Laplace transform and the idea of frequency domain analysis

Question

I have studied control theory in depth in my undergrad and still continue to use it on many occasions in my PhD work, but i still don't fully understand why we call Laplace transformation of system a "frequency domain analysis".

I know that Laplace transform is a mathematical tool to move from the time domain to the s-domaine to substitute differential equations to algebraic equations which makes the mathematical analysis a lot easier. And of course S-domaine is linked to the frequency domain with the relation S= alpha + JW.

But take for example the Laplace transform of the step function u(t) = 1 ; t>=0 , which is 1/s, the step function has a frequency of 0 and i don't see how 1/s represent a "frequency domain equivalent of the function".

what makes this more incomprehensible to me the fact that based on the S-plane analysis of the system, we draw bode plots which shows gain and phase shift based on the frequency of the input !!

My question is kind of an attempt to fully understand the mathematical idea behind the Laplace transform and how it relates to the actual physical properties of the control systems and the signals that drives them and i appreciate anybody sharing his/her prospective. Thanks

Answer 1

Applying a sinusoidal input to, e.g., a stable, SISO, LTI system will produce an output with a transient component and a steady-state component. The transient does what all transients do - it decays to zero after a certain time. The steady-state component is a sine wave at the same frequency as the input sinusoid but, generally, with a different amplitude and also a phase shift relative to the input

It turns out (see numerous references) that the steady-state sinusoidal response (or 'frequency response') can be obtained from the system transfer function by the simple algebraic substitution: \$\small s\rightarrow j\omega\$. And hence the frequency domain is extremely accessible from the s-domain. The time domain, for example, is not nearly so easy to access as it requires the inverse Laplace transform.

To use your example, if a system has TF \$\small G(s)=\large \frac{1}{s}\$, the frequency domain gain and phase angle would be: \$\frac{1}{\omega}\$ and \$\small -\frac{\pi}{2}\$, respectively.

From a signal, rather than system, perspective, the unit step signal, \$\small H(t)\$, has the same LT as the system, above, viz: \$\small H(s)=\large \frac{1}{s}\$, and in the frequency domain this maps to the positive spectrum \$\small H(j\omega)=\large\frac{1}{j\omega}\$, and has amplitude \$\frac{1}{\omega}\$, and a phase angle of \$\small -\frac{\pi}{2}\$ relative to a notional sine wave, \$\small sin(\omega t)\$

Answer 2

No hard argumentation but an attempt to make it plausible:

The Laplace Transform and Fourier Transform are very similar.
The difference is just the integration border and the additional factor \$i\$ (or if you prefere \$j\$). The Laplace Transform is just more general insofar as \$s\$ is complex.

For the Fourier Transform I assume it is obvious why the "other" domain is called the frequency domain. The little differences between the two do not justify why the \$s\$-domain of the Laplace Transform should not be called frequency domain too.

It is similar to the case of damped oscillation (e.g. RLC circuits) where it is also common to use the concept of a complex frequency that combines the "common" frequency (oscillation) and the damping.

Answer 3

The 2 main forms of representing a system in the frequency domain is by using 1) Foruier transform and 2) Laplace transform. Laplace is a bit more ahead than fourier , while foruier represents any signal in form of siusoids the laplace represents any signal in the form of damped sinusoids . The factor S = alpha + jw , while in fourier you only have jw!So in other words you could also say that when Xeta =0 you have got the fourier response from the laplace- This xeta adds a exponentil function to your already existing sinusoid (jw) . SO what happens when you add these 2, you get a damped sinusoid. What advantage does this posses ? Represting the magnitude response in Laplace gives you the poles whihc are also the natural response of the system ! It gives you a guess also at the transient response of the system along with the frequency response. But for your other question on why the step response is 1/S , F(s) =(Integral limit 0 to infinity) f(t)e^(-st) dt. So substitute f(t) = 1 as that is the case for a step function you will get the asnwer as 1/S where S=alpha + Jw . It still does not say that the step has a frequency in it it just say that it is represented in this form. So if you put w=0 then it does not have a frequency. This is what I think, it is justa form of representation of the step.

Answer 4

But take for example the Laplace transform of the step function u(t) = 1 ; t>=0 , which is 1/s, the step function has a frequency of 0 and i don't see how 1/s represent a "frequency domain equivalent of the function".

1/s is telling you that spectrum is infinite but falls as the reciprocal of s i.e. at S = 10 the amplitude might be 0.1 but at s = 100 the amplitude is 0.01. Here are a few other common examples (the step change scenario is 4th one down): -

My question is kind of an attempt to fully understand the mathematical idea behind the Laplace transform and how it relates to the actual physical properties of the control systems and the signals that drives them and i appreciate anybody sharing his/her prospective.

I've forgotten most of it but what I remember is that it's too much to cover on this site.

Answer 5

The answer is hidden in s=jw where w strands for omega i.e about frequency, j is the imaginary number and s is the laplace s