Fourier Transform vs. Laplace Transform?

Question

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.

Answer 1

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\$.

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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:

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