I just learned transient circuits and we solved them both in the time domain and the Laplace domain and I can't understand why we would use the Laplace domain as it seems it is a lot more simple to use the time domain especially in series RLC circuits.

  • 2
    \$\begingroup\$ For many circuits I'm more interested in the frequency domain behavior, filters and amplifiers are examples of this. Those circuit are much easier to analyze and design in the Laplace domain. \$\endgroup\$ – Bimpelrekkie Jun 25 '18 at 13:15
  • \$\begingroup\$ See en.wikipedia.org/wiki/Integral_transform#Usage_example Look especially at the second paragraph under the "Usage example" heading. \$\endgroup\$ – Solomon Slow Jun 25 '18 at 13:43
  • 2
    \$\begingroup\$ Algebra is easier than differential equations. Algebra is easier than convolution. \$\endgroup\$ – Chu Jun 25 '18 at 15:59

It's easy to solve simple filter circuits in the time domain but there becomes a tipping-point where most engineers would prefer to solve problems in the frequency domain and apply (for example) a step-function. Finding the inverse Laplace is fairly straightforward because of Laplace tables.

As an example of an RLC low pass filter, engineers become accustomed to the transfer function: -

$$H(s) = \dfrac{\omega_n^2}{s^2+2\zeta\omega_n^2 s + \omega_n^2}$$

And, applying (for example) a step function is as simple as multiplying by 1/s: -

$$\dfrac{1}{s}\cdot\dfrac{\omega_n^2}{s^2+2\zeta\omega_n^2 s + \omega_n^2}$$

Engineers who are familiar with this get to recognize that this converts to a standard form and the next step is just using the tables to derive the transient response.

In addition, the term \$\omega_n\$ (the natural resonant frequency) can be is factored out a lot of the time so that solving the step-function formula for \$\omega_n=1\$ begins with: -

$$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s + 1}$$

This is rearranged to a standard form such as this (underdamped resonance): -

$$\dfrac{1}{s[(s+a)^2 + b^2]}$$

Where \$a =\zeta\$ and \$b=\sqrt{1-\zeta^2}\$

The Laplace tables give us this: -

$$H(t) = 1+\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot \sin(t\cdot\sqrt{1-\zeta^2}+\phi)$$

Where \$\phi = \arccos(\zeta)\$

But you will probably only be convinced when you are dealing with slightly more complex situations or really do need to analyse the frequency spectrum.

| improve this answer | |

In this context, the Laplace transform is simply a tool that we use to deal with differential equations. I think it provides a slightly more intuitive feel for how responses change with frequency, but really, it's just a tool for dealing with differential equations.

The transform really starts to shine when you're cascading signals. A variety of properties of the transform really become very useful then. For example, convolutions in the time domain become simple multiplications in the frequency domain. There are also a variety of frequency domain tools that make control easier.

You start using the transforms for simple systems, so you're facile when you start using them for more advanced situations -- not because they're so intuitively easy. In fact, the understanding is very quantal. You'll be able to "do" problems successfully, but it may take many iterations, even years, before you grok the transforms.

| improve this answer | |

Some day you will need to understand a system design requirement for both frequency and time domain response.

Laplace Transforms are useful for many applications in the frequency domain with order of polynominal giving standard slopes of 6dB/octave per or 20 dB/decade. But the skirts can be made sharp or smooth as seen by this Bandpass filter at 50Hz +/-10%.

The time domain helps you understand why the small ringing comes from the step response or abrupt change in phase and voltage of this sweep generator from 200Hz to the start at 20 Hz.

Some day you may need this and the Math will help you understand the relationships.

enter image description here

| improve this answer | |
  • \$\begingroup\$ Does the OP understand why the small 50Hz ringing starts from the Laplace transform of a step function contains all frequencies with amplitude(f) integrated? When there is no 50Hz before or during the smaller envelope? Laplace (or Fourier) explains why. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 25 '18 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.