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The question arose to me earlier, and I never really understood it..

What is the difference between fourier transform and laplace transform in terms of analyzing an overall circuit? I don't quite understand.

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  • \$\begingroup\$ I don't trust my mathematical foundations to give a full answer but in brief: for practical purposes there is little difference, aside from the the Laplace transform being applicable to a wider range of input signals. In theory, they also have different Regions of Convergence and there are some subtle details regarding differing on the imaginary axis, or some point like that, but I'm not very familiar with these. \$\endgroup\$
    – jramsay42
    Dec 3 '18 at 2:49
  • \$\begingroup\$ Fourier is all time, Laplace is T> zero? \$\endgroup\$ Dec 3 '18 at 3:15
  • \$\begingroup\$ There are books and WIki pages to define the similarities and differences in mathematical characteristics and structure and both have conversions between time and frequency, but for practical use, Laplace transforms are more often to make it easier to solve analog circuits and filter equations to look like Bode Plots of gain and phase while Fourier transforms make it easier to solve problems on signals from time to spectral domain then add a design then you can convert back to time domain if you like. \$\endgroup\$ Dec 3 '18 at 3:50
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    \$\begingroup\$ Don't quote me on this, but I'm pretty sure the Fourier transform is just the Laplace transform evaluated along the imaginary axis. \$\endgroup\$
    – Hearth
    Dec 3 '18 at 4:05
  • \$\begingroup\$ This is a short summary. Though it is directed at a specific question, it covers a broad range. \$\endgroup\$
    – jonk
    Dec 3 '18 at 5:09
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The differences can be found in the definition. A Fourier transform:

$$\mathcal{F}\{f(t)\} = F(j\omega) = \int_{-\infty}^{+\infty} e^{-j\omega t}f(t)dt$$

While an ordinary Laplace transform is given by:

$$\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{+\infty} e^{-st}f(t)dt$$

There are two differences:

  1. \$j\omega t\$ is replaced by \$st\$. \$s\$ can be anywhere on the complex plain. In this sense, the Fourier transform can in some sense be thought of as a subset of the Laplace transform.
  2. The lower limit is different. This complicates things a little bit.

The two transformations become exactly equal if

$$\mathcal{F}\{f(t)\} \stackrel{!}{=} \left.\mathcal{L}\{f(t)\}\right|_{s=j\omega} \Leftrightarrow f(t < 0) = 0$$

Or in words: If \$f(t) = 0\$ for \$t < 0\$, then the Fourier transform is exactly the Laplace transform by following the imaginary axis, or \$s = j\omega\$.

Both have very similar properties. In particular, for a Linear and Time-Invariant (LTI) system with impulse reponse \$g(t)\$ (ie. not nonlinear and it doesn't matter when you start your input signal, the output will remain the same) you have the property that

$$y(t) = \int_{-\infty}^{+\infty} u(t)g(t-\tau)d\tau = u(t)*g(t)$$

$$\begin{align} Y(j\omega) &= U(j\omega)\cdot G(j\omega)\\ Y(s) &= U(s)\cdot G(s) \end{align}$$

This property works for any input \$u(t)\$, including the ones where \$u(t<0) = 0\$ in the Fourier transform. If the system \$g\$ is now also causal (ie. the system can't look into the future, which is always the case for analog electronics), then you can guarantee that \$y(t<0) = 0\$, and you can immediately state that \$G(j\omega) = \left.G(s)\right|_{s=j\omega}\$.

So when you calculate the Laplace transform of the impulse response of an LTI causal system, you can also immediately find the Fourier transform by replacing \$s\$ by \$j\omega\$.


Example

This can become an issue in the following example. Let's say we have the following transfer function (a regular RC low-pass filter):

$$G(s) = \frac{1}{1 + RC\cdot s}$$

If we wish to know the transient behavior when a sine wave starts at \$t=0\$, we would have to use the Laplace transform.

$$U(s) = \mathcal{L}\{\cos(\omega_0t)\} = \frac{s}{s^2 + \omega_0^2}$$

$$\begin{align} Y(s) &= \frac{s}{s^2 + \omega_0^2}\cdot\frac{1}{1 + RC\cdot s}\\ &= \frac{1}{1 + (\omega_0RC)^2}\left(\frac{s + RC\omega_0^2}{s^2 + \omega_0^2} - \frac{RC}{1 + RC\cdot s}\right)\\ y(t) &= \frac{1}{1 + (\omega_0RC)^2}\left[\cos(\omega_0t) + \omega_0RC\sin(\omega_0t) - e^{-\frac{t}{RC}}\right] \end{align}$$

However, if we wish to know the steady-state solution, assuming the input has always been a sine wave, then we need to use the Fourier transform. The Laplace transform will fall short now, because \$u(t < 0) \neq 0\$. We can however still reuse the transfer function \$G(j\omega) = \left.G(s)\right|_{s=j\omega}\$.

$$u(t) = \cos(\omega_0t) \Rightarrow U(j\omega)=\frac{1}{2}\left(\delta(\omega+\omega_0) + \delta(\omega-\omega_0)\right)$$

This is a bit annoying to use, so we will instead use phasors:

$$\begin{align} \underline{Y} &= \underline{U}\cdot G(j\omega_0)\\ &= 1\cdot\frac{1}{1 + RC\cdot j\omega_0}\\ &= \frac{1 - j\omega_0RC}{1 + (\omega_0RC)^2}\\ y(t) &= \mathcal{Re}\left\{\left|\underline{Y}\right|e^{j\omega_0t+\angle{\underline{Y}}}\right\}\\ &= \frac{1}{1 + (\omega_0RC)^2}\left[\cos(\omega_0t) + \omega_0RC\sin(\omega_0t)\right] \end{align}$$

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