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COrrected top integral limit
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Peter K.
Peter K.
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You can think of $\int_{-\infty}^{+infty} f(t) g(t;s) dt$$\int_{-\infty}^{+\infty} f(t) g(t;s) dt$ as a decomposition of $f(t)$ in terms of the basis functions $g(t)$.

There are several nice things about choosing $g(t;s) = e^{-st}$ as the set of basis functions to use, the prime one being that $\frac{dg(t;s)}{ds} = -se^{-st}$, which is a neat property if you're dealing with derivatives.

You can think of $\int_{-\infty}^{+infty} f(t) g(t;s) dt$ as a decomposition of $f(t)$ in terms of the basis functions $g(t)$.

There are several nice things about choosing $g(t;s) = e^{-st}$ as the set of basis functions to use, the prime one being that $\frac{dg(t;s)}{ds} = -se^{-st}$, which is a neat property if you're dealing with derivatives.

You can think of $\int_{-\infty}^{+\infty} f(t) g(t;s) dt$ as a decomposition of $f(t)$ in terms of the basis functions $g(t)$.

There are several nice things about choosing $g(t;s) = e^{-st}$ as the set of basis functions to use, the prime one being that $\frac{dg(t;s)}{ds} = -se^{-st}$, which is a neat property if you're dealing with derivatives.

Source Link
Peter K.
Peter K.
  • 221
  • 4
  • 9

You can think of $\int_{-\infty}^{+infty} f(t) g(t;s) dt$ as a decomposition of $f(t)$ in terms of the basis functions $g(t)$.

There are several nice things about choosing $g(t;s) = e^{-st}$ as the set of basis functions to use, the prime one being that $\frac{dg(t;s)}{ds} = -se^{-st}$, which is a neat property if you're dealing with derivatives.