5 ms·

I don't have any real understanding over the Laplace transform, but I understand Fourier transform well enough that it makes sense to me. Back then, I saw an cl

I don't have any real understanding over the Laplace transform, but I understand Fourier transform well enough that it makes sense to me. Back then, I saw an claim that Laplace transform is a generalization of Fourer transform in the sense that it transforms a function not only to a space of frequencies and phases of sine waves, but to a larger space of parameters of exponentials. Note that the parameter space of the sine waves is subset of the (complex) parameter space of exponentials. Is this claim correct?

beagle3 6y ago
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....
bollu 6y ago
If you understand fourier transform well, then perhaps this viewpoint will help. The fourier transform is 'just' a change of basis, with the basis being the sinusoidal functions. Why these? Well, because if we take a look at the discrete fourier transform, the matrix that changes the basis is both unitary [which is has to be as a change of basis] and vandermonde. So we can think of it as both a 'change of basis' and as a 'evaluation of polynomial'. This is where most of the power of the fourier transform comes from. Similarly, the laplace transform is also a change of basis. But the basis it chooses is a very special one --- it's the eigenvectors of the differential operator. Note that d/dx(e^(ax)) = ae^ax So e^(ax) literally an eigenvector of `(d/dx)`. And as we all know, going to the eigenbasis of a given operator/linear transform/matrix makes it easier to manipulate. The laplace transform is a change of basis that digonalizes the differential operator. This makes it easy to solve differentials.
- chombier 6y ago
  Exactly what I needed, thanks!
- hnra 6y ago
  Do you have any good text book where this is included? Seems like something very interesting to read about formally.
  - pokoleo 6y ago
    Math 213 at uwaterloo has publicly available notes in spades: https://www.gabrielwong.net/notes/math213_notes.pdf#page9 https://www.gabrielwong.net/notes/math213_notes.pdf#page9
  - novaRom 6y ago
    search for "Introduction to Signals & Systems"
  - abecedarius 6y ago
    I'm not sure this is what you're looking for, but the Feynman lectures give the basic idea in an elementary way: https://www.feynmanlectures.caltech.edu/I_23.html https://www.feynmanlectures.caltech.edu/I_23.html > This idea of using exponentials in linear differential equations is almost as great as the invention of logarithms, in which multiplication is replaced by addition. Here differentiation is replaced by multiplication. . . . See how simple it is! Differential equations are immediately converted, by sight, into mere algebraic equations
teleforce 6y ago
Please check this video presentation for a simple overview of Laplace Transform [1]. Basically Fourier Transform (FT) is a special case (subset) of Laplace Transform (LT) where the signal waveform revolves around a unit circle (real power of exponents) . Similarly, LT is basically a generalization (superset) of FT that the signal waveforms revolve off the unit circle (complex power of exponents). The discrete FT or DFT, however, as the name clearly implied, is the discrete version of FT and similarly discrete Laplace Transform (DLT) is the discrete version of LT. The main difference is that DFT covers finite sum but DLT covers infinite sum. The faster version of DFT (without compromising the resolution accuracy) is called FFT and it is probably the most useful and important algorithm in the 21st century! The inverse FFT is called IFFT and it was discovered around the same time of FFT. The faster version of DLT is interestingly called Chirp-Z Transform (CZT) and somehow its inverse (ICZT) discovery is at a much later date as has been reported recently [2] and also featured in HN [3]. This much later date of discovery is mainly due to the complexity of complex power exponents (pardon the pun but cannot resist). Fun fact, CT was discovered by Lawrence Rebinar who was working at AT&T's speech processing lab (SPL) [4]. The lab is so well funded that Kernighan and Ritchie who were belong to the other lab has to scrap by the older computer of the SDL (the infamous PDP-7) where Unix was originally developed on when Multics project got canceled. [1]https://youtu.be/n2y7n6jw5d0 https://youtu.be/n2y7n6jw5d0 [2]https://www.electronicsweekly.com/news/research-news/dsp-inverse-chirp-z-transform-works-doesnt-work-2020-03/ https://www.electronicsweekly.com/news/research-news/dsp-inv... [3]https://news.ycombinator.com/item?id=21230757 https://news.ycombinator.com/item?id=21230757 [4]https://ieeexplore.ieee.org/document/1276120 https://ieeexplore.ieee.org/document/1276120

beagle3 6y ago

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....

