0
$\begingroup$

I'm learning about transfer functions in control theory. I'm struggling to find a physical interpretation for the input and output of a transfer function, both of which may be complex numbers. In the time domain, the physical interpretation of the system solution is clear: the input is time and the output is a vector of physical state variables.

Is there a physical interpretation for the complex input and output of a transfer function?

$\endgroup$
2
$\begingroup$

Unlike Fourier transform*, Laplace transforms bears very little practical or physical insight. Actually the only reason you learn partial fraction expansion in the undergrad courses is because of Laplace Transforms.

The original insight of taking the Laplace transforms was to solve differential equations much more efficiently and in a more humane manner. But this got stuck and we are using it extremely blindly. 80% of control engineering students learn it as replace the number of dots with powers of s. It is very ill-taught and frankly you need to know lots of mathematics to start this type of queries if you want to do justice. For example, you stop worrying about the domain of convergence, analytic continuation and many many interesting features of this transform just after you learn about the definitions. If we are interested in stable systems and if s is not defined on the left half plane how is it that we discuss negative real part poles etc. See it immediately gets confusing because of the terrible motivation of such concepts. It is the same problem with Dirac's delta function. It only makes sense under the integral sine but we keep on multiplying with time functions etc. as if it is a real function. Hence, things become pretty tricky if you are not exposed to these concepts.

Another shortcoming of such thinking is that it forces control engineers think in terms of artificial causalities that bears no value in the original system. Take a clamped mass-spring-damper system, the equations that govern the motion is:

$$ m\ddot x(t) + b\dot x(t) + kx(t) = F(t) $$ now, if you follow the typical route, you get $$ \frac{X(s)}{F(s)} = \frac{1}{ms^2+bs+k} $$

here you must think that because of force we obtained a position output and hence the causality is position due to force. But obviously there is no such thing, the position and the force satisfy the same differential equation simultaneously. There is no such thing as force comes in and creates displacement.

My suggestion is that you keep this transform as a tool for solving/modeling differential equations as systems. Also, laplace transforms (or other integral kernels) are a much richer elements of mathematics. Hence, one should not overload them with artificial physical meanings.

* Fourier Transforms at least give some (and emphasis on some towards little) understanding of the steady state(!) behavior when the input is assumed to be pure sine. They are by no means indicative if you have a problematic transient regime.

$\endgroup$
  • $\begingroup$ Actually both the Fourier and the Laplace transforms give a lot more "physical" insight into the nature of signals than you give credit. The transforms represent signals in the frequency domain, a representation that often more useful and more practical than the time-domain description. The reason must be that systems in general have very different responses in different frequency ranges. One clear example is light. We think of light in terms of colors and frequencies. Time-domain signals are almost never mentioned. Music may be an even better example. $\endgroup$ – Pait May 27 '15 at 0:00
  • $\begingroup$ @Pait They are all examples of Fourier transform, a kernel evaluated on the imaginary axis, did you even see my footnote *? There is no particular insight Laplace transform gives in that regard other than converting a differential relation into an algebraic one. And again no transients only steady state. So you cannot guess the behavior of your system other than the final state. $\endgroup$ – percusse May 27 '15 at 11:18
  • 1
    $\begingroup$ @Pait We try to make Jan Willems' behavioral approach separated to bitesize chunks so that an undergrad can take it with a healthy dose of ss and tf). Again, laplace transforms are exceedingly difficult if you want to do justice. I don't buy control textbook way of teaching it. Not only they are superficial but also usually wrong (without an extensive discussion on analytic continuation and $L_p$f spaces). They confuse much more that what they intend to demonstrate and again for MIMO there is practically no insight plus there is no reliable tool. matlab converts all to ss behind the scenes. $\endgroup$ – percusse May 29 '15 at 16:27
  • 1
    $\begingroup$ An engineering course shouldn't pretend to be rigorous in math if it is not possible. Hence, we always say these are all mathematically wrong arguments, therefore don't overload with physical intuiton. The second case where it causes is the MIMO transmission zeros, especially, a MIMO system can have a noncancelling poles and zeros at the same location and don't appear in the rational entries is a mindboggling issue to explain. The only way to go about it is state space formulation. Frequency domain techniques are of course useful if we don't overload them with importance that they undeserve $\endgroup$ – percusse May 30 '15 at 13:16
  • 1
    $\begingroup$ @Pait I usually convert this narrative to a less formal and more pragmatic one. Then hopefully the students see that transfer functions are only another tool to solve the involved differentials but not a source of insight per se. Behaviors are not the holy grail anyways. I'm not saying this is the way to go. But this is at least general enough to show that state space transfer functions signal flows etc are all inside a one big universum of modeling. $\endgroup$ – percusse May 31 '15 at 10:37
0
$\begingroup$

They are the Laplace transforms of the time-domain input and output signals. In the frequency domain, the input and output signals are complex functions of a complex variable $s$. In the time domain they are real functions of a real variable $t$, different ways of representating the same signals.

It may be easier to think in terms of the Fourier transform, which is a generalization of the Fourier series to signals that are not necessarily periodic. There exist the sine series, the cosine series, and the complex exponential series, which is most convenient and leads to the complex Fourier transform. The Fourier transform represents a time-domain signal in terms of its frequency components. The transform is defined to be a function of a complex variable, to take advantage of the properties of the complex exponential. The transfer function relates the Fourier transforms of the input and the output.

The transfer function concept is the same, whether you use the Fourier or the Laplace transform. The latter is more practical in the study of control systems, because it takes care of initial conditions problems more easily, and is convenient in stability analysis.

The transfer function maps a complex function of the complex variable $s$ into another complex function - not a complex variable into another. The functions represent the input and the output in the frequency domain - as compared to the more usual representation of the input and output as functions of the time $t$.

$\endgroup$
  • 1
    $\begingroup$ I think the question is about how to interpret the complex character. For instance, suppose I fix $b \in \mathbb{R}$; what does the function $a \mapsto F(a+bi)$ tell me? $\endgroup$ – Ian May 13 '15 at 15:19
  • 1
    $\begingroup$ @Ian is exactly right. I'm trying to get some intuition about the physical meaning of the transfer function. $\endgroup$ – BmoreDaniel May 13 '15 at 16:23
  • $\begingroup$ See expanded answer. $\endgroup$ – Pait May 15 '15 at 0:41
  • $\begingroup$ @Pait I still don't see the physical interpretation of the input and output of the transfer function. Is it just a mathematical tool that has no physical interpretation? If the transfer function maps $a + bi$ to $c + di$, what does that tell me about the physical dynamics of the system? How can I physically interpret these complex numbers? $\endgroup$ – BmoreDaniel May 15 '15 at 19:44
  • $\begingroup$ The input and the output are not complex variables - they are complex functions, see above. $\endgroup$ – Pait May 17 '15 at 2:11

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.