Electronic – How does Laplace transform include the transient response

circuit analysiscontrol systemfrequencyimpedancelaplace transform

When we provide a sinsusoidal input, the evaluation of the solution of the differential equation head-on becomes horrendous. So we give an imaginary, complex exponential input and solve the differential equation and finally take the imaginary part as the solution (superposition theorem). So once we find the complex amplitude we can just multiply it with \$e^{iωt}\$, take imaginary part and arrive at the solution.

We can find the sinusoidal steady-state response for the given system, with this neat trick. When we use Laplace transform, as far as I can understand, it also gives us a complex amplitude, not only for the steady-state but also includes the transient response, for any input !. The idea behind the precious trick of modelling a sinusoidal input with complex numbers was perfectly intuitive. But I've got no idea how Laplace transform does all this, I don't even understand, why it demands the use complex numbers? rather how does it find this alternative model for any given input, as we did in the previous "trick" only for sinusoidal input.

Best Answer

My Laplace transform math is pretty rusty, but the neat trick behind a Laplace transform is that the test waveform that the transform compares your input function to changes amplitude over time.

Examine the Fourier transform:

\$\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t)\ e^{-i \omega t}\,dx,\$

Essentially, it takes a test sine wave (\$e^{-i \omega }\$) with frequency \$\omega\$ and determines how similar that sine wave is to your input function 1. It determines this similarity, or "cross-correlation", by multiplying that sine wave with your input function across all time and integrates. If there is little similarity between this frequency \$\omega\$ and your input, then this integration will be equal to zero.

As you've observed, this test signal never changes amplitude and can only be used to examine the steady state behavior of your input.

Now compare to the Laplace transform:

\$F(s) =\int_0^\infty f(t)e^{-st} \, dt\$

Very similar, except now \$s\$ is complex and replaces the real (not complex) \$\omega\$. In addition, the integration is now not for all time, but only the future.

The neat thing about raising to a complex power \$s\$ is that there is both a real and an imaginary component, so that if \$s= a + i b\$, then \$e^{-st} = e^{-at}e^{-ibt}\$. So now we have a sine wave that either grows exponentially, shrinks exponentially, or maintains a constant amplitude depending on the value of \$a\$.

This test signal that can change over time, plus the fact that the integration now only starts at time 0 means that we can now extract the transient response from the input signal.

\$a\$ and \$b\$ together determine a point on the complex plane (known here as the s-plane), whereas \$\omega\$ will always be on the real line. This is the extra degree of freedom that @hotpaw2 mentioned. This point defines the test signal used by the Laplace transform, and you can see the effects of moving the point around the complex s-plane in this image:

s-plane for Laplace Transform signals

Diagrams depicting the Fourier series decomposition of a signal are common (like the one here: https://en.wikipedia.org/wiki/Fourier_series#/media/File:Fourier_series_and_transform.gif)

Less common are diagrams showing a Laplace transform decomposition, but I found this example analyzing an RLC circuit: Transient response of RLC circuit

On the top left plot you can see the inductor current over time, with a large decaying transient from 0 to 5 seconds. On the bottom left plot you can see the end of the initial transient and the final steady state behavior (note the large change in plot scale!).

And the corresponding Laplace decomposition: Laplace decomposition of RLC circuit

Here you can see that the total response in the first plot is the sum of the two terms plotted here, and that the impact of the initial transient decays to zero over time.

See this page for much more information:


1: In reality, \$e^{-i \omega } = \cos x + i \sin x\$, so it's actually using both a cosine and a sine wave simultaneously (a rotating complex point) as the test function. This is how the Fourier transform gets phase information. That's also why the image above shows curly 3D spirals, since the Laplace transform also has a rotating complex point. But it's easier to just think of it as a single sine wave.

Image sources: https://www.dsprelated.com/freebooks/mdft/Comparing_Analog_Digital_Complex.html https://www.mathworks.com/help/symbolic/solve-differential-equations-using-laplace-transform.html