bollu 6y ago

If you understand fourier transform well, then perhaps this viewpoint will help. The fourier transform is 'just' a change of basis, with the basis being the sinusoidal functions. Why these? Well, because if we take a look at the discrete fourier transform, the matrix that changes the basis is both unitary [which is has to be as a change of basis] and vandermonde. So we can think of it as both a 'change of basis' and as a 'evaluation of polynomial'. This is where most of the power of the fourier transform comes from. Similarly, the laplace transform is also a change of basis. But the basis it chooses is a very special one --- it's the eigenvectors of the differential operator. Note that d/dx(e^(ax)) = ae^ax So e^(ax) literally an eigenvector of `(d/dx)`. And as we all know, going to the eigenbasis of a given operator/linear transform/matrix makes it easier to manipulate. The laplace transform is a change of basis that digonalizes the differential operator. This makes it easy to solve differentials.

chombier 6y ago
Exactly what I needed, thanks!
hnra 6y ago
Do you have any good text book where this is included? Seems like something very interesting to read about formally.
- pokoleo 6y ago
  Math 213 at uwaterloo has publicly available notes in spades: https://www.gabrielwong.net/notes/math213_notes.pdf#page9 https://www.gabrielwong.net/notes/math213_notes.pdf#page9
- novaRom 6y ago
  search for "Introduction to Signals & Systems"
- abecedarius 6y ago
  I'm not sure this is what you're looking for, but the Feynman lectures give the basic idea in an elementary way: https://www.feynmanlectures.caltech.edu/I_23.html https://www.feynmanlectures.caltech.edu/I_23.html > This idea of using exponentials in linear differential equations is almost as great as the invention of logarithms, in which multiplication is replaced by addition. Here differentiation is replaced by multiplication. . . . See how simple it is! Differential equations are immediately converted, by sight, into mere algebraic equations

teleforce 6y ago

Please check this video presentation for a simple overview of Laplace Transform [1]. Basically Fourier Transform (FT) is a special case (subset) of Laplace Transform (LT) where the signal waveform revolves around a unit circle (real power of exponents) . Similarly, LT is basically a generalization (superset) of FT that the signal waveforms revolve off the unit circle (complex power of exponents). The discrete FT or DFT, however, as the name clearly implied, is the discrete version of FT and similarly discrete Laplace Transform (DLT) is the discrete version of LT. The main difference is that DFT covers finite sum but DLT covers infinite sum. The faster version of DFT (without compromising the resolution accuracy) is called FFT and it is probably the most useful and important algorithm in the 21st century! The inverse FFT is called IFFT and it was discovered around the same time of FFT. The faster version of DLT is interestingly called Chirp-Z Transform (CZT) and somehow its inverse (ICZT) discovery is at a much later date as has been reported recently [2] and also featured in HN [3]. This much later date of discovery is mainly due to the complexity of complex power exponents (pardon the pun but cannot resist). Fun fact, CT was discovered by Lawrence Rebinar who was working at AT&T's speech processing lab (SPL) [4]. The lab is so well funded that Kernighan and Ritchie who were belong to the other lab has to scrap by the older computer of the SDL (the infamous PDP-7) where Unix was originally developed on when Multics project got canceled. [1]https://youtu.be/n2y7n6jw5d0 https://youtu.be/n2y7n6jw5d0 [2]https://www.electronicsweekly.com/news/research-news/dsp-inverse-chirp-z-transform-works-doesnt-work-2020-03/ https://www.electronicsweekly.com/news/research-news/dsp-inv... [3]https://news.ycombinator.com/item?id=21230757 https://news.ycombinator.com/item?id=21230757 [4]https://ieeexplore.ieee.org/document/1276120 https://ieeexplore.ieee.org/document/1276120