hat{\omega} t) x(t)$ for which the time "dependent frequency response" that you are suggesting would simply be $H(\omega,t)=\exp(\hat{\omega} t)$. Depending on your choice of $\hat{\omega}$ you get an arbitrary large shift betwee input and output frequencies. At which point would you say the time dependent amplification becomes a detuning? $\endgroup$ – Jazzmaniac Jan 28 at 23:22
  • $\begingroup$ i think $$h(t,u) = e^{\hat{\omega}u}\delta(t-u)$$. Then $$ H(f,u)=e^{j 2 \pi f u} \int\limits_{-\infty}^{\infty} h(t,u) \, e^{-j 2 \pi f t} \, \mathrm{d}t $$ it comes out to be something. $\endgroup$ – robert bristow-johnson Jan 28 at 23:32
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LTI means Linear Time-Invariant systems. The system has to satisfy two conditions. (1) Linear and (2) Time-Invariant. (1) Linear means, if the response of the system due to load Px and Py is Rx and Ry respectively, then for the load (Px+Py), the response of the system will be (Rx+Ry). (2) Time-Invariant means, the parameters of the system does not vary with time.

If the frequency response of the system is Time-Independent, then it suggests that the parameters of the systems are not changing over time, i.e., the system is time-invariant.

But it does not say anything about nonlinearity. Hence, in a nutshell, if the frequency response of a system is time-independent, this system is NOT an LTI?

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I think the answer to your question is "yes" and this is why:

A Linear Time-Variant (or possibly time-variant) system is fully described (from the POV of output $y(t)$ in terms of input $x(t)$) by this convolution integral:

$$ y(t) = \int\limits_{-\infty}^{\infty} h(t,u) \, x(u) \, \mathrm{d}u $$

$h(t,u)$ is the (possibly time-variant) impulse response function, evaluated at time $t$ of an impulse that was applied to the input at time $u$. That is what $y(t)$ is when $x(t)=\delta(t-u)$. In general, impulse responses of impulses applied at different times are different.

However, if the Linear system is also Time-Invariant (LTI) then

$$ h(t,u) = h(t-u) $$

which means that the only difference between impulse responses of impulses applied to the input at different times, $u$, is that they are delayed by the same time that the input impulse is delayed, otherwise exactly the same.

If the system is LTI, this impulse response that was dependent on the time that the impulse is applied, and is a function of two variables, $h(t,u)$, becomes a function of one variable $h(t)$. It doesn't matter what time, $u$, the impulse is applied other than the impulse response happens the same time later: $h(t-u)$. Knowing the impulse response for an impulse applied at time $0$ which is $h(t,0)$ tells you exactly what the impulse response will be for an impulse applied at any other time.

The LTI frequency response, $H(f)$ is the Fourier Transform of the impulse response of an impulse applied at time $0$:

$$ \begin{align} H(f) &\triangleq \mathscr{F} \big\{ h(t) \big\} \\ &= \int\limits_{-\infty}^{\infty} h(t) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\ &= \int\limits_{-\infty}^{\infty} h(t,0) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\ &= e^{j 2 \pi f 0} \int\limits_{-\infty}^{\infty} h(t,0) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\ \end{align}$$

That factor $e^{j 2 \pi f 0} = e^0 = 1$ is in there for a reason. This is the Fourier Transform of an impulse response of an input impulse that is not delayed by any time. But if the impulse response is of a delayed impulse, everything is delayed by that time and you get a linear phase shift factor because of that, which I want to remove, for the purpose of discussing time independence of the frequency response.

$$ \begin{align} H(f,u) &\triangleq \mathscr{F} \big\{ h(t+u,u) \big\} \\ &= \int\limits_{-\infty}^{\infty} h(t+u,u) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\ &= e^{j 2 \pi f u} \int\limits_{-\infty}^{\infty} h(t,u) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\ \end{align}$$

$H(f,u)$ is the time-dependent frequency response of the linear system around time $u$. You can tell that if the linear system is time-invariant (LTI) then $h(t,u)=h(t-u)$ and consequentially $H(f,u)$ does not vary with time $u$. So, at least, the converse is true: If the system is LTI, the frequency response is independent of time. But I cannot see how any other case of a Linear system where $H(f,u)=H(f)$ that can allow $h(t,u)$ to be dependent on $u$.

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  • $\begingroup$ and i think i have a pretty good idea whom the downvoter is. $\endgroup$ – robert bristow-johnson Jan 28 at 19:39
  • $\begingroup$ after taking a look at my old Linear Systems book by Chen (from the 1980s), i continue to stand by this answer, despite the downvote. it's a solid answer. $\endgroup$ – robert bristow-johnson Jan 31 at 17:18

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