4 ms·
Yeah, they're only equal when your f(t) = 0 for all t < 0. Otherwise they can be quite different, because the integral limits differ. In an undergrad engineerin
by hamiltonkibbe 6y ago
Yeah, they're only equal when your f(t) = 0 for all t < 0. Otherwise they can be quite different, because the integral limits differ. In an undergrad engineering context you're often evaluating the response of a system to some input and it's common to have "at t=0 the switch is closed/mass is released/etc." where the function is assumed to have been 0 previously.
- btrettel 6y agoSeems obvious in retrospect but wasn't obvious to me. Thanks for the insight!
- hamiltonkibbe 6y agoFor a little more intuition about why you can do that replacement: laplace is a representation of a system as a sum of sinusoids * exponentials, which are your two axis in the laplace plane. Frequency on the iw axis and exponential on the a axis. If you think of that replacement as s = iw + a | a = 0, you'll see the exponential terms go away and you're left with just the sinusoidal parts: f(t) * e^(iw + 0)t = f(t) * e^(iwt) * e(at) = f(t) * e(iwt) * e^(0t) = f(t) * e(iwt) integrated over time, which is your fourier transform subject to the condition above. It's just the laplace transform along the Y axis, or, the frequency response at steady state when not growing/decaying exponentially.