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Suppose I have an RLC network in a black box, and I bang it hard in the lab to get the impulse response. I have two options now, I can take the Fourier transform or I can take the Laplace transform to get the frequency response. How do I know which one to choose and what is the physical difference between each?

I have been told that the Laplace transform also gives you the transient response or the decay whereas the Fourier transform does not. Is this true? If I suddenly apply a sinusoidal signal at the input, then there should be a transient response for a brief period of time where the output is not a sinusoid until the system settles. Can someone give me a practical example in terms of an RLC network to show how this is true?

Also, often in circuits class, we take the Laplace transform of a circuit where the real part of \$s = \sigma + j\omega\$ is assumed to be zero anyway, so when we use \$\frac{1}{Cs}\$ to denote the Laplace transform of the capacitor, it is assumed that this is equivalent to \$\frac{1}{j\omega C}\$. I believe the real part is zero since the current through the capacitor is 90 degrees out of phase with the voltage across - is this correct? I thought Fourier transform was the same as Laplace transform with \$\sigma = 0\$. However, that does not seem to be true - consider \$x(t) = u(t)\$:

$$\mathcal{F}\{x(t)\} = \int_{-\infty}^\infty{u(t)e^{-j\omega t}}dt = \pi\delta(\omega) + \frac{1}{j\omega} \neq \mathcal{L}\{x(t)\} = \int_0^\infty{e^{-st}dt} = \frac{1}{s}$$

We can see that even if I substitute \$s = j \omega\$ with no real part at the output of the Laplace transform, they are still not equal. How come the Fourier transform has an extra impulse component but Laplace does not? When can I substitute \$s = j\omega\$ and expect the Fourier transform to equal the Laplace transform?

Edit: the latter part of my question has answers here and here.

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The Fourier and the Laplace transform are not the same. First of all, note that when we talk about the Laplace transform, we very often mean the unilateral Laplace transform, where the transformation integrals starts at \$t=0\$ (and not at \$t=-\infty\$), i.e. with the Laplace transform we usually analyze causal signals and systems. With the Fourier transform this is not always the case.

In order to understand the differences between the two, it is important to look at the region of convergence (ROC) of the Laplace transform. For causal signals, the ROC is always a right-half plane, i.e. there are no poles (of a rational function in \$s\$) to the right of some value \$\sigma_0\$ (where \$\sigma\$ denotes the real part of the complex variable \$s\$). Now if \$\sigma_0<0\$, i.e. if the \$j\omega\$ axis is inside the ROC, then you simply obtain the Fourier transform by setting \$s=j\omega\$. If \$\sigma_0>0\$ then the Fourier transform does not exist (because the corresponding system is unstable). The third case (\$\sigma_0=0\$) is interesting because here the Fourier transform does exist but it cannot be obtained from the Laplace transform by setting \$s=j\omega\$. Your example is of this type. The Laplace transform of the step function has a pole at \$s=0\$, which lies on the \$j\omega\$ axis. In all such cases the Fourier transform has additional \$\delta\$ impulses at the pole locations on the \$j\omega\$ axis.

Note that it is not true that the Fourier transform cannot deal with transients. This is just a misunderstanding which probably comes from the fact that we often use the Fourier transform to analyze the steady-state behavior of systems by applying sinusoidal input signals that are defined for \$-\infty<t<\infty\$. Please also see this answer to a similar question.

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  • \$\begingroup\$ Could you explain why in circuit analysis usually Laplace transform is used but finally the real part of s is set to 0? \$\endgroup\$ – anhnha Jul 6 '17 at 12:10
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Ok, so you bang a black box made of RLC components and you measure the response - the impulse response. Now you want to know the frequency response, meaning the response to any sinusoidal.

First of all, you can't really excite your system with a pure sinusoidal. It's too late, you should have started at the big bang. The best you can do is use a causal sinusoidal, which has extra frequency components.

But let's say that what you want to know is the response of the system to an arbitrary input in the time domain. You don't really need Fourier or Laplace to know this. A convolution will do.

What do you have in hand, really? You measured the impulse response. Somehow you plotted it out, let's say continuously, as opposed to an ADC that sampled the signal - which is usually what happens, and you'd be asking about the Z-transform vs FFT instead. Let's also assume that the bang you gave it was a good delta: strong but short.

Since your system is RLC, it is linear, so superposition principles work (we wouldn't be talking about this otherwise anyway). Any input can be constructed by adding attenuated impulses offset in time (sort of - it's a limit thing). So the total response is just adding all these individual responses together. This addition is exactly what the convolution input(t)*impulseResponse(t) does. You could consider the RLC system as a "hardware convoluter". This is probably the most accurate way of predicting a response to an arbitrary input.

Now I want to clarify something, which is how Laplace relates to Fourier. Our domain is causal functions, since it doesn't make sense to compare the unilateral Laplace with Fourier otherwise. Besides, all real signals are causal. Mathematically, the Laplace transform is just the Fourier transform of the function pre-multiplied by a decaying exponential. It is that simple. So if a Fourier transform doesn't exist because the integrals are infinite, Laplace may still exist if the decaying exponential is strong enough, because the intergral of the 'attenuated' function would converge. From a mathematical standpoint, this can be extremely useful in certain cases.

But what you really may want is to make a control system for your plant. In that case, what you do is inspect the response and then approximate it with a 1st or 2nd order model plus group delay. So it won't be exact, but by doing this you ditch all the little details of the actual response, and gain the enormous advantage of being able to plug this model onto control equations and algorithms and dozens of books' worth of control theory knowledge and design and simulate your control system. In that case, you'd use a Laplace model, since immediately get poles and zeroes that can be used for stability analysis.

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    \$\begingroup\$ Good answer. However, your statement "Laplace is more general than Fourier" is not true. In system theory it can be very useful, also for practical purposes, to study ideal systems and/or ideal signals. In these cases it is usually the Fourier transform that does exist, whereas the Laplace transform doesn't. Consider as an example the impulse response of ideal brick-wall filters. Their Laplace transform doesn't exist, but their Fourier transform does. The same is of course true for the transform of ideal signals, such as sinusoids (switched on at the big bang ...). \$\endgroup\$ – Matt L. Apr 6 '14 at 12:07
  • \$\begingroup\$ @apalopohapa: why "you can't really excite your system with a pure sinusoidal"? \$\endgroup\$ – anhnha Jul 12 '17 at 5:36

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