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Yes. The Fourier transform characterizes on a circle (usually the unit circle in the complex domain) at different frequencies. It is properly defined for period

Yes. The Fourier transform characterizes on a circle (usually the unit circle in the complex domain) at different frequencies. It is properly defined for periodic signals. The Laplace transform takes any exponential spiral in the complex plane, and reduces to the Fourier transform if you only care about the unit circle. I appreciate that doesn’t make things clearer unless you already have some understanding of integral transforms in the complex plane (in which case, you probably know this already). However, I have never met a simple intuitive explanation of the Laplace transform, - and actually no meaningful explanation that doesn’t involve integrals.

durbat 6y ago
With Fourier you can analyze oscillatory characteristics of function (frequency and phase). With Laplace you can also analyze amplification/attenuation.
- tilt_error 6y ago
  The Fourier transform, well, transforms a time-based phenomenon such as an alternating current sine wave into a frequency spectrum where you can observe the frequency spectrum components of the signal. A pure nice sine becomes a spike (delta function) located at a specific frequency. Music, as we observe it through our ears and can view it on an oscilloscope becomes moving spikes (lots of them :) in the frequency spectrum where the "amplitude" at a given frequency relates to the "amount" of that frequency in the music. The behaviour of a filter is much easier to describe in the frequency spectral domain than it would be in the time domain. Now to the direct current (DC) view. This cannot be handled by the Fourier transform -- at least the DC-part of the signal cannot be transformed to the frequency domain. As shown in the article, there were "steps", "ramps" and such. A typical scenario would be to describe what happens in your amplifier during startup, to describe how electrical circuits are behaving during startup before reaching the "running" state. The Laplace transform will handle these types of scenarios, and can thus be used to study (or describe) systems during other types of transitions than the "steady state" when you are up and running. Regarding filters, the Fourier transform describes things going on at the unit circle, while the Laplace transform can be used to study both the interior and exterior of the plane. In this sense, creating filters relates to locate "poles" and "zeros" in the plane (amplification and attenuation) which can be observed on the unit circle as the behaviour on periodic signals.
  - andi999 6y ago
    Why can DC not get fouriertransformed? Usually the f(p=0) is the DC component.
btrettel 6y ago
I recall from some undergraduate classes I took (in mechanical engineering) that you can "convert" between a Laplace transform and a Fourier transform by saying that s = i * omega. This has some appeal from a purely algebraic standpoint but doesn't seem rigorous to me. For one, the limits of the integral aren't the same. How valid is this? I always assumed it was an approximation. Is the Laplace transform in some sense similar to a one-sided/semi-infinite Fourier transform, provided that change of variables is made? Years ago in a complex analysis class I worked out the contour integration for a few Fourier transforms as I recall, but I've had no similar training for the Laplace transform and have forgotten many details.
- 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 ago
    Seems obvious in retrospect but wasn't obvious to me. Thanks for the insight!
    - hamiltonkibbe 6y ago
      For 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.
qppo 6y ago
I believe you're confusing discrete and continuous time, Fourier in continuous time is Laplace evaluated along s = jw or the vertical axis and not e^jw, the unit circle
- beagle3 6y ago
  You are correct. I mixed how I tend to visualize and think of them with what actually happens....

durbat 6y ago

With Fourier you can analyze oscillatory characteristics of function (frequency and phase). With Laplace you can also analyze amplification/attenuation.

tilt_error 6y ago
The Fourier transform, well, transforms a time-based phenomenon such as an alternating current sine wave into a frequency spectrum where you can observe the frequency spectrum components of the signal. A pure nice sine becomes a spike (delta function) located at a specific frequency. Music, as we observe it through our ears and can view it on an oscilloscope becomes moving spikes (lots of them :) in the frequency spectrum where the "amplitude" at a given frequency relates to the "amount" of that frequency in the music. The behaviour of a filter is much easier to describe in the frequency spectral domain than it would be in the time domain. Now to the direct current (DC) view. This cannot be handled by the Fourier transform -- at least the DC-part of the signal cannot be transformed to the frequency domain. As shown in the article, there were "steps", "ramps" and such. A typical scenario would be to describe what happens in your amplifier during startup, to describe how electrical circuits are behaving during startup before reaching the "running" state. The Laplace transform will handle these types of scenarios, and can thus be used to study (or describe) systems during other types of transitions than the "steady state" when you are up and running. Regarding filters, the Fourier transform describes things going on at the unit circle, while the Laplace transform can be used to study both the interior and exterior of the plane. In this sense, creating filters relates to locate "poles" and "zeros" in the plane (amplification and attenuation) which can be observed on the unit circle as the behaviour on periodic signals.
- andi999 6y ago
  Why can DC not get fouriertransformed? Usually the f(p=0) is the DC component.

btrettel 6y ago

I recall from some undergraduate classes I took (in mechanical engineering) that you can "convert" between a Laplace transform and a Fourier transform by saying that s = i * omega. This has some appeal from a purely algebraic standpoint but doesn't seem rigorous to me. For one, the limits of the integral aren't the same. How valid is this? I always assumed it was an approximation. Is the Laplace transform in some sense similar to a one-sided/semi-infinite Fourier transform, provided that change of variables is made? Years ago in a complex analysis class I worked out the contour integration for a few Fourier transforms as I recall, but I've had no similar training for the Laplace transform and have forgotten many details.

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 ago
  Seems obvious in retrospect but wasn't obvious to me. Thanks for the insight!
  - hamiltonkibbe 6y ago
    For 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.

qppo 6y ago

I believe you're confusing discrete and continuous time, Fourier in continuous time is Laplace evaluated along s = jw or the vertical axis and not e^jw, the unit circle

beagle3 6y ago
You are correct. I mixed how I tend to visualize and think of them with what actually happens....