What is the interpretation of $s$ compared to $t$?
Why is each Laplace transform only defined for some values of $s$?
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You can interpret $s$ as a frequency and $t$ as time, for example. The integral defining the transform may only make sense for certain $s$, usually something like $\operatorname{re} s > \sigma$. – copper.hat May 25 '15 at 22:38
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1The first part of this lecture gives a very nice interpretation of $s$: http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/ – Simon S May 25 '15 at 22:39
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Remember that $s$ can be complex valued and expand $e^{st}$ with Euler's identity. Notice how the imaginary part of $s$ shows up as the frequencies of sinusoids while the real part gives them an exponential amplitude envelope. If the real part of $s$ is zero, then $e^{st}$ is strictly sinusoidal and the Laplace transform becomes the Fourier transform.
Put another way, "frequency response" is the cross-section of the Laplace transform on the imaginary axis of the domain $s$. In that case, you can physically interpret $t$ as time and $s$ as frequency, but in general when considering "complex frequency" just imagine the exponentially enveloped sinusoids. The Laplace transform projects your function of $t$ onto a basis of those exponentially enveloped sinusoids. See here for more details.
It might also help to consider that, in some sense, $s$ has the reciprocal of whatever units $t$ has. If $t$ is time in seconds, then $s$ is frequency in (radians)/s.
