Convolution integral and fourier transform in linear response theory

  • Thread starter kgz
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  • #1
kgz
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Hello,

Consider I have a linear time-invariant (LTI) system, with ##x(t)##, ##y(t)##, and ##h(t)##, as input, output, and impulse response functions, respectively.
I have two choices to write the convolution integral to get ##y(t)##:
$$ 1)\ \ \ y(t) = \int_{0}^{t} h(t-t')x(t')dt' $$
and
$$ 2)\ \ \ y(t) = \int_{-\infty}^{t} h(t-t')x(t')dt' .$$
What are the differences between these two, and are initial conditions important factors in the decision of choosing one of these? Is it related to causality of the system?

Also, suppose I am just given the frequency response of the system,
$$ Y(\omega)=H(\omega)X(\omega) . $$
Using Fourier transform and convolution integral theory, I want to change from frequency domain to time domain. Which one of the convolution integrals above should I pick?
Because I need to work in the freuqency domain, I am trying not to use Laplace transform.

Thank you for sharing your ideas.
 

Answers and Replies

  • #2
marcusl
Science Advisor
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No, there is only one definition of convolution integral and it extends from -∞ to ∞. Specific times like 0 only have meaning in relation to the specifics of your problem (for instance, the input might be turned on at t=0).
 
  • #3
WannabeNewton
Science Advisor
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As marcus noted, for your linear time-shift invariant system the convolution integral in generality must be ##y(t) = \int_{-\infty}^{\infty}h(t - t')x(t')dt'##. Now if your system happened to be causal and you wanted the response at ##t## then you can write ##y(t) = \int_{-\infty}^{t}h(t - t')x(t')dt'## only because for a causal system, the response at ##t## will only depend on past contributions up to ##t## i.e. ##h(t - t')## vanishes for negative arguments. marcus explained the ##t = 0## part so I'll leave it at that.
 
  • #4
kgz
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Thank you guys for your responses.
 

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