# Physical realization of a system

Hi!

I have a simple question regarding the realization of systems. In my university course books, it's mentioned multiple times that a system with a transfer function G(s) is realizable, if the denominator's degree is at least that of the numerator's degree. I found the same statement on the internet as well: http://en.wikibooks.org/wiki/Control_Systems/Realizations

I don't understand what exactly this means, and it's always without explanation.

For example, suppose I have a transfer function G(s) = Y(s)/U(s) = Cs, which so shouldn't be realizable physically.
Doesn't a simple capacitor "realize" this transfer functions? If I consider the voltage of the capacitor my input, and the current through it the output? Then i = C*du/dt.
What do we mean by a system being realizable?

Thanks for any help!

-Tusike

I'm sorry, I don't understand how that helps... That's the same link I posted, and it doesn't provide an explanation as to why the statements it makes are true. What am I missing? Why wouldn't my capacitor example be considered a realization?

Stephen Tashi
Why wouldn't my capacitor example be considered a realization?

I'm not an expert on control systems, but I'll make a guess based on the "Implications" section of
http://en.wikipedia.org/wiki/Proper_transfer_function
[Edit: I edited the post to give the correct link.]

If a transfer function reduces to a form like $k_0 + k_1 s + k_2 s^2$ the terms in powers of $s$ have the effect of differentiating the input function. So if you set the the input to be a high frequency wave or rapidly wigglling curve you would get a high amplitude output. Apparently the ability to get an arbitrarily high amplitude output by picking a suitable high frequency input is considered not physically realizable.

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Simon Bridge
Homework Helper
It is not clear where you are having difficulty.

There are three "realization conditions" listed ... go through the one at a time and consider the converse.
Explain, in your own words, how the converse situation could be realized (or cannot be).

For your specific example - describe, in your own words, how you would physically realize the model.

OK, so I understand how the idea is that you shouldn't get an infinite response from an infinite frequency input, because a physical system wouldn't have enough energy to produce that.

1) However, why is an infinite frequency input considered physically realizable? Is it possible to generate such a thing? If not, then it shouldn't be used as a definition for why something isn't physically realizable...

2) Just because my capacitor wouldn't work at infinite frequency, it would still work in an operating range of e.g. 1kHz-1MHz. So, for such an application, I would describe it as I(s)/U(s)=G(s)=Cs, and I would use that transfer function for all my calculations, and it would work in practice just as I have calculated. Still, it's not considered physically realizable because it wouldn't work outside my planned operating range? Isn't this the same as saying a resistor isn't physically realizable, because it will break down at a very high U and won't follow it's supposed characteristic?

-Tusike

Stephen Tashi
1) However, why is an infinite frequency input considered physically realizable? Is it possible to generate such a thing? If not, then it shouldn't be used as a definition for why something isn't physically realizable...

Giving the Wikipedia article it's due, it defined a "proper" transfer functions. Even if we call them "realizable" transfer functions, there is no law of culture that says the use of words as technical terms must match their use in common speech. Why the terms "realizable" and "unrealizable" came into use mght have a historical explanation - perhaps it makes sense for people trying to design particular types of devices.

As I said, I'm not an expert in control theory. ( I once read a SAMS book about it.) Refresh my mind on how one evaluates if a given control system is "stable". Does that have anything to do with proper transfer functions?

Simon Bridge
Homework Helper
"realizable" is commonly said to mean that the model can be turned into a physical device - but that is not a definition in the strict mathematical sense. The definition is the list of conditions.
The question then goes to how that list got to be chosen.
The reason nobody derives the conditions is because they are usually considered self-evident when thought about.
So it is not possible to help you without you telling us more about how you are thinking re post #5.

FactChecker
Gold Member
For example, suppose I have a transfer function G(s) = Y(s)/U(s) = Cs, which so shouldn't be realizable physically.
Doesn't a simple capacitor "realize" this transfer functions?
I think that your problem is that you have the wrong Laplace transform for a capacitor. It is 1/(C*s). So the original statement that the order of the denominator must be at least the order of the numerator would say this is realizable..

@stephen: I guess that's the same conclusion I'm arriving at, that the naming is for historical reasons, or is very strict to avoid any confusion. Perhaps a better way of saying would be that the system is realizable "in a given frequency domain", and the formal definition requires it to be realizable at all domains. The requirement for a certain stability, say the output approaches zero after a dirac impulse, is that lim(sG(s)) be zero as 's' approaches zero. Whether or not the numerator has a higher polynomial in s than the denominator doesn't really effect this.

@simon: I'm not really sure how my first post isn't enough of an explanation for how to physically realize the example system. I can create a G(s)=Cs transfer function by connecting a capacitor to a signal generator. I consider the e.g. sinusoidal signal my input, and the current through the capacitor my output. Then I(s)=G(s)*U(s).

@FactChecker: the transfer function depends on what I consider my input and output, so Cs is okay.

Stephen Tashi
@stephen: The requirement for a certain stability, say the output approaches zero after a dirac impulse, is that lim(sG(s)) be zero as 's' approaches zero. Whether or not the numerator has a higher polynomial in s than the denominator doesn't really effect this.

Just from a mathematical point of view, doesn't the degree of the numerator and denominator of $G(s)$ have an important bearing on $lim_{s \rightarrow 0} s G(s)$ ?

FactChecker